CHAPTER 5 GASES AND THE KINETIC-MOLECULAR THEORY · 5-1 CHAPTER 5 GASES AND THE KINETIC-MOLECULAR THEORY FOLLOW–UP PROBLEMS . 5.1A . Plan: Use the equation for gas pressure in an

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CHAPTER 5 GASES AND THE KINETIC-MOLECULAR THEORY FOLLOW–UP PROBLEMS 5.1A Plan: Use the equation for gas pressure in an open-end manometer to calculate the pressure of the gas. Use

conversion factors to convert pressure in mmHg to units of torr, pascals and lb/in2. Solution: Because Pgas < Patm, Pgas = Patm – Δh Pgas = 753.6 mm Hg – 174.0 mm Hg = 579.6 mm Hg

Pressure (torr) = ( ) 1 torr579.6 mm Hg1 mm Hg

= 579.6 torr

Pressure (Pa) = ( )51 atm 1.01325 x 10 Pa579.6 mm Hg

760 mm Hg 1atm

= 7.727364 x 104 = 7.727 x 104 Pa

Pressure (lb/in2) = ( )21 atm 14.7 lb/in579.6 mm Hg

760 mm Hg 1 atm

= 11.21068 = 11.2 lb/in2

5.1B Plan: Convert the atmospheric pressure to torr. Use the equation for gas pressure in an open-end manometer to

calculate the pressure of the gas. Use conversion factors to convert pressure in torr to units of mmHg, pascals and lb/in2.

Solution: Because Pgas > Patm, Pgas = Patm + Δh

Pgas = (0.9475 atm)�760 torr1 atm

� + 25.8 torr = 745.9 torr

Pressure (mmHg) = (745.9 torr)�1 mmHg1 torr

� = 745.9 mmHg

Pressure (Pa) = (745.9 mmHg)� 1 atm760 mmHg

� �1.01325 x 105 Pa1 atm

� = 9.945 x 104 Pa

Pressure (lb/in2) = (745.9 mmHg)� 1 atm760 mmHg

� �14.7 lb/in2

1 atm� = 14.4 lb/in2

5.2A Plan: Given in the problem is an initial volume, initial pressure, and final volume for the argon gas. The final

pressure is to be calculated. The temperature and amount of gas are fixed. Rearrange the ideal gas law to the appropriate form and solve for P2. Once solved for, P2 must be converted from atm units to kPa units.

Solution: P1 = 0.871 atm; V1 = 105 mL P2 = unknown V2 = 352 mL

1 1 2 2

1 1 2 2 =

PV P Vn T n T

At fixed n and T:

1 1 2 2 = PV P V

P2 (atm) = 1 1

2

PVV

= (0.871 atm)(105 mL)

(352 mL) = 0.260 atm

P2 (kPa) = (0.260 atm)�101.325 kPa1 atm

� = 26.3 kPa

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5.2B Plan: Given in the problem is an initial volume, initial pressure, and final pressure for the oxygen gas. The final volume is to be calculated. The temperature and amount of gas are fixed. Convert the final pressure to atm units. Rearrange the ideal gas law to the appropriate form and solve for V2.

Solution: P1 = 122 atm; V1 = 651 L P2 = 745 mmHg V2 = unknown

1 1 2 2

1 1 2 2 =

PV P Vn T n T

At fixed n and T:

1 1 2 2 = PV P V

P2 (atm) = (745 mmHg)� 1 atm760 mmHg

� = 0.980 atm

V2 (atm) = 1 1

2

PVP

= (122 atm)(651 L)

(0.980 atm) = 8.10 x 104 L

5.3A Plan: Convert the temperatures to kelvin units and the initial pressure to units of torr. Examine the ideal gas law,

noting the fixed variables and those variables that change. R is always constant so 1 1 2 2

1 1 2 2 =

PV P Vn T n T

. In this problem,

P and T are changing, while n and V remain fixed. Solution: T1 = 23oC T2 = 100oC P1 = 0.991 atm P2 = unknown n and V remain constant Converting T1 from oC to K: 23oC + 273.15 = 296 K Converting T2 from oC to K: 100oC + 273.15 = 373 K

P1 (torr) = (0.991 atm)�760 torr1 atm

� = 753 torr

Arranging the ideal gas law and solving for P2: P1V1n1T1

= P2V2n2T2

or P1T1

= P2T2

P2 (torr) = P1 𝑇2𝑇1

= (753 torr)�373 K296 K

� = 949 torr

Because the pressure in the tank (949 torr) is less than the pressure at which the safety valve will open (1.00 x 103 torr), the safety valve will not open.

5.3B Plan: This is Charles’s law: at constant pressure and with a fixed amount of gas, the volume of a gas is directly

proportional to the absolute temperature of the gas. The temperature must be lowered to reduce the volume of a gas. Arrange the ideal gas law, solving for T2 at fixed n and P. Temperature must be converted to kelvin units.

Solution: V1 = 32.5 L V2 = 28.6 L T1 = 40°C (convert to K) T2 = unknown n and P remain constant Converting T from °C to K: T1 = 40 °C + 273 = 313K Arranging the ideal gas law and solving for T2:

1 1 2 2

1 1 2 2 =

PV P Vn T n T

or 1 2

1 2 =

V VT T

22 1

1 =

VT TV

= (313 K)�28.6 L32.5 L

� = 275 K – 273.15 = 2°C

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5.4A Plan: In this problem, the amount of gas is decreasing. Since the container is rigid, the volume of the gas will not change with the decrease in moles of gas. The temperature is also constant. So, the only change will be that the pressure of the gas will decrease since fewer moles of gas will be present after removal of the 5.0 g of ethylene. Rearrange the ideal gas law to the appropriate form and solve for P2. Since the ratio of moles of ethylene is equal to the ratio of grams of ethylene, there is no need to convert the grams to moles. (This is illustrated in the solution by listing the molar mass conversion twice.)

Solution: P1 = 793 torr; P2 = ? mass1 = 35.0 g; mass2 = 35.0 – 5.0 = 30.0 g

1 1 2 2

1 1 2 2 =

PV P Vn T n T

At fixed V and T:

1 2

1 2 =

P Pn n

P2 = 1 2

1

Pnn

= ( )( )

( )

2 42 4

2 4

2 42 4

2 4

1 mol C H30.0 g C H28.05 g C H

793 torr1 mol C H35.0 g C H

28.05 g C H

= 679.714 = 680. torr

5.4B Plan: Examine the ideal gas law, noting the fixed variables and those variables that change. R is always constant

so 1 1 2 2

1 1 2 2 =

PV P Vn T n T

. In this problem, n and V are changing, while P and T remain fixed.

Solution: m1 = 1.26 g N2 m2 = 1.26 g N2 + 1.26 g He V1 = 1.12 L V2 = unknown P and T remain constant

Converting m1 (mass) to n1 (moles): (1.26 g N2)� 1 mol N2

28.02 g N2� = 0.0450 mol N2 = n1

Converting m2 (mass) to n2 (moles): 0.0450 mol N2 + (1.26 g He)� 1 mol He4.003 g He

� = 0.0450 mol N2 + 0.315 mol He = 0.360 mol gas = n2

Arranging the ideal gas law and solving for V2: P1V1n1T1

= P2V2n2T2

or V1n1

= V2n2

V2 = V1 n2n1

= (1.12 L)� 0.360 mol0.0450 mol

� = 8.96 L

5.5A Plan: Convert the temperatures to kelvin. Examine the ideal gas law, noting the fixed variables and those

variables that change. R is always constant so 1 1 2 2

1 1 2 2 =

PV P Vn T n T

. In this problem, P, V, and T are changing, while n

remains fixed. Solution: T1 = 23oC T2 = 18oC P1 = 755 mmHg P2 = unknown

V1 = 2.55 L V2 = 4.10 L n remains constant Converting T1 from oC to K: 23oC + 273.15 = 296 K Converting T2 from oC to K: 18oC + 273.15 = 291 K


= P2V2n2T2

or P1V1

T1 =

P2V2T2

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P2 (mmHg) = P1 V1T2V2T1

= (755 mmHg)�(2.55 L)(291 K)(4.10 L)(296 K)

� = 462 mmHg

5.5B Plan: Convert the temperatures to kelvin. Examine the ideal gas law, noting the fixed variables and those

variables that change. R is always constant so 1 1 2 2

1 1 2 2 =

PV P Vn T n T

. In this problem, P, V, and T are changing, while n

remains fixed. Solution: T1 = 28oC T2 = 21oC P1 = 0.980 atm P2 = 1.40 atm

V1 = 2.2 L V2 = unknown n remains constant Converting T1 from oC to K: 28oC + 273.15 = 301 K Converting T2 from oC to K: 21oC + 273.15 = 294 K


= P2V2n2T2

or P1V1

T1 =

P2V2T2

V2 (L) = V1 P1T2P2T1

= (2.2 L)�(0.980 atm )(294 K)(1.40 atm)(301 K)

� = 1.5 L

5.6A Plan: From Sample Problem 5.6 the temperature of 21°C and volume of 438 L are given. The pressure is 1.37 atm

and the unknown is the moles of oxygen gas. Use the ideal gas equation PV = nRT to calculate the number of moles of gas. Multiply moles by molar mass to obtain mass.

Solution: PV = nRT

n = PVRT

= ( ) ( )

( )( )1.37 atm 438 L

0.0821 atm • L 273.15 21 Kmol • K

+

= 24.9 mol O2

Mass (g) of O2 = (24.9 mol O2) �32.00 g O2

1 mol O2� = 796.8 = 797 g O2

5.6B Plan: Convert the mass of helium to moles, the temperature to kelvin units, and the pressure to atm units. Use the

ideal gas equation PV = nRT to calculate the volume of the gas. Solution: P = 731 mmHg V = unknown m = 3950 kg He T = 20oC

Converting m (mass) to n (moles): (3950 kg He)�1000 g1 kg

� � 1 mol He4.003 g He

� = 9.87 x 105 mol = n

Converting T from oC to K: 20oC + 273.15 = 293 K

Converting P from mmHg to atm: (731 mmHg) � 1 atm760 mmHg

� = 0.962 atm

PV = nRT

V = nRT

P =

(9.87 x 105 mol)�0.0821 atm • Lmol • K�(293 K)

(0.962 atm) = 2.47 x 107 L

5.7A Plan: Balance the chemical equation. The pressure is constant and, according to the picture, the volume

approximately doubles. The volume change may be due to the temperature and/or a change in moles. Examine the

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balanced reaction for a possible change in number of moles. Rearrange the ideal gas law to the appropriate form and solve for the variable that changes.

Solution: The balanced chemical equation must be 2CD → C2 + D2

Thus, the number of mole of gas does not change (2 moles both before and after the reaction). Only the temperature remains as a variable to cause the volume change. Let V1 = the initial volume and 2V1 = the final volume V2. T1 = (–73 + 273.15) K = 200.15 K

1 1 2 2

1 1 2 2 =

PV P Vn T n T

At fixed n and P:

1 2

1 2 =

V VT T

2 12

1 = =

V TTV

( )( )( )

1

1

2 200.15 KVV

= 400.30 K – 273.15 = 127.15 = 127°C

5.7B Plan: The pressure is constant and, according to the picture, the volume approximately decreases by a factor of 2

(the final volume is approximately one half the original volume). The volume change may be due to the temperature change and/or a change in moles. Consider the change in temperature. Examine the balanced reactions for a possible change in number of moles. Think about the relationships between the variables in the ideal gas law in order to determine the effect of temperature and moles on gas volume.

Solution: Converting T1 from oC to K: 199oC + 273.15 = 472 K Converting T2 from oC to K: –155oC + 273.15 = 118 K According to the ideal gas law, temperature and volume are directly proportional. The temperature decreases by a

factor of 4, which should cause the volume to also decrease by a factor of 4. Because the volume only decreases by a factor of 2, the number of moles of gas must have increased by a factor of 2 (moles of gas and volume are also directly proportional).

1/4 (decrease in V from the decrease in T) x 2 (increase in V from the increase in n) = 1/2 (a decrease in V by a factor of 2)

Thus, we need to find a reaction in which the number of moles of gas increases by a factor of 2. In equation (1), 3 moles of gas yield 2 moles of gas. In equation (2), 2 moles of gas yield 4 moles of gas. In equation (3), 1 mole of gas yields 3 moles of gas. In equation (4), 2 moles of gas yield 2 moles of gas.

Because the number of moles of gas doubles in equation (2), that equation best describes the reaction in the figure in this problem.

5.8A Plan: Density of a gas can be calculated using a version of the ideal gas equation,PdRT

=M

. Two calculations are

required, one with T = 0°C = 273 K and P = 380 torr and the other at STP which is defined as T = 273 K and P = 1 atm.

Solution: Density at T = 273 K and P = 380 torr:

d = ( )( )

( )

380 torr 44.01 g/mol 1 atm0.0821 atm • L 760 torr273 K

mol • K

= 0.981783 = 0.982 g/L

Density at T = 273 K and P = 1 atm. (Note: The 1 atm is an exact number and does not affect the significant figures in the answer.)

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d = ( ) ( )

( )

44.01 g/mol 1 atm0.0821 atm • L 273 K

mol • K

= 1.9638566 = 1.96 g/L

The density of a gas increases proportionally to the increase in pressure.

5.8B Plan: Density of a gas can be calculated using a version of the ideal gas equation,PdRT

=M

Solution: Density of NO2 at T = 297 K (24oC + 273.15) and P = 0.950 atm:

d = (0.950 atm)(46.01 g/mol)

�0.0821 atm • Lmol • K�(297 K)

= 1.7926 = 1.79 g/L

Nitrogen dioxide is more dense than dry air at the same conditions (density of dry air = 1.13 g/L). 5.9A Plan: Calculate the mass of the gas by subtracting the mass of the empty flask from the mass of the flask

containing the condensed gas. The volume, pressure, and temperature of the gas are known.

The relationship PdRT

=M

is rearranged to give M = dRT

P or M = mRT

PV

Solution: Mass (g) of gas = mass of flask + vapor – mass of flask = 68.697 – 68.322 = 0.375 g T = 95.0°C + 273 = 368 K

P = ( ) 1 atm740. torr760 torr

= 0.973684 atm

V = 149 mL = 0.149 L

M = mRTPV

= ( ) ( )

( )( )

0.0821 atm • L0.375 g 368 Kmol • K

0.973684 atm 0.149 L

= 78.094 = 78.1 g

5.9B Plan: Calculate the mass of the gas by subtracting the mass of the empty glass bulb from the mass of the bulb

containing the gas. The volume, pressure, and temperature of the gas are known. The relationship PdRT

=M

is

rearranged to give M = dRT

P or M = mRT

PV. Use the molar mass of the gas to determine its identity.

Solution: Mass (g) of gas = mass of bulb + gas – mass of bulb = 82.786 – 82.561 = 0.225 g T = 22°C + 273.15 = 295 K

P = (733 mmHg) � 1 atm760 mmHg

� = 0.965 atm

V = 350. mL = 0.350 L

M = mRTPV

= (0.225 g)�0.0821 atm • L

mol • K�(295 K)

(0.965 atm )(0.350 L) = 16.1 g/mol

Methane has a molar mass of 16.04 g/mol. Nitrogen monoxide has a molar mass of 30.01 g/mol. The gas that has a molar mass that matches the calculated value is methane.

5.10A Plan: Calculate the number of moles of each gas present and then the mole fraction of each gas. The partial

pressure of each gas equals the mole fraction times the total pressure. Total pressure equals 1 atm since the problem specifies STP. This pressure is an exact number, and will not affect the significant figures in the answer

Solution:

Moles of He = ( ) 1 mol He5.50 g He4.003 g He

= 1.373970 mol He

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Moles of Ne = ( ) 1 mol Ne15.0 g Ne20.18 g Ne

= 0.743310 mol Ne

Moles of Kr = ( ) 1 mol Kr35.0 g Kr83.80 g Kr

= 0.417661 mol Ke

Total number of moles of gas = 1.373970 + 0.743310 + 0.417661 = 2.534941 mol PA = XA x P total

PHe = ( )1.37397 mol He 1 atm2.534941 mol

= 0.54201 = 0.542 atm He

PNe = ( )0.74331 mol Ne 1 atm2.534941 mol

= 0.29323 = 0.293 atm Ne

PKr = ( )0.41766 mol Kr 1 atm2.534941 mol

= 0.16476 = 0.165 atm Kr

5.10B Plan: Use the formula PA = XA x P total to calculate the mole fraction of He. Multiply the mole fraction by 100% to

calculate the mole percent of He. Solution: PHe = XHe x P total

Mole percent He = XHe (100%) = � PHe

Ptotal� (100%) = �143 atm

204 atm� (100%) = 70.1%

5.11A Plan: The gas collected over the water will consist of H2 and H2O gas molecules. The partial pressure of the water

can be found from the vapor pressure of water at the given temperature given in the text. Subtracting this partial pressure of water from total pressure gives the partial pressure of hydrogen gas collected over the water. Calculate the moles of hydrogen gas using the ideal gas equation. The mass of hydrogen can then be calculated by converting the moles of hydrogen from the ideal gas equation to grams.

Solution: From the table in the text, the partial pressure of water is 13.6 torr at 16°C.

P = 752 torr – 13.6 torr = 738.4 = 738 torr H2 The unrounded partial pressure (738.4 torr) will be used to avoid rounding error.

Moles of hydrogen = n = PVRT

= ( ) ( )

( )( )

3738.4 torr 1495 mL 1 atm 10 L0.0821 atm • L 760 torr 1 mL273.15 16 K

mol • K

− +

= 0.061186 mol H2

Mass (g) of hydrogen = ( ) 22

2

2.016 g H0.061186 mol H

1 mol H

= 0.123351 = 0.123 g H2

5.11B Plan: The gas collected over the water will consist of O2 and H2O gas molecules. The partial pressure of the water

can be found from the vapor pressure of water at the given temperature given in the text. Subtracting this partial pressure of water from total pressure gives the partial pressure of oxygen gas collected over the water. Calculate the moles of oxygen gas using the ideal gas equation. The mass of oxygen can then be calculated by converting the moles of oxygen from the ideal gas equation to grams.

Solution: From the table in the text, the partial pressure of water is 17.5 torr at 20°C.

P = 748 torr – 17.5 torr = 730.5 = 730. torr O2

Moles of oxygen = n = PVRT

= (730. torr)(307 mL)

�0.0821 atm • Lmol • K�(293 K)

� 1 atm760 torr

� � 1 L1000 mL

�

= 0.012258 mol O2

Mass (g) of oxygen = (0.012258 mol O2) �32.00 g O2

1 mol O2� = 0.3923 = 0.392 g O2

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5.12A Plan: Write a balanced equation for the reaction. Calculate the moles of HCl(g) from the starting amount of

sodium chloride using the stoichiometric ratio from the balanced equation. Find the volume of the HCl(g) from the molar volume at STP. Solution:

The balanced equation is H2SO4(aq) + 2NaCl(aq) → Na2SO4(aq) + 2HCl(g).

Moles of HCl = ( )310 g 1 mol NaCl 2 mol HCl0.117 kg NaCl

1 kg 58.44 g NaCl 2 mol NaCl

= 2.00205 mol HCl

Volume (mL) of HCl = ( ) 322.4 L 1 mL2.00205 mol HCl

1 mol HCl 10 L−

= 4.4846 x 104 = 4.48 x 104 mL HCl 5.12B Plan: Write a balanced equation for the reaction. Use the ideal gas law to calculate the moles of CO2(g) scrubbed.

Use the molar ratios from the balanced equation to calculate the moles of lithium hydroxide needed to scrub that amount of CO2. Finally, use the molar mass of lithium hydroxide to calculate the mass of lithium hydroxide required. Solution:

The balanced equation is 2LiOH(s) + CO2(g) → Li2CO3(s) + H2O(l).

Amount (mol) of CO2 scrubbed = n = PVRT

= (0.942 atm)(215 L)

�0.0821 atm • Lmol • K�(296 K)

= 8.3340 = 8.33 mol CO2

Mass (g) of LiOH = 8.33 mol CO2 �2 mol LiOH1 mol CO2

� �23.95 g LiOH1 mol LiOH

� = 399.0070 = 399 g LiOH

5.13A Plan: Balance the equation for the reaction. Determine the limiting reactant by finding the moles of each reactant

from the ideal gas equation, and comparing the values. Calculate the moles of remaining excess reactant. This is the only gas left in the flask, so it is used to calculate the pressure inside the flask.

Solution: The balanced equation is NH3(g) + HCl(g) → NH4Cl(s).

The stoichiometric ratio of NH3 to HCl is 1:1, so the reactant present in the lower quantity of moles is the limiting reactant.

Moles of ammonia = PVRT

= ( ) ( )

( )( )0.452 atm 10.0 L

0.0821 atm • L 273.15 22 Kmol • K

+

= 0.18653 mol NH3

Moles of hydrogen chloride =PVRT

= ( )( )

( )

37.50 atm 155 mL 10 L0.0821 atm • L 1 mL271 K

mol • K

−

= 0.052249 mol HCl

The HCl is limiting so the moles of ammonia gas left after the reaction would be 0.18653 – 0.052249 = 0.134281 mol NH3.

Pressure (atm) of ammonia = nRTV

= ( ) ( )( )

( )

0.0821 atm • L0.134281 mol 273.15 22 Kmol • K

10.0L

+

= 0.325387 = 0.325 atm NH3

5.13B Plan: Balance the equation for the reaction. Use the ideal gas law to calculate the moles of fluorine that react.

Determine the limiting reactant by determining the moles of product that can be produced from each of the reactants and comparing the values. Use the moles of IF5 produced and the ideal gas law to calculate the volume of gas produced.

Solution: The balanced equation is I2(s) + 5F2 (g) → 2IF5(g).

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Amount (mol) of F2 that reacts = n = PVRT

= (0.974 atm)(2.48 L)

�0.0821 atm • Lmol • K�(291 K)

= 0.1011 = 0.101 mol F2

Amount (mol) of IF5 produced from F2 = 0.101 mol F2 �2 mol IF5

5 mol F2� = 0.0404 mol IF5

Amount (mol) of IF5 produced from I2 = 4.16 g I2 � 1 mol I2

253.8 g I2� �2 mol IF5

1 mol I2� = 0.0328 mol IF5

Because a smaller number of moles is produced from the I2, I2 is limiting and 0.0328 mol of IF5 are produced.

Volume (L) of IF5 = nRT

P =

(0.0328 mol)�0.0821 atm • Lmol • K�(378 K)

(0.935 atm) = 1.08867 = 1.09 L

5.14A Plan: Graham’s law can be used to solve for the effusion rate of the ethane since the rate and molar mass of

helium are known, along with the molar mass of ethane. In the same way that running slower increases the time to go from one point to another, so the rate of effusion decreases as the time increases. The rate can be expressed as 1/time.

Solution:

2 6C H

2 6 He

Rate HeRateC H

=M

M

( )( )

2 6

2 6

C H

0.010 mol He30.07 g/mol1.25 min4.003 g/mol0.010 mol C H

=

t

0.800 t = 2.74078 t = 3.42597 = 3.43 min 5.14B Plan: Graham’s law can be used to solve for the molar mass of the unknown gas since the rates of both gases and

the molar mass of argon are known. Rate can be expressed as the volume of gas that effuses per unit time. Solution: Rate of Ar = 13.8 mL/time Rate of unknown gas = 7.23 mL/time Mass of Ar = 39.95 g/mol

Rate of Ar

Rate of unknown gas = ��

Munknown gas

MAr�

� Rate of ArRate of unknown gas

�2

= �Munknown gas

MAr�

Munknown gas = (MAr) �Rate of Ar

Rate of unknown gas�

2

Munknown gas = (39.95 g/mol) �13.8 mL/time7.23 mL/time

�2 = 146 g/mol

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CHEMICAL CONNECTIONS BOXED READING PROBLEMS B5.1 Plan: Examine the change in density of the atmosphere as altitude changes. Solution: The density of the atmosphere decreases with increasing altitude. High density causes more drag on the aircraft. At high altitudes, low density means that there are relatively few gas particles present to collide with the aircraft. B5.2 Plan: The conditions that result in deviations from ideal behavior are high pressure and low temperature. At

high pressure, the volume of the gas decreases, the distance between particles decreases, and attractive forces between gas particles have a greater effect. A low temperature slows the gas particles, also increasing the affect of attractive forces between particles.

Solution: Since the pressure on Saturn is significantly higher and the temperature significantly lower than that on Venus, atmospheric gases would deviate more from ideal gas behavior on Saturn. B5.3 Plan: To find the volume percent of argon, multiply its mole fraction by 100. The partial pressure of argon gas can

be found by using the relationship PAr = XAr x P total. The mole fraction of argon is given in Table B5.1. Solution: Volume percent = mole fraction x 100 = 0.00934 x 100 = 0.934 % The total pressure at sea level is 1.00 atm = 760 torr. PAr = XAr x P total = 0.00934 x 760 torr = 7.0984 = 7.10 torr B5.4 Plan: To find the moles of gas, convert the mass of the atmosphere from t to g and divide by the molar mass of air. Knowing the moles of air, the volume can be calculated at the specified pressure and temperature by using the ideal gas law. Solution:

a) Moles of gas = ( )15 1000 kg 1000 g 1 mol5.14x10 t1 t 1 kg 28.8 g

= 1.78472x1020 = 1.78x1020 mol b) PV = nRT

V = nRT

P =

( ) ( )( )( )

20 L•atm1.78472 x10 mol 0.0821 273 25 Kmol•K

1 atm

+ = 4.36646x1021 = 4 x 1021 L

END–OF–CHAPTER PROBLEMS 5.1 Plan: Review the behavior of the gas phase vs. the liquid phase.

Solution: a) The volume of the liquid remains constant, but the volume of the gas increases to the volume of the larger

container. b) The volume of the container holding the gas sample increases when heated, but the volume of the container holding the liquid sample remains essentially constant when heated. c) The volume of the liquid remains essentially constant, but the volume of the gas is reduced. 5.2 The particles in a gas are further apart than those are in a liquid. a) The greater empty space between gas molecules allows gases to be more compressible than liquids. b) The greater empty space between gas molecules allows gases to flow with less resistance (hindrance) than liquids. c) The large empty space between gas molecules limits their interaction, allowing all mixtures of gases to be solutions. d) The large empty space between gas molecules increases the volume of the gas, therefore decreasing the density.

5-11

5.3 The mercury column in the mercury barometer stays up due to the force exerted by the atmosphere on the mercury in the outer reservoir just balancing the gravitational force on the mercury in the tube. Its height adjusts according to the air pressure on the reservoir. The column of mercury is shorter on a mountaintop as there is less atmosphere to exert a force on the mercury reservoir. On a mountaintop, the air pressure is less, so the height of mercury it balances in the barometer is shorter than at sea level where there is more air pressure.

5.4 The pressure of mercury is its weight (force) per unit area. The weight, and thus the pressure, of the mercury column is directly proportional to its height. 5.5 When the mercury level in the arm attached to the flask is higher than the level in the other arm, the pressure in the flask is less than the pressure exerted in the other arm. This is an impossible situation for a closed-end manometer as the flask pressure cannot be less than the vacuum in the other arm. 5.6 Plan: The ratio of the heights of columns of mercury and water are inversely proportional to the ratio of the

densities of the two liquids. Convert the height in mm to height in cm. Solution:

2

2

H O Hg

Hg H O

h dh d

=

2

2

HgH O Hg

H O x =

dh h

d = ( )

3

213.5 g/mL 10 m 1 cm730 mmHg1.00 g/mL 1 mm 10 m

−

−

= 985.5 = 990 cm H2O

5.7 Plan: The ratio of the heights of columns of mercury and water are inversely proportional to the ratio of the

densities of the two liquids. Solution:

2

2

H O Hg

Hg H O

h dh d

=

2

2

HgH O Hg

H O x =

dh h

d = ( )13.5 g/mL 755 mmHg

1.00 g/mL

= 10,192.5 = 1.02x104 mm H2O

5.8 Plan: Use the conversion factors between pressure units:

1 atm = 760 mmHg = 760 torr = 101.325 kPa = 1.01325 bar Solution:

a) Converting from atm to mmHg: P(mmHg) = ( ) 760 mmHg0.745 atm1 atm

= 566.2 = 566 mmHg

b) Converting from torr to bar: P(bar) = ( ) 1.01325 bar992 torr760 torr

= 1.32256 = 1.32 bar

c) Converting from kPa to atm: P(atm) = ( ) 1 atm365 kPa101.325 kPa

= 3.60227 = 3.60 atm

d) Converting from mmHg to kPa: P(kPa) = ( ) 101.325 kPa804 mmHg760 mmHg

= 107.191 = 107 kPa


1 atm = 760 mmHg = 760 torr = 101.325 kPa = 1.01325 bar Solution: a) Converting from cmHg to atm:

P(atm) = ( )2

310 m 1 mm 1 atm76.8 cmHg1 cm 760 mmHg10 m

−

−

= 1.01053 = 1.01 atm

5-12

b) Converting from atm to kPa: P(kPa) = ( ) 101.325 kPa27.5 atm1 atm

= 2.786x103 = 2.79x103 kPa

c) Converting from atm to bar: P(bar) = ( ) 1.01325 bar6.50 atm1 atm

= 6.5861 = 6.59 bar

d) Converting from kPa to torr: P(torr) = ( ) 760 torr0.937 kPa101.325 kPa

= 7.02808 = 7.03 torr

5.10 Plan: This is an open-end manometer. Since the height of the mercury column in contact with the gas is higher

than the column in contact with the air, the gas is exerting less pressure on the mercury than the air. Therefore the pressure corresponding to the height difference (Δh) between the two arms is subtracted from the atmospheric pressure. Since the height difference is in units of cm and the barometric pressure is given in units of torr, cm must be converted to mm and then torr before the subtraction is performed. The overall pressure is then given in units of atm. Solution:

( )2

310 m 1 mm 1 torr2.35 cm1 cm 1 mmHg10 m

−

−

= 23.5 torr

738.5 torr – 23.5 torr = 715.0 torr

P(atm) = ( ) 1 atm715.0 torr760 torr

= 0.940789 = 0.9408 atm

5.11 Plan: This is an open-end manometer. Since the height of the mercury column in contact with the gas is higher

than the column in contact with the air, the gas is exerting less pressure on the mercury than the air. Therefore the pressure corresponding to the height difference (Δh) between the two arms is subtracted from the atmospheric pressure. Since the height difference is in units of cm and the barometric pressure is given in units of mmHg, cm must be converted to mm before the subtraction is performed. The overall pressure is then given in units of kPa. Solution:

( )2

310 m 1 mm1.30 cm1 cm 10 m

−

−

= 13.0 mmHg

765.2 mmHg – 13.0 mmHg = 752.2 mmHg

P(kPa) = ( ) 101.325 kPa752.2 torr760 torr

= 100.285 = 100.3 kPa

5.12 Plan: This is a closed-end manometer. The difference in the height of the Hg (Δh) equals the gas pressure. The

height difference is given in units of m and must be converted to mmHg and then to atm. Solution:

P(atm) = ( ) 31 mmHg 1 atm0.734 mHg

760 mmHg10 mHg−

= 0.965789 = 0.966 atm

5.13 Plan: This is a closed-end manometer. The difference in the height of the Hg (Δh) equals the gas pressure.

The height difference is given in units of cm and must be converted to mmHg and then to Pa. Solution:

P(Pa) = ( )2 5

310 mHg 1 mmHg 1.01325x10 Pa3.56 cm1 cmHg 760 mmHg10 m

−

−

= 4746.276 = 4.75x103 Pa


1 atm = 760 mmHg = 760 torr = 1.01325x105 Pa = 14.7 psi Solution:

5-13

a) Converting from mmHg to atm: P(atm) = ( )2 1 atm2.75x10 mmHg760 mmHg

= 0.361842 = 0.362 atm

b) Converting from psi to atm: P(atm) = ( ) 1 atm86 psi14.7 psi

= 5.85034 = 5.9 atm

c) Converting from Pa to atm: P(atm) = ( )65

1 atm9.15x10 Pa1.01325x10 Pa

= 90.303 = 90.3 atm

d) Converting from torr to atm: P(atm) = ( )4 1 atm2.54x10 torr760 torr

= 33.42105 = 33.4 atm

5.15 Plan: 1 atm = 1.01325x105 Pa = 1.01325x105 N/m2. So the force on 1 m2 of ocean is 1.01325x105 N where

1 N = 1 2kg•m

s. Use F = mg to find the mass of the atmosphere in kg/m2 for part a). For part b), convert this mass

to g/cm2 and use the density of osmium to find the height of this mass of osmium. Solution:

a) F = mg 1.01325x105 N = mg

52

kg•m1.01325 x 10s

= (mass) (9.81 m/s2)

mass = 1.03287x104 = 1.03x104 kg

b) 23 2

42

kg 10 g 10 m1.03287x101 kg 1 cmm

−

= 1.03287x103 g/cm2 (unrounded)

Height = 3

32

g 1 mL 1 cm1.03287x1022.6 g 1 mLcm

= 45.702 = 45.7 cm Os

5.16 The statement is incomplete with respect to temperature and mass of sample. The correct statement is: At constant

temperature and moles of gas, the volume of gas is inversely proportional to the pressure. 5.17 a) Charles’s law: At constant pressure, the volume of a fixed amount of gas is directly proportional to its Kelvin

temperature. Variable: volume and temperature; Fixed: pressure and moles b) Avogadro’s law: At fixed temperature and pressure, the volume occupied by a gas is directly proportional to the moles of gas. Variable: volume and moles; Fixed: temperature and pressure c) Amontons’s law: At constant volume, the pressure exerted by a fixed amount of gas is directly proportional to the Kelvin temperature. Variable: pressure and temperature; Fixed: volume and moles

5.18 Plan: Examine the ideal gas law; volume and temperature are constant and pressure and moles are variable.

Solution:

PV = nRT RTP = nV

R, T, and V are constant

P = n x constant At constant temperature and volume, the pressure of the gas is directly proportional to the amount of gas in moles.

5.19 Plan: Examine the ideal gas law, noting the fixed variables and those variables that change. R is always constant

so 1 1 2 2

1 1 2 2 =

PV P Vn T n T

.

Solution:

5-14

a) P is fixed; both V and T double: 1 1 2 2

1 1 2 2 =

PV P Vn T n T

or 1 2

1 1 2 2 =

V Vn T n T

T can double as V doubles only if n is fixed.

b) T and n are both fixed and V doubles: 1 1 2 2

1 1 2 2 =

PV P Vn T n T

or P1V1 = P2V2

P and V are inversely proportional; as V doubles, P is halved. c) T is fixed and V doubles. n doubles since one mole of reactant gas produces a total of 2 moles of product gas.

1 1 2 2

1 1 2 2 =

PV P Vn T n T

or 1 1 2 2

1 2 =

PV P Vn n

V and n can both double only if P is fixed. d) P is fixed and V doubles. n is fixed since 2 moles of reactant gas produce 2 moles of product gas.

1 1 2 2

1 1 2 2 =

PV P Vn T n T

or 1 2

1 2 =

V VT T

V and T are directly proportional so as V is doubled, T is doubled.

5.20 Plan: Use the relationship 1 1 2 2

1 1 2 2 =

PV P Vn T n T

or 1 1 2 22

2 1 1 =

PV n TVP n T

.

Solution: a) As the pressure on a fixed amount of gas (n is fixed) increases at constant temperature (T is fixed), the molecules move closer together, decreasing the volume. When the pressure is tripled, the volume decreases to one-third of the original volume at constant temperature (Boyle’s law).

1 1 2 2 1 12

2 1 1 1

( )( )(1)(1) = =

(3 )(1)(1)PV n T P VVP n T P

V2 = ⅓V1

b) As the temperature of a fixed amount of gas (n is fixed) increases at constant pressure (P is fixed), the gas molecules gain kinetic energy. With higher energy, the gas molecules collide with the walls of the container with greater force, which increases the size (volume) of the container. If the temperature is increased by a factor of 3.0 (at constant pressure) then the volume will increase by a factor of 3.0 (Charles’s law).

1 1 2 2 1 12

2 1 1 1

(1)( )(1)(3 ) = =

(1)(1)( )PV n T V TVP n T T

V2 = 3V1

c) As the number of molecules of gas increases at constant pressure and temperature (P and T are fixed), the force they exert on the container increases. This results in an increase in the volume of the container. Adding 3 moles of gas to 1 mole increases the number of moles by a factor of 4, thus the volume increases by a factor of 4(Avogadro’s law).

1 1 2 2 1 12

2 1 1 1

(1)( )(4 )(1) = =

(1)( )(1)PV n T V nVP n T n

V2 = 4V1


1 2 =

PV P VT T

or 1 1 22

2 1 =

PV TVP T

. R and n are fixed.

Solution: a) As the pressure on a fixed amount of gas (n is fixed) doubles from 101 kPa to 202 kPa at constant temperature, the volume decreases by a factor of ½. As the temperature of a fixed amount of gas (n is fixed) decreases by a factor of ½ (from 310 K to 155 K) at constant pressure, the volume decreases by a factor of ½. The changes in pressure and temperature combine to decrease the volume by a factor of 4. P1 = 760 torr = 101 kPa T1 = 37°C + 273 = 310 K

1 1 2 12

2 1

(101 kPa)( )(155 K) = =

(202 kPa)(310 K)PV T VVP T

V2 = 14 V1

5-15

b) As the pressure on a fixed amount of gas (n is fixed) decreases at constant temperature (T is fixed), the molecules move farther together, increasing the volume. When the pressure is reduced by a factor of 2, the volume increases by a factor of 2 at constant temperature (Boyle’s law). T2 = 32°C + 273 = 305 K P2 = 101 kPa = 1 atm

1 1 2 12

2 1

(2 atm)( )(305 K) = =

(1 atm)(305 K)PV T VVP T

V2 = 2V1

c) As the pressure on a fixed amount of gas (n is fixed) decreases at constant temperature (T is fixed), the molecules move farther together, increasing the volume. When the pressure is reduced by a factor of 4, the volume increases by a factor of 4 at constant temperature (Boyle’s law).

1 1 2 1 12

2 1 1

( )( )(1) = =

(1/ 4 )(1)PV T P VVP T P

V2 =4V1


1 2 =

PV P VT T

or 1 1 22

2 1 =

PV TVP T

. R and n are fixed.

Solution: a) The temperature is decreased by a factor of 2, so the volume is decreased by a factor of 2 (Charles’s law).

1 1 2 12

2 1

(1)( )(400 K) = =

(1)(800 K)PV T VVP T

V2 = ½ V1

b) T1 = 250°C + 273 = 523 K T2 = 500°C + 273 = 773 K The temperature increases by a factor of 773/523 = 1.48, so the volume is increased by a factor of 1.48

(Charles’s law). 1 1 2 12

2 1

(1)( )(773 K) = =


V2 = 1.48V1

c) The pressure is increased by a factor of 3, so the volume decreases by a factor of 3 (Boyle’s law).

1 1 2 12

2 1

(2 atm)( )(1) = =

(6 atm)(1)PV T VVP T

V2 = ⅓V1


1 1 2 2 =

PV P Vn T n T

or 1 1 2 22

2 1 1 =

PV n TVP n T

.

Solution:

a) P1 = ( ) 1 atm722 torr760 torr

= 0.950 atm

T1 = 59

[T (in °F) – 32] = 59

[32°F – 32] = 0°C T1 = 0°C + 273 = 273 K

Both P and T are fixed: P1 = P2 = 0.950 atm; T1 = T2 = 273 K, so the volume remains constant. 1 1 2 2 1

22 1 1

(1)( )(1)(1) = =

(1)(1)(1)PV n T VVP n T

V2 =V1

b) Since the number of moles of gas is decreased by a factor of 2, the volume would be decreased by a factor of 2 (Avogadro’s law).

1 11 1 2 22

2 1 1 1

12

(1)( )( )(1) = =

(1)( )(1)

V nPV n TVP n T n

V2 = ½V1

c) If the pressure is decreased by a factor of 4, the volume will increase by a factor of 4 (Boyle’s law). If the temperature is decreased by a factor of 4, the volume will decrease by a factor of 4 (Charles’s law). These two effects offset one another and the volume remains constant.

1 1 11 1 2 22

2 1 1 1 1

14

14

( )( )(1)( ) = =

( )(1)( )P V TPV n TV

P n T P T V2 =V1

5-16


so 1 1 2 2

1 1 2 2 =

PV P Vn T n T

. In this problem, P and V are changing, while n and T remain fixed.

Solution: V1 = 1.61 L V2 = unknown P1 = 734 torr P2 = 0.844 atm n and T remain constant

Converting P1 from torr to atm: (734 torr)� 1 atm760 torr

� = 0.966 atm

Arranging the ideal gas law and solving for V2: 1 1 2 2

1 1 2 2 =

PV P Vn T n T

or P1V1 = P2V2

V2 = V1 𝑃1𝑃2

= (1.61 L)�0.966 atm0.844 atm

� = 1.84 L


so 1 1 2 2

1 1 2 2 =

PV P Vn T n T

. In this problem, P and V are changing, while n and T remain fixed.

Solution: V1 = 10.0 L V2 = 7.50 L P1 = 725 mmHg P2 = unknown n and T remain constant

Arranging the ideal gas law and solving for P2: 1 1 2 2

1 1 2 2 =

PV P Vn T n T

or P1V1 = P2V2

P2 = P1 𝑉1𝑉2

= (725 mmHg)�10.0 L7.50 L

� = 967 mmHg

5.26 Plan: This is Charles’s law: at constant pressure and with a fixed amount of gas, the volume of a gas is directly

proportional to the absolute temperature of the gas. The temperature must be lowered to reduce the volume of a gas. Arrange the ideal gas law, solving for T2 at fixed n and P. Temperature must be converted to kelvin.

Solution: V1 = 9.10 L V2 = 2.50 L T1 = 198°C (convert to K) T2 = unknown n and P remain constant Converting T from °C to K: T1 = 198°C + 273 = 471K Arranging the ideal gas law and solving for T2:

1 1 2 2

1 1 2 2 =

PV P Vn T n T

or 1 2

1 2 =

V VT T

22 1

1 =

VT TV

= 2.50 L471 K 9.10 L

= 129.396 K – 273 = –143.604 = –144°C

5.27 Plan: This is Charles’s law: at constant pressure and with a fixed amount of gas, the volume of a gas is directly

proportional to the absolute temperature of the gas. If temperature is reduced, the volume of gas will also be reduced. Arrange the ideal gas law, solving for V2 at fixed n and P. Temperature must be converted to kelvins.

Solution: V1 = 93 L V2 = unknown

T1 = 145°C (convert to K) T2 = –22°C n and P remain constant Converting T from °C to K: T1 = 145°C + 273 = 418 K T2 = –22°C + 273 = 251 K

5-17

Arranging the ideal gas law and solving for V2:

1 1 2 2

1 1 2 2 =

PV P Vn T n T

or 1 2

1 2 =

V VT T

22 1

1 =

TV VT

= 251 K93 L418 K

= 55.844 = 56 L


so 1 1 2 2

1 1 2 2 =

PV P Vn T n T

. In this problem, P and T are changing, while n and V remain fixed.

Solution: T1 = 25 oC T2 = 195 oC P1 = 177 atm P2 = unknown n and V remain constant Converting T1 from oC to K: 25 oC + 273.15 = 298 K Converting T2 from oC to K: 195 oC + 273.15 = 468 K


= P2V2n2T2

or P1T1

= P2T2

P2 = P1 𝑇2𝑇1

= (177 atm)�468 K298 K

� = 278 atm


so 1 1 2 2

1 1 2 2 =

PV P Vn T n T

. In this problem, P and T are changing, while n and V remain fixed.

Solution: T1 = 30.0 oC T2 = unknown P1 = 110. psi P2 = 105 psi n and V remain constant Converting T1 from oC to K: 30.0 oC + 273.15 = 303.2 K

Arranging the ideal gas law and solving for T2: P1V1n1T1

= P2V2n2T2

or P1T1

= P2T2

T2 = T1 𝑃2𝑃1

= (303.2 K)�105 psi110. psi

� = 289 K

Converting T2 from K to oC: 289 K - 273.15 = 16 oC


so 1 1 2 2

1 1 2 2 =

PV P Vn T n T


Solution: m1 = 1.92 g He m2 = 1.92 g – 0.850 g = 1.07 g He V1 = 12.5 L V2 = unknown P and T remain constant

Converting m1 (mass) to n1 (moles): (1.92 g He)� 1 mol He4.003 g He

� = 0.480 mol He = n1

Converting m2 (mass) to n2 (moles): (1.07 g He)� 1 mol He4.003 g He

� = 0.267 mol He = n2

5-18


= P2V2n2T2

or V1n1

= V2n2

V2 = V1 𝑛2𝑛1

= (12.5 L)�0.267 mol He0.480 mol He

� = 6.95 L


so 1 1 2 2

1 1 2 2 =

PV P Vn T n T


Solution: n1 = 1 x 1022 molecules of air* n2 = unknown V1 = 500 mL V2 = 350 mL P and T remain constant

*The number of molecules of any substance is directly proportional to the moles of that substance, so we can use number of molecules in place of n in this problem.

Arranging the ideal gas law and solving for n2: P1V1n1T1

= P2V2n2T2

or V1n1

= V2n2

n2 = n1 𝑉2𝑉1

= (1 x 1022 molecules of air)�350 mL500 mL

� = 7 x 1021 molecules of air

5.32 Plan: Since the volume, temperature, and pressure of the gas are changing, use the combined gas law. Arrange the ideal gas law, solving for V2 at fixed n. STP is 0°C (273 K) and 1 atm (101.325 kPa) Solution: P1 = 153.3 kPa P2 = 101.325 kPa

V1 = 25.5 L V2 = unknown T1 = 298 K T2 = 273 K n remains constant Arranging the ideal gas law and solving for V2:

1 1 2 2

1 1 2 2 =

nPV P Vn T T

or 1 1 2 2

1 2 =

PV P VT T

V2 = 2 11

1 2

T PV

T P = ( ) 273 K 153.3 kPa25.5 L

298 K 101.325 kPa

= 35.3437 = 35.3 L

5.33 Plan: Since the volume, temperature, and pressure of the gas are changing, use the combined gas law. Arrange the ideal gas law, solving for V2 at fixed n. Temperature must be converted to kelvins. Solution:

P1 = 745 torr P2 = 367 torr V1 = 3.65 L V2 = unknown T1 = 298 K T2 = –14°C + 273 = 259 K n remains constant Arranging the ideal gas law and solving for V2:

1 1 2 2

1 1 2 2 =

nPV P Vn T T

or 1 1 2 2

1 2 =

PV P VT T

V2 = 2 11

1 2

T PV

T P = ( )

torr367 torr745

K 298K 259 L 65.3 = 6.4397 = 6.44 L

5-19

5.34 Plan: Given the volume, pressure, and temperature of a gas, the number of moles of the gas can be calculated using the ideal gas law, solving for n. The gas constant, R = 0.0821 L•atm/mol•K, gives pressure in atmospheres and temperature in Kelvin. The given pressure in torr must be converted to atmospheres and the temperature converted to kelvins. Solution:

P = 328 torr (convert to atm) V = 5.0 L T = 37°C n = unknown

Converting P from torr to atm: P = ( ) 1 atm328 torr760 torr

= 0.43158 atm

Converting T from °C to K: T = 37°C + 273 = 310 K PV = nRT Solving for n:

(0.43158 atm)(5.0 L)L•atm0.0821 (310 K)mol•K

PVn =RT

=

= 0.08479 = 0.085 mol chlorine

5.35 Plan: Given the volume, moles, and temperature of a gas, the pressure of the gas can be calculated using the ideal

gas law, solving for P. The gas constant, R = 0.0821 L•atm/mol•K, gives volume in liters and temperature in Kelvin. The given volume in mL must be converted to L and the temperature converted to kelvins. Solution:

V = 75.0 mL T = 26°C n = 1.47 x 10–3 mol P = unknown

Converting V from mL to L: V = ( )310 L75.0 mL

1 mL

−

= 0.0750 L

Converting T from °C to K: T = 26°C + 273 = 299 K PV = nRT Solving for P:

P = nRTV

= ( ) ( )3 L•atm1.47x10 mol 0.0821 299 K

mol•K0.0750 L

− = 0.48114 atm

Convert P to units of torr: ( ) 760 torr0.48114 atm1 atm

= 365.6664 = 366 torr

5.36 Plan: Solve the ideal gas law for moles and convert to mass using the molar mass of ClF3.

The gas constant, R = 0.0821 L•atm/mol•K, gives volume in liters, pressure in atmospheres, and temperature in Kelvin so volume must be converted to L, pressure to atm, and temperature to K.

Solution: V = 357 mL T = 45°C

P = 699 mmHg n = unknown

Converting V from mL to L: V = ( )310 L357 mL

1 mL

−

= 0.357 L

Converting T from °C to K: T = 45°C + 273 = 318 K

Converting P from mmHg to atm: P = ( ) 1 atm699 mmHg760 mmHg

= 0.91974 atm

PV = nRT Solving for n:

( ) ( )

( )

0.91974 atm 0.357 LL•atm0.0821 318 Kmol•K

PVn =RT

=

= 0.01258 mol ClF3

5-20

Mass ClF3 = ( ) 33

3

92.45 g ClF0.01258 mol ClF

1 mol ClF

= 1.163021 = 1.16 g ClF3

5.37 Plan: Solve the ideal gas law for pressure; convert mass to moles using the molar mass of N2O.

The gas constant, R = 0.0821 L•atm/mol•K, gives temperature in Kelvin so the temperature must be converted to units of kelvins.

Solution: V = 3.1 L T = 115°C

n = 75.0 g (convert to moles) P = unknown Converting T from °C to K: T = 115°C + 273 = 388 K

Converting from mass of N2O to moles: n = ( ) 22

2

1 mol N O75.0 g N O

44.02 g N O

= 1.70377 mol N2O

PV = nRT Solving for P:

P = ( ) ( )

( )

L•atm1.70377 mol 0.0821 388 Kmol•K

3.1 LnRTV

= = 17.5075 = 18 atm N2O

5.38 Plan: Solve the ideal gas law for moles. The gas constant, R = 0.0821 L•atm/mol•K, gives pressure in

atmospheres, and temperature in Kelvin so pressure must be converted to atm and temperature to K. Solution: V = 1.5 L T = 23°C

P = 85 + 14.7 = 99.7 psi n = unknown Converting T from °C to K: T = 23°C + 273 = 296 K

Converting P from psi to atm: P = ( ) 1 atm99.7 psi14.7 psi

= 6.7823 atm


( ) ( )

( )

6.7823 atm 1.5 LL•atm0.0821 296 Kmol•K

PVn =RT

=

= 0.41863 = 0.42 mol SO2

5.39 Plan: Assuming that while rising in the atmosphere the balloon will neither gain nor lose gas molecules, the

number of moles of gas calculated at sea level will be the same as the number of moles of gas at the higher altitude (n is fixed). Volume, temperature, and pressure of the gas are changing. Arrange the ideal gas law, solving for V2 at fixed n. Given the sea-level conditions of volume, pressure, and temperature, and the temperature and pressure at the higher altitude for the gas in the balloon, we can set up an equation to solve for the volume at the higher altitude. Comparing the calculated volume to the given maximum volume of 835 L will tell us if the balloon has reached its maximum volume at this altitude. Temperature must be converted to kelvins and pressure in torr must be converted to atm for unit agreement.

Solution: P1 = 745 torr P2 = 0.066 atm V1 = 65 L V2 = unknown T1 = 25°C + 273 = 298 K T2 = –5°C + 273 = 268 K n remains constant


= 0.98026 atm


1 1 2 2 =

nPV P Vn T T

or 1 1 2 2

1 2 =

PV P VT T

5-21

V2 = 2 11

1 2

T PV

T P = ( ) 268 K 0.98026 atm65 L

298 K 0.066 atm

= 868.219 = 870 L

The calculated volume of the gas at the higher altitude is more than the maximum volume of the balloon. Yes, the balloon will reach its maximum volume. Check: Should we expect that the volume of the gas in the balloon should increase? At the higher altitude, the pressure decreases; this increases the volume of the gas. At the higher altitude, the temperature decreases, this decreases the volume of the gas. Which of these will dominate? The pressure decreases by a factor of

0.98/0.066 = 15. If we label the initial volume V1, then the resulting volume is 15V1. The temperature decreases by a factor of 298/268 = 1.1, so the resulting volume is V1/1.1 or 0.91V1. The increase in volume due to the change in pressure is greater than the decrease in volume due to change in temperature, so the volume of gas at the higher altitude should be greater than the volume at sea level.

5.40 Air is mostly N2 (28.02 g/mol), O2 (32.00 g/mol), and argon (39.95 g/mol). These “heavy” gases dominate the density of dry air. Moist air contains H2O (18.02 g/mol). The relatively light water molecules lower the density of the moist air. 5.41 The molar mass of H2 is less than the average molar mass of air (mostly N2, O2, and Ar), so air is denser. To collect a beaker of H2(g), invert the beaker so that the air will be replaced by the lighter H2. The molar mass of CO2 is greater than the average molar mass of air, so CO2(g) is more dense. Collect the CO2 holding the beaker upright, so the lighter air will be displaced out the top of the beaker. 5.42 Gases mix to form a solution and each gas in the solution behaves as if it were the only gas present. 5.43 PA = XA PT The partial pressure of a gas (PA) in a mixture is directly proportional to its mole fraction (XA). 5.44 Plan: Calculate the mole fraction of each gas; the partial pressure of each gas is directly proportional to its mole fraction so the gas with the highest mole fraction has the highest partial pressure. Use the relationship between partial pressure and mole fraction to calculate the partial pressure of gas D2. Solution:

a) AA

total =

nXn

= 4 A particles

16 total particles = 0.25 XB = B

total

nn

= 3 B particles

16 total particles = 0.1875

XC = C

total

nn

= 5 C particles

16 total particles= 0.3125

2DX = 2D

total

nn

= 24 D particles16 total particles

= = 0.25

Gas C has the highest mole fraction and thus the highest partial pressure. b) Gas B has the lowest mole fraction and thus the lowest partial pressure. c)

2 2D D total = x P X P 2D = P 0.25 x 0.75 atm = 0.1875 = 0.19 atm

5.45 Plan: Rearrange the ideal gas law to calculate the density of xenon from its molar mass at STP. Standard

temperature is 0°C (273 K) and standard pressure is 1 atm. Do not forget that the pressure at STP is exact and will not affect the significant figures.

Solution: P = 1 atm T = 273 K

M of Xe = 131.3 g/mol d = unknown PV = nRT Rearranging to solve for density:

( ) ( )

( )

1 atm 131.3 g/molL•atm0.0821 273 Kmol•K

Pd =RT

=

M = 5.8581 = 5.86 g/L

5.46 Plan: Rearrange the ideal gas law to calculate the density of CFCl3 from its molar mass. Temperature must

be converted to kelvins.

5-22

Solution: P = 1.5 atm T = 120°C + 273 = 393 K

M of CFCl3 = 137.4 g/mol d = unknown PV = nRT Rearranging to solve for density:

( ) ( )

( )

1.5 atm 137.4 g/molL•atm0.0821 393 Kmol•K

Pd =RT

=

M = 6.385807663 = 6.4 g/L

5.47 Plan: Solve the ideal gas law for moles. Convert moles to mass using the molar mass of AsH3 and divide this

mass by the volume to obtain density in g/L. Standard temperature is 0°C (273 K) and standard pressure is 1 atm. Do not forget that the pressure at STP is exact and will not affect the significant figures.

Solution: V = 0.0400 L T = 0°C + 273 = 273 K

P = 1 atm n = unknown M of AsH3 = 77.94 g/mol PV = nRT Solving for n:

=PVn =RT

( ) ( )

( )

1 atm 0.0400 LL•atm0.0821 273 Kmol•K

= 1.78465x10–3 = 1.78x10–3 mol AsH3

Converting moles of AsH3 to mass of AsH3:

Mass (g) of AsH3 = ( )3 33

3

77.94 g AsH1.78465x10 mol AsH1 mol AsH

−

= 0.1391 g AsH3

d = volumemass = ( )

( )0.1391 g0.0400L

= 3.4775 = 3.48 g/L

5.48 Plan: Solve the density form of the ideal gas law for molar mass. Temperature must be converted to kelvins.

Compare the calculated molar mass to the molar mass values of the noble gases to determine the identity of the gas. Solution: P = 3.00 atm T = 0°C + 273 = 273 K d = 2.71 g/L M = unknown

Pd =RTM

Rearranging to solve for molar mass:

( ) ( )

( )

L•atm2.71 g/L 0.0821 273 Kmol•K = =

3.00 atmdRT

P

M = 20.24668 = 20.2 g/mol

Therefore, the gas is Ne. 5.49 Plan: Rearrange the formula PV = (m/M)RT to solve for molar mass. Convert the mass

in ng to g and volume in µL to L. Temperature must be in Kelvin and pressure in atm. Solution: V = 0.206 μL T = 45°C + 273 = 318 K

P = 380 torr m = 206 ng M = unknown


= 0.510526 atm

5-23

Converting V from μL to L: V = ( )610 L0.206 L

1 L

− µ µ

= 2.06x10–7 L

Converting m from ng to g: m = ( )910 g206 ng

1 ng

−

= 2.06x10–7 g

=

mPV RTM

Solving for molar mass, M:

( ) ( )

( ) ( )

7

7

L•atm2.06x10 g 0.0821 318 Kmol•K =

0.510526 atm 2.06x10 LmRTPV

−

−

=M = 51.1390 = 51.1 g/mol

5.50 Plan: Rearrange the formula PV = (m/M)RT to solve for molar mass. Compare the calculated molar mass

to that of N2, Ne, and Ar to determine the identity of the gas. Convert volume to liters, pressure to atm, and temperature to Kelvin. Solution:

V = 63.8 mL T = 22°C + 273 = 295 K P = 747 mm Hg m = 0.103 g M = unknown


= 0.982895 atm


1 mL

−

= 0.0638 L

=

mPV RTM


( ) ( )

( ) ( )

L•atm0.103 g 0.0821 295 Kmol•K =

0.982895 atm 0.0638 LmRTPV

=M = 39.7809 = 39.8 g/mol

The molar masses are N2 = 28 g/mol, Ne = 20 g/mol, and Ar = 40 g/mol. Therefore, the gas is Ar. 5.51 Plan: Use the ideal gas law to determine the number of moles of Ar and of O2. The gases are combined

(n total = nAr + n 2O ) into a 400 mL flask (V) at 27°C (T). Use the ideal gas law again to determine the total pressure from n total, V, and T. Pressure must be in units of atm, volume in units of L and temperature in K.

Solution: For Ar: V = 0.600 L T = 227°C + 273 = 500. K

P = 1.20 atm n = unknown PV = nRT Solving for n:

( ) ( )

( )

1.20 atm 0.600 LL•atm0.0821 500. Kmol•K

PVn =RT

=

= 0.017539586 mol Ar

For O2: V = 0.200 L T = 127°C + 273 = 400. K

P = 501 torr n = unknown

5-24


= 0.6592105 atm


( ) ( )

( )

0.6592105 atm 0.200 LL•atm0.0821 400. Kmol•K

PVn =RT

=

= 0.004014680 mol O2

n total = nAr + n 2O = 0.017539586 mol + 0.004014680 mol = 0.021554266 mol For the mixture of Ar and O2:

V = 400 mL T = 27°C + 273 = 300. K P = unknownn n = 0.021554265 mol


1 mL

−

= 0.400 L


Pmixture = ( ) ( )

( )

L•atm0.021554266 mol 0.0821 300 Kmol•K

0.400 LnRTV

= = 1.32720 = 1.33 atm

5.52 Plan: Use the ideal gas law, solving for n to find the total moles of gas. Convert the mass of Ne to moles and subtract moles of Ne from the total number of moles to find moles of Ar. Volume must be in units of liters, pressure in units of atm, and temperature in kelvins. Solution: V = 355 mL T = 35°C + 273 = 308 K

P = 626 mmHg n total = unknown


= 0.823684 atm


1 mL

−

= 0.355 L

PV = nRT Solving for n total:

( ) ( )

( )total

0.823684 atm 0.355 LL•atm0.0821 308 Kmol•K

PVn =RT

=

= 0.011563655 mol Ne + mol Ar

Moles Ne = ( ) 1 mol Ne0.146 g Ne20.18 g Ne

= 0.007234886 mol Ne

Moles Ar = n total – nNe = (0.011563655 – 0.007234886) mol = 0.004328769 = 0.0043 mol Ar 5.53 Plan: Use the ideal gas law, solving for n to find the moles of O2. Use the molar ratio from the balanced equation to determine the moles (and then mass) of phosphorus that will react with the oxygen. Standard temperature is 0°C (273 K) and standard pressure is 1 atm. Solution: V = 35.5 L T = 0°C + 273 = 273 K

P = 1 atm n = unknown PV = nRT

5-25

Solving for n: ( ) ( )

( )

1 atm 35.5 LL•atm0.0821 273 Kmol•K

PVn =RT

=

= 1.583881 mol O2

P4(s) + 5O2(g) → P4O10(s)

Mass P4 = ( ) 4 42

2 4

1 mol P 123.88 g P1.583881 mol O

5 mol O 1 mol P

= 39.24224 = 39.2 g P4

5.54 Plan: Use the ideal gas law, solving for n to find the moles of O2 produced. Volume must be in units of liters, pressure in atm, and temperature in kelvins. Use the molar ratio from the balanced equation to determine the moles (and then mass) of potassium chlorate that reacts.

Solution: V = 638 mL T = 128°C + 273 = 401 K

P = 752 torr n = unknown


= 0.9894737 atm


1 mL

−

= 0.638 L


( ) ( )

( )

0.9894737 atm 638 LL•atm0.0821 401 Kmol•K

PVn =RT

=

= 0.0191751 mol O2

2KClO3(s) → 2KCl(s) + 3O2(g)

Mass (g) of KClO3 = ( ) 3 32

2 3

2 mol KClO 122.55 g KClO0.0191751 mol O

3 mol O 1 mol KClO

= 1.5666 = 1.57 g KClO3

5.55 Plan: Since the amounts of two reactants are given, this is a limiting reactant problem. To find the mass of PH3,

write the balanced equation and use molar ratios to find the number of moles of PH3 produced by each reactant. The smaller number of moles of product indicates the limiting reagent. Solve for moles of H2 using the ideal gas law.

Solution: Moles of hydrogen: V = 83.0 L T = 0°C + 273 = 273 K

P = 1 atm n = unknown PV = nRT Solving for n:

( ) ( )

( )

1 atm 83.0 LL•atm0.0821 273 Kmol•K

PVn =RT

=

= 3.7031584 mol H2

P4(s) + 6H2(g) → 4PH3(g)

PH3 from P4 = ( ) 344

4 4

4 mol PH1 mol P37.5 g P

123.88 g P 1 mol P

= 1.21085 mol PH3

PH3 from H2 = ( ) 32

2

4 mol PH3.7031584 mol H

6 mol H

= 2.4687723 mol PH3

P4 is the limiting reactant because it forms less PH3.

5-26

Mass PH3 = ( ) 3 344

4 4 3

4 mol PH 33.99 g PH1 mol P37.5 g P

123.88 g P 1 mol P 1 mol PH

= 41.15676 = 41.2 g PH3

5.56 Plan: Since the amounts of two reactants are given, this is a limiting reactant problem. To find the mass of NO,

write the balanced equation and use molar ratios to find the number of moles of NO produced by each reactant. Since the moles of gas are directly proportional to the volumes of the gases at the same temperature and pressure,

the limiting reactant may be found by comparing the volumes of the gases. The smaller volume of product indicates the limiting reagent. Then use the ideal gas law to convert the volume of NO produced to moles and then to mass. Solution: 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l)

Mol NO from NH3 = ( )33

4 L NO35.6 L NH4 L NH

= 35.6 L NO

Mol NO from O2 = ( )22

4 L NO40.5 L O5 L O

= 32.4 L NO

O2 is the limiting reactant since it forms less NO. Converting volume of NO to moles and then mass:

V = 32.4 L T = 0°C + 273 = 273 K P = 1 atm n = unknown PV = nRT Solving for n:

( ) ( )

( )

1 atm 32.4 LL•atm0.0821 273 Kmol•K

PVn =RT

=

= 1.44557 mol NO

Mass (g) of NO = ( ) 30.01 g NO1.44557 mol NO1 mol NO

= 43.38156 = 43.4 g NO

5.57 Plan: First, write the balanced equation. The moles of hydrogen produced can be calculated from the ideal gas

law. The problem specifies that the hydrogen gas is collected over water, so the partial pressure of water vapor must be subtracted from the overall pressure given. Table 5.2 reports pressure at 26°C (25.2 torr) and 28°C (28.3 torr), so take the average of the two values to obtain the partial pressure of water at 27oC. Volume must be in units of liters, pressure in atm, and temperature in kelvins. Once the moles of hydrogen produced are known, the molar ratio from the balanced equation is used to determine the moles of aluminum that reacted.

Solution: V = 35.8 mL T = 27°C + 273 = 300 K

P total = 751 mmHg n = unknown Pwater vapor = (28.3 + 25.2) torr/2 = 26.75 torr = 26.75 mmHg Phydrogen = P total – Pwater vapor = 751 mmHg – 26.75 mmHg = 724.25 mmHg

Converting P from mmHg to atm: P = ( ) 1 atm724.25 mmHg760 mmHg

= 0.952960526 atm


1 mL

−

= 0.0358 L


( ) ( )

( )

0.952960526 atm 0.0358 LL•atm0.0821 300. Kmol•K

PVn =RT

=

= 0.0013851395 mol H2

5-27

2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g)

Mass (g) of Al = ( )22

2 mol Al 26.98 g Al0.0013851395 mol H3 mol H 1 mol Al

= 0.024914 = 0.0249 g Al

5.58 Plan: First, write the balanced equation. Convert mass of lithium to moles and use the molar ratio from the

balanced equation to find the moles of hydrogen gas produced. Use the ideal gas law to find the volume of that amount of hydrogen. The problem specifies that the hydrogen gas is collected over water, so the partial pressure of water vapor must be subtracted from the overall pressure given. Table 5.2 reports the vapor pressure of water at 18°C (15.5 torr). Pressure must be in units of atm and temperature in kelvins. Solution:

2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g)

Moles H2 = ( ) 21 mol H1 mol Li0.84 g Li6.941 g Li 2 mol Li

= 0.0605100 mol H2

Finding the volume of H2: V = unknown T = 18°C + 273 = 291 K

P total = 725 mmHg n = 0.0605100 mol Pwater vapor = 15.5 torr = 15.5 mmHg Phydrogen = P total – Pwater vapor = 725 mmHg – 15.5 mmHg = 709.5 mmHg


= 0.933552631 atm

PV = nRT Solving for V:

V = ( ) ( )

( )

L•atm0.0605100 mol 0.0821 291 Kmol•K =

0.933552631 atmnRT

P

= 1.5485 = 1.5 L H2

5.59 Plan: Rearrange the ideal gas law to calculate the density of the air from its molar mass. Temperature must

be converted to kelvins and pressure to atmospheres. Solution: P = 744 torr T = 17°C + 273 = 290 K or T = 60°C + 273 = 333 K

M of air = 28.8 g/mol d = unknown


= 0.978947368 atm

PV = nRT Rearranging to solve for density: At 17°C

( ) ( )

( )


Pd =RT

=

M = 1.18416 = 1.18 g/L

At 60.0°C ( ) ( )0.978947368 atm 28.8 g/mol

L•atm0.0821 (333 K)mol•K

Pd =RT

=

M = 1.03125 = 1.03 g/L

5.60 Plan: Solve the ideal gas law for molar volume, n/V. Pressure must be converted to atm and temperature to K. Solution:

P = 650. torr T = –25°C + 273 = 248 K n/V = unknown

5-28

Converting P from torr to atm: P = ( ) 1 atm650. torr760 torr

= 0.855263157 atm

PV = nRT Solving for n/V:

( )

( )

0.855263157 atmL•atm0.0821 248 Kmol•K

n P=V RT

=

= 0.042005 = 0.0420 mol/L

5.61 Plan: The problem gives the mass, volume, temperature, and pressure of a gas; rearrange the formula

PV = (m/M)RT to solve for the molar mass of the gas. Temperature must be in Kelvin and pressure in atm. The problem also states that the gas is a hydrocarbon, which by, definition, contains only carbon and hydrogen atoms. We are also told that each molecule of the gas contains five carbon atoms so we can use this information and the calculated molar mass to find out how many hydrogen atoms are present and the formula of the compound. Solution:

V = 0.204 L T = 101°C + 273 = 374 K P = 767 torr m = 0.482 g M = unknown


= 1.009210526 atm

=

mPV RTM


( ) ( )

( ) ( )

L•atm0.482 g 0.0821 374 Kmol•K =

1.009210526 atm 0.204 LmRTPV

=M = 71.8869 g/mol (unrounded)

The mass of the five carbon atoms accounts for [5(12 g/mol)] = 60 g/mol; thus, the hydrogen atoms must make up the difference (72 – 60) = 12 g/mol. A value of 12 g/mol corresponds to 12 H atoms. (Since fractional atoms are not possible, rounding is acceptable.) Therefore, the molecular formula is C5H12.

5.62 Plan: Solve the ideal gas law for moles of air. Temperature must be in units of kelvins. Use Avogadro’s

number to convert moles of air to molecules of air. The percent composition can be used to find the number of molecules (or atoms) of each gas in that total number of molecules. Solution: V = 1.00 L T = 25°C + 273 = 298 K P = 1.00 atm n = unknown PV = nRT Solving for n:

Moles of air = =PVn =RT

( ) ( )

( )

1.00 atm 1.00 LL•atm0.0821 298 Kmol•K

= 0.040873382 mol

Converting moles of air to molecules of air:

Molecules of air = ( )236.022x10 molecules0.040873382 mol

1 mol

= 2.461395x1022 molecules

Molecules of N2 = ( )22 2 78.08% N molecules2.461395x10 air molecules100% air

= 1.921857x1022 = 1.92x1022 molecules N2

Molecules of O2 = ( )22 2 20.94% O molecules2.461395x10 air molecules100% air

= 5.154161x1021 = 5.15x1021 molecules O2

5-29

Molecules of CO2 = ( )22 2 0.05% CO molecules2.461395x10 air molecules100% air

= 1.2306975x1019 = 1x1019 molecules CO2

Molecules of Ar = ( )22 0.93% Ar molecules2.461395x10 air molecules100% air

= 2.289097x1020 = 2.3x1020 molecules Ar 5.63 Plan: Since you have the pressure, volume, and temperature, use the ideal gas law to solve for n, the total

moles of gas. Pressure must be in units of atmospheres and temperature in units of kelvins. The partial pressure of SO2 can be found by multiplying the total pressure by the volume fraction of SO2. Solution: a) V = 21 L T = 45°C + 273 = 318 K P = 850 torr n = unknown


= 1.118421053 atm

PV = nRT

Moles of gas = ( ) ( )

( )

1.118421053 atm 21 LL•atm0.0821 318 Kmol•K

PVn =RT

=

= 0.89961 = 0.90 mol gas

b) The equation 2 2SO SO total = x P X P can be used to find partial pressure. The information given in ppm is a way

of expressing the proportion, or fraction, of SO2 present in the mixture. Since n is directly proportional to V, the volume fraction can be used in place of the mole fraction,

2SO X . There are 7.95x103 parts SO2 in a million parts of mixture, so volume fraction = (7.95x103/1x106) = 7.95x10–3.

2D total = volume fraction x P P = (7.95x10–3) (850. torr) = 6.7575 = 6.76 torr

5.64 Plan: First, write the balanced equation. Convert mass of P4S3 to moles and use the molar ratio from the balanced

equation to find the moles of SO2 gas produced. Use the ideal gas law to find the volume of that amount of SO2. Pressure must be in units of atm and temperature in kelvins. Solution:

P4S3(s) + 8O2(g) → P4O10(s) + 3SO2(g)

Moles SO2 = ( ) 4 3 24 3

4 3 4 3

1 mol P S 3 mol SO0.800 g P S

220.06 g P S 1 mol P S

= 0.010906 mol SO2

Finding the volume of SO2: V = unknown T = 32°C + 273 = 305 K

P = 725 torr n = 0.010905 mol


= 0.953947368 atm


V = ( ) ( )

( )

L•atm0.010906 mol 0.0821 305 Kmol•K =

0.953947368 atmnRT

P

= 0.28627543 L

Converting V from L to mL:

V = ( ) 31 mL0.28627543 L

10 L−

= 286.275 = 286 mL SO2

5.65 Plan: The moles of Freon-12 produced can be calculated from the ideal gas law. Volume must be in units of L,

pressure in atm, and temperature in kelvins. Then, write the balanced equation. Once the moles of Freon-12

5-30

produced is known, the molar ratio from the balanced equation is used to determine the moles and then grams of CCl4 that reacted.

Solution: V = 16.0 dm3 T = 27°C + 273 = 300 K

P total = 1.20 atm n = unknown

Converting V from dm3 to L: V = ( )33

1 L16.0 dm1 dm

= 16.0 L


Moles of Freon-12 = ( ) ( )

( )

1.20 atm 16.0 LL•atm0.0821 300. Kmol•K

PVn =RT

=

= 0.779537149 mol Freon-12

CCl4(g) + 2HF(g) → CF2Cl2(g) + 2HCl(g)

Mass of Freon-12 (CF2Cl2) = ( ) 4 42 2

2 2 4

1 mol CCl 153.81 g CCl0.779537149 mol CF Cl

1 mol CF Cl 1 mol CCl

= 119.9006 = 1.20x102 g CCl4 5.66 Plan: First, write the balanced equation. Given the amount of xenon hexafluoride that reacts, we can find the

number of moles of silicon tetrafluoride gas formed by using the molar ratio in the balanced equation. Then, using the ideal gas law with the moles of gas, the temperature, and the volume, we can calculate the pressure of the silicon tetrafluoride gas. Temperature must be in units of kelvins. Solution:

2XeF6(s) + SiO2(s) → 2XeOF4(l) + SiF4(g)

Moles SiF4 = n = ( ) 6 46

6 6

1 mol XeF 1 mol SiF2.00 g XeF

245.3 g XeF 2 mol XeF

= 0.0040766 mol SiF4

Finding the pressure of SiF4: V = 1.00 L T = 25°C + 273 = 298 K

P = unknown n = 0.0040766 mol PV = nRT Solving for P:

Pressure SiF4 = P = nRTV

= ( ) ( )4

L•atm0.0040766 mol SiF 0.0821 298 Kmol•K

1.00 L

= 0.099737 = 0.0997 atm SiF4

5.67 Plan: Use the ideal gas law with T and P constant; then volume is directly proportional to moles.

Solution: PV = nRT. At constant T and P, V α n. Since the volume of the products has been decreased to ½ the original volume, the moles (and molecules) must have been decreased by a factor of ½ as well. Cylinder A best represents the products as there are 2 product molecules (there were 4 reactant molecules).

5.68 Plan: Write the balanced equation. Since the amounts of 2 reactants are given, this is a limiting reactant problem.

To find the volume of SO2, use the molar ratios from the balanced equation to find the number of moles of SO2 produced by each reactant. The smaller number of moles of product indicates the limiting reagent. Solve for moles of SO2 using the ideal gas law.

Solution: Moles of oxygen: V = 228 L T = 220°C + 273 = 493 K

P = 2 atm n = unknown PV = nRT Solving for n:

5-31

Moles of O2 =( ) ( )

( )

2 atm 228 LL•atm0.0821 493 Kmol•K

PVn =RT

=

= 11.266 mol O2

2PbS(s) + 3O2(g) → 2PbO(g) + 2SO2(g)

Moles SO2 from O2 = ( ) 22

2

2 mol SO11.266 mol O

3 mol O

= 7.5107mol SO2

Moles SO2 from PbS = ( )3

22 mol SO10 g 1 mol PbS3.75 kg PbS1 kg 239.3 g PbS 2 mol PbS

= 15.6707 mol SO2 (unrounded)

O2 is the limiting reagent because it forms less SO2. Finding the volume of SO2: V = unknown T = 0°C + 273 = 273 K

P total = 1 atm n = 7.5107 mol PV = nRT

Solving for V:

V = ( ) ( )

( )

L•atm7.5107 mol 0.0821 273 Kmol • K =

1 atmnRT

P

= 168.34 = 1.7x102 L SO2

5.69 Plan: First, write the balanced equation. Given the amount of xenon HgO that reacts (20.0% of the given amount),

we can find the number of moles of oxygen gas formed by using the molar ratio in the balanced equation. Then, using the ideal gas law with the moles of gas, the temperature, and the volume, we can calculate the pressure of the silicon tetrafluoride gas. Temperature must be in units of kelvins. Solution: 2HgO(s) → 2Hg(l) + O2(g)

Mole O2 = n = ( ) 21 mol O20.0% 1 mol HgO40.0 g HgO100% 216.6 g HgO 2 mol HgO

= 0.01846722 mol O2

Finding the pressure of O2: V = 502 mL T = 25°C + 273 = 298 K

P = unknown n = 0.01846722 mol


1 mL

−

= 0.502 L


Pressure O2 = P = nRTV

= ( ) ( )2

L•atm0.01846722 mol O 0.0821 298 Kmol•K

0.502 L

= 0.9000305 = 0.900 atm O2

5.70 As the temperature of the gas sample increases, the most probable speed increases. This will increase both the number of collisions per unit time and the force of each collision with the sample walls. Thus, the gas pressure increases. 5.71 At STP (or any identical temperature and pressure), the volume occupied by a mole of any gas will be identical.

One mole of krypton has the same number of particles as one mole of helium and, at the same temperature, all of the gas particles have the same average kinetic energy, resulting in the same pressure and volume.

5.72 The rate of effusion is much higher for a gas than its rate of diffusion. Effusion occurs into an evacuated space,

whereas diffusion occurs into another gas. It is reasonable to expect that a gas will escape faster into a vacuum than it will into a space already occupied by another gas. The ratio of the rates of effusion and diffusion for two gases will be the same since both are inversely proportional to the square root of their molar masses.

5.73 a) PV = nRT Since the pressure, volume, and temperature of the two gases are identical, n must be the same

5-32

for the two gases. Since the molar mass of O2 (32.0 g/mol) is greater than the molar mass of H2 (2.02 g/mol), a given number of moles of O2 has a greater mass than the same number of moles of H2. mass O2 > mass H2

b) Pd =RTM

The pressure and temperature are identical and density is directly proportional to molar

mass M. Since the molar mass of O2 (32.0 g/mol) is greater than the molar mass of H2 (2.02 g/mol), the density of O2 is greater than that of H2. O H2 2

>d d

c) The mean free path is dependent on pressure. Since the two gases have the same pressure, their mean free paths are identical.

d) Kinetic energy is directly proportional to temperature. Since the two gases have the same temperature, their average moelcular kinetic energies are identical. e) Kinetic energy = ½mass x speed2. O2 and H2 have the same average kinetic energy at the same temperature and mass and speed are inversely proportional. The lighter H2 molecules have a higher speed than the heavier O2 molecules. average speed H2 > average speed O2

f) Rate of effusion 1 ∝M

H2 molecules with the lower molar mass have a faster effusion time than O2

molecules with a larger molar mass. effusion time H2 < effusion time O2 5.74 Plan: The molar masses of the three gases are 2.016 for H2 (Flask A), 4.003 for He (Flask B), and 16.04 for CH4

(Flask C). Since hydrogen has the smallest molar mass of the three gases, 4 g of H2 will contain more gas molecules (about 2 mole’s worth) than 4 g of He or 4 g of CH4. Since helium has a smaller molar mass than methane, 4 g of He will contain more gas molecules (about 1 mole’s worth) than 4 g of CH4 (about 0.25 mole’s worth). Solution: a) PA > PB > PC The pressure of a gas is proportional to the number of gas molecules (PV = nRT). So, the gas sample with more gas molecules will have a greater pressure.

b) EA = EB = EC Average kinetic energy depends only on temperature. The temperature of each gas sample is 273 K, so they all have the same average kinetic energy.

c) rateA > rateB > rateC When comparing the speed of two gas molecules, the one with the lower mass travels faster.

d) total EA > total EB > total EC Since the average kinetic energy for each gas is the same (part b) of this problem), the total kinetic energy would equal the average times the number of molecules. Since the hydrogen flask contains the most molecules, its total kinetic energy will be the greatest.

e) dA = dB = dC Under the conditions stated in this problem, each sample has the same volume, 5 L, and the same mass, 4 g. Thus, the density of each is 4 g/5 L = 0.8 g/L. f) Collision frequency (A) > collision frequency (B) > collision frequency (C) The number of collisions depends on both the speed and the distance between gas molecules. Since hydrogen is the lightest molecule it has the greatest speed and the 5 L flask of hydrogen also contains the most molecules, so collisions will occur more frequently between hydrogen molecules than between helium molecules. By the same reasoning, collisions will occur more frequently between helium molecules than between methane molecules. 5.75 Plan: To find the ratio of effusion rates, calculate the inverse of the ratio of the square roots of the molar masses

(Graham’s law). Solution:

2

6

Rate HRate UF

= 6

2

molar mass UFmolar mass H

= 352.0 g/mol2.016 g/mol

= 13.2137 = 13.21

5.76 Plan: To find the ratio of effusion rates, calculate the inverse of the ratio of the square roots of the molar masses

(Graham’s law). Solution:

2Rate ORate Kr

= 2

molar mass Krmolar mass O

= 83.80 g/mol32.00 g/mol

= 1.618255 = 1.618

5.77 Plan: Recall that the heavier the gas, the slower the molecular speed. The molar mass of Ar is 39.95 g/mol while

5-33

the molar mass of He is 4.003 g/mol. Solution: a) The gases have the same average kinetic energy because they are at the same temperature. The heavier Ar atoms are moving more slowly than the lighter He atoms to maintain the same average kinetic energy. Therefore, Curve 1 with the lower average molecular speed, better represents the behavior of Ar.

b) A gas that has a slower molecular speed would effuse more slowly, so Curve 1 is the better choice. c) Fluorine gas exists as a diatomic molecule, F2, with M = 38.00 g/mol. Therefore, F2 is much closer in mass to Ar (39.95 g/mol) than He (4.003 g/mol), so Curve 1 more closely represents the behavior of F2. 5.78 Plan: Recall that the lower the temperature, the lower the average kinetic energy and the slower the molecular

speed. Solution: a) At the lower temperature, the average molecular speed is lower so Curve 1 represents the gas at the lower temperature.

b) When a gas has a higher kinetic energy, the molecules have a higher molecular speed. Curve 2 with the larger average molecular speed represents the gas when it has a higher kinetic energy.

c) If a gas has a higher diffusion rate, then the gas molecules are moving with a higher molecular speed as in Curve 2.

5.79 Plan: To find the ratio of effusion rates, calculate the inverse of the ratio of the square roots of the molar masses (Graham’s law). Then use the ratio of effusion rates to find the time for the F2 effusion. Effusion rate and time required for the effusion are inversely proportional.

Solution: M of He = 4.003 g/mol M of F2 = 38.00 g/mol

2

Rate HeRate F

= 2molar mass Fmolar mass He

= 38.00 g/mol4.003 g/mol

= 3.08105 (unrounded)

2

Rate HeRate F

= 2time Ftime He

23.08105 time F = 1.00 4.85 min He

Time F2 = 14.9431 = 14.9 min

5.80 Plan: Effusion rate and time required for the effusion are inversely proportional. Therefore, time of effusion for

a gas is directly proportional to the square root of its molar mass. The ratio of effusion times and the molar mass of H2 are used to find the molar mass of the unknown gas. Solution: M of H2 = 2.016 g/mol Time of effusion of H2 = 2.42 min Time of effusion of unknown = 11.1

min

2rate Hrate unknown

= 2

time unknowntime H

= 2

molar mass unknownmolar mass H

11.1 min2.42 min

= molar mass unknown2.016 g/mol

4.586777 = molar mass unknown2.016 g/mol

21.03852196 = molar mass unknown2.016 g/mol

Molar mass unknown = 42.41366 = 42.4 g/mol 5.81 Plan: White phosphorus is a molecular form of the element phosphorus consisting of some number, x, of

phosphorus atoms; the number of atoms in a molecule determines the molar mass of the phosphorus molecule. Use the relative rates of effusion of white phosphorus and neon (Graham’s law) to determine the molar mass of white phosphorus. From the molar mass of white phosphorus, determine the number of phosphorus atoms, x, in one molecule of white phosphorus.

5-34

Solution: M of Ne = 20.18 g/mol

xRate PRate Ne

= 0.404 = x

molar mass Nemolar mass P

0.404 = x

20.18 g/molmolar mass P

(0.404)2 = x


0.163216 = x


Molar mass Px = 123.6398 g/mol

x

123.6398 g 1 mol Pmol P 30.97 g P

= 3.992244 = 4 mol P/mol Px or 4 atoms P/molecule Px

Thus, 4 atoms per molecule, so Px = P4. 5.82 Plan: Use the equation for root mean speed (u rms) to find this value for He at 0.°C and 30.°C and for Xe at 30.°C. The calculated root mean speed is then used in the kinetic energy equation to find the average kinetic energy for the two gases at 30.°C. Molar mass values must be in units of kg/mol and temperature in kelvins. Solution:

a) 0°C = 273 K 30°C + 273 = 303 K M of He = 3

4.003 g He 1 kgmol 10 g

= 0.004003 kg/mol

R = 8.314 J/mol•K 1 J = kg•m2/s2

u rms = 3RTM

u rmsHe (at 0°C) = ( ) 2 2

J3 8.314 273 Kkg•m /smol•K

0.004003 kg/mol J

= 1.3042x103 = 1.30x103 m/s

u rmsHe (at 30°C) = ( ) 2 2

J3 8.314 303 Kkg•m /smol•K

0.004003 kg/mol J

= 1.3740x103 = 1.37x103 m /s

b) 30°C + 273 = 303 K M of Xe = 3

131.3 g Xe 1 kgmol 10 g

= 0.1313 kg/mol

R = 8.314 J/mol•K 1 J = kg•m2/s2

u rms = 3RTM

u rmsXe (at 30°C) = ( ) 2 2

J3 8.314 303 Kkg•m /smol•K

0.1313 kg/mol J

= 239.913 m/s (unrounded)

Rate He/Rate Xe = (1.3740x103 m/s)/(239.913 m/s) = 5.727076 = 5.73 He molecules travel at almost 6 times the speed of Xe molecules. c) 2

k12

= E mu

EHe = ( ) ( ) ( )2 2 231

20.004003 kg/mol 1.3740x10 m/s 1 J/kg•m /s = 3778.58 = 3.78x103 J/mol

EXe = ( ) ( ) ( )12

2 2 2 0.1313 kg/mol 239.913 m/s 1 J/kg•m /s = 3778.70 = 3.78x103 J/mol

5-35

d) 233778.58 J 1 mol

mol 6.022x10 atoms

= 6.2746x10–21 = 6.27x10–21 J/He atom

5.83 Plan: Use Graham’s law: the rate of effusion of a gas is inversely proportional to the square root of the molar

mass. When comparing the speed of gas molecules, the one with the lowest mass travels the fastest. Solution:

a) M of S2F2 = 102.12 g/mol; M of N2F4 = 104.02 g/mol; M of SF4 = 108.06 g/mol SF4 has the largest molar mass and S2F2 has the smallest molar mass:

4SFrate < 2 4N Frate <

2 2S Frate

b) 2 2

2 4

S F

N F

RateRate

= 2 4

2 2

molar mass N F 104.02 g/mol = molar mass S F 102.12 g/mol

= 1.009260 = 1.0093:1

c) 4

Rate XRate SF

= 0.935 = 4molar mass SFmolar mass X

0.935 = 108.06 g/molmolar mass X

(0.935)2 = 108.06 g/molmolar mass X

0.874225 = 108.06 g/molmolar mass X

Molar mass X = 123.60662 = 124 g/mol 5.84 Interparticle attractions cause the real pressure to be less than ideal pressure, so it causes a negative deviation. The size of the interparticle attraction is related to the constant a. According to Table 5.4, N2

a = 1.39,

Kra = 2.32, and CO2a = 3.59. Therefore, CO2 experiences a greater negative deviation in pressure than the other

two gases: N2 < Kr < CO2. 5.85 Particle volume causes a positive deviation from ideal behavior. Thus, VReal Gases > VIdeal Gases. The

particle volume is related to the constant b. According to Table 5.4, H2b = 0.0266, O2

b = 0.0318, and

Cl2b = 0.0562. Therefore, the order is H2 < O2 < Cl2.

5.86 Nitrogen gas behaves more ideally at 1 atm than at 500 atm because at lower pressures the gas molecules are farther apart. An ideal gas is defined as consisting of gas molecules that act independently of the other gas molecules. When gas molecules are far apart they act more ideally, because intermolecular attractions are less important and the volume of the molecules is a smaller fraction of the container volume. 5.87 SF6 behaves more ideally at 150°C. At higher temperatures, intermolecular attractions become less important and

the volume occupied by the molecules becomes less important. 5.88 Plan: To find the total force, the total surface area of the can is needed. Use the dimensions of the can to find the

surface area of each side of the can. Do not forget to multiply the area of each side by two. The surface area of the can in cm2 must be converted to units of in2. Solution: Surface area of can = 2(40.0 cm)(15.0 cm) + 2(40.0 cm)(12.5 cm) + 2(15.0 cm)(12.5 cm) = 2.575x103 cm2

Total force = ( )2

3 22

1 in 14.7 lb2.575x10 cm2.54 cm 1 in

= 5.8671x103 = 5.87x103 lbs

5-36

5.89 Plan: Use the ideal gas law to find the number of moles of O2. Moles of O2 is divided by 4 to find moles of Hb since O2 combines with Hb in a 4:1 ratio. Divide the given mass of Hb by the number of moles of Hb to obtain molar mass, g/mol. Temperature must be in units of kelvins, pressure in atm, and volume in L. Solution:

V = 1.53 mL T = 37°C + 273 = 310 K P = 743 torr n = unknown


1 mL

−

= 1.53x10–3 L


= 0.977631578 atm


Moles of O2 = ( ) ( )

( )

30.977631578 atm 1.53x10 L

L•atm0.0821 310 Kmol•K

PVn =RT

−

=

= 5.87708x10–5 mol O2

Moles Hb = ( )52

2

1 mol Hb5.87708x10 mol O4 mol O

−

= 1.46927x10–5 mol Hb (unrounded)

Molar mass hemoglobin = 51.00 g Hb

1.46927x10 Hb− = 6.806098x104 = 6.81x104 g/mol

5.90 Plan: First, write the balanced equations. Convert mass of NaHCO3 to moles and use the molar ratio from each

balanced equation to find the moles of CO2 gas produced. Use the ideal gas law to find the volume of that amount of CO2. Temperature must be in kelvins. Solution: Reaction 1: 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)

Moles CO2 = ( ) 3 23

3 3

1 mol NaHCO 1 mol CO1.00 g NaHCO

84.01 g NaHCO 2 mol NaHCO

= 5.95167x10–3 mol CO2

Finding the volume of CO2: V = unknown T = 200.°C + 273 = 473 K

P = 0.975 atm n = 5.95167x10–3 mol PV = nRT

Solving for V:

Volume of CO2 = V = ( ) ( )

( )

3 L•atm5.95167x10 mol 0.0821 473 Kmol•K =

0.975 atmnRT

P

− = 0.237049 L


V = ( ) 31 mL0.237049 L

10 L−

= 237.049 = 237 mL CO2 in Reaction 1

Reaction 2: NaHCO3(s) + H+(aq) → H2O(l) + CO2(g) + Na+(aq)

Moles CO2 = ( ) 3 23

3 3

1 mol NaHCO 1 mol CO1.00 g NaHCO

84.01 g NaHCO 1 mol NaHCO

= 1.1903x10–2 mol CO2

Finding the volume of CO2: V = unknown T = 200.°C + 273 = 473 K

P = 0.975 atm n = 1.1903x10–2 mol PV = nRT

5-37

Solving for V:

Volume of CO2 = V = ( ) ( )

( )

2 L•atm1.1903x10 mol 0.0821 473 Kmol•K =

0.975 atmnRT

P

− = 0.4740986 L


V = ( ) 31 mL0.4740986 L

10 L−

= 474.0986 = 474 mL CO2 in Reaction 2


1 1 2 2 =

PV P Vn T n T

or 1 1 2 22

2 1 1 =

PV n TVP n T

; P is fixed while V and T change. n2 is 0.75n1 since

one-fourth of the gas leaks out. Only the initial and final conditions need to be considered. Solution: P1 = 1.01 atm P2 = 1.01 atm (Thus, P has no effect, and does not need to be included.) T1 = 305 K T2 = 250 K n1 = n1 n2 = 0.75n1 V1 = 600. L V2 = ?

1 2 22

1 1 =

V n TVn T

= ( )( )( )( )( )

1

1

600. L 0.75 250 K305 K

nn

= 368.852 = 369 L

5.92 Plan: Convert the mass of Cl2 to moles and use the ideal gas law and van der Waals equation to

find the pressure of the gas. Solution:

a) Moles Cl2: ( )3

22

2

1 mol Cl10 g0.5950 kg Cl1 kg 70.90 g Cl

= 8.3921016 mol

V = 15.50 L T = 225°C + 273 = 498 K

n = 8.3921016 mol P = unknown Ideal gas law: PV = nRT Solving for P:

P IGL = nRTV

= ( ) ( )L•atm8.3921016mol 0.0821 498 K

mol•K15.50 L

= 22.1366 = 22.1 atm

b) van der Waals equation: ( )2

2

+ − =

n aP V nb nRTV

Solving for P:

PVDW = 2

2nRT n a

V nb V−

− From Table 5.4: a =

2

2atm•L6.49mol

; b = 0.0562 Lmol

n = 8.3921016 mol from part a)

PVDW = ( ) ( )

( )

( )

( )

22

2 22

2

2

atm•LL•atm 8.3921016 mol Cl 6.498.3921016 mol Cl 0.0821 498 K molmol•KL 15.50 L15.50 L 8.3921016 mol Cl 0.0562

mol

−

−

= 20.928855 = 20.9 atm 5.93 Plan: Rearrange the formula PV = (m/M)RT to solve for molar mass. Convert the volume in mL to L.

Temperature must be in Kelvin. To find the molecular formulas of I, II, III, and IV, assume 100 g of each sample so the percentages are numerically equivalent to the masses of each element. Convert each of the masses to moles by using the molar mass of each element involved. Divide all moles by the lowest number of moles and convert to

5-38

whole numbers to determine the empirical formula. The empirical formula mass and the calculated molar mass will then relate the empirical formula to the molecular formula. For gas IV, use Graham’s law to find the molar mass

Solution: a) V = 750.0 mL T = 70.00°C + 273.15 = 343.15 K

m = 0.1000 g P = 0.05951 atm (I); 0.07045 atm (II); 0.05767 atm (III) M = unknown


1 mL

−

= 0.7500 L

=

mPV RTM


Molar mass I = ( ) ( )

( ) ( )

L•atm0.1000 g 0.08206 343.15 Kmol•K =

0.05951 atm 0.7500 LmRTPV

=M = 63.0905 = 63.09 g I/mol

Molar mass II = ( ) ( )

( ) ( )

L•atm0.1000 g 0.08206 343.15 Kmol•K =

0.07045 atm 0.7500 LmRTPV

=M = 53.293 = 53.29 g II/mol

Molar mass III = ( ) ( )

( ) ( )

L•atm0.1000 g 0.08206 343.15 Kmol•K =

0.05767 atm 0.7500 LmRTPV

=M = 65.10349 = 65.10 g III/mol

b) % H in I = 100% – 85.63% = 14.37% H % H in II = 100% – 81.10% = 18.90% H % H in III = 100% – 82.98% = 17.02% H Assume 100 g of each so the mass percentages are also the grams of the element. I

Moles B = ( ) 1 mol B85.63 g B10.81 g B

= 7.921369 mol B (unrounded)

Moles H = ( ) 1 mol H14.37 g H1.008 g H

= 14.25595 mol H (unrounded)

7.921369 mol B7.921369 mol B

= 1.00 14.25595 mol H7.921369 mol B

= 1.7997

The hydrogen value is not close enough to a whole number to round. Thus, both amounts need to be multiplied by the smallest value to get near whole numbers. This value is 5. Multiplying each value by 5 gives (1.00 x 5) = 5 for B and (1.7997 x 5) = 9 for H. The empirical formula is B5H9, which has a formula mass of 63.12 g/mol. The empirical formula mass is near the molecular mass from part a) (63.09 g/mol). Therefore, the empirical and molecular formulas are both B5H9.

II

Moles B = ( ) 1 mol B81.10 g B10.81 g B


Moles H = ( ) 1 mol H18.90 g H1.008 g H


7.50231 mol B7.50231 mol B

= 1.00 18.750 mol H7.50231 mol B

= 2.4992

The hydrogen value is not close enough to a whole number to round. Thus, both amounts need to be multiplied by the smallest value to get near whole numbers. This value is 2. Multiplying each value by 2 gives (1.00 x 2) = 2 for B and (2.4992 x 2) = 5 for H. The empirical formula is B2H5, which has a

5-39

formula mass of 26.66 g/mol. Dividing the molecular formula mass from part a) by the empirical formula mass gives the relationship between the formulas: (53.29 g/mol)/(26.66 g/mol) = 2. The molecular formula is two times the empirical formula, or B4H10.

III

Moles B = ( ) 1 mol B82.98 g B10.81 g B


Moles H = ( ) 1 mol H17.02 g H1.008 g H


7.6762 mol B7.6762 mol B

= 1.00 16.8849 mol H7.6762 mol B

= 2.2

The hydrogen value is not close enough to a whole number to round. Thus, both amounts need to be multiplied by the smallest value to get near whole numbers. This value is 5. Multiplying each value by 5 gives (1.00 x 5) = 5 for B and (2.2 x 5) = 11 for H. The empirical formula is B5H11, which has a formula mass of 65.14 g/mol. The empirical formula mass is near the molecular mass from part a). Therefore, the empirical and molecular formulas are both B5H11.

c) 2Rate SORate IV

= 2

molar mass IVmolar mass SO

250.0 mL13.04 min350.0 mL12.00 min

= 0.657318 = molar mass IV64.06 g/mol

0.6573182 = molar mass IV64.06 g/mol

Molar mass IV = 27.6782 = 27.68 g/mol % H in IV = 100% – 78.14% = 21.86% H

Moles B = ( ) 1 mol B78.14 g B10.81 g B


Moles H = ( ) 1 mol H21.86 g H1.008 g H


7.22849 mol B7.22849 mol B

= 1.00 21.6865 mol H7.22849 mol B

= 3.00

The empirical formula is BH3, which has a formula mass of 13.83 g/mol. Dividing the molecular formula mass by the empirical formula mass gives the relationship between the formulas: (27.68 g/mol)/(13.83 g/mol) = 2. The molecular formula is two times the empirical formula, or B2H6. 5.94 Plan: Calculate the mole fraction of each gas; the partial pressure of each gas is directly proportional to its mole fraction so the gas with the highest mole fraction has the highest partial pressure. Remember that kinetic energy is directly proportional to Kelvin temperature. Solution:

a) AA

total =

nXn

I. XA = 3 A particles

9 total particles = 0.33; II. XA =

4 A particles12 total particles

= 0.33; III. XA = 5 A particles


The partial pressure of A is the same in all 3 samples since the mole fraction of A is the same in all samples.

5-40

b) I. XB = 3 B particles

9 total particles = 0.33; II. XB =

3 B particles12 total particles

= 0.25; III. XB = 3 B particles


The partial pressure of B is lowest in Sample III since the mole fraction of B is the smallest in that sample. c) All samples are at the same temperature, T, so all have the same average kinetic energy.


1 2 =

PV P VT T

or 1 1 22

2 1 =

PV TVP T

. R and n are fixed. Temperatures must be converted to

Kelvin. Solution: a) T1 = 200°C + 273 = 473 K; T2 = 100°C + 273 = 373 K

1 1 2 12

2 1

(2 atm)( )(373 K) = =


= 1.577 V1 Increase

b) T1 = 100°C + 273 = 373 K; T2 = 300°C + 273 = 573 K 1 1 2 1

22 1

(1 atm)( )(573 K) = =


= 0.51206 V1 Decrease

c) T1 = –73°C + 273 = 200 K; T2 = 127°C + 273 = 400 K 1 1 2 1

22 1

(3 atm)( )(400 K) = =


= V1 Unchanged

d) T1 = 300°C + 273 = 573 K; T2 = 150°C + 273 = 423 K

1 1 2 12

2 1

(0.2 atm)( )(423 K) = =

(0.4 atm)(573 K)PV T VVP T

= 0.3691 V1 Decrease

5.96 Plan: Partial pressures and mole fractions are calculated from Dalton’s law of partial pressures: PA = XA x P total.

Remember that 1 atm = 760 torr. Solve the ideal gas law for moles and then convert to molecules using Avogadro’s number to calculate the number of O2 molecules in the volume of an average breath. Solution:

a) Convert each mole percent to a mole fraction by dividing by 100%. P total = 1 atm = 760 torr PNitrogen = XNitrogen x P total = 0.786 x 760 torr = 597.36 = 597 torr N2 POxygen = XOxygen x P total = 0.209 x 760 torr = 158.84 = 159 torr O2 PCarbon Dioxide = XCarbon Dioxide x P total = 0.0004 x 760 torr = 0.304 = 0.3 torr CO2 PWater = XWater x P total = 0.0046 x 760 torr = 3.496 = 3.5 torr H2O b) Mole fractions can be calculated by rearranging Dalton’s law of partial pressures:

XA = A

total

PP

and multiply by 100 to express mole fraction as percent

P total = 1 atm = 760 torr

XNitrogen = 569 torr760 torr

x 100% = 74.8684 = 74.9 mol% N2

XOxygen = 104 torr760 torr

x 100% = 13.6842 = 13.7 mol% O2

X Carbon Dioxide = 40 torr760 torr

x 100% = 5.263 = 5.3 mol% CO2

XWater =47 torr760 torr

x 100% = 6.1842 = 6.2 mol% H2O

c) V = 0.50 L T = 37°C + 273 = 310 K P = 104 torr n = unknown

5-41


= 0.136842105 atm


( ) ( )

( )

0.136842105 atm 0.50 LL•atm0.0821 310 Kmol•K

PVn =RT

=

= 0.0026883 mol O2

Molecules of O2 = ( )23

22

2

6.022x10 molecules O0.0026883 mol O1 mol O

= 1.6189x1021 = 1.6x1021 molecules O2 5.97 Plan: Convert the mass of Ra to moles and then atoms using Avogadro’s number. Convert from number of Ra atoms to Rn atoms produced per second and then to Rn atoms produced per day. The number of Rn atoms is converted to moles and then the ideal gas law is used to find the volume of this amount of Rn. Solution:

Atoms Ra = ( )231 mol Ra 6.022 x10 Ra atoms1.0 g Ra

226 g Ra 1 mol Ra

= 2.664602x1021 Ra atoms

Atoms Rn produced/s = ( )4

2115

1.373x10 Rn atoms2.664602x10 Ra atoms1.0x10 Ra atoms

= 3.65849855x1010 Rn atoms/s

Moles Rn produced/day = 10

233.65849855x10 Rn atoms 3600 s 24 h 1 mol Rn

s h day 6.022x10 Rn atoms

= 5.248992x10–9 mole Rn/day PV = nRT Solving for V (at STP):

Volume of Rn = V = ( ) ( )

( )

9 L•atm5.248992x10 mol 0.0821 273 Kmol•K =

1 atmnRT

P

−

= 1.17647x10–7 = 1.2x10–7 L Rn 5.98 Plan: For part a), since the volume, temperature, and pressure of the gas are changing, use the combined gas law.

For part b), use the ideal gas law to solve for moles of air and then moles of N2. Solution:

a) P1 = 1450. mmHg P2 = 1 atm V1 = 208 mL V2 = unknown T1 = 286 K T2 = 298 K

Converting P1 from mmHg to atm: P1 = ( ) 1 atm1450. mmHg760 mmHg

= 1.9079 atm


1 2 =

PV P VT T

V2 = 2 11

1 2

T PV

T P = ( ) 298 K 1.9079 atm208 L

286 K 1 atm

= 413.494 mL = 4x102 mL

b) V = 208 mL T = 286 K

P = 1450 mmHg = 1.9079 atm n = unknown

5-42


1 mL

−

= 0.208 L


Moles of air = ( ) ( )

( )

1.9079 atm 0.208 LL•atm0.0821 286 Kmol•K

PVn =RT

=

= 0.016901 mol air

Mole of N2 = ( ) 277% N0.016901 mol

100%

= 0.01301 = 0.013 mol N2

5.99 Plan: The amounts of both reactants are given, so the first step is to identify the limiting reactant.

Write the balanced equation and use molar ratios to find the number of moles of NO2 produced by each reactant. The smaller number of moles of product indicates the limiting reagent. Solve for volume of NO2 using the ideal gas law. Solution: Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)

Moles NO2 from Cu = ( )3 23

2 mol NO8.95 g Cu 1 mol Cu4.95 cm63.55 g Cu 1 mol Cucm

= 1.394256 mol NO2

Moles NO2 from HNO3 = ( )3

3 3 23

3

68.0% HNO 1 mol HNO 2 mol NO1 cm 1.42 g230.0 mL100% 1 mL 63.02 g 4 mol HNOcm

= 1.7620 mol NO2 Since less product can be made from the copper, it is the limiting reactant and excess nitric acid will be left after the reaction goes to completion. Use the calculated number of moles of NO2 and the given temperature and pressure in the ideal gas law to find the volume of nitrogen dioxide produced. Note that nitrogen dioxide is the only gas involved in the reaction. V = unknown T = 28.2°C + 273.2 = 301.4 K P = 735 torr n = 1.394256 mol NO2


= 0.967105 atm


( ) ( )

( )

L•atm1.394256 mol 0.0821 301.4 Kmol•K

0.967105 atmnRTV =

P

= = 35.67429 = 35.7 L NO2

5.100 Plan: Solve the ideal gas law for moles of air and then convert to molecules using Avogadro’s number. Volume

must be converted to liters and temperature to kelvins. Solution:

a) V = 1200 mL T = 37°C + 273 = 310 K P = 1 atm n = unknown


1 mL

−

= 1.2 L


5-43

Moles of air =( ) ( )

( )

1 atm 1.2 LL•atm0.0821 310 Kmol•K

PVn =RT

=

= 0.047149 = 0.047 mol air

b) Molecules of air = ( )236.022 x10 molecules0.047149 mol air

1 molair

= 2.83931x1022 = 2.8x1022 molecules air

5.101 Plan: The amounts of two reactants are given, so the first step is to identify the limiting reactant.

Write the balanced equation and use molar ratios to find the number of moles of Br2 produced by each reactant. The smaller number of moles of product indicates the limiting reagent. Solve for volume of Br2 using the ideal gas law. Solution: 5NaBr(aq) + NaBrO3(aq) + 3H2SO4(aq) → 3Br2(g) + 3Na2SO4(aq) + 3H2O(g)

Moles Br2 from NaBr = ( ) 23 mol Br1 mol NaBr275 g NaBr102.89 g NaBr 5 mol NaBr

= 1.60365 mol Br2 (unrounded)

Moles Br2 from NaBrO3 = ( ) 3 23

3 3

1 mol NaBrO 3 mol Br175.6 g NaBrO

150.89 g NaBrO 1 mol NaBrO

= 3.491285 mol Br2

The NaBr is limiting since it produces a smaller amount of Br2. Use the ideal gas law to find the volume of Br2:

V = unknown T = 300°C + 273 = 573 K P = 0.855 atm n = 1.60365 mol Br2 PV = nRT Solving for V:

Volume (L) of Br2 = ( ) ( )

( )

L•atm1.60365 mol 0.0821 573 Kmol•K

0.855 atmnRTV =

P

= = 88.235 = 88.2 L Br2

5.102 Plan: First, write the balanced equation. Convert mass of NaN3 to moles and use the molar ratio from the

balanced equation to find the moles of nitrogen gas produced. Use the ideal gas law to find the volume of that amount of nitrogen. The problem specifies that the nitrogen gas is collected over water, so the partial pressure of water vapor must be subtracted from the overall pressure given. Table 5.2 reports the vapor pressure of water at 26°C (25.2 torr). Pressure must be in units of atm and temperature in kelvins. Solution: 2NaN3(s) → 2Na(s) + 3N2(g)

Moles N2 = ( ) 3 23

3 3

1 mol NaN 3 mol N50.0 g NaN

65.02 g NaN 2 mol NaN

= 1.15349 mol N2

Finding the volume of N2: V = unknown T = 26°C + 273 = 299 K

P total = 745.5 mmHg n = 1.15319 mol Pwater vapor = 25.2 torr = 25.2 mmHg Pnitrogen = P total – Pwater vapor = 745.5 mmHg – 25.2 mmHg = 720.3 mmHg


= 0.9477632 atm


V = ( ) ( )

( )

L•atm1.15349 mol 0.0821 299 Kmol•K =

0.9477632 atmnRT

P

= 29.8764 = 29.9 L N2

5-44

5.103 Plan: Use the percent composition information to find the empirical formula of the compound. Assume 100 g of sample so the percentages are numerically equivalent to the masses of each element. Convert each of the masses to moles by using the molar mass of each element involved. Divide all moles by the lowest number of moles and convert to whole numbers to determine the empirical formula. Rearrange the formula PV = (m/M)RT to solve for molar mass.The empirical formula mass and the calculated molar mass will then relate the empirical formula to the molecular formula.

Solution: Empirical formula: Assume 100 g of each so the mass percentages are also the grams of the element.

Moles C = ( ) 1 mol C64.81 g C12.01 g C

= 5.39634 mol C

Moles H = ( ) 1 mol H13.60 g H1.008 g H

= 13.49206 mol H

Moles O = ( ) 1 mol O21.59 g O16.00 g O

= 1.349375 mol O

5.39634 mol C1.349375 mol O

= 4 13.749206 mol H1.349375 mol O

= 10 1.349375 mol O1.349375 mol O

= 1.00

Empirical formula = C4H10O (empirical formula mass = 74.12 g/mol) Molecular formula: V = 2.00 mL T = 25°C + 273 = 298 K

m = 2.57 g P = 0.420 atm M = unknown

=

mPV RTM


Molar mass = ( ) ( )

( ) ( )

L•atm2.57 g 0.0821 298 Kmol•K =

0.420 atm 2.00 LmRTPV

=M = 74.85 g/mol

Since the molar mass (74.85 g/mol ) and the empirical formula mass (74.12 g/mol) are similar, the empirical and molecular formulas must both be: C4H10O

5.104 Plan: The empirical formula for aluminum chloride is AlCl3 (Al3+ and Cl–). The empirical formula mass is

(133.33 g/mol). Calculate the molar mass of the gaseous species from the ratio of effusion rates (Graham’s law). This molar mass, divided by the empirical weight, should give a whole-number multiple that will yield the molecular formula. Solution:

Rate unknownRate He

= 0.122 = molar mass Hemolar mass unknown

0.122 = 4.003 g/molmolar mass unknown

0.014884 = 4.003 g/molmolar mass unknown

Molar mass unknown = 268.9465 g/mol The whole-number multiple is 268.9465/133.33, which is about 2. Therefore, the molecular formula of the gaseous species is 2 x (AlCl3) = Al2Cl6.

5-45

5.105 Plan: First, write the balanced equation. Convert mass of C8H18 to moles and use the molar ratio from the

balanced equation to find the total moles of gas produced. Use the ideal gas law to find the volume of that amount of gas. For part b), use the molar ratio from the balanced equation to find the moles of oxygen that react with the C8H18. Use the composition of air to calculate the amount of N2 and Ar that remains after the O2 is consumed and use the ideal gas law to find the volume of that amount of gas. The volume of the gaseous exhaust is the sum of the volume of gaseous products and the residual air (N2 and Ar) that does not react. Solution: 2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O (g)

a) Moles gaseous products = ( ) 8 188 18

8 18 8 18

1 mol C H 34 mol gas100. g C H114.22 g C H 2 mol C H

= 14.883558 mol gas

Finding the volume of gaseous product: V = unknown T = 350°C + 273 = 623 K

P total = 735 torr n = 14.883558 mol


= 0.9671053 atm


Volume gas = V = ( ) ( )

( )

L•atm14.883558 mol 0.0821 623 Kmol•K =

0.9671053 atmnRT

P

= 787.162 = 787 L gas

b) Moles O2 = ( ) 8 18 28 18

8 18 8 18

1 mol C H 25 molO100. g C H

114.22 g C H 2 mol C H

= 10.94379 mol O2

Moles other gases = ( ) 22

2

78% + 1% N and Ar10.94379 mol O

21% O

= 41.1695 mol Ar + N2

Finding the volume of Ar + N2: V = unknown T = 350°C + 273 = 623 K

P total = 735 torr = 0.9671053 atm n = 41.1695 mol PV = nRT

Solving for V:

Volume gas = V = ( ) ( )

( )

L•atm41.1695 mol 0.0821 623 Kmol•K =

0.9671053 atmnRT

P

= 2177.37 L residual air

Total volume of gaseous exhaust = 787.162 L + 2177.37 L = 2964.53 = 2.96x103 L 5.106 Plan: First, write the balanced equation for the reaction: 2SO2 + O2 → 2SO3. The total number of moles of gas will change as the reaction occurs since 3 moles of reactant gas forms 2 moles of product gas. From

the volume, temperature, and pressures given, we can calculate the number of moles of gas before and after the reaction using the ideal gas law. For each mole of SO3 formed, the total number of moles of gas decreases by 1/2 mole. Thus, twice the decrease in moles of gas equals the moles of SO3 formed. Solution:

Moles of gas before and after reaction: V = 2.00 L T = 800. K

P total = 1.90 atm n = unknown PV = nRT

Initial moles = n = PVRT

= ( ) ( )

( )

1.90 atm 2.00 LL•atm0.0821 800. Kmol•K

= 0.05785627 mol

5-46

Final moles = n = PVRT

= ( ) ( )

( )

1.65 atm 2.00 LL•atm0.0821 800. Kmol•K

= 0.050243605 mol

Moles of SO3 produced = 2 x decrease in the total number of moles = 2 x (0.05785627 mol – 0.050243605 mol) = 0.01522533 = 1.52x10–2 mol

Check: If the starting amount is 0.0578 total moles of SO2 and O2, then x + y = 0.0578 mol, where x = mol of SO2 and y = mol of O2. After the reaction: (x – z) + (y – 0.5z) + z = 0.0502 mol Where z = mol of SO3 formed = mol of SO2 reacted = 2(mol of O2 reacted).

Subtracting the two equations gives: x – (x – z) + y – (y – 0.5z) – z = 0.0578 – 0.0502 z = 0.0152 mol SO3 The approach of setting up two equations and solving them gives the same result as above. 5.107 Plan: Use the density of C2H4 to find the volume of one mole of gas. Then use the van der Waals equation with

1.00 mol of gas to find the pressure of the gas (the mole ratio is 1:1, so the number of moles of gas remains the same). Solution:

a) ( )3

2 42 4

2 4

28.05 g C H 1 mL 10 L1 mole C H1 mole C H 0.215 g 1 mL

−

= 0.130465 L = 0.130 L

V = 0.130 L T = 10°C + 273 = 283 K + 950 K = 1233 K P total = unknown n = 1.00 mol

From Table 5.4 for CH4: a = 2

2atm•L2.25mol

; b = 0.0428 Lmol

( )2

2

+ − =

n aP V nb nRTV

Pressure of CH4 = PVDW = 2

2−−

nRT n aV nb V

PVDW = ( ) ( )

( )

( )

( )

22

2

2

atm•LL•atm 1.00 mol 2.251.00 mol 0.0821 1233 K molmol•K0.130 L 1.00 mol 0.0428 L/mol 0.130 L

−

− = 1027.7504 = 1028 atm

b) ( ) ( )

( )

1028 atm 0.130 L =

L•atm0.0821 1233 Kmol•K

PVRT

= 1.32

This value is smaller than that shown in Figure 5.23 for CH4. The temperature in this situation is very high (1233 K). At high temperatures, the gas particles have high kinetic energy. Thus the gas particles have the energy to overcome the effects of intermolecular attraction and the gas behaves more ideally.

5.108 Plan: First, write the balanced equation. Convert mass of ammonium nitrate to moles and use the molar ratio from

the balanced equation to find the moles of gas produced. Use the ideal gas law to find the volume of that amount of gas. Solution:

Moles gas = ( )3

4 34 3

4 3 4 3

1 mol NH NO10 g 7 mol gas15.0 kg NH NO1 kg 80.05 g NH NO 2 mol NH NO

= 655.840 mol gas

5-47

Finding the volume of gas: V = unknown T = 307°C + 273 = 580 K

P = 1.00 atm n = 655.840 mol PV = nRT

Solving for V:

V = ( ) ( )

( )

L•atm655.840 mol 0.0821 580 Kmol•K =

1.00 atmnRT

P

= 3.122979x104 = 3.12x104 L

5.109 Plan: Multiply the molarity and volume of the I2 solution to find initial moles of I2 added. Multiply molarity and volume of the S2O3

2– solution to obtain moles of that solution and use the molar ratio in the balanced equation to find moles of excess I2. Subtract the excess I2 from the initial moles of added I2 to find the moles of I2 reacted with the SO2; the molar ratio betweeen SO2 and I2 gives the moles of SO2 present. Use the ideal gas law to find the moles of air which is compared to the moles of SO2 present. Solution:

The balanced equation is: I2(aq) + 2S2O32−(aq) → 2I−(aq) + S4O6

2−(aq).

Initial moles of I2 = ( )3

20.01017 mol I10 L20.00 mL1 mL L

−

= 2.034x10–4 mol I2 initial

Moles I2 reacting with S2O32– = ( )

232 3 2

22 3

0.0105 mol S O 1 mol I10 L11.37 mL1 mL L 2 mol S O

−−

−

= 5.96925x10–5 mol I2 reacting with S2O32– (not reacting with SO2)

Moles I2 reacting with SO2 = 2.034x10–4 mol – 5.96925x10–5 mol = 1.437075x10–4 mol I2 reacted with SO2 The balanced equation is: SO2(aq) + I2(aq) + 2H2O(l) → HSO4

−(aq) + 2I−(aq) + 3H+(aq).

Moles SO2 = ( )4 22

2

1 mol SO1.437075x10 mol I1 mol I

−

= 1.437075x10–4 mol SO2

Moles of air: V = 500 mL = 0.500 L T = 38°C + 273 = 311 K

P total = 700. torr n = unknown


= 0.921052631 atm

PV = nRT

Moles air = n = PVRT

= ( ) ( )

( )

0.921052632 atm 0.500 LL•atm0.0821 311 Kmol•K

= 0.018036 mol air (unrounded)

Volume % SO2 = mol % SO2 = ( )4

21.437075x10 mol SO 1000.018036 mol air

−= 0.796781 = 0.797%

5.110 Plan: First, write the balanced equation. The moles of CO that react in part a) can be calculated from the ideal gas

law. Volume must be in units of L, pressure in atm, and temperature in kelvins. Once the moles of CO that react are known, the molar ratio from the balanced equation is used to determine the mass of nickel that will react with the CO. For part b), assume the volume is 1 m3. Use the ideal gas law to solve for moles of Ni(CO)4, which equals the moles of Ni, and convert moles to grams using the molar mass. For part c), the mass of Ni obtained from 1 m3 (part b)) can be used to calculate the amount of CO released. Use the ideal gas law to calculate the volume of CO. The vapor pressure of water at 35°C (42.2 torr) must be subtracted from the overall pressure (see Table 5.2). Solution:

a) Ni(s) + 4CO(g) → Ni(CO)4(g) V = 3.55 m3 T = 50°C + 273 = 323 K

P = 100.7 kPa n = unknown

5-48

Converting V from m3 to L: V = ( )33 3

1 L3.55 m10 m−

= 3550 L

Converting P from kPa to atm: P = ( ) 1 atm100.7 kPa101.325 kPa

= 0.993831729 atm


Moles of CO = ( ) ( )

( )

0.993831729 atm 3550 LL•atm0.0821 323 Kmol•K

PVn =RT

=

= 133.044073 mol CO

Mass Ni = ( ) 1 mol Ni 58.69g Ni133.044073 mol CO4 mol CO 1mol Ni

= 1952.089 = 1.95x103 g Ni

b) Ni(s) + 4 CO(g) → Ni(CO)4(g) V = 1 m3 T = 155°C + 273 = 428 K

P = 21 atm n = unknown

Converting V from m3 to L: V = ( )33 3

1 L1 m10 m−

= 1000 L


Moles of Ni(CO)4 = ( ) ( )

( )

21 atm 1000 LL•atm0.0821 428 Kmol•K

PVn =RT

=

= 597.62997 mol Ni(CO)4

Mass Ni = ( )44

1 mol Ni 58.69 g Ni597.62997 mol Ni(CO)1 mol Ni(CO) 1 mol Ni

= 3.50749x104 = 3.5x104 g Ni The pressure limits the significant figures.

c) Moles CO = ( )4 1 mol Ni 4 mol CO3.50749x10 g Ni58.69 g Ni 1 mol Ni

= 2390.51968 mol CO

Finding the volume of CO: V = unknown T = 35°C + 273 = 308 K

P total = 769 torr n = 2390.51968 mol Pwater vapor = 42.2 torr PCO = P total – Pwater vapor = 769 torr – 42.2 torr = 726.8 torr

Converting P from torr to atm: P = ( ) 1 atm726.8 torr760 torr

= 0.956315789 atm


V = ( ) ( )

( )

L•atm2390.51968 mol 0.0821 308 Kmol•K =

0.956315789 atmnRT

P

= 63209.86614 L CO

Converting V from L to m3: V = ( )3 310 m63209.86614 L

1 L

−

= 63.209866 = 63 m3 CO

The answer is limited to two significant figures because the mass of Ni comes from part b).

5-49

5.111 Plan: Use the percent composition information to find the empirical formula of the compound. Assume

100 g of sample so the percentages are numerically equivalent to the masses of each element. Convert each of the masses to moles by using the molar mass of each element involved. Divide all moles by the lowest number of moles and convert to whole numbers to determine the empirical formula. Rearrange the formula PV = (m/M)RT to solve for molar mass.The empirical formula mass and the calculated molar mass will then relate the empirical formula to the molecular formula.

Solution: Empirical formula: Assume 100 g of each so the mass percentages are also the grams of the element.

Moles Si = ( ) 1 mol Si33.01 g Si28.09 g Si

= 1.17515 mol Si

Moles F = ( ) 1 mol F66.99 g F19.00 g F

= 3.525789 mol F

1.17515 mol Si1.17515 mol Si

= 1 3.525789 mol F1.17515 mol Si

= 3

Empirical formula = SiF3 (empirical formula mass = 85.1 g/mol) Molecular formula: V = 0.250 L T = 27°C + 273 = 300 K

m = 2.60 g P = 1.50 atm M = unknown

=

mPV RTM


Molar mass = ( ) ( )

( ) ( )

L•atm2.60 g 0.0821 300 Kmol•K =

1.50 atm 0.250 LmRTPV

=M = 170.768 g/mol

The molar mass (170.768 g/mol ) is twice the empirical formula mass (85.1 g/mol), so the molecular formula must be twice the empirical formula, or 2 x SiF3 = Si2F6.

5.112 a) A preliminary equation for this reaction is 4CxHyNz + nO2 → 4CO2 + 2N2 + 10H2O. Since the organic compound does not contain oxygen, the only source of oxygen as a reactant is oxygen gas. To form 4 volumes of CO2 would require 4 volumes of O2 and to form 10 volumes of H2O would require 5 volumes of O2. Thus, 9 volumes of O2 was required. b) Since the volume of a gas is proportional to the number of moles of the gas we can equate volume and moles. From a volume ratio of 4CO2:2N2:10H2O we deduce a mole ratio of 4C:4N:20H or 1C:1N:5H for an empirical formula of CH5N. 5.113 a) There is a total of 6x106 blue particles and 6x106 black particles. When equilibrium is reached after

opening the stopcocks, the particles will be evenly distributed among the three containers. Therefore, container B will have 2x106 blue particles and 2x106 black particles. b) The particles are evenly distributed so container A has 2x106 blue particles and 2x106 black particles.

c) There are 2x106 blue particles and 2x106 black particles in C for a total of 4x106 particles.

Final pressure in C = ( )66

750 torr4x10 particles6x10 particles

= 500 torr

d) There are 2x106 blue particles and 2x106 black particles in B for a total of 4x106 particles.

5-50

Final pressure in B = ( )66

750 torr4x10 particles6x10 particles

= 500 torr

5.114 Plan: Write the balanced equation for the combustion of n-hexane. For part a), assuming a 1.00 L sample of air at STP, use the molar ratio in the balanced equation to find the volume of n-hexane required to react with the oxygen in 1.00 L of air. Convert the volume n-hexane to volume % and divide by 2 to obtain the LFL. For part b), use the LFL calculated in part a) to find the volume of n-hexane required to produce a flammable mixture and then use the ideal gas law to find moles of n-hexane. Convert moles oft-hexane to mass and then to volume using the density.

Solution: a) 2C6H14(l) + 19O2(g) →12CO2(g) + 14H2O(g) For a 1.00 L sample of air at STP:

Volume of C6H14 vapor needed = ( ) 6 142

2

2 L C H20.9 L O1.00 L air

100 L air 19 L O

= 0.0220 L C6H14

Volume % of C6H14 = ( ) ( )6 14 6 14C H volume 0.0220 L C H100 = 100

air volume 1.00 L air = 2.2% C6H14

LFL = 0.5(2.2%) = 1.1% C6H14

b) Volume of C6H14 vapor = ( )3 6 143 3

1.1% C H1 L1.000 m air100% air10 m−

= 11.0 L C6H14


Moles of C6H14 = ( ) ( )

( )

1 atm 11.0 LL•atm0.0821 273 Kmol•K

PVn =RT

=

= 0.490780 mol C6H14

Volume of C6H14 liquid = ( ) 6 146 14

6 14 6 14

86.17 g C H 1 mL0.490780 mol C H1 mol C H 0.660 g C H

= 64.0765 = 64 mL

C6H14 5.115 Plan: To find the factor by which a diver’s lungs would expand, find the factor by which P changes from 125 ft to

the surface, and apply Boyle’s law. To find that factor, calculate Pseawater at 125 ft by converting the given depth from ft-seawater to mmHg to atm and adding the surface pressure (1.00 atm). Solution:

P(H2O) = ( )2

312 in 2.54 cm 10 m 1 mm125 ft1 ft 1 in 1 cm 10 m

−

−

= 3.81x104 mmH2O

P(Hg): H O Hg2

Hg H O2

= h dh d

4

2

Hg

3.81x10 mmH O 13.5 g/mL = 1.04 g/mLh

hHg = 2935.1111 mmHg

P(Hg) = ( ) 1 atm2935.11111 mmHg760 mm Hg

= 3.861988 atm (unrounded)

P total = (1.00 atm) + (3.861988 atm) = 4.861988 atm (unrounded) Use Boyle’s law to find the volume change of the diver’s lungs:

P1V1 = P2V2

2 1

1 2

= V PV P

2

1

4.861988 atm = 1 atm

VV

= 4.86

5-51

To find the depth to which the diver could ascend safely, use the given safe expansion factor (1.5) and the pressure at 125 ft, P125, to find the safest ascended pressure, Psafe.

P125/Psafe = 1.5 Psafe = P125/1.5 = (4.861988 atm)/1.5 = 3.241325 atm (unrounded) Convert the pressure in atm to pressure in ft of seawater using the conversion factors above. Subtract this distance from the initial depth to find how far the diver could ascend.

h(Hg): ( ) 760 mmHg4.861988 3.241325 atm1 atm

−

= 1231.7039 mmHg

H O Hg2

Hg H O2

= h dh d

H O2 13.5 g/mL = 1231.7039 mmHg 1.04 g/mL

h H O2

h = 15988.464 mm

( )3

210 m 1.094 yd 3 ft15988.464 mmH O1 mm 1 m 1 yd

−

= 52.4741 ft

Therefore, the diver can safely ascend 52.5 ft to a depth of (125 – 52.4741) = 72.5259 = 73 ft. 5.116 Plan: Write a balanced equation. Convert mass of CaF2 to moles and use the molar ratio from the balanced

equation to find the moles of gas produced. Use the ideal gas law to find the temperature required to store that amount of HF gas at the given conditions of temperature and pressure. Solution: CaF2(s) + H2SO4(aq) → 2HF(g) + CaSO4(s)

Moles HF gas = ( ) 22

2 2

1 mol CaF 2 mol HF15.0 g CaF78.08 g CaF 1 mol CaF

= 0.3842213 mol HF

Finding the temperature: V = 8.63 L T = unknown

P = 875 torr n = 0.3842213 mol


= 1.151312789 atm

PV = nRT Solving for T:

= PVTnR

= ( ) ( )

( )

1.151315789 atm 8.63 LL•atm0.3842213 mol HF 0.0821mol•K

= 314.9783 K

The gas must be heated to 315 K. 5.117 Plan: First, write the balanced equation. According to the description in the problem, a given volume of peroxide

solution (0.100 L) will release a certain number of “volumes of oxygen gas” (20). Assume that 20 is exact. A 0.100 L solution will produce (20 x 0.100 L) = 2.00 L O2 gas. Use the ideal gas law to convert this volume of

O2 gas to moles of O2 gas and convert to moles and then mass of H2O2 using the molar ratio in the balanced equation.

Solution: 2H2O2(aq) → 2H2O(l) + O2(g)


Moles of O2 = ( ) ( )

( )

1 atm 2.00 LL•atm0.0821 273 Kmol•K

PVn =RT

=

= 8.92327x10–2 mol O2

5-52

Mass H2O2 = ( )2 2 2 2 22

2 2 2

2 mol H O 34.02 g H O8.92327x10 mol O1 mol O 1 mol H O

−

= 6.071395 = 6.07 g H2O2

5.118 Plan: The moles of gas may be found using the ideal gas law. Multiply moles of gas by Avogadro’s number to obtain the number of molecules.

Solution: V = 1 mL = 0.001 L T = 500 K P = 10–8 mmHg n = unknown

Converting P from mmHg to atm: P = ( )8 1 atm10 mmHg760 mmHg

−

= 1.315789x10–11 atm


Moles of gas = ( ) ( )

( )

111.315789x10 atm 0.001 L


PVn =RT

−

=

= 3.2053337x10–16 mol gas

Molecules = ( )23

16 6.022 x10 molecules3.2053337x10 mol1 mol

−

= 1.93025x108 = 108 molecules

(The 10–8 mmHg limits the significant figures.) 5.119 Plan: Use the equation for root mean speed (u rms) to find this value for O2 at 0.°C. Molar mass values must be

in units of kg/mol and temperature in kelvins. Divide the root mean speed by the mean free path to obtain the collision frequency.

Solution:

a) 0°C = 273 K M of O2 = 23

32.00 g O 1 kgmol 10 g

= 0.03200 kg/mol

R = 8.314 J/mol•K 1 J = kg•m2/s2

u rms = 3RTM

u rms = ( ) 2 2

J3 8.314 273 Kkg•m /smol•K

0.03200 kg/mol J

= 461.2878 = 461 m/s

b) Collision frequency = rms8

461.2878 m/s = mean free path 6.33x10 m

u−

= 7.2873x109 = 7.29x109 s–1

5.120 Plan: Convert the volume of the tubes from ft3 to L. Use the ideal gas law to find the moles of gas

that will occupy that volume at the given conditions of pressure and temperature. From the mole fraction of propylene and the total moles of gas, the moles of propylene can be obtained. Convert moles to mass in grams and then pounds using the molar mass and scale the amount added in 1.8 s to the amount in 1 h. Solution:

Volume of tubes = ( ) ( )( )

( )( )

( )( )

323 33

3 3 3 3 3

10 m12 in 2.54 cm 1 L100 ft10 m1 ft 1 in 1 cm

−

−

= 2831.685 L

V = 2831.685 L T = 330°C + 273 = 603 K P = 2.5 atm n = unknown PV = nRT

5-53

Moles of gas = n = ( )( )

( )

2.5 atm 2831.685 L =


PVRT

= 142.996 mol gas

Xpropylene = moles of propylene moles of mixture

0.07 = x moles of propylene142.996 moles of mixture

Moles of propylene = 10.00972 moles

Pounds of propylene = ( ) 3 6 3 63 6

3 6 3 6

42.08 g C H 1 lb C H 1 3600 s10.00972 moles C H1 mole C H 453.6 g C H 1.8 s

= 1857.183 = 1900 lb C3H6 5.121 Plan: Use the ideal gas law to calculate the molar volume, the volume of exactly one mole of gas, at the temperature and pressure given in the problem.

Solution: V = unknown T = 730. K P = 90 atm n = 1.00 mol PV = nRT

Solving for V:

V = ( ) ( )

( )

L•atm1.00 mol 0.0821 730. Kmol•K =

90 atmnRT

P

= 0.66592 = 0.67 L/mol

5.122 Plan: The diagram below describes the two Hg height levels within the barometer. First, find the pressure of the

N2. The 2NP is directly related to the change in column height of Hg. Then find the volume occupied by the N2.

The volume of the space occupied by the N2(g) is calculated from the length and cross-sectional area of the barometer. To find the mass of N2, use these values of P and V (T is given) in the ideal gas law to find moles which is converted to mass using the molar mass of nitrogen.

vacuum with N2(74.0 cm) (64.0 cm)

Solution:

Pressure of the nitrogen = ( )2

310 m 1 mm 1 atm74.0 cm 64.0 cm

1 cm 760 atm10 m

−

−

− = 0.1315789 atm

Volume of the nitrogen = ( )( )3

2 23

1 mL 10 L1.00 x 10 cm 64.0 cm 1.20 cm1 mL1 cm

− − = 0.0432 L

V = 0.0432 L T = 24°C + 273 = 297 K P = 0.1315789 atm n = unknown PV = nRT Solving for n:

Moles of N2 = n = ( )( )

( )

0.1315789 atm 0.0432 L =


PVRT

= 2.331151x10–4 mol N2

Mass of N2 = ( )4 22

2

28.02 g N2.331151x10 mol N1 mol N

−

= 6.5319x10–3 = 6.53x10–3 g N2

5-54

5.123 Plan: Use the ideal gas law to find the moles of ammonia gas in 10.0 L at this pressure and temperature.

Molarity is moles per liter. Use the moles of ammonia and the final volume of solution (0.750 L) to get the molarity. Solution: V = 10.0 L T = 33°C + 273 = 306 K P = 735 torr n = unknown


= 0.9671053 atm


Moles of ammonia = ( ) ( )

( )

0.9671053 atm 10.0 LL•atm0.0821 306 Kmol•K

PVn =RT

=

= 0.384954 mol

Molarity = M = mol ammonia 0.384954 mol = liters of solution 0.750 L

= 0.513272 = 0.513 M

5.124 Plan: Use the ideal gas law to determine the total moles of gas produced. The total moles multiplied by the

fraction of each gas gives the moles of that gas which may be converted to metric tons. Solution:

V = 1.5x103 m3 T = 298 K P = 1 atm n = unknown

Converting V from m3 to L: V = ( )3 33 3

1 L1.5x10 m10 m−

= 1.5x106 L


Moles of gas/day = ( ) ( )

( )

61 atm 1.5x10 L


PVn =RT

=

= 6.13101x105 mol/day

Moles of gas/yr = 56.13101x10 mol 365.25 day

day 1 yr

= 2.23935x107 mol/yr

Mass CO2 = ( )7

23 3

2

2.23935x10 mol 44.01 g CO 1 kg 1 t0.4896yr 1 mol CO 10 g 10 kg

= 482.519 = 4.83x102 t CO2/yr

Mass CO = ( )7

3 32.23935x10 mol 28.01 g CO 1 kg 1 t0.0146

yr 1 mol CO 10 g 10 kg

= 9.15773 = 9.16 t CO/yr

Mass H2O = ( )7

23 3

2

2.23935x10 mol 18.02 g H O 1 kg 1 t0.3710yr 1 mol H O 10 g 10 kg

= 149.70995 = 1.50x102 t H2O/yr

Mass SO2 = ( )7

23 3

2

64.06 g SO2.23935x10 mol 1 kg 1 t0.1185yr 1 mol SO 10 g 10 kg

= 169.992 = 1.70x102 t SO2/yr

Mass S2 = ( )7

23 3

2

64.12 g S2.23935x10 mol 1 kg 1 t0.0003yr 1 mol S 10 g 10 kg

= 0.4307614 = 4x10–1 t S2/yr

Mass H2 = ( )7

23 3

2

2.23935x10 mol 2.016 g H 1 kg 1 t0.0047yr 1 mol H 10 g 10 kg

= 0.21218 = 2.1x10–1 t H2/yr

5-55

Mass HCl = ( )7

3 32.23935x10 mol 36.46 g HCl 1 kg 1 t0.0008

yr 1 mol HCl 10 g 10 kg

= 0.6531736 = 6x10–1 t HCl/yr

Mass H2S = ( )7

23 3

2

34.08 g H S2.23935x10 mol 1 kg 1 t0.0003yr 1 mol H S 10 g 10 kg

= 0.228951 = 2x10–1 t H2S/yr

5.125 Plan: Use the molar ratio from the balanced equation to find the moles of H2 and O2 required to form 28.0 moles

of water. Then use the ideal gas law in part a) and van der Waals equation in part b) to find the pressure needed to provide that number of moles of each gas.

Solution:

a) The balanced chemical equation is: 2H2(g) + O2(g) → 2H2O(l)

Moles H2 = ( ) 22

2

2 mol H28.0 mol H O

2 mol H O

= 28.0 mol H2

Moles O2 = ( ) 22

2

1 mol O28.0 mol H O

2 mol H O

= 14.0 mol O2

V = 20.0 L T = 23.8°C + 273.2 =297 K P = unknown n = 28.0 mol H2; 14.0 mol O2 PV = nRT Solving for P:

PIGL of H2 = nRTV

= ( ) ( )L•atm28.0mol 0.0821 297 K

mol•K20.0 L

= 34.137 = 34.1 atm H2

PIGL of O2 = nRTV

= ( ) ( )L•atm14.0mol 0.0821 297 K

mol•K20.0 L

= 17.06859 = 17.1 atm O2

b) V = 20.0 L T =23.8°C + 273.2 =297 K P = unknown n = 28.0 mol H2; 14.0 mol O2 Van der Waals constants from Table 5.4:

H2: a = 2

2atm•L0.244mol

; b = 0.0266 Lmol

O2: a = 2

2atm•L1.36mol

; b = 0.0318 Lmol

( )2

2

+ − =

n aP V nb nRTV

PVDW = 2

2−−

nRT n aV nb V

PVDW of H2 = ( ) ( )

( )

( )

( )

22

2

2


−

− = 34.9631 = 35.0 atm H2

PVDW of O2 = ( ) ( )

( )

( )

( )

22

2

2


−

− = 16.78228 = 16.8 atm O2

c) The van der Waals value for hydrogen is slightly higher than the value from the ideal gas law. The van der Waals value for oxygen is slightly lower than the value from the ideal gas law.

5-56

5.126 Plan: Deviations from ideal gas behavior are due to attractive forces between particles which reduce the pressure

of the real gas and due to the size of the particle which affects the volume. Compare the size and/or attractive forces between the substances. The greater the size and/or the stronger the attractive forces, the greater the deviation from ideal gas behavior. Solution: a) Xenon would show greater deviation from ideal behavior than argon since xenon is a larger atom than argon.

The electron cloud of Xe is more easily distorted so intermolecular attractions are greater. Xe’s larger size also means that the volume the gas occupies becomes a greater proportion of the container’s volume at high pressures.

b) Water vapor would show greater deviation from ideal behavior than neon gas since the attractive forces

between water molecules are greater than the attractive forces between neon atoms. We know the attractive forces are greater for water molecules because it remains a liquid at a higher temperature than neon (water is a liquid at room temperature while neon is a gas at room temperature). c) Mercury vapor would show greater deviation from ideal behavior than radon gas since the attractive forces between mercury atoms is greater than that between radon atoms. We know that the attractive forces for mercury are greater because it is a liquid at room temperature while radon is a gas.

d) Water is a liquid at room temperature; methane is a gas at room temperature. Therefore, water molecules have stronger attractive forces than methane molecules and should deviate from ideal behavior to a greater extent than methane molecules.

5.127 Plan: Use the molarity and volume of the solution to find the moles of HBr needed to make the solution. Then use the ideal gas law to find the volume of that number of moles of HBr gas at the given conditions. Solution:

Moles of HBr in the hydrobromic acid: ( )1.20 mol HBr 3.50 LL

= 4.20 mol HBr

V = unknown T = 29°C + 273 =302 K P = 0.965 atm n = 4.20 mol PV = nRT

Solving for V:

V = ( ) ( )

( )

L•atm4.20 mol 0.0821 302 Kmol•K =

0.965 atmnRT

P

= 107.9126 = 108 L HBr

5.128 Plan: Convert the mass of each gas to moles using the molar mass. Calculate the mole fraction of CO. Use the

relationship between partial pressure and mole fraction (PA = XA x P total) to calculate the partial pressure of CO. Solution:

Moles CO = ( ) 1 mol CO7.0 g CO28.01 g CO

= 0.24991 mol CO

Moles SO2 = ( ) 22

2

1 mol SO10.0 g SO

64.06 g SO

= 0.156104 mol SO2

XCO = 2

mol COmol CO + mol SO

= 2

0.24991 mol CO 0.24991 mol CO + 0.156104 mol SO

= 0.615521

PCO = XCO x Ptotal = 0.615521 x 0.33 atm = 0.20312 = 0.20 atm CO 5.129 Plan: First, balance the equation. The pressure of N2 is found by subtracting the pressure of O2 from the total pressure. The pressure of the remaining 15% of O2 is found by taking 15% of the original O2 pressure. The molar ratio between O2 and SO2 in the balanced equation is used to find the pressure of the SO2 that is produced. Since pressure is directly proportional to moles of gas, the molar ratio may be expressed as a pressure ratio. Solution:

5-57

4FeS2(s) + 11O2(g) → 8SO2(g) + 2Fe2O3(s) Pressure of N2 = 1.05 atm – 0.64 atm O2 = 0.41 atm N2 Pressure of unreacted O2 = (0.64 atm O2) [(100 – 85)%/100%] = 0.096 atm O2

Pressure of SO2 produced = ( ) 22

2

8 atm SO0.64 atm O

11 atm O

= 0.46545 = 0.47 atm SO2

Total Pressure = (0.41 atm) + (0.096 atm) + (0.46545 atm) = 0.97145 = 0.97 atm total 5.130 Plan: V and T are not given, so the ideal gas law cannot be used. The total pressure of the mixture is given.

Use PA = XA x P total to find the mole fraction of each gas and then the mass fraction. The total mass of the two gases is 35.0 g. Solution:

P total = Pkrypton + Pcarbon dioxide = 0.708 atm The NaOH absorbed the CO2 leaving the Kr, thus Pkrypton = 0.250 atm. Pcarbon dioxide = P total – Pkrypton = 0.708 atm – 0.250 atm = 0.458 atm Determining mole fractions: PA = XA x P total

Carbon dioxide: CO2

total

= P

XP

= 0.458 atm0.708 atm

= 0.64689

Krypton: Kr

total

= PXP

= 0.250 atm0.708 atm

= 0.353107

Relative mass fraction = ( )

( ) 2

83.80 g Kr0.353107mol

44.01 g CO0.64689mol

= 1.039366

35.0 g = x g CO2 + (1.039366 x) g Kr 35.0 g = 2.039366 x Grams CO2 = x = (35.0 g)/(2.039366) = 17.16219581 = 17.2 g CO2 Grams Kr = 35.0 g – 17.162 g CO2 = 17.83780419 = 17.8 g Kr 5.131 As the car accelerates, the air in the car is pressed toward the back. The helium balloon floats on “top” of this denser air, which pushes it toward the front of the car. 5.132 Plan: Convert molecules of •OH to moles and use the ideal gas law to find the pressure of this

number of moles of •OH in 1 m3 of air. The mole percent is the same as the pressure percent as long as the other conditions are the same. Solution:

Moles of •OH = 12 3 3

3 232.5x10 molecules 1 mol 10 m

1 Lm 6.022 x10 molecules

−

= 4.151445x10–15 mol/L

V = 1.00 L T = 22°C + 273 =295 K P = unknown n = 4.151445x10–15 mol PV = nRT Solving for P:

Pressure of •OH = P = nRTV

= ( ) ( )15 L•atm4.151445x10 mol 0.0821 295 K

mol•K1.00 L

−

=1.005459x10–13 = 1.0x10–13 atm •OH

Mole percent = ( )131.005459x10 atm 100

1.00 atm

− = 1.005459x10–11 = 1.0x10–11%

5.133 Plan: Write the balanced equations. Use the ideal gas law to find the moles of SO2 gas and then use the molar

5-58

ratio between SO2 and NaOH to find moles and then molarity of the NaOH solution. Solution:

The balanced chemical equations are: SO2(g) + H2O(l) → H2SO3(aq) H2SO3(aq) + 2NaOH(aq) → Na2SO3(aq) + 2H2O(l) Combining these equations gives: SO2(g) + 2NaOH(aq) → Na2SO3(aq) + H2O(l)

V = 0.200 L T = 19°C + 273 =292 K P = 745 mmHg n = unknown


= 0.980263 atm


Moles of SO2 = ( ) ( )

( )

0.980263 atm 0.200 LL•atm0.0821 292 Kmol•K

PVn =RT

=

= 8.17799x10–3 mol SO2

Moles of NaOH = ( )32

2

2 mol NaOH8.17799x10 mol SO1 mol SO

−

= 0.01635598 mol NaOH

M NaOH = 3mol NaOH 0.01635598 mol NaOH 1 mL =

volume of NaOH 10.0 mL 10 L−

= 1.635598 = 1.64 M NaOH

5.134 Plan: Rearrange the formula PV = (m/M)RT to solve for molar mass. The mass of the flask and water and the density of water are used to find the volume of the flask and thus the gas. The mass of the condensed liquid

equals the mass of the gas in the flask. Pressure must be in units of atmospheres, volume in liters, and temperature in kelvins.

Solution: Mass of water = mass of flask filled with water – mass of flask = 327.4 g – 65.347 g = 262.053 g

Volume of flask = volume of water = ( )31 mL 10 L262.053 g water

0.997 g water 1 mL

−

= 0.2628415 L

Mass of condensed liquid = mass of flask and condensed liquid – mass of flask = 65.739 g – 65.347 g = 0.392 g V = 0.2628415 L T = 99.8°C + 273.2 = 373.0 K

P = 101.2 kPa m = 0.392 g M = unknown


= 0.998766 atm

=

mPV RTM


( ) ( )

( ) ( )

L•atm0.392 g 0.0821 373.0 Kmol•K =

0.998766 atm 0.2628415 LmRTPV

=M = 45.72781 = 45.7 g/mol

5.135 Plan: Write the balanced chemical equations. Convert mass of hydride to moles and use the molar ratio from the

balanced equation to find the moles of hydrogen gas produced. Use the ideal gas law to find the volume of that amount of hydrogen. Pressure must be in units of atm and temperature in kelvins. Solution:

LiH(s) + H2O(l) → LiOH(aq) + H2(g) MgH2(s) + 2H2O(l) → Mg(OH)2(s) + 2H2(g) LiOH is water soluble, however, Mg(OH)2 is not water soluble.

5-59

Lithium hydride LiH:

Moles H2 = ( )3

21 mol H1 kg 10 g 1 mol LiH1.00 lb LiH2.205 lb 1 kg 7.946 g LiH 1 mol LiH

= 57.0746 mol H2


P = 750. torr n = 57.0746 mol


= 0.98684 atm


V = ( ) ( )

( )

L•atm57.0746 mol 0.0821 300 Kmol•K =

0.98684 atmnRT

P

= 1424.493736 = 1420 L H2 from LiH

Magnesium hydride, MgH2

Moles H2 = ( )3

2 22

2 2

1 mol MgH 2 mol H1 kg 10 g1.00 lb MgH2.205 lb 1 kg 26.33 g MgH 1 mol MgH

= 34.44852 mol H2


P = 750. Torr = 0.98684 atm n = 34.44852 mol PV = nRT

Solving for V:

V = ( ) ( )

( )

L•atm34.44852 mol 0.0821 300 Kmol•K =

0.98684 atmnRT

P

= 859.7817758 = 8.60x102 L H2 from MgH2

5.136 Plan: Use the equation for root mean speed (u rms). Molar mass values must be in units of kg/mol and temperature

in kelvins. Solution:

u rms = 3 RTM

u rms Ne = ( ) ( )

( )3 2 23 8.314J/mol•K 370 K 10 g kg•m /s

20.18 g/mol kg J

= 676.24788 = 676 m/s Ne

u rms Ar = ( ) ( )

( )3 2 23 8.314J/mol•K 370 K 10 g kg•m /s

39.95 g/mol kg J

= 480.6269 = 481 m/s Ar

u rms He = ( ) ( )

( )3 2 23 8.314 J/mol•K 370 K 10 g kg•m /s

4.003 g/mol kg J

= 1518.356 = 1.52x103 m/s He

5.137 Plan: Use the ideal gas law to find the number of moles of CO2 and H2O in part a). The molar mass is then

used to convert moles to mass. Temperature must be in units of kelvins, pressure in atm, and volume in L. For part b), use the molar ratio in the balanced equation to find the moles and then mass of C6H12O6 that produces the number of moles of CO2 exhaled during 8 h. Solution:

a) V = 300 L T = 37.0°C + 273.2 = 310.2 K P = 30.0 torr n = unknown

Converting P from torr to atm: P = ( ) 1 atm30.0 torr760 torr

= 0.0394737 atm

5-60


Moles of CO2 = moles of H2O = ( ) ( )

( )

0.0394737 atm 300 LL•atm0.0821 310.2 Kmol•K

PVn =RT

=

= 0.464991 mol

Mass (g) of CO2 = ( ) 22

2

44.01 g CO0.464991 mol CO

1 mol CO

= 20.4643 = 20.5 g CO2

Mass (g) of H2O = ( ) 22

2

18.02 g H O0.464991 mol H O

1 mol H O

= 8.3791 = 8.38 g H2O

b) C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(g)

Moles of CO2 exhaled in 8 h = ( )20.464991 mol CO8 h

h

= 3.719928 mol CO2

Mass (g) of C6H12O6 = ( ) 6 12 6 6 12 62

2 6 12 6

1 mol C H O 180.16 g C H O3.719928 mol CO

6 mol CO 1 mol C H O

= 111.6970 = 1x102 g C6H12O6 (= body mass lost) (This assumes the significant figures are limited by the 8 h.) 5.138 Plan: For part a), the number of moles of water vapor can be found using the ideal gas law. Convert moles of

water to mass using the molar mass and adjust to the 1.6% water content of the kernel. For part b), use the ideal gas law to find the volume of water vapor at the stated set of condition.

Solution:

a) Volume of water in kernel = ( ) 275% H O0.25 mL kernel

100 % kernel

= 0.1875 mL = 1.875x10–4 L

V = 1.875x10–4 L T = 170°C + 273.2 = 443 K P = 9.0 atm n = unknown PV = nRT Solving for n:

Moles of H2O =( ) ( )

( )

49.0 atm 1.875 10 L


PVn =RT

−

=

= 4.639775x10–5 mol

Mass (g) of = ( )5 22

2

18.02 g H O 100%4.639775x10 mol H O1 mol H O 1.6%

−

= 0.052255 = 0.052 g

b) V = unknown T = 25°C + 273 = 298 K P = 1.00 atm n = 4.639775x10–5 mol PV = nRT

Solving for V:

V = ( ) ( )

( )

4 L•atm4.639775x10 mol 0.0821 298 Kmol•K =

1.00 atmnRT

P

− = 0.00113516 L = 1.13516 mL = 1.1 mL

5.139 Plan: For part a), find the SO2 volume in 4 GL of flue gas and take 95% of that as the volume of SO2 removed.

The ideal gas law is used to find the number of moles of SO2 in that volume. The molar ratio in the balanced equation is used to find the moles and then mass of CaSO4 produced. For part b), use the molar ratio in the balanced equation to calculate the moles of O2 needed to produce the amount of CaSO4 found in part a). Use the ideal gas law to obtain the volume of that amount of O2 and adjust for the mole fraction of O2 in air. Solution: a) Mole fraction SO2 = 1000(2x10–10) = 2x10–7

5-61

Volume of SO2 removed = ( ) ( )9

710 L 95%4 GL 2x101 GL 100%

−

= 760 L

V = 760 L T = 25°C + 273 = 298 K P = 1.00 atm n = unknown PV = nRT Solving for n:

Moles of SO2 =( ) ( )

( )

1.00 atm 760 LL•atm0.0821 298 Kmol•K

PVn =RT

=

= 31.063771mol

The balanced chemical equations are: CaCO3(s) + SO2(g) → CaSO3(s) + CO2(g) 2CaSO3(s) + O2(g) → 2CaSO4(s)

Mass (kg) of CaSO4 = ( ) 4 42 3

2 4

1 mol CaSO 136.14 g CaSO 1 kg31.063771 mol SO1 mol SO 1 mol CaSO 10 g

= 4.2290 = 4 kg CaSO4

b) Moles of O2 =

( )3

4 24

4 4

1 mol CaSO 1 mol O10 g4.2290 kg CaSO1 kg 136.14 g CaSO 2 mol CaSO

= 15.531805 mol O2

V = unknown T = 25°C + 273 = 298 K P = 1.00 atm n = 15.531766 mol PV = nRT

Solving for V:

Volume of O2 = V = ( ) ( )

( )

L•atm15.531805 mol 0.0821 298 Kmol•K =

1.00 atmnRT

P

= 379.998 L O2

Volume of air = ( )21379.998 L O

0.209

= 1818.1722 = 2000 L air

5.140 Plan: Use the ideal gas law to find the moles of gas occupying the tank at 85% of the 85.0 atm ranking. Then use van der Waals equation to find the pressure of this number of moles of gas. Solution:

a) V = 850. L T = 298 K

P = ( ) 80%85.0 atm100%

= 68.0 atm n = unknown


Moles of Cl2 =( ) ( )

( )

68.0 atm 850. LL•atm0.0821 298 Kmol•K

PVn =RT

=

= 2.36248x103 = 2.36x103 mol Cl2

b) V = 850. L T =298 K P = unknown n = 2.36248x103 mol Cl2 Van der Waals constants from Table 5.4:

a = 2

2atm•L6.49mol

; b = 0.0562 Lmol

( )2

2

+ − =

n aP V nb nRTV

5-62

PVDW = 2

2−−

nRT n aV nb V

PVDW = ( ) ( )

( )

( )( )

22332 22

23

2

atm•LL•atm 2.36248x10 mol Cl 6.492.36248x10 mol Cl 0.08206 298 K molmol•K L 850. L850. L 2.36248x10 molCl 0.0562

mol

−

−

= 30.4134 = 30.4 atm c) The engineer did not completely fill the tank. She should have filled it to (80.0%/100%)(85.0 atm) = 68 atm, but only filled it to 30.4 atm. 5.141 Plan: Solve the density form of the ideal gas law for molar mass. Temperature must be converted to kelvins and

pressure to atmospheres. Solution: P = 102.5 kPa T = 10.0°C + 273.2 = 283.2 K d = 1.26 g/L M = unknown


= 1.011596 atm

Pd =RTM

Rearranging to solve for molar mass:

( ) ( )

( )

L•atm1.26 g/L 0.0821 283.2 Kmol•K = =

1.011596 atmdRT

P

M = 28.9601 = 29.0 g/mol


1 1 2 2 =

PV P Vn T n T

or 1 1 2 22

2 1 1 =

PV n TVP n T

. R is fixed.

Solution: a) As the pressure on a fixed amount of gas (n is fixed) increases at constant temperature (T is fixed), the molecules move closer together, decreasing the volume. When the pressure is increased by a factor of 2, the volume decreases by a factor of 2 at constant temperature (Boyle’s law).

1 1 2 1 12

2 1 1

( )( )(1) = =

(2 )(1)PV T P VVP T P

V2 = ½V1

Cylinder B has half the volume of the original cylinder. b) The temperature is decreased by a factor of 2, so the volume is decreased by a factor of 2 (Charles’s law).

1 1 2 12

2 1

(1)( )(200 K) = =


V2 = ½ V1

Cylinder B has half the volume of the original cylinder. c) T1 = 100°C + 273 = 373 K T2 = 200°C + 273 = 473 K The temperature increases by a factor of 473/373 = 1.27, so the volume is increased by a factor of 1.27 (Charles’s law).

1 1 2 12

2 1

(1)( )(473 K) = =


V2 = 1.27V1

None of the cylinders show a volume increase of 1.27. d) As the number of molecules of gas increases at constant pressure and temperature (P and T are fixed), the force they exert on the container increases. This results in an increase in the volume of the container. Adding 0.1 mole of gas to 0.1 mole increases the number of moles by a factor of 2, thus the volume increases by a factor of 2 (Avogadro’s law).

5-63

1 1 2 2 12

2 1 1

(1)( )(0.2)(1) = =

(1)(0.1)(1)PV n T VVP n T

V2 = 2V1

Cylinder C has a volume that is twice as great as the original cylinder. e) Adding 0.1 mole of gas to 0.1 mole increases the number of moles by a factor of 2, thus increasing the volume by a factor of 2. Increasing the pressure by a factor of 2 results in the volume decreasing by a factor of ½. The two volume changes cancel out so that the volume does not change.

1 1 2 2 1 12

2 1 1 1

( )( )(0.2)(1) = =

(2 )(0.1)(1)PV n T P VVP n T P

V2 = V1

Cylinder D has the same volume as the original cylinder. 5.143 Plan: Use both the ideal gas law and van der Waals equation to solve for pressure. Convert mass of

ammonia to moles and temperature to Kelvin. Solution:

Ideal gas law: V = 3.000 L T = 0°C + 273 =273 K or 400.°C + 273 =673 K

P = unknown n = ( ) 33

3

1 mol NH51.1 g NH

17.03 g NH

= 3.0005872 mol


PIGL of NH3 at 0°C = nRTV

= ( ) ( )L•atm3.0005872 mol 0.0821 273 K

mol•K3.000 L

= 22.417687 = 22.4 atm

PIGL of NH3 at 400.°C = nRTV

= ( ) ( )L•atm3.0005872 mol 0.0821 673 K

mol•K3.000 L

= 55.264115 = 55.3 atm

van der Waals equation: V = 3.000 L T = 0°C + 273 =273 K or 400.°C + 273 =673 K

P = unknown n = ( ) 33

3

1 mol NH51.1 g NH

17.03 g NH

= 3.0005872 mol

van der Waals constants from Table 5.4:

a = 2

2atm•L4.17mol

; b = 0.0371 Lmol

( )2

2

+ − =

n aP V nb nRTV

PVDW = 2

2−−

nRT n aV nb V

PVDW of of NH3 at 0°C = ( ) ( )

( )

( )

( )

22

2

2


−

−

= 19.10997 = 19.1 atm NH3

5-64

PVDW of of NH3 at 400.°C = ( ) ( )

( )

( )

( )

22

2

2


−

−

= 53.2222 = 53.2 atm NH3 5.144 Plan: Since the mole fractions of the three gases must add to 1, the mole fraction of methane is found by subtracting the sum of the mole fractions of helium and argon from 1. Pmethane = Xmethane P total is used to calculate the pressure of methane and then the ideal gas law is used to find moles of gas. Avogadro’s number is needed to convert moles of methane to molecules of methane. Solution:

Xmethane = 1.00 – (Xargon + Xhelium) = 1.00 – (0.35 + 0.25) = 0.40 Pmethane = Xmethane P total = (0.40)(1.75 atm) = 0.70 atm CH4

V = 6.0 L T = 45°C + 273 =318 K P = 0.70 atm n = unknown PV = nRT Solving for n:

Moles of CH4 =( ) ( )

( )

0.70 atm 6.0 LL•atm0.0821 318 Kmol•K

PVn =RT

=

= 0.1608715 mol

Molecules of CH4 = ( )23

44

4

6.022x10 CH molecules0.1608715 mol CH1 mol CH

= 9.68768x1022 = 9.7x1022 molecules CH4 5.145 Plan: For part a), convert mass of glucose to moles and use the molar ratio from the balanced equation to find the

moles of CO2 gas produced. Use the ideal gas law to find the volume of that amount of CO2. Pressure must be in units of atm and temperature in kelvins. For part b), use the molar ratios in the balanced equation to calculate the moles of each gas and then use Dalton’s law of partial pressures to determine the pressure of each gas. Solution: a) C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(g)

Moles CO2: ( ) 6 12 6 26 12 6

6 12 6 6 12 6

1 mol C H O 6 mol CO20.0 g C H O

180.16 g C H O 1 mol C H O

= 0.666075 mol CO2

Finding the volume of CO2: V = unknown T = 37°C + 273 = 310 K

P = 780. torr n = 0.666075 mol


= 1.0263158 atm


V = ( ) ( )

( )

L•atm0.666075 mol 0.0821 310 Kmol•K =

1.0263158 atmnRT

P

= 16.5176 = 16.5 L CO2

This solution assumes that partial pressure of O2 does not interfere with the reaction conditions.

b) Moles CO2 = moles O2 = ( ) 6 12 66 12 6

6 12 6 6 12 6

1 mol C H O 6 mol 10.0 g C H O180.16 g C H O 1 mol C H O

= 0.333037 mol CO2 = mol O2 At 37°C, the vapor pressure of water is 48.8 torr. No matter how much water is produced, the partial pressure of H2O will still be 48.8 torr. The remaining pressure, 780 torr – 48.8 torr = 731.2 torr is the sum of partial pressures

5-65

for O2 and CO2. Since the mole fractions of O2 and CO2 are equal, their pressures must be equal, and must be one-half of 731.2 torr.

Pwater = 48.8 torr (731.2 torr)/2 = 365.6 = 3.7x102 torr Poxygen = Pcarbon dioxide 5.146 Plan: Use the equations for average kinetic energy and root mean speed (u rms) to find these values for N2 and H2.

Molar mass values must be in units of kg/mol and temperature in kelvins. Solution:

The average kinetic energies are the same for any gases at the same temperature: Average kinetic energy = 3/2RT = (3/2)(8.314 J/mol•K) (273 K) = 3.40458x103 = 3.40x103 J for both gases

rms speed:

T = 0°C + 273 = 273 K M of N2 = 23

28.02 g N 1 kgmol 10 g

= 0.02802 kg/mol

M of H2 = 23

2.016 g H 1 kgmol 10 g

= 0.002016 kg/mol

R = 8.314 J/mol•K 1 J = kg•m2/s2

u rms = 3RTM

urmsN2 =( ) ( )

( )2 23 8.314J/mol•K 273 K kg•m /s

0.02802 kg/mol J

= 4.92961x102 = 4.93x102 m/s N2

urms H2 = ( ) ( )( )

2 23 8.314J/mol•K 273K kg•m /s0.002016 kg/mol J

= 1.83781x103 = 1.84x103 m/s H2

5.147 Plan: Use the relationship between mole fraction and partial pressure, PA = XA P total, to find the mole fraction of

each gas in parts a) and b). For parts c) and d), use the ideal gas law to find the moles of air in 1000 L of air at these conditions and compare the moles of each gas to the moles of air. Mass and molecules must be converted to moles. Solution: a) Assuming the total pressure is 1 atm = 760 torr. PA = XA P total

22

BrBr

total

0.2 torr = = 760 torr

PX

P = 2.6315789x10–4x(106) = 263.15789 = 300 ppmv Unsafe

b) 22

COCO

total

0.2 torr = = 760 torr

PX

P = 2.6315789x10–4x(106) = 263.15789 = 300 ppmv Safe

(0.2 torr CO2/760 torr)(106) = 263.15789 = 300 ppmv CO2 Safe

c) Moles Br2 = ( ) 22

2

1 mol Br0.0004 g Br

159.80 g Br

= 2.5031x10–6 mol Br2 (unrounded)

Finding the moles of air: V = 1000 L T = 0°C + 273 =273 K P = 1.00 atm n = unknown PV = nRT

Moles of air =( ) ( )

( )

1.00 atm 1000 LL•atm0.0821 273 Kmol•K

PVn =RT

=

= 44.616 mol air (unrounded)

Concentration of Br2 = mol Br2/mol air(106) = [(2.5031x10–6 mol)/(44.616 mol)] (106) = 0.056103 = 0.06 ppmv Br2 Safe

5-66

d) Moles CO2 = ( )22 22 23

2

1 mol CO2.8x10 molecules CO6.022x10 molecules CO

= 0.046496 mol CO2

Concentration of CO2 = mol CO2/mol air(106) = [(0.046496 mol)/(44.616 mol)] (106) = 1042.1 = 1.0x103 ppmv CO2 Safe 5.148 Plan: For part a), use the ideal gas law to find the moles of NO in the flue gas. The moles of NO are converted

to moles of NH3 using the molar ratio in the balanced equation and the moles of NH3 are converted to volume using the ideal gas law. For part b), the moles of NO in 1 kL of flue gas is found using the ideal gas law; the molar ratio in the balanced equation is used to convert moles of NO to moles and then mass of NH3.

Solution: a) 4NH3(g) + 4NO(g) + O2(g) → 4N2(g) + 6H2O(g) Finding the moles of NO in 1.00 L of flue gas: V = 1.00 L T = 365°C + 273 =638 K P = 4.5x10–5 atm n = unknown PV = nRT Solving for n:

Moles of NO =( ) ( )

( )

54.5x10 atm 1.00 L


PVn =RT

−

=

= 8.5911x10–7 mol NO

Moles of NH3 = ( )7 34 mol NH8.5911x10 mol NO4 mol NO

−

= 8.5911x10–7 mol NH3

Volume of NH3: V = unknown T =365°C + 273 =638 K P = 1.00 atm n = 8.5911x10–7 mol PV = nRT

Solving for V:

V = ( ) ( )

( )

7 L•atm8.5911x10 mol 0.0821 638 Kmol•K =

1.00 atmnRT

P

− = 4.5000x10–5 = 4.5x10–5 L NH3

b) Finding the moles of NO in 1.00 kL of flue gas: V = 1.00 kL = 1000 L T = 365°C + 273 =638 K P = 4.5 x 10–5 atm n = unknown PV = nRT Solving for n:

Moles of NO =( ) ( )

( )

54.5x10 atm 1000 L


PVn =RT

−

=

= 8.5911x10–4 mol NO

Moles of NH3 = ( )4 34 mol NH8.5911x10 mol NO4 mol NO

−

= 8.5911x10–4 mol NH3

Mass of NH3 = ( )4 33

3

1 mol NH8.59x10 mol NH17.03 g NH

−

= 0.014631 = 0.015 g NH3

5.149 Plan: Use Graham’s law to compare effusion rates. Solution:

Rate NeRate Xe

= molar mass Xemolar mass Ne

= 131.3 g/mol20.18 g/mol

= 2.55077

1 enrichment factor (unrounded)

5-67

Thus XNe = moles of Ne 2.55077 mol = moles of Ne + moles of Xe 2.55077 mol + 1 mol

= 0.71837 = 0.7184

5.150 Plan: To find the number of steps through the membrane, calculate the molar masses to find the ratio of effusion rates. This ratio is the enrichment factor for each step.

Solution:

235 UF6

238 UF6

Rate

Rate =

2386

2356

molar mass UFmolar mass UF

= 352.04 g/mol349.03 g/mol

= 1.004302694 enrichment factor Therefore, the abundance of 235UF6 after one membrane is 0.72% x 1.004302694 Abundance of 235UF6 after “N” membranes = 0.72% x (1.004302694)N Desired abundance of 235UF6 = 3.0% = 0.72% x (1.004302694)N

Solving for N: 3.0% = 0.72% x (1.004302694)N 4.16667 = (1.004302694)N ln 4.16667 = ln (1.004302694)N ln 4.16667 = N x ln (1.004302694) N = (ln 4.16667)/(ln 1.004302694)

N = 1.4271164/0.004293464 = 332.39277 = 332 steps 5.151 Plan: Use van der Waals equation to calculate the pressure of the gas at the given conditions. Solution:

V = 22.414 L T = 273.15 K P = unknown n = 1.000 mol Van der Waals constants from Table 5.4:

a = 2

2atm•L1.45mol

; b = 0.0395 Lmol

( )2

2

+ − =

n aP V nb nRTV

PVDW = 2

2−−

nRT n aV nb V

PVDW = ( ) ( )

( )

( )

( )

22

2

2

atm•LL•atm 1.000 mol CO 1.451.000 mol CO 0.08206 273.15 K molmol•K L 22.414 L22.414 L 1.000 mol CO 0.0395

mol

−

−

= 0.998909977 = 0.9989 atm 5.152 Plan: The amount of each gas that leaks from the balloon is proportional to its effusion rate. Using 35% as the rate for H2, the rate for O2 can be determined from Graham’s law.

Solution:

2

2

Rate ORate H

= 2

2

molar mass Hmolar mass O

= 2.016 g/mol32.00 g/mol

= 2rate O35

0.250998008 = 2rate O35

Rate O2 = 8.78493 Amount of H2 that leaks = 35%; 100–35 = 65% H2 remains

5-68

Amount of O2 that leaks = 8.78493%; 100–8.78493 = 91.21507% O2 remains

2

2

O 91.21507 = H 65

= 1.40331 = 1.4

5.153 Plan: For part a), put together the various combinations of the two isotopes of Cl with P and add the masses.

Multiply the abundances of the isotopes in each combination to find the most abundant for part b). For part c), use Graham’s law to find the effusion rates.

Solution: a) Options for PCl3:

All values are g/mol P First Cl Second Cl Third Cl Total 31 35 35 35 136 31 37 35 35 138 31 37 37 35 140 31 37 37 37 142 b) The fraction abundances are 35Cl = 75%/100% = 0.75, and 37Cl = 25%/100% = 0.25. The relative amount of each mass comes from the product of the relative abundances of each Cl isotope. Mass 136 = (0.75) (0.75) (0.75) = 0.421875 = 0.42 (most abundant) Mass 138 = (0.25) (0.75) (0.75) = 0.140625 = 0.14 Mass 140 = (0.25) (0.25) (0.75) = 0.046875 = 0.047 Mass 142 = (0.25) (0.25) (0.25) = 0.015625 = 0.016

c) 37

335

3

Rate P ClRate P Cl

= 35

337

3

molar mass P Clmolar mass P Cl

= 136 g/mol142 g/mol

= 0.978645 = 0.979 5.154 Plan: Use the combined gas law for parts a) and b). For part c), use the ideal gas law to find the moles of air represented by the pressure decrease and convert moles to mass using molar mass. Solution:

a) 1 1 2 2

1 2 =

PV P VT T

( )( )1 1 2 1 22

2 1 1

35.0 psi 318 K= = =

295 KP V T PTPV T T

= 37.7288 = 37.7 psi

b)New tire volume = V2 = V1(102%/100%) = 1.02V1 1 1 2 2

1 2 =

PV P VT T

( )( )( )( )( )

11 1 2 1 22

2 1 1 1

35.0 psi 318 K= = =

1.02 295 KVPV T PTP

V T T V= 36.9890 = 37.0 psi

c) Pressure decrease = 36.9890 – 35.0 psi = 1.989 psi; ( ) 1 atm1.989 psi14.7 psi

= 0.135306 atm

V = 218 L T = 295 K P = 0.135306 atm n = unknown PV = nRT Solving for n:

Moles of air lost =( ) ( )

( )

0.135306 atm 218 LL•atm0.0821 295 Kmol•K

PVn =RT

=

= 1.217891 mol

5-69

Time = ( ) 28.8 g 1 min1.217891 mol1 mol 1.5

= 23.384 = 23 min

5.155 Plan: Rearrange the ideal gas law to calculate the density of O2 and O3 from their molar masses. Temperature

must be converted to kelvins and pressure to atm. Solution: P = 760 torr = 1.00 atm T = 0°C + 273 = 273 K

M of O2 = 32.00 g/mol d = unknown M of O3 = 48.00 g/mol PV = nRT Rearranging to solve for density:

d of O2 =( ) ( )

( )


PRT

=

M = 1.42772 = 1.43 g O2/L

d of O3 =( ) ( )

( )


PRT

=

M = 2.141585576 = 2.14 g O3/L

b) dozone/doxygen = (2.141585576)/(1.42772) = 1.5 The ratio of the density is the same as the ratio of the molar masses. 5.156 Plan: Since the amounts of two reactants are given, this is a limiting reactant problem. Write the balanced

equation and use molar ratios to find the number of moles of IF7 produced by each reactant. The mass of I2 is converted to moles using its molar mass and the moles of F2 is found using the ideal gas law. The smaller number of moles of product indicates the limiting reagent. Determine the moles of excess reactant gas and the moles of product gas and use the ideal gas law to solve for the total pressure.

Solution: Moles of F2: V = 2.50 L T = 250. K

P = 350. torr n = unknown


= 0.460526315 atm


( ) ( )

( )

0.460526315 atm 2.50 LL•atm0.0821 250. Kmol•K

PVn =RT

=

= 0.056093339 mol F2

7F2(g) + I2(s) → 2IF7(g

Moles IF7 from F2 = ( ) 72

2

2 mol IF0.056093339 mol F

7 mol F

= 0.016026668 mol IF7 (unrounded)

Moles IF7 from I2 = ( ) 722

2 2

2 mol IF1 mol I2.50 g I

253.8 g I 1 mol I

= 0.019700551 mol IF7 (unrounded)

F2 is limiting. All of the F2 is consumed. Mole I2 remaining = original amount of moles of I2 – number of I2 moles reacting with F2

Mole I2 remaining = ( ) 22

2

1 mol I2.50 g I

253.8 g I

– ( ) 22

2

1 mol I0.056093339 mol F

7 mol F

= 1.83694x10–3 mol I2

Total moles of gas = (0 mol F2) + (0.016026668 mol IF7) + (1.83694x10–3 mol I2) = 0.0178636 mol gas V = 2.50 L T = 550. K

P = unknown n = 0.0178636 mol

5-70


P (atm) = nRTV

= ( ) ( )L•atm0.0178636 mol 0.0821 550. K

mol•K2.50 L

= 0.322652 atm

P (torr) = ( ) 760 torr0.322652 atm1 atm

= 245.21552 = 245 torr

P iodine (torr) = X iodine P total = [(1.83694x10–3 mol I2)/(0.0178636 mol)] (245.21552 torr) = 25.215869 = 25.2 torr

Documents

CHAPTER 5 GASES AND THE KINETIC-MOLECULAR THEORY · 5-1 CHAPTER 5 GASES AND THE KINETIC-MOLECULAR THEORY FOLLOW–UP PROBLEMS . 5.1A . Plan: Use the equation for gas pressure in an