The Gas Laws-Part II Applications

Avogadro’s LawAmedeo Avogadro (1776–1856)

“Equal volumes of gases, under the same conditions

of temperature and pressure, contain the same

number of molecules.”

Avogadro’s LawAmedeo Avogadro (1776–1856)

Volume is directly proportional to the number of gas molecules (constant P and T)

More gas molecules = larger volume

Equal volumes of gases contain equal numbers of molecules.

The gas doesn’t matter.

V ∝ n

V = kn

Avogadro’s Law At constant T and P, the volume of a gas increases proportionately as its molar amount increases. If the molar amount is doubled, the volume is doubled.

add gas

remove gas

P= 1.0 atm

V= 44.8 LV= 22.4 L

P= 1.0 atm

n = 1.0 mol n = 2.0 mol

V vs nVo

lum

e(L)

0

100

200

300

400

number of moles (n)0 2 4 6 8 10

Avogadro’s Law

V ∝ nV = kn

6. If 1.00 mole of a gas occupies 22.4 L and the temperature and pressure are not changed, what volume would

0.750 moles occupy?

V1 =22.4 L, n1 = 1.00 mol, n2 = 0.750 mol V2 = ?

V1, n1, n2 V2

V1n1

V2n2

=

(V1)(n2)(n1)

V2 = = 16.8 L(22.4 L)(0.750 mol)(1.00 mol)=

Standard Conditions

Because the volume of a gas varies with pressure and temperature, chemists have agreed on a set of conditions to report our measurements so that comparison is easy – we call these standard conditions - STP

Standard pressure = 1 atm Standard temperature = 273 K, 0°C

(1.00 atm)(22.414 L) (1.00 mole)(273.15 K)

V ∝ 1 P V ∝ T V ∝ n

V ∝ nT P V =(k)nT P

k= = R =PV nT

R = 0.0821(atm)(L)/(mol)(K)

Ideal Gas LawBy combining the gas laws we can write a general equation:

R is called the gas constant.

The value of R depends on the units of P and V.

We will use 0.0821 and convert P to atm and V to L. atm · L mol· K

= nRTP

V V= nRTP

V =R TP

nV

= RT

Pn

V= T

RPn

(L)(K)

(atm)(mole)

“R”

7. Calculate the volume occupied by 637 g of SO2 (MM 64.07) at 6.08 x 104 mmHg and –23 °C.

mSO2 = 637 g, P = 6.08 x 104 mmHg, t = −23 °C, V = ?

K = 273+ºC = 273+(-23) = 250

g n

1.00 mol64.07 g SO2

P, n, T, R V

(n)(R)(T)PV =

(9.942 mol)(0.0821 )(250 K)80.0 atm

1.00 mol SO264.07 g SO2

637 g SO2 x = 9.942 mol SO2

1.00 atm760 mm Hg6.08 x 104 mmHg x = 80.0 atm

atm Lmol K

= 2.55L

8. A gas occupies 10.0 L at 44.1 psi and 27 °C. What volume will it occupy at standard conditions?

V1 = 10.0L, P1 = 44.1 psi, t1 = 27 °C, P2 = 1.00 atm, t2 = 0 °C, V2 = ?

P1, V1, T1, R n

n = PVRT V = nRT

P

P2, n, T2, R V2

1.00 atm14.7 psi

44.1 psi x = 3.00 atm

T1(K) = 27ºC + 273.15 = 300. K

n = PVRT

= 1.218 mol

atm Lmol K

(3.00 atm)(10.0 L)(0.0821 )(300. K)=

= 1.22 mol

8. A gas occupies 10.0 L at 44.1 psi and 27 °C. What volume will it occupy at standard conditions?

n =1.218 mol

P1, V1, T1, R n

n = PVRT V = nRT

P

P2, n, T2, R V2

V = (1.22 mol)(.0821 )(273 K)1.00 atm

atm Lmol K

= 27.3 L

8. A gas occupies 10.0 L at 44.1 psi and 27 °C. What volume will it occupy at standard conditions?

P1 = 3.00 atmV1 = 10.0 LT1 = 300 KP2 = 1.00 atmT2 = 273 KV2 = ?

P1V1 T1

P2V2 T2

=

P1V1T2 T1P2

V2 =

(3.00 atm)(10.1L)(273 K) (300 K)(1.00 atm)V2 =

V2 = 27.3 L

This is a comparison of two sets of conditions; you can use the combined gas law.

Molar Mass of a Gas

One of the methods chemists use to determine the molar mass of an unknown substance is to heat a weighed sample until it becomes a gas,

measure the temperature, pressure, and volume, and use the ideal gas law.

Determining the Molar Mass of an Unknown Volatile Liquid.

JBA Dumas (1800-1884)

n = PVRT

molar mass = gramsmole (n)

Mixtures of Gases

When gases are mixed together, their molecules behave independently, but

All the gases in the mixture have the same volume.

All gases in the mixture are at the same temperature.

In certain applications, the mixture can be thought of as one gas

We can calculate the total moles of molecules in an air sample, knowing P, V, and T, even though they are different molecules.

Partial PressureThe pressure of a single gas in a mixture of gases is

called its partial pressure.

We can calculate the partial pressure of a gas if

we know what fraction of the mixture it composes and the total pressure,or

we know the number of moles of the gas in a container of known volume and temperature.

Dalton’s Law of Partial Pressures - The sum of the partial pressures of all the gases in the mixture equals the total pressure

Dalton's Law of Partial Pressures

+Maintain volume at 5.00 L

Ptot = 3.90 atm, V = 8.70 L, T = 598 K, nXe = 0.175 mol , nNe = ?

9. A mixture of neon and xenon has a volume of 8.70 L at temperature of 598 K and 3.90 atm pressure. If the mixture contains 0.175 mol of xenon, how many moles of neon are present ?

nXe,V, T, R PXe

PXe =(0.175 mol)(.0821 )(598 K)8.70 atm

atm Lmol K

PXe = 0.988 atm

PXe = (nXe)(R)(T)V

Ptotal, PXe PNe

PNe = Ptotal - PXe

PNe = 3.90 atm-0.988 atm

PNe = 2.91 atm

nNe = (PXe)(V)(R)(T)

Ptot = 3.90 atm, V = 8.70 L, T = 598 K, nXe = 0.175 mol , nNe = ?

9. A mixture of neon and xenon has a volume of 8.70 L at temperature of 598 K and 3.90 atm pressure. If the mixture contains 0.175 mol of xenon, how many moles of neon are present ?

nXe,V, T, R PXe Ptotal, PXe PNe

nNe = (PNe)(V)(R)(T)

PXe = 0.988 atm PNe = 2.91 atm

PNe,V, T, R nNe

= (2.91 atm)(8.70 L)(0.0821 )(598 K)atm L

mol K = 0.516 mol Ne

Collecting Gas by Water Displacement

A chemical process

generates a gas.

Collection of a Gas Over Water

It is necessary to adjust the position of the bottle so that the water levels inside and outside the bottle are the same.

This makes the total pressure of the gaseous mixture in the bottle equal to

the atmospheric pressure as measured with a barometer.

Pgas = Pbar - Pwater

Pgas + Pwater

A chemical process

generates a gas.

Collection of a Gas Over Water

It is necessary to adjust the position of the bottle so that the water levels inside and outside the bottle are the same.

This makes the total pressure of the gaseous mixture in the bottle equal to

the atmospheric pressure as measured with a barometer.

Pgas = Pbar - Pwater

Pgas + Pwater

Vapor Pressure of Water vs Temperature

If you collect a gas sample with a total pressure of 758.2 mmHg* at 25 °C, the partial pressure of the water vapor will be 23.78 mmHg – so the partial pressure of the dry gas will be 734.4 mmHg. This is the pressure you would use to calculate the amount of gas present.

10. 1.02 L of O2 collected over water at 20ºC with a total pressure of 755.2 mmHg. Find the mass of O2 collected.

V=1.02 L, P=755.2 mmHg, T=293 K, mass O2 = ?

Ptotal = PO2 + PH2O @20ºC

PO2 = Ptotal - PH2O @20ºC

PO2 = 755.2 mmHg - 17.55 mmHg

PO2 = 737.7 mmHg1 atm

760.0 mmHg PO2 = 737.7 mmHg x

PO2 = 0.9706 atm

n = (P)(V)(R)(T)

n = (0.9706 atm)(1.02 L)(0.0821 )(293 K)atm L

mol K

n = 0.0412 mol

0.0412 mol x 32.00 g O21.00 mol O2

= 1.32 g O2

Ptotal , PH2O PO2 PO2,V, T nO2 gO2

Reactions Involving Gases

The principles stoichiometry can be combined with the gas laws for reactions involving gases.

In reactions of gases, the amount of a gas is often given as a volume instead of moles.

The ideal gas law allows us to convert from the volume of the gas to moles; then we can use the coefficients in the equation as a mole ratio.

P,V,T of Gas A mol of Gas A mol of Gas B P,V,T of Gas B

The Big Picture of Stoichiometry

Moles of A

Moles of B

Mole to

Mole Ratio*

Grams of A

Grams of B

Molar Mass

Particles of A

Particles of B

Avogadro’s Number

Liters of a Solution of A

Liters of a Solution of B

MolarityMolarity

grams or L of gas A

grams or L of gas B

Gas Laws Gas Laws

10.0 g HgO X X

11. What volume of O2 at 0.750 atm and 313 K is generated by the thermolysis of 10.0 g of HgO? 2 HgO(s) → 2 Hg(l) + O2(g)

mHgO = 10.0g; P = 0.750 atm; T = 313 K VO2 = ?

g HgO mol HgO mol O2

1 mol HgO216.59 g

1 mol O22 mol HgO

1 mol HgO216.59 g HgO

1 mol O22 mol HgO

= 0.0231 mol O2

V = (0.0231 mol)(.0821 )(313 K)0.750 atm

atm Lmol K

= 0.791 L O2

V = (n)(R)(T)P

P,n,T,R V

12. What volume of H2 is needed to make 37.5 g of CH3OH at 738 mmHg and 355 K? CO(g) + 2 H2(g) → CH3OH(g)

mCH3OH = 37.5g, P = 738 mmHg, T=355 K , VH2 = ?

g CH3OH mol CH3OH mol H2 P,n,T,R V

1 mol CH3OH32.04 g CH3OH

2 mol H21 mol CH3OH V = (n)(R)(T)

P

37.5 g CH3OH X

X

= 2.23 mol H2

1 mol CH3OH32.04 g CH3OH

2 mol H21 mol CH3OH V = (2.23 mol)(.0821 )(355 K)

0.971 atm

atm Lmol K

= 66.9 L H2

738 mm Hg X 1 atm760 mm Hg

= 0.971 atm

Molar Volume

Solving the ideal gas equation for the volume of 1 mol of gas at STP gives 22.4 L

for 6.022 x 1023 molecules of any gas

We call the volume of 1 mole of gas at STP the molar volume.

It is important to recognize that one mole measures of different gases have different masses, even though they have the same volume.

Molar Volume

13. How many liters of O2 @ STP can be made from the decomposition of 100.0 g of PbO2? 2 PbO2(s) → 2 PbO(s) + O2(g) (PbO2 , 239.2 g/mol)

100.0 g PbO2, LO2 = ?g PbO2 mol PbO2 mol O2 L O2

1 mol PbO2239.2 g PbO2

1 mol O22 mol PbO2

22.4 L O21 mol O2

= 4.68 L O2

14. How many liters of H2O form when 1.24 L H2 reacts completely with O2 @ STP? O2(g) + 2 H2(g) → 2 H2O(g)

VH2 = 1.24 L, P = 1.00 atm, T = 273 K L H2O (g) = ?

L H2 mol H2 mol H2O L H2O

1 mol H222.4 L H2

2 mol H2O2 mol H2

22.4 L H21 mol H2O

1.24 L H2 X X X 1 mol H2 (g)22.4 L H2 (g)

2 mol H2O (g) 2 mol H2 (g)

22.4 L H2O (g)1 mol H2O (g)

= 1.24 L H2O (g)

4 NH3 (g) + 5 O2 (g) 4 NO (g) + 6 H2O (g)

15. Consider the reaction:

How many liters of NO are produced from 10 L of NH3 at STP?

How many liters of NO are produced from 10 L of O2 at STP?

How many liters of H2O are produced from 10 L of NH3 at STP?

How many liters of H2O are produced from 10 L of O2 at STP?

Does the “STP” matter?