When you can measure what you are speaking about and

2/3/2003 OFB Chapter 5 1

When you can measure what you are speaking about and express it in numbers, you know something about it; but when you cannot measure it, when you cannot express it in numbers, you knowledge is of a meager and unsatisfactory kind; it may be the beginning of knowledge but you have scarcely, in your thoughts advanced to the stage of science, whatever the matter may be.

--Lord Kelvin, May 3rd, 1883


OFB Chapter 5

The Gaseous State

5-1 The Chemistry of Gases

5-2 Pressure and Boyle’s Law

5-3 Temperature and Charles’s Law

5-4 The Ideal Gas Law

5-5 Chemical Calculations for Gases

5-6 Mixtures of Gases

5-7 Real Gases



OFB Chapter 5

The Gaseous StateEarly discoveries of gases formed by chemical reactions:

2 HgO(s) → 2 Hg(l) + O2(g)

Lavoisier used this to establish the conservation of mass theory

heat

Marble: CaCO3(s) → CaO(s) + CO2(g)heat

NH4Cl(s) → HCl(g) + NH3(g)heat

Nitroglycerin: 4 C3H5(NO3)3(l) →

6 N2(g) + 12 CO2(g) + O2(g) + 10 H2O(g)

CaCO3(s) + HCl(aq) →

CaCl2(aq) + H2O(g) + CO2(g)


Pressure and Boyle’s Law• In physics, a force (F) is a simple push

exerted by one object on another.

• Applying a force to a stationary object sets it in motion, unless the object pushes back with an equal force.

• Applying a force to an object that pushes back creates a pressure (P) on the object.

• Applying a force to an object that pushes back creates a pressure (P) on the object.

• The pressure equals the force divided by the area (A) over which the force is applied:


Pressure and Boyle’s Law

force (F) = mass * acceleration = Newton (N) = kg m s-2

area (A) = m2

acceleration (a) = velocity per unit of time [m s-2]

mass (m) = quantity of matter [kg]





P = gdh

g = acceleration of gravity at the surface of the Earth

= 9.80665 m s-2

d = density of the liquid = for Hg at 0ºC

= 13.5951 g cm-3 = 13.5951 kg m-3

h = height of mercury in the column

= 76 cm = 760 mm = 0.76 m

P = gdh = (9.80665 m s-2)(13.5951 kg m-3)

(0.76 m)


A pressure of 101.325 kPa is need to raise the column of

Hg 760 mm or 76 cm

Called standard pressure


Boyle’s Law: The Effect of Pressure on Gas Volume

The product of the pressure and volume, PV, of a sample of gas is a constant at a constant temperature:


Boyle’s Law: The Effect of Pressure on Gas Volume

STP

For 1 mole of any gas

(i.e., 32.0 g of O2; 28.0 g N2; 2.02 g H2),

STP = standard temperature and pressure= 0oC and 1 atm


Boyle’s Law: The Effect of Pressure on Gas VolumeExercise 5-3The long cylinder of a bicycle pump has a volume of 1131 cm3 and is filled with air at a pressure of 1.02 atm. The outlet valve is sealed shut, and the pump handle is pushed down until the volume of the air is 517 cm3. The temperature of the air trapped inside does not change. Compute the pressure inside the pump.


Charles’ Law: The Effect of Temperature on Gas Volume

Temperature and Charles’ Law

−°=

°1)

VV C273.15t



Absolute Temperature

V = Vo ( 1 + )t273.15oC

Kelvin temperature scale




Exercise 5-4

The gas in a gas thermometer that has been placed in a furnace has a volume that is 2.56 times larger than the volume that it occupies at 100oC. Determine the temperature in the furnace (in degrees Celsius).


V1 / V2 = T1 / T2(at a fixed pressure)

P1V1 = P2V2(at a fixed

temperature)

Boyle’s Law

Charles’ Law

(at a fixed pressure and temperature)Avogadro


The Ideal Gas Law

V ∝ nTP-1


The Ideal Gas Law

11

11

TnVPR =

22

22

TnVPR =


Exercise 5-5At one point during its ascent, a weather balloon filled with helium at a volume of 1.0 × 104 L at 1.00 atm and 30oC reaches an altitude at which the temperature is -10oC yet the volume is unchanged. Compare the pressure at that point.


The Ideal Gas Law

R universal gas constant ?for fixed V, P, and T, the number of n is fixed as well, and independent of the particular gas studied

nTPVR =

5K)mol)(273.1 (1.00)m N 10)(101.325xm 10 x (22.414 R

-233-3

=

K) 5mol)(273.1 (1.001atm)(22.414L)(

=


Exercise 5-6

What mass of Hydrogen gas is needed to fill a weather balloon to a volume of 10,000 L at 1.00 atm and 30°C?


Exercise 5-6

What mass of Hydrogen gas is needed to fill a weather balloon to a volume of 10,000 L at 1.00 atm and 30°C?


Gas Density and Molar Mass

RTMmPV

nRTPV

=

=


Gas Density and Molar Mass

Exercise 5-7Calculate the density of gaseous hydrogen at a pressure of 1.32 atm and a temperature of -45oC.


Molar Mass

RTMmPV

nRTPV

=

=


Exercise 5-8Fluorocarbons are compounds of fluorine and carbon. A 45.60 g sample of a gaseous fluorocarbon contains 7.94 g of carbon and 37.66 g of fluorine and occupies 7.40 L at STP (P = 1.00 atm and T = 273.15 K). Determine the approximate molar mass of the fluorocarbon and give its molecular formula.


Exercise 5-8Fluorocarbons are compounds of fluorine and carbon. A 45.60 g sample of a gaseous fluorocarbon contains 7.94 g of carbon and 37.66 g of fluorine and occupies 7.40 L at STP (P = 1.00 atm and T = 273.15 K). Determine the approximate molar mass of the fluorocarbon and give its molecular formula.

1

11

mol g 138M

1atm273K x Kmol atm L 0.082

7.40L45.60g

−

−−

=

PRTdM =

F mol 1.982F g 19F 1mol x F g 37.66n

C mol 0.661C g 12C 1mol x C g 7.94n

F

C

=

=

=

=

Cpart 1mol 0.661 =÷

F parts 3mol 0.661 =÷


Why use Volume for gases in chemical reaction calculations?The volume of a gas is easier to

measure than the mass of a gas.Exercise 5-9

Ethylene burns in oxygen:

C2H4(g) + 3 O2(g) → 2 CO2(g) + 2H2O(g)

A volume of 3.51 L of C2H4(g) at a temperature of 25oC and a pressure of 4.63 atm reacts completely with O2(g). The water vapor is collected at a temperature of 130oC and a pressure of 0.955 atm. Calculate the volume of the water vapor.

Chemical Calculations for Gases


Exercise 5-9

Ethylene burns in oxygen:

C2H4(g) + 3 O2(g) → 2 CO2(g) + 2H2O(g)

A volume of 3.51 L of C2H4(g) at a temperature of 25oC and a pressure of 4.63 atm reacts completely with O2(g). The water vapor is collected at a temperature of 130oC and a pressure of 0.955 atm. Calculate the volume of the water vapor.

22

22

11

11

TnVP

TnVP

=

xV

403KT

atm 0.955P

298KT

3.51LV

4.63atmP

n 2nOH parts 2for HC1part

OH is 2Conditon andHC is 1Condition

OH

OH

OH

HC

HC

HC

0HHC

242

2

42

2

2

2

42

42

42

242

=

=

=

=

=

=

= L 46.0V OH2=

OHOH

OHOH

HCHC

HCHC

TnVP

TnVP

22

22

4242

4242 =


Exercise 5-10

Hydrazine (N2H4), a rocket fuel, is prepared by the reaction of ammonia with a solution of sodium hypochlorite:

2 NH3(g) + NaOCl(aq) →

N2H4(aq) + NaCl(aq) + H2O(g)

What volume of gaseous ammonia at a temperature of 10oC and a pressure of 3.63 atm is required to produce 15.0 kg of hydrazine according to this equation.


Exercise 5-10Hydrazine (N2H4), a rocket fuel, is prepared by the reaction of ammonia with a solution of sodium hypochlorite:

2 NH3(g) + NaOCl(aq) →

N2H4(aq) + NaCl(aq) + H2O(g)

What volume of gaseous ammonia at a temperature of 10oC and a pressure of 3.63 atm is required to produce 15.0 kg of hydrazineaccording to this equation.

342 NH mol 2HN mol 1nRTPV

==

( )

L 10 x 6.4V

3.63atm283K x 0.082

HN mol g 30.04HN g 10 x 152V

PTR

Mm

2V

PTR2n

PTRn

V

3NH

421-

423

NH

NH

NHNH

HN

HNNH

NH

NHNHHN

NH

NHNHNHNH

3

3

3

33

42

42

3

3

3342

3

333

3

=

=

=

==

PRT

Mm

PnRTV

==


Mixtures of Gases

Dalton’s Law of Partial Pressures

The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases.


Mole Fractions and Partial PressuresThe mole faction of a component in a mixture is define as the number of moles of the components that are in the mixture divided by the total number of moles present.

NBA

A

tot

AA

A

n...nnn

nnX

X A ofFraction Mole

+++==

=

totAA

tottot

AA

tot

A

tot

A

tottot

AA

tottot

AA

PXP

PnnPor

nn

PPor

RTnVPRTnVP

equations divideRTnVP

RTnVP

=

====

==


Exercise 5-11

A solid hydrocarbon is burned in air in a closed container, producing a mixture of gases having a total pressure of 3.34 atm. Analysis of the mixture shows it to contain 0.340 g of water vapor, 0.792 g of carbon dioxide, 0.288 g of oxygen, 3.790 g of nitrogen, and no other gases. Calculate the mole fraction and partial pressure of carbon dioxide in this mixture.


The Kinetic Theory of Gases

1. A pure gas consists of a large number of identical molecules separated by distances that are large compared with their size.2. The molecules of a gas are constantly moving in random directions with a distribution of speeds.

3. The molecules of a gas exert no forces on one another except during collisions, so that between collisions they move in straight lines with constant velocities.4. The collisions of the molecules with each other and with the walls of the container are elastic; no energy is lost during a collision.


Temperature and Molecular Motion

Pressure ∝ (impulse per collision) x (rate of collisions with the walls)

•impulse pre collision

∝ momentum (m × u)•rate of collisions

∝ number of molecules per unit volume (N/V)•rate of collisions

∝ speed of molecules (u)

P ∝ (m × u) × [(N/V) × u]


The Kinetic Theory of Gases

PV ∝ Nmu2

mean-square speed

M3RTu

mNM

RTumN31

NnNnumber sAvogadro'N if

nRTuNm31

nRTPV

uNm31PV

2

2

2

2

=

=

=∴

==

=∴

=

=

o

o

o

o


Distribution of Molecular Speeds

M3RTu2 =


Exercise 5-12

At a certain speed, the root-mean-square-speed of the molecules of hydrogen in a sample of gas is 1055 ms-1. Compute the root-mean square speed of molecules of oxygen at the same temperature.


Exercise 5-12



Exercise 5-12


Strategy

1. Find T for the H2 gas with a urms =1055 ms-1

2. Find urms of O2 at the same temperature

( )RMu

T HHrms

32

22

=

12

Orms

O

H2HO

rms

O

H2H

Orms

O

Orms

264.8ms32

(2)(1005)u

MMu

u

M3RMu

3Ru

M3RTu

2

2

222

2

22

2

2

2

−==

=

=

=


Speed Distribution Curves

Maxwell-Boltzmann speed distribution

Temperature is a measure of the average kinetic energy of molecules when their speeds have Maxwell Boltzmann distribution. I.e., the molecules come to thermal equilibrium.


Gaseous DiffusionRate of effusion of A ∝ uA

rms

factor enrichmentMM

M3RTM3RT

uu

B Eff of RateA Eff of Rate

A

B

B

A

Brms

Arms

==

=

=

Rate of effusion of B ∝ uBrms


Exercise 5-13

A gas mixture contains equal numbers of molecules of N2and SF6. A small portion of it is passed through a gaseous diffusion apparatus. Calculate how many molecules of N2are present in the product of gas for every 100 molecules of SF6.


Exercise 5-13

A gas mixture contains equal numbers of molecules of N2 and SF6. A small portion of it is passed through a gaseous diffusion apparatus. Calculate how many molecules of N2 are present in the product of gas for every 100 molecules of SF6.

A

B

MMfactor enrichment =

2.2814 x 2

19) x (632=

+=

2

6

N of molecules 2282.28 x SF molecules 100 =


SkipSection 5-8Real Gases



Chapter 5The Gaseous State

Examples / Exercises– 5-1 thru 5-13

Problems– 34, 38, 48, 62, 70, 81

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When you can measure what you are speaking about and