Physical Chemistry II Chem 402 Spring 2011

Chapter 1 (10, 11, 13, 20, 31, 35, 37)

P1.10: A typical diver inhales 0.500 liters of air per breath and carries a 20. L breathing tank containing

air at a pressure of 300. bar. As she dives deeper, the pressure increases by 1 bar for every 10.08 m. How

many breaths can the diver take from this tank at a depth of 25 m? Assume that the temperature remains

constant.

325

251

325

20. L 300. bar1.72 10 [L]

25 m1

10.08 bar m

number breaths = 3.5 10

surface surfacem

m

m

breath

V PV

P

V

V

P1.11: Use the ideal gas and van der Waals equations to calculate the pressure when 3.00 mol H2 are

confined to a volume of 1.00 L at 298 K. Is the gas in the repulsive or attractive region of the molecule-

molecule potential?

2

2

2 6 22 3 1 1

23 3 1 3

3 mol 0.2452 bar dm mol3 mol 8.314×10 bar dm mol K ×298 K

1.00 dm 3 mol 0.0265 dm mol 1.00dm

78.5 bar

nRT n aP

V nb V

2 1 13 mol 8.3145 10 ×L bar mol K ×298 K74.3 [bar]

1.00 Lideal

nRTP

V

Because P > Pideal, the repulsive part of the potential dominates.

P1.13: A mixture of oxygen and hydrogen is analyzed by passing it over hot copper oxide and through a

drying tube. Hydrogen reduces the CuO according to the reaction CuO(s) + H2(g) Cu(s) + H2O(l), and

oxygen reoxidizes the copper formed according to Cu(s) + 1/2 O2(g) CuO(s). At 25°C and 750. Torr,

125.0 cm3 of the mixture yields 82.5 cm3 of dry oxygen measured at 25°C and 750. Torr after passage

over CuO and the drying agent. What is the original composition of the mixture?

Two equilibria must be considered:

idealP

CuO(s) + H2(g) → H2O(l) + Cu(s)

at equilibrium 2Hn α–β

Cu(s) + ½ O2(g) → CuO(s)

at equilibrium 2

1

2On

In the final state, only O2 is present. Therefore 2.Nn In an excess of O2, all the copper is oxidized.

Therefore α–β = 0 or2.Hn We conclude that

2 2 2

1.

2O O Hn n n

Let V1 and V2 be the initial and final volumes.

2 2 2 2 2 21 2

1 ; [ ( )] 0

2H O H O O H

RT RT RTV n n V n n n n

P P P

Dividing the second equation by the first yields

2 2

2 2 2 2 2

2 2 2 2

2 2 2

2

1

32

31

1 1 1 31 1

2 2 2 2

2 2 82.5 cm1 1 0.227; 1 = 0.773

3 3 125 cm

O HO H H H H

H O H O

H O H

n nVx x x x x

V n n n n

Vx x x

V

P1.20) In the absence of turbulent mixing, the partial pressure of each constituent of air would fall off

with height above sea level in the Earth’s atmosphere as 0 iM gz RTi iP P e where Pi is the partial pressure at

the height z, 0iP is the partial pressure of component i at sea level, g is the acceleration of gravity, R is the

gas constant, T is the absolute temperature, and Mi is the molecular mass of the gas. As a result of

turbulent mixing, the composition of the Earth’s atmosphere is constant below an altitude of 100 km, but

the total pressure decreases with altitude as 0 aveM gz RTP P e where Mave is the mean molecular weight of

air. At sea level, 20.78084 and 0.00000524N Hex x

and T = 300 K.

a) Calculate the total pressure at 10.0 km assuming a mean molecular mass of 28.9 g mol-1 and that T =

300 K throughout this altitude range

b) Calculate the value that2N Hex x would have at 10.0 km in the absence of turbulent mixing. Compare

your answer with the correct value.

2

2 2

3 2 30 5 4

1 1

0

3 2 35

28.9 10 kg 9.81 m s 10.0 10 m1.0125 10 Pa exp 3.25 10 [Pa]

8.314 J mol K 300. K

28.02 10 kg 9.81 m s 10.0 10 m0.78084 1.0125 10 Pa exp

8.314 J mo

i

i

M gz

RTtot N

M gz

RTN N

P P e

P P e

2

2 2

41 1

0

3 2 36 5

1 1

4

2.63 10 [Pa]l K 300. K

4.003 10 kg 9.81 m s 10.0 10 m5.24 10 1.0125 10 Pa exp 0.453 [Pa]

8.314 J mol K 300. K

5.79 10 without mixing

0

iM gz

RTHe He

N

He

N N

He He

P P e

P

P

P x

P x

56

.780841.49 10 with mixing

5.24 10

P1.31: The total pressure of a mixture of oxygen and hydrogen is 1.50 atm. The mixture is ignited and the

water is removed. The remaining gas is pure hydrogen and exerts a pressure of 0.250 atm when measured

at the same values of T and V as the original mixture. What was the composition of the original mixture in

mole percent?

2H2(g) + O2(g) → 2H2O(l)

initial moles 2Hn

2On 0

at equilibrium 2

2Hn 2On 2

If the O2 is completely consumed, 2 2

0 or .O On n The number of moles of H2 remaining is

2 2 22 2 .H H On n n

Let P1 be the initial total pressure and P2 be the total pressure after all the O2 is consumed.

2 2 2 2

2 2

2 2 2 2 2

2 2 2 2

2 2

1 2

2

1

2

1

and 2

Dividing the second equation by the first

2 2 1 2 1 3

1 1 0.250 atm1 1 0.278; 0.722

3 3 1.50 atm

H O H O

H OH O O O O

H O H O

O H

RT RTP n n P n n

V V

n nPx x x x x

P n n n n

Px x

P

P1.35: As a result of photosynthesis, an acre of forest (1 acre = 4047 square meter) can take up 1000. kg

of CO2. Assuming air is 0.0314% CO2 by volume, what volume of air is required to provide 1000. kg of

CO2? Assume T = 298 K and P = 1.00 atm.

2

2

2

2 1 13 1

5

9 6 3 3 3

1000. kg8.206 10 L atm mol K 298 K

44.01 10 kg mol1.00 atm

5.55 10 L

1.77 10 L = 1.77 10 m = 1.77 10 km0.000314

COCO

COair

n RTV

P

VV

I am not sure of the time unit, perhaps per hr, i.e. [hr-1].

P1.37) Use L’Hôpital’s rule, 0

0

lim limx

x

df x dxf x g x

dg x dx

to show that the expression

derived for Pf in part b of Example Problem 1.1 has the correct limit as → 0.

2

22

22

0

2

22

0

1;

1

1 0

1 1 4

2

1 1 4lim

2

2 1 4

2 1lim

f i f ii i

i f

i f f i f i

f i f i i i f

i i i i i f i

fi

i i i i i f i

i

i i i i fi i

i i

P V P PPV

T T

PT P T P P

P T P T P PT

T P T P TT PP

T

d T P T P TT P ddf d

dg d d T d

T P P TT PT P

T P

22 44 2

2 2

2

2

i i i fi i

i f i i

i i

i i i i i f i f

i i

T P TT PT P

TT P T

T T

T P PT T P T P

T T