Homework 1 Solutions

Homework Assignment # 1 (Due February 2, 2010) Chapter 1 1) Consider a gas mixture in a 2.00-dm3 flask at 27.0C. For each of the following mixtures, calculate the partial pressure of each gas, the total pressure, and the composition of the mixture in mole percent. a) 1.00 g H2 and 1.00 g O2 b) 1.00 g N2 and 1.00 g O2 c) 1.00 g CH4 and 1.00 g NH3 a)

2

2

2

2

1 15

3 3

1 14

3 3

5

1.00 2.016 mol 8.314 J mol K 300 K 6.24 10 Pa2.00 10 m

1.00 32.00 mol 8.314 J mol K 300 K 3.90 10 Pa2.00 10 m

6.57 10 Pa

HH

OO

total

n RTP

Vn RT

PV

P

= = =

= = =

=

22

2 2

22

2 2

mol H 1.00 2.016mol % H 100 100 94.1%mol H mol O 1.00 2.016 1.00 32.00

mol O 1.00 32.00mol % O 100 100 5.9%mol H mol O 1.00 2.016 1.00 32.00

= = =+ +

= = =+ +

b)

2

2

2

2

1 14

3 3

1 14

3 3

4

1.00 28.02 mol 8.314 J mol K 300 K 4.45 10 Pa2.00 10 m

1.00 32.00 mol 8.314 J mol K 300 K 3.90 10 Pa2.00 10 m

8.35 10 Pa

NN

OO

total

n RTP

Vn RT

PV

P

= = =

= = =

=

22

2 2

22

2 2

mol N 1.00 28.02mol % N 100 100 53.3%mol N mol O 1.00 28.02 1.00 32.00

mol O 1.00 32.00mol % O 100 100 46.7%mol N mol O 1.00 28.02 1.00 32.00

= = =+ +

= = =+ +

c)

3

3

4

4

1 14

3 3

1 14

3 3

5

1.00 17.03 mol 8.314 J mol K 300 K 7.32 10 Pa2.00 10 m

1.00 16.04 mol 8.314 J mol K 300 K 7.77 10 Pa2.00 10 m

1.51 10 Pa

NHNH

CHCH

total

n RTP

Vn RT

PV

P

= = =

= = =

=

6.19 x105 Pa

33

3 4

42

3 4

mol NH 1.00 17.03mol % NH 100 100 48.5%mol NH mol CH 1.00 17.03 1.00 16.04

mol CH 1.00 16.04mol % O 100 100 51.5%mol NH mol CH 1.00 17.03 1.00 16.04

= = =+ +

= = =+ +

2) Consider a 20.0-L sample of moist air at 60C and 1 atm in which the partial pressure of water vapor is 0.120 atm. Assume that dry air has the composition 78.0 mole percent N2, 21.0 mole percent O2, and 1.00 mole percent Ar. a) What are the mole percentages of each of the gases in the sample?

b) The percent relative humidity is defined as % RH = 22

*H O

H O

PP

where 2H O

P is the partial pressure of

water in the sample and 2H O

P = 0.197 atm is the equilibrium vapor pressure of water at 60C. The gas is compressed at 60C until the relative humidity is 100%. What volume does the mixture now occupy? c) What fraction of the water will be condensed if the total pressure of the mixture is isothermally increased to 200 atm? a)

2

2

2

2

2

2

0.78 0.88 atmmol % N = 100 100 68.6%1 atm

0.21 0.88 atmmol % O = 100 100 18.5%1 atm

0.01 0.88 atmmol % Ar = 100 100 0.9%1 atm

0.12 atmmol % H O = 100 1001 atm

N

total

O

total

Ar

total

H O

total

PPPPPP

PP

= =

= =

= =

= 12.0%=

b) 2

2

2 2

2

2

where the primed quantities refer to 100% RH

0.12 atm 20.0 L 12.2 L0.197 atm

H OH O

H O H O

H O

H O

n RTP V

VP V P V

P VV

P

=

=

= = =

c) If all the water remained in the gas phase, the partial pressure of water at a total pressure of 200 atm would be

2 2mol fraction H O = 200 atm 0.12 = 24.0 atmH O totalP P=

However, the partial pressure of water cannot be greater than 0.197 atm, and the excess will condense. The fraction that condenses is given by

24.0 atm 0.197 atmfraction condensed = 0.99224.0 atm

=

3) The total pressure of a mixture of oxygen and hydrogen is 1.00 atm. The mixture is ignited and the water is removed. The remaining gas is pure hydrogen and exerts a pressure of 0.400 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

2Hno

2Ono 0

at equilibrium 2

2Hn o

2On o 2

If the O2 is completely consumed, 2 20 or .O On n = =

o o The number of moles of H2 remaining

is 2 2 2

2 2 .H H On n n = o o o

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 1

and 2

Dividing the second equation by the first

2 2 1 2 1 3

1 1 0.400 atm1 1 0.20; 0.803 3 1.00 atm

H O H O

H OH O O O O

H O H O

O H

RT RTP n n P n nV V

n nP x x x x xP n n n n

Px xP

= + =

= = = = + +

= = = =

o o o o

o o

o o o o

4) The barometric pressure falls off with height above sea level in the Earth's atmosphere as

0iM 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, and T is the absolute temperature. Consider an atmosphere that has the composition

2 20.600 and 0.400N COx x= = and that T = 300 K. Near sea level, the total pressure is 1.00 bar.

Calculate the mole fractions of the two components at a height of 50.0 km. Why is the composition different from its value at sea level? Mi is the molecular mass of the gas.

Mok Wei ChiangSticky Note


2 2

2 2

3 2 30 5

1 1

3 2 30 5

1

28.04 10 kg 9.81 m s 50 10 m0.600 1.0125 10 Pa exp8.314 J mol K 300 K

242 Pa

44.04 10 kg 9.81 m s 50 10 m0.400 1.0125 10 Pa exp8.314 J mol K

i

i

M gzRT

N N

M gzRT

CO CO

P P e

P P e

= = =

= =

2

2 2 2

2 2

1 300 K 6.93 Pa

6.93 0.028 1 0.9726.93 242

COCO N CO

CO N

Px x x

P P

=

= = = = =+ +

The mole fraction of CO2 at the high altitude is much reduced relative to its value at sea level because it has a larger molecular mass than N2. 5) Calculate the pressure exerted by benzene for a molar volume 1.42 L at 790 K using the

Redlich-Kwong equation of state: ( ) ( )

21 1

m m m

RT a nRT n aPV b V V b V nb V V nbT T

= = + +

. The

Redlich-Kwong parameters a and b for benzene are 452.0 bar dm6 mol2 K1/2 and 0.08271 dm3 mol1, respectively. Is the attractive or repulsive portion of the potential dominant under these conditions?

( )

( )

2 3 1 1

3 1 3 1

16 2 2

3 1 3 1 3 1

1

8.31410 bar dm mol K 790 K 1.42dm mol 0.08271dm mol

452.0bar dm mol K 1 1.42dm mol 1.42dm mol + 0.08271dm mol790 K

41.6 bar

m m m

RT aPV b V V bT

P

= +

=

=2 1 18.314510 L bar mol K 790 K 46.3 bar

1.42 LidealRTPV

= = =

Because P < Pideal, the attractive part of the potential dominates.

Chapter 2

1) 3.00 moles of a gas are compressed isothermally from 60.0 L to 20.0 L using a constant external pressure of 5.00 atm. Calculate q, w, U, and H.

( )5 3 3 4

4

5 1.013 10 Pa 60 10 L 20 10 L 2.03 10 J

0 and 0 because 02.03 10 J

externalw P V

U H Tq w

=

= =

= = =

= =

2) An automobile tire contains air at 320 103 Pa at 20C. The stem valve is removed and the air is allowed to expand adiabatically against the constant external pressure of 100 103 Pa until P = Pexternal. For air, CV,m = 5/2 R. Calculate the final temperature of the gas in the tire. Assume ideal gas behavior.

( ) ( )

( )

,

,

because 0,

V m f i ext f i

f iV m f i ext

f i

q U w

nC T T P V V

nRT nRTnC T T PP P

= =

=

=

The factor n cancels out. Rearranging the equation

, ,

,

,

-1 1 5-1 -1

5

-1 1 5-1 1

5

8.314 J mol K 10 Pa2.5 8.314 J mol K3.20 10 Pa

8.314 J mol K 10 Pa2.5 8.314 J mol K10 Pa

0.804 235

ext extV m f V m i

f i

extV m

f i

extiV m

f

f i f

RP RPC T C TP P

RPCT P

RPT CP

T T T

+ = +

+=

+

+

=

+

= = K

3) An ideal gas described by Ti = 300 K, Pi = 1.00 bar and Vi = 10.0 L is heated at constant volume until P = 10.0 bar. It then undergoes a reversible isothermal expansion until P = 1.00 bar. It is then restored to its original state by the extraction of heat at constant pressure. Depict this closed cycle process in a P-V diagram. Calculate w for each step and for the total process. What values for w would you calculate if the cycle were traversed in the opposite direction?

(20x10-3L 60x10-3L) = +2.03 x 104J

-2.03 x104 J

-2 -1 -1

1.00 bar 10.0 L 0.401 mol8.314510 L bar mol K 300 K

i i

i

PVnRT

= = =

The process can be described by step 1: Pi,Vi,Ti P1 = 10.0 bar,Vi, T1 step 2: P1,Vi, T1 Pi,V2 T1 step 3: Pi, V2, T1 Pi,Vi,Ti. In step 1, Pi,Vi,Ti P1,Vi, T1, w = 0 because V is constant. In step 2, P1,Vi, T1 Pi, V2, T1 Before calculating the work in step 2, we first calculate T1.

11

10.0 bar300 K 3000 K1.00 bari i

PT TP

= = =

1 1

-1 -1 3

ln ln

10.0 bar 0.401 mol 8.314 J mol K 3000 Kln 23.010 J1.00 bar

f i

i f

V Pw nRT nRTV P

= =

= =

In step 3,

( )

11 2 2

5 -3 33

; 10 100 L

10 Pa 10 m1.00 bar 10 L 100 L 9.0010 Jbar L

ii i i

i

external

PVPV PV V VP

w P V

= = = =

= = =

3 3 30 23.0 10 J 9.00 10 J 14.0 10 Jcyclew = + =

If the cycle were traversed in the opposite direction, the magnitude of each work term would be unchanged, but all signs would change.

4) A pellet of Zn of mass 10.0 g is dropped into a flask containing dilute H2SO4 at a pressure P = 1.00 bar and temperature T = 298 K. What is the reaction that occurs? Calculate w for the process. Zn(s) + H2SO4(aq) Zn2+(aq) + SO42-(aq) +H2(g) The volume of H2 produced is given by

( )

-1 -1-3 32

-1 5

25 -3 3

1mol H10.0 g 8.314 J mol K 298 K 3.79 10 m1mol Zn 110 Pa65.39 g mol Zn

volume of H produced.

110 Pa 3.7910 m 379 J

external

V

w P VV

w

= =

=

= =

5) One mole of an ideal gas with CV,m = 3/2R initially at 298 K and 1.00 105 Pa undergoes a reversible adiabatic compression. At the end of the process, the pressure is 1.00 106 Pa. Calculate the final temperature of the gas. Calculate q, w, U and H for this process.

( )

11 11 1

513

5 5 0.43

6

; ;

1.0010 Pa 0.100 2.511.0010 Pa

2.51298 K 749 K

f f f f fi i i

i i i f i f i f

f

i

f

T V T T TP P PT V T P T P T P

TTT

= = = = =

= = =

= =

q = 0 because the process is adiabatic.

( )

( ) ( )

-1 -13

,

3 -1 -1

3

38.314 J mol K1 mol 749 K 298 K 5.62 10 J2

5.62 10 J 8.314 J mol K 749 K 298 K

9.37 10 J

V mw U nC T

H U PV U R T

H

= = = =

= + = + = +

=



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Homework 1 Solutions