1

Homework 1 Answer

(Due 1/21/2009 Wednesday)

P1.2) Consider a gas mixture in a 2.00-dm3 flask at 27.0°C. 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

To calculate the partial pressures we use the ideal gas law:

a)

( ) ( ) ( )( ) ( ) Pa 10 6.18

mol kg10 2.02 m 10 2.00K 300.15mol K J 8.314472 kg 0.001

VMT R m

VT R n

p 513-33-

11

2

22

2×=

×××××

=== −

−−

H

HHH

( ) ( ) ( )( ) ( ) Pa 10 3.90

mol kg10 32.0 m 10 2.00K 300.15mol K J 8.314472 kg 0.001

VMT R m

VT R n

p 413-33-

11

2

22

2×=

×××××

=== −

−−

O

OOO

Pa 10 57.6ppp 5total 22

×=+= OH

( ) % 4.19

mol kg10 32.0kg 0.001

mol kg10 2.02kg 0.001

mol kg10 2.02kg 0.001

100O molH mol

H mol100H % mol

13-13-

13-

22

22 =

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛×

+⎟⎟⎠

⎞⎜⎜⎝

⎛×

⎟⎟⎠

⎞⎜⎜⎝

⎛×

×=+

×=

−−

−

( ) % .95

mol kg10 32.0kg 0.001

mol kg10 2.02kg 0.001

mol kg10 32.0kg 0.001

100O molH mol

O mol100O % mol

13-13-

13-

22

22 =

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛×

+⎟⎟⎠

⎞⎜⎜⎝

⎛×

⎟⎟⎠

⎞⎜⎜⎝

⎛×

×=+

×=

−−

−

2

b)

( ) ( ) ( )( ) ( ) Pa 10 4.45

mol kg10 28.02 m 10 2.00K 300.15mol K J 8.314472 kg 0.001

VMT R m

VT R n

p 413-33-

11

2

22

2×=

×××××

=== −

−−

N

NNN

( ) ( ) ( )( ) ( ) Pa 10 3.90

mol kg10 32.0 m 10 2.00K 300.15mol K J 8.314472 kg 0.001

VMT R m

VT R n

p 413-33-

11

2

22

2×=

×××××

=== −

−−

O

OOO

Pa 10 35.8ppp 4total 22

×=+= OH

( ) % 3.35

mol kg10 32.0kg 0.001

mol kg10 28.02kg 0.001

mol kg10 28.02kg 0.001

100O molN mol

N mol100N % mol

13-13-

13-

22

22 =

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛×

+⎟⎟⎠

⎞⎜⎜⎝

⎛×

⎟⎟⎠

⎞⎜⎜⎝

⎛×

×=+

×=

−−

−

( ) % 6.74

mol kg10 32.0kg 0.001

mol kg10 28.02kg 0.001

mol kg10 32.0kg 0.001

100O molN mol

O mol100O % mol

13-13-

13-

22

22 =

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛×

+⎟⎟⎠

⎞⎜⎜⎝

⎛×

⎟⎟⎠

⎞⎜⎜⎝

⎛×

×=+

×=

−−

−

c)

( ) ( ) ( )( ) ( ) Pa 10 7.32

mol kg10 17.03 m 10 2.00kg 300.15mol K J 8.314472 kg 0.001

VMT R m

VT R n

p 413-33-

11

3

33

3×=

×××××

=== −

−−

NH

NHNHNH

( ) ( ) ( )( ) ( ) Pa 10 7.77

mol kg10 16.04 m 10 2.00kg 300.15mol K J 8.314472 kg 0.001

VMT R m

VT R n

p 413-33-

11

4

44

4×=

×××××

=== −

−−

CH

CHCHCH

Pa 10 51.1ppp 5total 22

×=+= OH

( )

% 8.54

mol kg10 17.03kg 0.001

mol kg10 16.04kg 0.001

mol kg10 17.03kg 0.001

100

CH molNH molNH mol

100NH % mol

13-13-

13-

43

33

=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛×

+⎟⎟⎠

⎞⎜⎜⎝

⎛×

⎟⎟⎠

⎞⎜⎜⎝

⎛×

×=

+×=

−−

−

3

( )

% 1.55

mol kg10 17.03kg 0.001

mol kg10 16.04kg 0.001

mol kg10 16.04kg 0.001

100

CH molNH molCH mol100CH % mol

13-13-

13-

43

44

=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛×

+⎟⎟⎠

⎞⎜⎜⎝

⎛×

⎟⎟⎠

⎞⎜⎜⎝

⎛×

×=

+×=

−−

−

P1.4) In a normal breath, about 0.5 L of air at 1.0 atm and 293 K is inhaled. About 25.0% of the

oxygen in air is absorbed by the lungs and passes into the bloodstream. For a respiration rate of 18

breaths per minute, how many moles of oxygen per minute are absorbed by the body? Assume the

mole fraction of oxygen in air is 0.21. Compare this result with Example Problem 1.1.

We use the ideal gas law to calculate the number of moles of air inhaled every minute:

( ) ( )( ) ( ) mol 0208.0

mol K J 8.314472 K 293m 10 5.0 Pa 101325

T R V pn 11

3-4

air =×

××== −−

We then use the mole fraction to determine the number of moles of O2 for 18 breaths:

( ) ( ) ( ) ( ) 11O min mol 01966.0 min 180.25 0.21 mol .02080n

2

−− =×××=

P1.5) Suppose that you measured the product PV of 1 mol of a dilute gas and found that PV =

22.98 L atm at 0.00°C and 31.18 L atm at 100°C. Assume that the ideal gas law is valid, with T =

t(°C) + a, and that the value of R is not known. Determine R and a from the measurements

provided.

The ideal gas law and solving for R:

( )aT R V pn+

=

4

( )aTn V pR+

=

Using the information for the two different conditions and solving for a yields a:

( ) ( )aTn V p

aTn V p

2

22

1

11

+=

+

( ) ( )( )( )

( )( )

C2.280 1

atm L 31.18atm L 22.98

atm L 31.18atm L 22.98C100-C0

1V pV p

V pV p

T-Ta

22

11

22

1121

o

oo

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

⎟⎟⎠

⎞⎜⎜⎝

⎛

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

⎟⎟⎠

⎞⎜⎜⎝

⎛

=

R is then:

( )( )

( ) ( ) ( )( )1-1-

1

11 K mol atm L 0.0820 C280.2C0mol 1

atm L 22.98aTn V pR =

+×=

+=

oo

P1.6) Devise a temperature scale, abbreviated G, for which the magnitude of the ideal gas constant

is 1.00 J G–1 mol–1.

The G-temperature scale is given by:

( ) -1KG 8.314472KTG ×=

So that on the Kelvin scale:

( ) ( ) -1-1-1 mol J 14.2436K 293K mol J 8.314472T R =×=

And on the G scale:

( ) ( ) ( ) -1-1-1-1G mol J 14.2436KG 8.314472K 293G mol J 1.0T R =××=

P1.8) In normal respiration, an adult exhales about 500. L of air per hour. The exhaled air is

saturated with water vapor at body temperature T = 310. K. At this temperature water vapor in

5

equilibrium with liquid water has a pressure of P = 0.062 atm. Assume water vapor behaves ideally

under these conditions. What mass of water vapor is exhaled in an hour?

Using the ideal gas law, and solving for m yields:

T RMmT Rn V p ==

( ) ( ) ( ) ( )( ) ( ) ( )

1-11

3-11-3

h kg .02200 h 1 mol K J 8.314472 K 310

m 0.5 atm Pa 101325 atm 0.062mol kg 1018.02T RV p Mm =

××××××

== −−

−

P1.15) Consider a 20.0-L sample of moist air at 60.°C and 1 atm in which the partial pressure of

water vapor is 0.120 atm. Assume that dry air has the composition 78.0 mol % N2, 21.0 mol % O2,

and 1.00 mol % Ar.

a. What are the mole percentages of each of the gases in the sample?

b. The percent relative humidity is defined as %RH = PH2O PH2O* where

PH2O is the partial

pressure of water in the sample and PH2O* = 0.197 atm is the equilibrium vapor pressure of

water at 60.°C. The gas is compressed at 60.°C until the relative humidity is 100%. What

volume does the mixture contain now?

c. What fraction of the water will be condensed if the total pressure of the mixture is isothermally

increased to 200. atm?

a) ( )( ) % 68.6

atm 00.1atm 0.880.78100

p p

100N % moltotal

N2

2 =×

×=×=

( )( ) % 18.5

atm 00.1atm 0.880.21100

p p

100O % moltotal

O2

2 =×

×=×=

( )( ) % 0.9

atm 00.1atm 0.880.01100

p p

100Ar % moltotal

Ar =×

×=×=

6

( )( ) % 12.0

atm 00.1atm 0.12100

p p

100OH % moltotal

OH2

2 =×=×=

b)

V T R n

V p OHOH

2

2=

V p V p OHOH 22=′′ , where the prime refers to 100% RH

( ) ( )( ) L 12.2

atm 0.197L 20.0atm 0.12

p V p

V OH

OH

2

2 =×

=′

=′

c)

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

would be:

atm 24.0 0.12 atm 200 OHfraction molp p 2totalOH2=×=×=

However, the partial pressure of water cannot be greater than 0.197 atm, and the excess will

condense. PH2O is maintained at 0.197atm, thus

P’gas= P’N2+P’O2+P’Ar = 200atm-P’H2O = 200atm-0.197atm = 199.803atm

P’gasV’ = PgasV

0.08809L803.199

0.20880.0P'

VPV'

gas

gas =×

==atm

Latm

Total amount of H2O: mol7880.0)15.27360(134.8

100.20101325120.0RT

VPn 11

331

OH2=

+××××

== −−

−−

KmolJKmPaatmatm

After condense, amount of H2O stay as water vapor:

molKmolJK

mPaatmatm 000635.0)15.27360(134.8

1008809.0101325197.0RT

V''Pn' 11

331

OH2=

+××××

== −−

−−

7

The fraction that condenses is given by:

( ) ( )( ) 0.993 0.0878mol

l0.000635mo0.0878molcondensedfrcation =−

=

P1.18) A normal adult inhales 0.500 L of air at T = 293 K and 1.00 atm. To explore the surface of

the moon, an astronaut requires a 25.0-L breathing tank containing air at a pressure of 200. atm.

How many breaths can the astronaut take from this tank?

We first need to calculate the number of moles inhaled at 1 atm using the ideal gas law:

( ) ( )( ) ( ) mol 0.0208

K 293 mol K J 8.314472 m 0.0005Pa 101325

T R V pn 11

3

breath =×

×==

−−

Next we determine how many moles of air are in the tank:

( ) ( )( ) ( )

3

tank 1 1

200 x 101325 Pa 0.025 m p V n 207.96 molR T 8.314472 J K mol 293 K− −

×= = =

×

Therefore, the number of breaths the astronaut can take is:

( )( ) 9998

mol 0.0208 mol 207.96

n n

breaths ofnumber breath

tank ===

P1.20) Yeast and other organisms can convert glucose (C6H12O6) to ethanol (CH3CH2OH) by a

process called alcoholic fermentation. The net reaction is

C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)

Calculate the mass of glucose required to produce 1.0 L of CO2 measured at P = 1.00 atm and T =

300. K.

First, we calculate the number of moles of CO2 under the conditions given:

8

( ) ( )( ) ( ) mol 0406.0

mol K J 8.314472 K 300Pa 101325 m 101

T Rp V n 11

3-3

CO2=

×××

== −−

From the equation of the reaction, we can see that the number of moles of glucose is half the

number of moles of CO2, and the mass of glucose necessary is:

( ) g 66.3 2

mol 0.0406mol g 180.18 2

nMn Mm 1CO

glucoseglucose2 =⎟

⎠⎞

⎜⎝⎛×=== −

P1.30) The barometric pressure falls off with height above sea level in the Earth’s atmosphere as

RTgzMoii

iePP /−= where Pi is the partial pressure at the height z, Pi0 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. Consider an atmosphere that has the

composition xN2

= 0.600 and xCO2

= 0.400 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?

PaKmolJK

mmsKgmolExpPa

ePP RTgzM

NN

N

242)300()314472.8(

)1050()81.9()1004.28()101((0.600) 11

32135

02

22

=⎥⎦

⎤⎢⎣

⎡×

××××−×××=

=

−−

−−−

−

PaKmolJK

mmsKgmolExpPa

ePP RTgzM

COCO

CO

94.6)300()314472.8(

)1050()81.9()1004.44()101((0.400) 11

32135

02

22

=⎥⎦

⎤⎢⎣

⎡×

××××−×××=

=

−−

−−−

−

028.0)24294.6(

)94.6()( 22

22 =

+=

+=

PaPaPa

PNPcoPcoXco

972.0 x1x22 CON =−=

9

N2 and CO2 have different composition a height of 50.0 km from sea level, because they have

different molecular mass, so their barometric pressure fall off at different rate with height.

10

1. For N2, Calculate (a) the molecular mass, (b) the molecular weight, (c) the mass of 1 molecule of N2, (d) the molar mass.

a) molecular mass of N2, 123123

1

10653.410022.6

02.28 −−−

−

×=×

moleculeglmoleculemo

gmol or 28.02 u (Da)

b) molecular weight of N2, 123123

1

10653.410022.6

02.28 −−−

−

×=×

moleculeglmoleculemo

gmol or 28.02 u (Da)

c) mass of 1molecule of N2, gmol

gmol 23123

1

10653.410022.6

02.28 −−

−

×=×

d) molar mass, 28.02gmol-1 2. Give the (a) weight % and (a) molar % of O2 in air at 25 °C. Assume air composition 78%N2, 21%O2, 1%Ar

a) weight% of O2 = %19.2395.39%100.32%2102.28%78

00.32%21100 111

1

=×+×+×

×× −−−

−

gmolgmolgmolgmol

b) molar% of O2 = 21% 3. What is the relationship of temperatures denoted in K (T) and °C (T’)? Use one equation. K=C+273.15 4. In. P1.15, what is the partial pressure of N2 for 1.00 atm of air at 60 °C? Assume air composition 78%N2, 21%O2, 1%Ar

Partial pressure of N2, atmatmPP totalN 78.000.1%78%782

=×=×=

5. Plot P (in bar) for a varied V (0 < V ≤ 1.0 L) for one mole of ideal gas at T = 300 K and 450 K. Use Excel or other software to draw graphs. Present the equations you used to draw the graphs and explain how you generated each equation.

PV=nRT

When T=300K, barVVdm

KmolbarKdmmolV

nRTP 9.24300083.013

113

=××

==−−

11

When T=450K, barVVdm

KmolbarKdmmolV

nRTP 35.37450083.013

113

=××

==−−

V (L)

0.0 .2 .4 .6 .8 1.0

P (b

ar)

0

1000

2000

3000

4000

T=300KT=450K

6. Plot P for a varied V (0< V ≤ 1.0 L) for one mole of non-ideal gas, CO2 at T = 300 K and 450 K using equation 1.12 in page 9 and values of a and b in Table 1.3. Present the equations you used to draw the graphs and explain how you generated each equation. What is the major difference from the graphs in 5.

Non-ideal gas, equation 1.12 2

2

Van

nbVnRTP −−

=

From table 1.3 in appendix, van der Waals parameters for CO2, a=3.658 dm3bar mol-2 b=0.0429 dm3mol-1

When T=300K,

barVVV

barmoldmmolmoldmmolVdm

KmolbarKdmmolP 22

232

133

113 658.30429.09.24658.31

0429.01300083.01

−−

=×

−×−

××=

−

−

−−

When T=450K,

barVVV

barmoldmmolmoldmmolVdm

KmolbarKdmmolP 22

232

133

113 658.30429.035.37658.31

0429.01450083.01

−−

=×

−×−

××=

−

−

−−

12

V (L)

0.0 .2 .4 .6 .8 1.0

Pres

sure

(bar

)

0

50

100

150

200

250

300

T=300KT=450K