chapter 1: fundamental concepts of thermodynamics · 2017-12-04 · chapter 1/fundemental concepts...

1

Chapter 1: Fundamental Concepts of Thermodynamics

Questions on Concepts

Q1.1) The location of the boundary between the system and the surroundings is a choice that

must be made by the thermodynamicist. Consider a beaker of boiling water in an airtight room.

Is the system open or closed if you place the boundary just outside the liquid water? Is the sys-

tem open or closed if you place the boundary just inside the walls of the room?

If the system boundaries are just outside of the liquid water, the system is open because water

can escape from the top surface. The system is closed if the boundary is just inside the walls, be-

cause the room is airtight.

Q1.2) Real walls are never totally adiabatic. Order the following walls in increasing order with

respect to their being diathermal: 1-cm-thick concrete, 1-cm-thick vacuum, 1-cm-thick copper,

1-cm-thick cork.

1-cm-thick vacuum < 1-cm-thick cork < 1-cm-thick concrete < 1-cm-thick copper

Q1.3) Why is the possibility of exchange of matter or energy appropriate to the variable of in-

terest a necessary condition for equilibrium between two systems?

Equilibrium is a dynamic process in which the rates of two opposing processes are equal. How-

ever, if the rate in each direction is zero, no exchange is possible, and the system cannot reach

equilibrium.

Q1.4) At sufficiently high temperatures, the van der Waals equation has the form P ≈ RT/(Vm –

b). Note that the attractive part of the potential has no influence in this expression. Justify this

behavior using the potential energy diagram of Figure 1.7.

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Chapter 1/Fundemental Concepts of Thermodynamics

2

In this case, the energy well depth is small compared to the total energy of the particle. There-

fore, the particle is unaffected by the attractive part of the potential.

Q1.5) Parameter a in the van der Waals equation is greater for H2O than for He. What does this

say about the form of the potential function in Figure 1.7 for the two gases?

This means that the depth of the attractive potential is greater for H2O than for He.

Problems

Problem numbers in RED indicate that the solution to the problem is given in the Student Solu-

tions Manual.

P1.1) A sealed flask with a capacity of 1.00 dm3 contains 5.00 g of ethane. The flask is so weak

that it will burst if the pressure exceeds 1.00 × 106 Pa. At what temperature will the pressure of

the gas exceed the bursting pressure?

With nRTpV = and M

mn = :

r

V(r) E

r

V(r) E

r

V(r) E





3

( ) ( ) ( )( ) ( ) K 723.6

mol K J 8.314472 kg 0.005

mol kg10 0.083 m 0.001 Pa 101.00

R m

M V pT

11

1-336

=×

××××== −−

−

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 2.00

kg 300.15mol K J 8.314472 kg 0.001

VM

T R m

V

T R np 5

13-33-

11

2

22

2×=

×××××=== −

−−

H

HHH

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

mol kg10 32.0 m 10 2.00

kg 300.15mol K J 8.314472 kg 0.001

VM

T R m

V

T R np 5

13-33-

11

2

22

2×=

×××××=== −

−−

O

OOO

Pa 10 57.6ppp 5total 22

×=+= OH

( ) % 4.19

mol kg10 32.0

kg 0.001

mol kg10 2.02

kg 0.001

mol kg10 2.02

kg 0.001

100O molH mol

H mol100H % mol

13-13-

13-

22

22 =

×

+

×

×

×=+

×=

−−

−

( ) % .95

mol kg10 32.0

kg 0.001

mol kg10 2.02

kg 0.001

mol kg10 32.0

kg 0.001

100O molH mol

O mol100O % mol

13-13-

13-

22

22 =

×

+

×

×

×=+

×=

−−

−

b)





4

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

mol kg10 28.02 m 10 2.00

kg 300.15mol K J 8.314472 kg 0.001

VM

T R m

V

T R np 4

13-33-

11

2

22

2×=

×××××=== −

−−

N

NNN

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

mol kg10 32.0 m 10 2.00

kg 300.15mol K J 8.314472 kg 0.001

VM

T R m

V

T R np 5

13-33-

11

2

22

2×=

×××××=== −

−−

O

OOO


×=+= OH

( ) % 3.35

mol kg10 32.0

kg 0.001

mol kg10 28.02

kg 0.001

mol kg10 28.02

kg 0.001

100O molN mol

N mol100N % mol

13-13-

13-

22

22 =

×

+

×

×

×=+

×=

−−

−

( ) % 6.74

mol kg10 32.0

kg 0.001

mol kg10 28.02

kg 0.001

mol kg10 32.0

kg 0.001

100O molN mol

O mol100O % mol

13-13-

13-

22

22 =

×

+

×

×

×=+

×=

−−

−

c)

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

mol kg10 17.03 m 10 2.00

kg 300.15mol K J 8.314472 kg 0.001

VM

T R m

V

T R np 4

13-33-

11

3

33

3×=

×××××=== −

−−

NH

NHNH

NH

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

mol kg10 16.04 m 10 2.00

kg 300.15mol K J 8.314472 kg 0.001

VM

T R m

V

T R np 4

13-33-

11

4

44

4×=

×××××=== −

−−

CH

CHCHCH


×=+= OH





5

( )

% 8.54

mol kg10 17.03

kg 0.001

mol kg10 16.04

kg 0.001

mol kg10 17.03

kg 0.001

100

CH molNH mol

NH mol100NH % mol

13-13-

13-

43

33

=

×

+

×

×

×=

+×=

−−

−

( )

% 1.55

mol kg10 17.03

kg 0.001

mol kg10 16.04

kg 0.001

mol kg10 17.03

kg 0.001

100

CH molNH mol

CH mol100CH % mol

13-13-

13-

43

44

=

×

+

×

×

×=

+×=

−−

−

P1.3) Approximately how many oxygen molecules arrive each second at the mitochondrion of

an active person? The following data are available: oxygen consumption is about 40. mL of O2

per minute per kilogram of body weight, measured at T = 300. K and P = 1.0 atm. An adult with

a body weight of 64 kg has about 1 × 1012 cells. Each cell contains about 800. mitochondria.

With nRTpV = the number of moles per minute and per kg of body weight is:

( ) ( )( ) ( ) ( )

13-11

3-5

min mol 10 1.6249min 1 mol K J 8.314472 K 300

m 10 4.0 Pa 101325

T R

V pn −

−− ×=××

××==

For a body weight of 64 kg and per second the number of moles is:

( ) ( ) mol 10 7332.1s 60

min1 kg 64 min mol 10 1.6249n 3-13- ×=

×××= −

Converting to molecules:

( ) ( ) molecules 10 1.04mol molecules 10 6.02214 mol 10 1.7332 O molecules 21-123-32 ×=×××= Wit

h 1 × 1012 cells in a 64 kg body, and 800 mitochondria in each cell:

( )( ) ( ) molecules 10 1.25

800 10 1.0

10 1.04 O molecules 6

12

21

2 ×=××

×=

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





6

18 breaths per minute, how many moles of oxygen per minute are absorbed by the body? As-

sume 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 2930

m 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 pro-

vided.

The ideal gas law and solving for R:

( )aT R

V pn

+=

( )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

1atm L 31.18

atm L 22.98

atm L 31.18

atm L 22.98C100-C0

1V p

V p

V p

V pT-T

a

22

11

22

1121

=

−

=

−

=





7

R is then:

( )( )

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

1

11 C mol atm L 0.0820 C280.2C0mol 1

atm L 22.98

aTn

V pR

=+×

=+

=

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

stant 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 =××=

P1.7) A rigid vessel of volume 0.500 m3 containing H2 at 20.5°C and a pressure of 611 × 103 Pa

is connected to a second rigid vessel of volume 0.750 m3 containing Ar at 31.2°C at a pressure of

433 × 103 Pa. A valve separating the two vessels is opened and both are cooled to a temperature

of 14.5°C. What is the final pressure in the vessels?

We need to first calculate the number of moles of H2 and Ar using the ideal gas law:

( ) ( )( ) ( ) mol 125

mol K J 8.314472 K 20.5K 273.15

m .5000 Pa 10611

T R

V pn

11

33

H2=

×+××== −−

( ) ( )( ) ( ) mol 128

mol K J 8.314472 K 2.13K 273.15

m .7500 Pa 10334

T R

V pn

11

33

Ar =×+

××== −−





8

Using the final number of moles, the final pressure at 14.5ºC is then:

( ) ( ) ( )( ) Pa 1084.4

m .7500m .5000

mol K J 8.314472 K 5.14K 273.15 mol 125mol 281

V

T R np 5

33

11

total

total ×=+

×+×+==−−

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

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

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

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

T RM

mT Rn V p ==

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

1-11

3-11-3

h kg .02270 h 1 mol K J 8.314472 K 300

m 0.5 atm Pa 101325 atm 0.062mol kg 1018.02

T R

V p Mm =

××××××== −−

−

P1.9) At T = 293 K and at 50.% relative humidity, the pressure of water vapor in equilibrium

with liquid water is 0.0115 atm. Using the information in Problem P1.8, determine what mass of

water is inhaled per hour and the net loss of water through respiration per hour.

Using the ideal gas law, and solving for m, with the volume from P1.8 yields:

T RM

mT Rn V p ==

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

1-3-11

3-11-3

h kg 10.152 0.5h 1 mol K J 8.314472 K 293

m 0.5 atm Pa 101325 atm 0.0115mol kg 1018.02

T R

V p Mm ×=×

××××××== −−

−





9

With the results from P1.8, the net loss of water is then:

-1-1-3-1 h kg 0.0205 h kg 102.15 -h kg 0.0227 lossnet =×=

P1.10) A compressed cylinder of gas contains 1.50 × 103 g of N2 gas at a pressure of 2.00 × 107

Pa and a temperature of 17.1°C. What volume of gas has been released into the atmosphere if the

final pressure in the cylinder is 1.80 × 105 Pa? Assume ideal behavior and that the gas tempera-

ture is unchanged.

Let ni and nf be the initial and final number of moles of N2 in the cylinder, respectively:

f

f

i

i

p

T R n

p

T R n=

Solving for nf yields:

( )( )

( )( ) mol 0.482

Pa 102.99

Pa 101.80

mol kg 1028.01

kg 1.50

p

p nn

5

5

13-i

fif =

×××

×== −

The initial number of moles is:

( )( ) mol 55.35

mol kg 1028.01

kg 1.50n

13-i =×

= −

The volume of gas that is released is:

( ) ( ) ( ) ( )( )

311

if m 1.26 Pa 101325

mol K J 8.314472 K 290.25mol 0.482 - mol 53.55

p

RTnnV =××=

−=

−−

P1.11) As a result of photosynthesis, 1.0 kg of carbon is fixed per square meter of forest. As-

suming air is 0.046% CO2 by weight, what volume of air is required to provide 1.0 kg of fixed

carbon? Assume T = 298 K and P = 1.00 atm. Also assume that air is approximately 20.% oxy-





10

gen and 80.% nitrogen by weight.

We first calculate the number of moles necessary to provide one 1 kg of CO2:

( ) ( )( )

( )( ) mol 22.722

mol kg 1044

kg 1

COM

COmCOn

1-3-2

22 =

×==

With the composition of air, a certain volume of air containing one kg of CO2 also contains:

( ) ( )( )

( )( ) mol 96.35861

mol kg 102300046.0

2.0kg 1

OM

OmOn

1-3-2

22 =

×

×

==

( ) ( )( )

( )( ) mol 8.62111

mol kg 102800046.0

8.0kg 1

NM

NmNn

1-3-2

22 =

×

×

==

The volume of air can then be obtained by considering the total number of moles:

( ) ( ) ( ){ }

( ) ( ) ( ){ } ( ) ( )( )

3

1-1-

222totair

m 6.8511

Pa 013251

K 298mol K J 8.314472mol 62111.8mol 13586.962mol 22.722

p

T R NnOnCOn

p

T R nV

=

××++=

++==

P1.12) A balloon filled with 10.50 L of Ar at 18.0°C and 1 atm rises to a height in the atmos-

phere where the pressure is 248 Torr and the temperature is –30.5°C. What is the final volume of

the balloon?

We first calculate the number of moles of Ar at 1 atm using the ideal gas law:

( ) ( ) ( )( ) ( )

-1 3

1 1

1 atm 101325 Pa atm 0.0105 mp Vn 0.4395 mol

R T 291.15 K 8.314472 J K mol− −

× ×= = =

×

We can then determine the volume after the balloon raised using the number of moles:





11

( ) ( ) ( )( ) ( ) L 26.8m 0.0268

Torr Pa 133.32 Torr 248

mol K J 8.314472 K 30.5K 291.150.4395mol

p

T Rn V 3

1-

11

==×

×−×==−−

P1.13) One liter of fully oxygenated blood can carry 0.20 L of O2 measured at T = 273 K and P

= 1.00 atm. Calculate the number of moles of O2 carried per liter of blood. Hemoglobin, the oxy-

gen transport protein in blood, has four oxygen-binding sites. How many hemoglobin molecules

are required to transport the O2 in 1.0 L of fully oxygenated blood?

With nRTpV = the number of moles of O2 in one liter of fully oxygenated blood is:

( ) ( ) ( )( ) ( ) mol10 9279.8

K 273 mol K J 8.314472

m 0.0002Pa 013251

T R

V pOn 3-

11

3

2 ×=×

×== −−

Converting to molecules:

( ) ( ) molecules 10 5.377mol molecules 10 6.02214 mol10 9279.8 O molecules 21-123-32 ×=×××= Fi-

nally, four binding sites per Hemoglobin molecule have to be considered, so that the number of

O2 molecules required is:

molecules 10 1.34 4 / molecules 10 5.377required O molecules 21212 ×=×=

P1.14) Myoglobin is a protein that stores oxygen in the tissues. Unlike hemoglobin, which has

four oxygen-binding sites, myoglobin has only a single oxygen-binding site. How many my-

oglobin molecules are required to transport the oxygen absorbed by the blood in Problem 1.13?

Since myoglobin can only bind ¼ of the amount of O2 that hemoglobin can bind the number of

myoglobin molecules is:

molecules 10 5.36 4molecules 10 1.344myoglobin moleculesmyoglobin molecules 2121 ×=××=×=

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?





12

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 isother-

mally increased to 200. atm?

a) ( )( ) % 68.6

atm 00.1

atm 0.880.78100

p

p100N % mol

total

N

22 =××=×=

( )( ) % 18.5

atm 00.1

atm 0.880.21100

p

p100O % mol

total

O

22 =××=×=

( )( ) % 0.9

atm 00.1

atm 0.880.01100

p

p100Ar % mol

total

Ar =××=×=

( )( ) % 12.0

atm 00.1

atm 0.12100

p

p100OH % mol

total

OH

22 =×=×=

b)

V

T R n V p

OH

OH2

2=

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

( ) ( )( ) L 12.2

atm 0.197

L 20.0atm 0.12

p

V p V

OH

OH

2

2 =×=′

=′

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

atm would be:





13

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 con-

dense. The fraction that condenses is given by:

( ) ( )( ) 0.992

atm 24.0

atm 0.197atm 24.0condensedfrcation =−=

P1.16) A mixture of 2.50 × 10–3 g of O2, 3.51 × 10–3 mol of N2, and 4.67 × 1020 molecules of

CO is placed into a vessel of volume 3.50 L at 5.20°C.

a. Calculate the total pressure in the vessel.

b. Calculate the mole fractions and partial pressures of each gas.

a) The pressure in the vessel can be calculated by using the total number of moles:

( ) mol 10 7.8125mol g 32.0

g10 2.5On 5

1-

-3

2−×=×= ( ) mol 10 7.7547

molmolecules106.02214

molecules 10 4.67COn 4

1-23

20−×=

××=

( ) ( ) ( )( )

bar 0.00288 Pa 2885.375

m 0.0035

K 278.35 mol K J 8.314472 10 3.51 mol 10 7.7547mol 10 7.8125

V

T R np

3

11345

tot

=

=×××+×+×

==

−−−−−

b) The mole fractions, tot

ii n

nx = , and partial pressures, pnp ii ×= , for the gases are:

( ) 0.0179 mol 10 4.3636

mol 10 7.8125Ox

3

5

2 =××= −

−

( ) bar 10 5.16 0.0179 bar 0.00288Op 42

−×=×=

( ) 0.8044 mol 10 4.3636

mol 10 3.51Nx

3

3

2 =×

×= −

−

( ) bar 10 2.32 0.8044 bar 0.00288Np 32

−×=×=

( ) 0.1777 mol 10 4.3636

mol 10 7.7547COx

3

4

=××= −

−

( ) bar 10 5.12 0.1777 bar 0.00288COp 4−×=×=

P1.17) Carbon monoxide (CO) competes with oxygen for binding sites on the transport protein

hemoglobin. CO can be poisonous if inhaled in large quantities. A safe level of CO in air is 50.





14

parts per million (ppm). When the CO level increases to 800. ppm, dizziness, nausea, and uncon-

sciousness occur, followed by death. Assuming the partial pressure of oxygen in air at sea level

is 0.20 atm, what ratio of O2 to CO is fatal?

Converting the partial pressure of O2 in the atmosphere to ppm using p

pii =x :

( ) ppm 200000%200.2atm 1

atm 0.2Ox 2 ====

Therefore, the fatal O2/CO ratio is:

( )( ) 250

800

200000

COx

Ox 2 ==

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 mol

R 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

nbreaths ofnumber

breath

tank ===

P1.19) Liquid N2 has a density of 875.4 kg m–3 at its normal boiling point. What volume does a

balloon occupy at 18.5°C and a pressure of 1.00 atm if 2.00 × 10–3 L of liquid N2 is injected into





15

it?

We can use the density of liquid N2 to calculate the mass of N2 that is injected into the balloon:

( ) ( )2

-3 -6 3 -3N LN2 LN2m d V 875.4 kg m 2 10 m 1.7508 10 kg= = × × = ×

We calculate the volume using the ideal gas law:

( ) ( ) ( )( ) ( )

-3 1 1

-3 3

-3 1

n R T m R TV

p M p

1.7508 10 kg 291.15 K 8.314472 J K mol 1.50 10 m 1.50 L

28 10 kg mol 101325 Pa

− −

−

= =

× × ×= = × =

× ×

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:

( ) ( )( ) ( ) mol 0406.0

mol K J 8.314472 K 300

Pa 101325 m 101

T R

p 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 =

×=== −

P1.21) A sample of propane (C3H8) is placed in a closed vessel together with an amount of O2

that is 3.00 times the amount needed to completely oxidize the propane to CO2 and H2O at con-





16

stant temperature. Calculate the mole fraction of each component in the resulting mixture after

oxidation assuming that the H2O is present as a gas.

The reaction we have to consider is:

( ) ( ) ( ) ( )g OH 4 g CO 3 g O 5 g HC 22283 +⎯→⎯+

If m moles of propane are present initially, there must be 15 m moles of O2. After the reaction is

complete, there are 3 m moles of CO2, 4 m moles of H2O, and 10 m moles of O2. Therefore:

.1760 m 17

m 3x

2CO == ; .2350 m17

m 4x OH2

== , .5880 m17

m 10x

2O ==

P1.22) Calculate the volume of all gases evolved by the complete oxidation of 0.25 g of the

amino acid alanine (NH2CHCH3COOH) if the products are liquid water, nitrogen gas, and car-

bon dioxide gas and the total pressure is 1.00 atm and T = 310. K.

The reaction equation is:

( ) ( ) ( ) ( ) ( )g CO 6 l OH 6 g N g O 9 s COOHCHCHNH 2 222232 ++⎯→⎯+

We use the mass of alanine to calculate the number of moles of alanine that were oxidized:

( )( ) mol 102.8387

mol g 88.07

g 0.25

M

mn 3-

1alanine

alaninealanine ×=== −

From the reaction equation we know that:

mol 101.4194 n 2

1n 3-

alanineN2×==

And the volume of N2 consumed in the reaction is:





17

( ) ( ) ( )( ) L 0.0361m 103.61

Pa 101325

mol K J 8.314472 K 310mol 101.4194

p

T Rn V 35-

11-3

N2=×=×××==

−−

mol 108.5161 n 3n -3alanineCO2

×==

( ) ( ) ( )( ) L 0.2166m 10166.2

Pa 101325

mol K J 8.314472 K 310mol 108.5161

p

T Rn V 34-

11-3

CO2=×=×××==

−−

Then the total volume of gases produced in the reaction is:

L 0.253 L 0.0361L 0.2166 VVV22 CONtotal =+=+=

P1.23) A gas sample is known to be a mixture of ethane and butane. A bulb having a 200.0-cm3

capacity is filled with the gas to a pressure of 100.0 × 103 Pa at 20.0°C. If the weight of the gas

in the bulb is 0.3846 g, what is the mole percent of butane in the mixture?

With nRTpV = the total number of moles of moles of the mixture is:

( ) ( )( ) ( ) mol10 2055.8

K 293.15 mol K J 8.314472

m 0.0002Pa 10 100

T R

V pn 3-

11

3-3

tot ×=×

××== −−

The total number of moles can also be expressed as:

butane

butane

ethane

ethanebutaneethanetot M

m

M

mnnn +=+=

With methane = mtot – mbutane:

( )( )butaneethane

ethanebutanebutanebutanetottot MM

MmMmmn

×+×−

=

Solving for mbutane and dividing by Mbutane yields after some rearrangement:

( )( )

( ) ( ) ( )( )( ) ( )( ) mol109101.4

mol g 58.14mol g 30.08

g 0.3846mol g 30.08mol108.2055

MM

mMnn 3

1-1-

-13

butaneethane

totethanetotbutane

−−

×=−

−××=−

−×= An

d finally:

( )( ) %9.59599.0

mol108.2055

mol109101.4

n

nx

3

3

total

butanebutane ==

××== −

−





18

P1.24) A glass bulb of volume 0.136 L contains 0.7031 g of gas at 759.0 Torr and 99.5°C. What

is the molar mass of the gas?

With n

mM = and nRTpV = the molar mass of the gas is:

( ) ( ) ( )( ) ( )-3 1 1

-3 -1 -1

3

0.7031 10 kg 8.314472 J K mol 372.65 Km R TM 0.1583 10 kg mol 158.3 g mol

p V 101191.68 Pa 0.000136 m

− −× × ×= = = × =

×

P1.25) The total pressure of a mixture of oxygen and hydrogen is 1.00 atm. The mixture is ig-

nited 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?

We start by constructing the following table:

2 H2 (g) + O2 (g) → 2 H2O ( )

initial moles 2Hn

2On 0

at equilibrium α2n2H − α−

2On α2

If O2 is consumed completely, 0n2O =−α , or α=

2On . The number of moles of H2 remaining is

222 OHH n2n2n −=− α . Let p1 be the initial total pressure and p2 be the total pressure after all O2 is

consumed.

( )V

T Rnnp

22 OH1 += and ( )

V

T Rn 2np

22 OH2 +=

Dividing the second equation by the first:

( ) ( ) 22222

22

2

22

2

OOOOHOH

O

OH

H

1

2 x31x2x1x2xnn

n2

nn

n

p

p −=−−=−=+

−+

=





19

( )( ) 20.0

atm 1.00

atm 0.4001

3

1

p

p1

3

1x

1

2O2

=

−=

−= , and 80.0 x

2H =

P1.26) The photosynthetic formation of glucose in spinach leaves via the Calvin cycle involves

the fixation of carbon dioxide with ribulose 1-5 diphosphate C5H8P2O114−(aq) to form 3-

phosphoglycerate C3H4PO7

3−(aq) :

C5H8P2O11

4−(aq) + H 2O(l) + CO2 (g)

→ 2C3H 4PO7

3− (aq) + 2H+ (aq)

If 1.00 L of carbon dioxide at T = 273 K and P = 1.00 atm is fixed by this reaction, what mass of

3-phosphoglycerate is formed?

First we need to calculate the number of moles of fixed CO2:

( ) ( )( ) ( )2

3

1 1

101191.68 Pa 0.001 mpVn 0.0446 mol

RT 8.314472 J K mol 273 KCO − −

×= = =

×

The number of moles of 3-phosphoglycerate formed is given by:

23n 2 n 0.0892 molphos CO− = =

And finally, the mass of 3-phosphoglycerate formed is:

( ) ( )1m n M 0.0928 atm 182.97 g mol 16.32 g−= = × =

P1.27) Calculate the pressure exerted by Ar for a molar volume of 1.42 L mol–1 at 300. K using

the van der Waals equation of state. The van der Waals parameters a and b for Ar are 1.355 bar

dm6 mol–2 and 0.0320 dm3 mol–1, respectively. Is the attractive or repulsive portion of the poten-

tial dominant under these conditions?





20

To determine what portion of the potential is dominant we need to compare the pressure pre-

dicted by the van der Waals equation of state with that predicted by the ideal gas law. The van

der Waals equation of state yields:

( )( ) ( )

( ) ( )( )( )

( ) bar 17.3 mol dm 1.42

mol dmbar 1.355

mol dm 0.0321 mol dm 1.42

K 300 mol K dmbar 108.314472

V

a

bV

T Rp 213

26

1313

113-2

2mm

vdW =−−

××=−−

=−

−

−−

−−

The ideal gas law gives:

( ) ( )( ) bar 17.6

mol dm 1.42

K 300 mol K dmbar 108.314472

V

T Rp

13

113-2

mideal =××== −

−−

Because pvdW < pideal, the attractive part of the potential dominates.

P1.28) Calculate the pressure exerted by benzene for a molar volume of 1.42 L at 790. K using

the Redlich–Kwong equation of state:

P =

RT

Vm − b−

a

T

1

Vm(Vm + b)=

nRT

V − nb−

n2a

T

1

V (V + nb)

The Redlich–Kwong parameters a and b for benzene are 452.0 bar dm6 mol–2 K1/2 and 0.08271

dm3 mol–1, respectively. Is the attractive or repulsive portion of the potential dominant under

these conditions?

The exerted benzene pressure is calculated using ( ) ( )bVV

1

T

a

bV

T Rp

mmm +×−

−= :

( ) ( )( )

( )( ) ( ) ( ) bar 41.6

mol L 0.08271mol L 1.42mol L 1.42

1

790K

mol Kbar L 452.0

mol L 0.08271mol L 1.42

K 790 mol Kbar L 10 8.314p

111

21/2-2

11

1-1-2

=−×

−−

××=

−−−

−

−−

−

To determine whether the attractive or repulsive portion of the potential is dominant we need to

compare the pressure from above with that obtained by using the ideal gas law:





21

( ) ( ) ( ) ( )( ) ( ) bar 6.34

Pa 10m 10 1.42

bar 1K 790.0mol K J 8.314472 mol 1

V

T Rn p

533-

11

=××

×××==−−

ideal

Since the pressure calculated with the ideal gas law is higher than the pressure calculated with

the Redlich–Kwong equation of state, the attractive forces are dominant.

P1.29) When Julius Caesar expired, his last exhalation had a volume of 500. cm3 and contained

1.00 mol % argon. Assume that T = 300. K and P = 1.00 atm at the location of his demise. As-

sume further that T and P currently have the same values throughout the Earth’s atmosphere. If

all of his exhaled Ar atoms are now uniformly distributed throughout the atmosphere (which for

our calculation is taken to have a thickness of 1.00 km), how many inhalations of 500. cm3 must

we make to inhale one of the Ar atoms exhaled in Caesar’s last breath? Assume the radius of the

Earth to be 6.37 × 106 m. (Hint: Calculate the number of Ar atoms in the atmosphere in the sim-

plified geometry of a plane of area equal to that of the Earth’s surface and a height equal to the

thickness of the atmosphere. See Problem P1.30 for the dependence of the barometric pressure

on the height above the Earth’s surface.)

The total number of Ar atoms in the atmosphere is:

dzA N~

N0

Ar∞

= Ar , where

ArN~

is the number Ar atoms per m3 at the surface of the earth. ArN

~ is

given by:

( ) ( )( ) ( )

3-2311

5-123ArA

Ar m 1041.2K 300mol K J 8.314472

Pa 1010100.0mol 10023.6

T R

pNN~ ×=

×××××== −−

The total number at Ar atoms in the atmosphere is:





22

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

411-3

11263-23

ArAr

0

T R

z g M

Ar

0

Ar

1085.7s m 9.81kg 109.39

K 300mol K J 8.314472m 1037.64m 1041.2

g M

T RA N

~dzA eN

~dzA N

~N

Ar

×=××

××××××=

===

−

−−

∞ −∞

π

Ar

The fraction of these atoms that came from Caesar’s last breath, f, is given by:

( ) ( )( )

2241

33-323Ar 1053.1

1085.7

m 10500.0m 1041.2

N

V N~

f −−

×=×

×××==Ar

The number of Ar atoms that we inhale with each breath is:

( ) ( ) ( )( ) ( )

2011

3351-23

A 1021.1K 300mol K J 8.314472

m 10500.0Pa 1010100.0mol 10023.6

T R

V pNN ×=

×××××××== −−

−

The number of these that came from Caesar’s last breath is f × N:

f×N = 1.53 × 10-22 × 1.21 × 1020 = 1.85 × 10-2

The reciprocal of this result, or 54, is the number of breaths needed to inhale one Ar atom that

Caesar exhaled in his last breath.

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

as Pi = Pi

0e− Mi g RT

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 com-

position 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?





23

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

N2

2 2

M g z

R TN N

3 -1 -1 3

5

1 1

p p e

28.04 10 kg mol 9.81 m s 50 10 m0.600 1.01325 10 Pa Exp 248 Pa

8.314472 J K mol 300 K

−

−

− −

=

× × × ×= × × × − =

×

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

CO2

2 2

M g z

R TCO CO

3 -1 -1 3

5

1 1

p p e

44.04 10 kg mol 9.81 m s 50 10 m0.400 1.01325 10 Pa Exp 7.02 Pa

8.314472 J K mol 300 K

−

−

− −

=

× × × ×= × × × − =

×

( )( )

( )2

2

2 2

COCO

CO N

p 7.02 Pax 0.028

7.02 Pa 248 Pap p= = =

++

972.0 x1x22 CON =−=

P1.31) Assume that air has a mean molar mass of 28.9 g mol–1 and that the atmosphere has a

uniform temperature of 25.0°C. Calculate the barometric pressure at Denver, for which z = 1600.

m. Use the information contained in Problem P1.30.

Using

−×=

T R

z g MExpPP i0

ii the barometric pressure at 1600 m is:

( ) ( ) ( )( ) ( ) Pa 10 8.44atm 0.833

K 298.15 mol K J 8.314472

m 1600 s m 9.80665mol kg10 28.9Expatm 1P 4

11

-2-13

i ×==

×

×××−×= −−

−

P1.32) 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 + H2 → Cu +

H2O, and oxygen reoxidizes the copper formed according to Cu + 1/2 O2 → CuO. At 25°C and

750. Torr, 100.0 cm3 of the mixture yields 84.5 cm3 of dry oxygen measured at 25°C and 750.

Torr after passage over CuO and the drying agent. What was the original composition of the

mixture?





24

We need to consider two equilibria:

CuO (s) + H2 (g) → H2O (l) + Cu (s)

at equilibrium α−2Hn βα −

Cu (s) + ½ O2 (g) → CuO (s)

at equilibrium βα − β2

1n

2O −

In the final state, only O2 is present. Therefore, α−2On . In an excess of O2, all the copper is oxi-

dized, and 0=− βα and β=2On . That means that

222 HOO n2

1nn −= .

Let V1 and V2 be the initial and final volumes:

( )p

T RnnV

22 OH1 += and

p

T Rn

2

1nV

22 HO2

+=

Dividing the second equation by the first yields:

( ) ( )

2222

22

2

22

2

HHHOOH

H

OH

O

1

2 x2

31x

2

11x

2

1x

nn

n

2

1

nn

n

V

V −=−=−=+

−+

=

( )( ) 103.0

cm 100

cm 84.51

3

2

V

V1

3

2x

3

3

1

2H2

=

−=

−= , and 897.0 x1x

22 HO =−=

P1.33) Aerobic cells metabolize glucose in the respiratory system. This reaction proceeds ac-

cording to the overall reaction

6O2(g) + C6H12O6(s)→6CO2(g) + 6H2O(l)

Calculate the volume of oxygen required at STP to metabolize 0.010 kg of glucose (C6H12O6).

STP refers to standard temperature and pressure, that is, T = 273 K and P = 1.00 atm. Assume





25

oxygen behaves ideally at STP.

The number of moles of O2 according to the stoichiometry of the equation is:

( ) ( ) ( )( ) mol 0.0555

mol g 180.18

g 106glucosen6On

12 =×=×= −

Therefore, the volume of O2 required is:

( ) ( ) ( )( ) L 7.47m10 7.47

Pa 101325

K 273 mol K J 8.314472 mol 0.0555

p

T R nV 33-

11O

O2

2=×=××==

−−

P1.34) Consider the oxidation of the amino acid glycine (NH2CH2COOH) to produce water,

carbon dioxide, and urea (NH2CONH2):

NH2CH2COOH(s) + 3O2(g) →NH2CONH2(s) +3CO2(g) + 3H2O(l)

Calculate the volume of carbon dioxide evolved at P = 1.00 atm and T = 310. K from the oxida-

tion of 0.0100 g of glycine.

The number of moles of CO2 according to the stoichiometry of the equation is:

( ) ( ) ( )( ) mol 10997.3

mol g 75.05

g 0.01003glycinen3COn 4

12−

− ×=×=×=

Therefore, the volume of CO2 evolved is:

( ) ( ) ( )( ) L 10 1.02m10 1.0168

Pa 101325

K 310 mol K J 8.314472 mol 10997.3

p

T R nV 2-35-

114CO

CO2

2×=×=×××==

−−−

P1.35) An initial step in the biosynthesis of glucose (C6H12O6) is the carboxylation of pyruvic

acid (CH3COCOOH) to form oxaloacetic acid (HOOCCOCH2COOH):

CH3COCOOH(s)+CO2(g)→HOOCCOCH2COOH(s)

If you knew nothing else about the intervening reactions involved in glucose biosynthesis other

than that no further carboxylations occur, what volume of CO2 is required to produce 0.50 g of

glucose? Assume P = 1 atm and T = 310. K.

The number of moles of CO according to the stoichiometry of the equations is:





26

( ) ( ) ( )( ) mol 10 2.775

mol g 180.18

g 0.50glucosenCOn 3

12−

− ×===

Therefore, the volume of CO2 required is:

( ) ( ) ( )( ) L 0706.0m10 7.059

Pa 101325

K 310 mol K J 8.314472 mol 10 2.775

p

T Rn V 35-

113

=×=×××==−−−




chapter 1: fundamental concepts of thermodynamics · 2017-12-04 · chapter 1/fundemental concepts...

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