Solution manual for physical chemistry for the life sciences 1st edition by engel

Solution Manual for Physical Chemistry for the Life Sciences 1st Edition by Engel
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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 system 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, because 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 interest a necessary condition for equilibrium between two systems?
Equilibrium is a dynamic process in which the rates of two opposing processes are equal. However, 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.
1

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E

V(r)

r

In this case, the energy well depth is small compared to the total energy of the particle. Therefore, 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 Solutions 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 pV = nRT and n =

m
:
M

2

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Solution Manual for Physical Chemistry for the Life Sciences 1st Edition by Engel
Full file at />T=

(

) (

Chapter 1/Fundemental Concepts of Thermodynamics

) (

)

p V M 1.00 × 10 6 Pa × 0.001 m 3 × 30.08 × 10 -3 kg mol −1
=
= 723.6 K
mR
(0.005 kg ) × 8.314472 J K −1 mol −1

(

)

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)
p H2 =

p O2 =

n H2 R T

V
n O2 R T
V

=

=

m H2 R T
M H2 V
m O2 R T
M O2 V

=

(0.001 kg ) × (8.314472 J K −1 mol −1 )× (300.15 kg ) = 6.18 × 10 5 Pa

(2.00 × 10

-3

) (

m 3 × 2.02 × 10 -3 kg mol −1

)

(
0.001 kg ) × (8.314472 J K −1 mol −1 )× (300.15 kg )
=

(2.00 × 10

-3

3

) (

-3

m × 32.0 × 10 kg mol

−1

)

= 3.90 × 10 5 Pa

p total = p H 2 + p O2 = 6.57 × 10 5 Pa



0.001 kg


-3
−1 
2.02 × 10 kg mol 
mol H 2


= 100 ×
= 94.1 %
mol % H 2 = 100 ×
(mol H 2 + mol O 2 )

 

0.001 kg
0.001 kg

 

  2.02 × 10 -3 kg mol −1  +  32.0 × 10 -3 kg mol −1  
 




0.001 kg


-3
−1 
mol O 2
 32.0 × 10 kg mol 
mol % O 2 = 100 ×
= 5.9 %
= 100 ×
(mol H 2 + mol O 2 )


 

0.001 kg
0.001 kg

 

  2.02 × 10 -3 kg mol −1  +  32.0 × 10 -3 kg mol −1  






b)
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p N2 =

p O2 =

n N2 R T
V
n O2 R T

V

=

=

m N2 R T
M N2 V
m O2 R T
M O2 V

=

(0.001 kg ) × (8.314472 J K −1 mol −1 )× (300.15 kg ) = 4.45 × 10 4 Pa

=

(0.001 kg ) × (8.314472 J K −1 mol −1 )× (300.15 kg )

(2.00 × 10

-3

(2.00 × 10

-3

) (

m 3 × 28.02 × 10 -3 kg mol −1

3

) (

-3

m × 32.0 × 10 kg mol

−1

)

)

= 3.90 × 10 5 Pa

p total = p H 2 + p O2 = 8.35 × 10 4 Pa



0.001 kg


−1 
-3
28.02 × 10 kg mol 
mol N 2

= 100 ×

= 53.3 %
mol % N 2 = 100 ×
(mol N 2 + mol O 2 )

 

0.001 kg
0.001 kg

 

  28.02 × 10 -3 kg mol −1  +  32.0 × 10 -3 kg mol −1  







0.001 kg


-3
−1 
mol O 2
 32.0 × 10 kg mol 
mol % O 2 = 100 ×
= 46.7 %
= 100 ×
(mol N 2 + mol O 2 )


 

0.001 kg
0.001 kg

 

  28.02 × 10 -3 kg mol −1  +  32.0 × 10 -3 kg mol −1  
 



c)
p NH 3 =

p CH 4 =

n NH 3 R T
V

n CH 4 R T
V

=

=

m NH 3 R T
M NH 3 V

m CH 4 R T
M CH 4 V

=

(0.001 kg ) × (8.314472 J K −1 mol −1 )× (300.15 kg ) = 7.32 × 10 4 Pa

=

(0.001 kg ) × (8.314472 J K −1 mol −1 )× (300.15 kg )

(2.00 × 10

(2.00 × 10

-3

-3

) (

m 3 × 17.03 × 10 -3 kg mol −1

3

m × 16.04 × 10 kg mol

p total = p H 2 + p O2 = 1.51 × 10 5 Pa

4

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) (

-3

−1

)

)

= 7.77 × 10 4 Pa


Solution Manual for Physical Chemistry for the Life Sciences 1st Edition by Engel
Full file at />mol % NH 3 = 100 ×

Chapter 1/Fundemental Concepts of Thermodynamics

mol NH 3
(mol NH 3 + mol CH 4 )



0.001 kg


-3

−1 
17.03 × 10 kg mol 

= 100 ×
= 48.5 %

 

0.001 kg
0.001 kg

 

  16.04 × 10 -3 kg mol −1  +  17.03 × 10 -3 kg mol −1  





mol % CH 4 = 100 ×

mol CH 4
(mol NH 3 + mol CH 4 )



0.001 kg


-3

−1 
17.03 × 10 kg mol 

= 100 ×
= 51.5 %

 

0.001 kg
0.001 kg

 

  16.04 × 10 -3 kg mol −1  +  17.03 × 10 -3 kg mol −1  






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 pV = nRT the number of moles per minute and per kg of body weight is:

(

)

(
pV
101325 Pa ) × 4.0 × 10 -5 m 3
n=
= 1.6249 × 10 -3 mol min −1
=
−1
−1
R T (300 K ) × 8.314472 J K mol × (1 min )

(

)

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

 1 min 
-3
n = 1.6249 × 10 -3 mol min −1 × (64 kg ) × 
 = 1.7332 × 10 mol
 60 s 

(

)

Converting to molecules:

(

) (

)

molecules O 2 = 1.7332 × 10 -3 mol × 6.02214 × 10 23 molecules mol -1 = 1.04 × 10 21 molecules Wit
h 1 × 1012 cells in a 64 kg body, and 800 mitochondria in each cell:

(1.04 × 10 ) = 1.25 × 10
=
(1.0 × 10 )× (800)
21

molecules O 2

12

6

molecules

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

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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:

n air

(

)

(
pV
101325 Pa ) × 5.0 × 10 -4 m 3
=
=
= 0.0208 mol
R T (2930 K ) × 8.314472 J K −1 mol −1

(

)

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

(

n O 2 = (0.0208 mol) × (0.21) × (0.25) × 18 min −1

)

= 0.01966 mol min −1

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:
n=

pV
R (T + a )

R=

pV
n (T + a )

Using the information for the two different conditions and solving for a yields a:
p1 V1
p 2 V2
=
n (T1 + a ) n (T2 + a )


p V 
 T1 - T2 1 1 
p 2 V2 

a=
=
 p1 V1


− 1
 p 2 V2


 
(22.98 L atm ) 
 0 C - 100  C
(31.18 L atm ) 

= 280.2  C
 (22.98 L atm ) 

− 1
 (31.18 L atm ) 

( )(

)

6

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Chapter 1/Fundemental Concepts of Thermodynamics

R is then:
R=

p1 V1
(22.98 L atm )
=
n (T1 + a ) (1 mol) × 0  C + 280.2  C

(( ) (

)) = 0.0820 L atm mol

-1 

C -1

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:
G = T(K ) × 8.314472 G K -1
So that on the Kelvin scale:

(

)

R T = 8.314472 J mol -1 K -1 × (293 K ) = 2436.14 J mol -1

And on the G scale:

(

)

(

)

R G T = 1.0 J mol -1 G -1 × (293 K ) × 8.314472 G K -1 = 2436.14 J mol -1
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:
n H2 =

(611× 10 3 Pa )× (0.500 m 3 )
pV
= 125 mol
=
R T (273.15 K + 20.5 K ) × (8.314472 J K −1 mol −1 )

n Ar =

(433 × 103 Pa )× (0.750 m 3 )
pV
= 128 mol

=
R T (273.15 K + 31.2 K ) × (8.314472 J K −1 mol −1 )

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

n total R T (128 mol + 125 mol) × (273.15 K + 14.5 K ) × (8.314472 J K −1 mol −1 )
=
= 4.84 × 10 5 Pa
3
3
Vtotal
(0.500 m + 0.750 m )

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 ideally under these conditions. What mass of water vapor is exhaled in an hour?
Using the ideal gas law, and solving for m yields:
pV =nRT =

m=

m
RT
M

M p V (18.02 × 10 -3 kg mol −1 )× (0.062 atm ) × (101325 Pa atm -1 )× (0.5 m 3 )
=
= 0.0227 kg h -1
−1
−1
RT
(300 K ) × (8.314472 J K mol )× (1 h )

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:
pV =nRT =

m=

m
RT
M

M p V (18.02 × 10 -3 kg mol −1 )× (0.0115 atm ) × (101325 Pa atm -1 )× (0.5 m 3 )
=
× 0.5 = 2.15 × 10 -3 kg h -1
−1
−1

RT
(293 K ) × (8.314472 J K mol )× (1 h )

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Chapter 1/Fundemental Concepts of Thermodynamics

With the results from P1.8, the net loss of water is then:
net loss = 0.0227 kg h -1 - 2.15 × 10 -3 kg h -1 = 0.0205 kg h -1

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 temperature is unchanged.
Let ni and nf be the initial and final number of moles of N2 in the cylinder, respectively:
ni R T
n RT
= f
pi
pf

Solving for nf yields:
nf = ni

pf
(1.50 kg )

(1.80 × 10 5 Pa ) = 0.482 mol
=
×
pi
(28.01× 10 -3 kg mol −1 ) (2.99 × 105 Pa )

The initial number of moles is:
ni =

(1.50 kg )

(28.01× 10

-3

kg mol −1 )

= 53.55 mol

The volume of gas that is released is:
V=

(n f

− n i )RT (53.55 mol - 0.482 mol) × (290.25 K ) × (8.314472 J K −1 mol −1 )
=
= 1.26 m 3
(101325 Pa )
p

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-

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Solution Manual for Physical Chemistry for the Life Sciences 1st Edition by Engel
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at />1/Fundemental Concepts of Thermodynamics
gen and 80.% nitrogen by weight.
We first calculate the number of moles necessary to provide one 1 kg of CO2:
n (CO 2 ) =

(1 kg )
m(CO 2 )
=
= 22.722 mol
M(CO 2 ) 44 × 10 -3 kg mol-1

(

)

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

n (O 2 ) =

n (N 2 ) =

m(O 2 )
M(O 2 )

m(N 2 )
M (N 2 )

(1 kg ) × 

0.2 

 0.00046  = 13586.96 mol
=
(32 × 10-3 kg mol-1 )

(1 kg ) × 

0.8 

0.00046 

=
(28 × 10-3 kg mol-1 ) = 62111.8 mol

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

n tot R T {n (CO 2 ) + n (O 2 ) + n (N 2 )} R T

=
p
p

{(22.722 mol) + (13586.962 mol) + (62111.8 mol)}× (8.314472 J K -1 mol-1 )× (298 K )
(101325 Pa )

= 1851.6 m 3
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:

n=

-1
3
p V (1 atm ) × (101325 Pa atm ) × ( 0.0105 m )
=
= 0.4395 mol
RT
( 291.15 K ) × (8.314472 J K −1 mol −1 )

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

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Chapter 1/Fundemental Concepts of Thermodynamics

(

)

n R T (0.4395mol) × (291.15 K − 30.5 K ) × 8.314472 J K −1 mol −1
=
= 0.0268 m 3 = 26.8 L
-1
p
(248 Torr ) × 133.32 Pa Torr

(

)

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 oxygen 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 pV = nRT the number of moles of O2 in one liter of fully oxygenated blood is:
n (O 2 ) =

(

)

(101325 Pa ) × 0.0002 m 3 = 8.9279 × 10 -3 mol
pV
=
R T 8.314472 J K −1 mol −1 × (273 K )

(

)

Converting to molecules:

(

) (

)

molecules O 2 = 8.9279 × 10 -3 mol × 6.02214 × 10 23 molecules mol -1 = 5.377 × 10 21 molecules Fi-

nally, four binding sites per Hemoglobin molecule have to be considered, so that the number of
O2 molecules required is:
molecules O 2 required = 5.377 × 10 21 molecules / 4 = 1.34 × 10 21 molecules

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 myoglobin 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 myoglobin = molecules myoglobin × 4 = 1.34 × 10 21 molecules × 4 = 5.36 × 10 21 molecules

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?
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at />1/Fundemental Concepts of Thermodynamics
b. The percent relative humidity is defined as %RH = PH

pressure of water in the sample and PH*

2O

2O

P*H

2O

where PH

2O

is the partial

= 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) mol % N 2 = 100 ×

mol % O 2 = 100 ×

mol % Ar = 100 ×

p N2
p total

p O2
p total

= 100 ×

= 100 ×

(0.78 × 0.88 atm ) = 68.6 %
(1.00 atm )

(0.21× 0.88 atm ) = 18.5 %
(1.00 atm )

p Ar
(0.01 × 0.88 atm) = 0.9 %
= 100 ×
(1.00 atm )
p total

mol % H 2 O = 100 ×

p H 2O
p total

= 100 ×

(0.12 atm) = 12.0 %
(1.00 atm )

b)
p H 2O V =

n H 2O R T
V

p ′H 2 O V ′ = p H 2O V , where the prime refers to 100% RH

V′ =

p H 2O V
p ′H 2 O

=

(0.12 atm )× (20.0 L )
(0.197 atm )

= 12.2 L

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

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Chapter 1/Fundemental Concepts of Thermodynamics

p H 2 O = p total × mol fraction H 2 O = 200 atm × 0.12 = 24.0 atm

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:
frcation condensed =

(24.0 atm ) − (0.197 atm )
(24.0 atm )

= 0.992

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:
n (O 2 ) =

4.67 × 10 20 molecules
2.5 × 10 -3 g
−5
(
)
n
CO
=
= 7.7547 × 10 − 4 mol
=
7.8125
×
10
mol
6.02214 × 10 23 molecules mol -1
32.0 g mol -1

n tot R T
=
V
7.8125 × 10 −5 mol + 7.7547 × 10 − 4 mol + 3.51 × 10 −3 × 8.314472 J K −1 mol −1 × (278.35 K )
=
0.0035 m 3
2885.375 Pa = 0.00288 bar

p=

(

(

b) The mole fractions, x i =

)

) (

)

ni
, and partial pressures, p i = n i × p , for the gases are:
n tot

x (O 2 ) =

7.8125 × 10 −5 mol
= 0.0179
4.3636 × 10 −3 mol

p(O 2 ) = 0.00288 bar × 0.0179 = 5.16 × 10 −4 bar

x (N 2 ) =

3.51 × 10 −3 mol

= 0.8044
4.3636 × 10 −3 mol

p(N 2 ) = 0.00288 bar × 0.8044 = 2.32 × 10 −3 bar

x (CO ) =

7.7547 × 10 −4 mol
= 0.1777
4.3636 × 10 −3 mol

p(CO ) = 0.00288 bar × 0.1777 = 5.12 × 10 −4 bar

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.

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parts per million (ppm). When the CO level increases to 800. ppm, dizziness, nausea, and unconsciousness 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 xi =
x (O 2 ) =

pi

:
p

0.2 atm
= 0.2 = 20% = 200000 ppm
1 atm

Therefore, the fatal O2/CO ratio is:
x (O 2 ) 200000
=
= 250
x (CO )
800

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:
n breath =

(

)

(101325 Pa ) × 0.0005 m 3
pV
=
= 0.0208 mol
R T 8.314472 J K −1 mol −1 × (293 K )

(

)

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

n tank

( 200 x 101325 Pa ) × ( 0.025 m3 )
pV
=
=
= 207.96 mol
R T ( 8.314472 J K −1 mol −1 ) × ( 293 K )

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

n tank
(207.96 mol) = 9998
=
n breath (0.0208 mol)

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

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it?
We can use the density of liquid N2 to calculate the mass of N2 that is injected into the balloon:
m N 2 = d LN2 VLN2 = ( 875.4 kg m -3 ) × ( 2 × 10-6 m3 ) = 1.7508 × 10-3 kg

We calculate the volume using the ideal gas law:
V=

nRT mRT
=
p
Mp

(1.7508 × 10 kg ) × ( 291.15 K ) × (8.314472 J K
=
( 28 × 10 kg mol ) × (101325 Pa )
-3

−1

-3

−1

mol −1 )

= 1.50 × 10-3 m3 = 1.50 L

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:
n CO 2

(

)

Vp
1 × 10 -3 m 3 × (101325 Pa )
=
=
R T (300 K ) × 8.314472 J K −1 mol −1

(

= 0.0406 mol

)

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:

m glucose = M n glucose = M

n CO 2

 0.0406 mol 
= 180.18 g mol −1 × 
 = 3.66 g
2
2



(

)

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 con15

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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:

C 3 H 8 (g ) + 5 O 2 (g ) ⎯

⎯→ 3 CO 2 (g ) + 4 H 2 O (g )
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:
x CO 2 =

3m
= 0.176 ;
17 m

x H 2O =

4m
= 0.235 ,
17 m

x O2 =

10 m
= 0.588
17 m

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 carbon dioxide gas and the total pressure is 1.00 atm and T = 310. K.
The reaction equation is:
2 NH 2CHCH 3COOH (s ) + 9 O 2 (g ) ⎯
⎯→ N 2 (g ) + 6 H 2O (l ) + 6 CO 2 (g )
We use the mass of alanine to calculate the number of moles of alanine that were oxidized:
n alanine =

(0.25 g ) = 2.8387 ×10-3 mol
m alanine
=
M alanine 88.07 g mol−1

(

)

From the reaction equation we know that:
n N2 =

1
n alanine = 1.4194 × 10-3 mol
2

And the volume of N2 consumed in the reaction is:

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Full file at />VN2 =

(

Chapter 1/Fundemental Concepts of Thermodynamics

)

(

)

n R T 1.4194 × 10-3 mol × (310 K )× 8.314472 J K −1 mol−1
=
= 3.61× 10-5 m 3 = 0.0361 L
(101325 Pa )
p

n CO2 = 3 n alanine = 8.5161× 10-3 mol

VCO2 =

(

)

(

)

n R T 8.5161× 10-3 mol × (310 K )× 8.314472 J K −1 mol−1
=
= 2.166 × 10-4 m 3 = 0.2166 L
(101325 Pa )
p

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

Vtotal = VN2 + VCO2 = 0.2166 L + 0.0361 L = 0.253 L
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 pV = nRT the total number of moles of moles of the mixture is:
n tot =

(

) (

)

pV
100 × 10 -3 Pa × 0.0002 m 3
=
= 8.2055 × 10 -3 mol
R T 8.314472 J K −1 mol −1 × (293.15 K )

(

)

The total number of moles can also be expressed as:
n tot = n ethane + n butane =

m ethane m butane
+
M ethane M butane

With methane = mtot – mbutane:
n tot =

((m tot − m butane ) × M butane + m butane M ethane )
M ethane × M butane

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

(n tot × M ethane − m tot ) ((8.2055 × 10 −3 mol) × (30.08 g mol -1 ) − (0.3846 g ))
=
= 4.9101 × 10 −3 mol An
-1
-1
(M ethane − M butane )
((30.08 g mol ) − (58.14 g mol ))

d finally:
x butane =

(
(

)
)

n butane
4.9101 × 10 −3 mol
=

= 0.599 = 59.9%
n total
8.2055 × 10 −3 mol

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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 M =

m
and pV = nRT the molar mass of the gas is:
n

-3
−1
−1
m R T ( 0.7031 ×10 kg ) × (8.314472 J K mol ) × ( 372.65 K)
M=
=
= 0.1583 ×10 -3 kg mol -1 = 158.3 g mol -1
3
pV
(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)

initial moles

n H2

n O2

at equilibrium

n H2 − 2α

n O2 − α

→

2 H2O (  )
0
2α

If O2 is consumed completely, n O − α = 0 , or n O = α . The number of moles of H2 remaining is
2

2

n H2 − 2α = n H2 − 2n O2 . Let p1 be the initial total pressure and p2 be the total pressure after all O2 is

consumed.

(

p1 = n H2 + n O2

) RVT and p = (n
2


H2

+ 2 n O2

) RVT

Dividing the second equation by the first:
n H2
n O2
p2
=
−2 

= x H2 − 2x O2 = 1 − x O2 − 2x O2 = 1 − 3x O2
p1 n H2 + n O2
n H2 + n O2

(

) (

)

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Chapter 1/Fundemental Concepts of Thermodynamics

1  p  1  (0.400atm) 
 = 0.20 , and x H2 = 0.80
x O2 = 1 − 2  = 1 −
(1.00 atm) 
3  p1  3 

P1.26) The photosynthetic formation of glucose in spinach leaves via the Calvin cycle involves
4−
the fixation of carbon dioxide with ribulose 1-5 diphosphate C5 H 8 P2 O11
(aq) to form 3-

phosphoglycerate C3 H 4 PO 3−

7 (aq) :
4−
(aq) + H 2O(l) + CO 2 (g)
C5 H 8 P2 O11

+
→ 2C 3H 4 PO 3−
7 (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:

nCO2 =

(101191.68 Pa ) × ( 0.001 m 3 )
pV
=
= 0.0446 mol
RT (8.314472 J K−1 mol −1 ) × ( 273 K )

The number of moles of 3-phosphoglycerate formed is given by:
n3− phos = 2 nCO2 = 0.0892 mol

And finally, the mass of 3-phosphoglycerate formed is:
m = n M = ( 0.0928 atm ) × (182.97 g mol −1 ) = 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 potential dominant under these conditions?
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To determine what portion of the potential is dominant we need to compare the pressure predicted by the van der Waals equation of state with that predicted by the ideal gas law. The van
der Waals equation of state yields:
p vdW =

(

)

(

)

RT
a
8.314472 × 10-2 bar dm 3 K −1 mol−1 × (300 K ) 1.355 bar dm 6 mol−2
− 2 =
−
= 17.3 bar
2
(Vm − b ) Vm
1.42 dm 3 mol−1 − 0.0321 dm 3 mol−1
1.42 dm 3 mol −1

((

) (

))

(

)

The ideal gas law gives:
p ideal =

(

)

R T 8.314472 × 10-2 bar dm 3 K −1 mol−1 × (300 K )
=
= 17.6 bar
Vm
1.42 dm 3 mol−1

(

)

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=

a
nRT
n2 a
1
1
RT
−
=
−
Vm − b
T Vm (Vm + b) V − nb
T 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 p =

(8.314 × 10 L bar K mol )× (790 K ) −
(1.42 L mol − 0.08271 L mol )
(452.0 L bar K mol )
1
(1.42 L mol )× (1.42 L mol
( 790K )
-2

p=

-1

−1

2

-1/2

RT
a
1
:
−
(Vm − b ) T Vm × (Vm + b )

−1

−1

−2

−1

−1

− 0.08271 L mol −1

)

= 41.6 bar

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:

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(

Chapter 1/Fundemental Concepts of Thermodynamics

)
) (

n R T (1 mol) × 8.314472 J K −1 mol −1 × (790.0 K ) × (1 bar )
= 46.3 bar
=
V
1.42 × 10 -3 m 3 × 10 5 Pa

(

)

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. Assume 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 simplified 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:
∞

~
~
~
N Ar =  N Ar A dz , where N Ar is the number Ar atoms per m3 at the surface of the earth. N Ar is
0

given by:

(

) (

)

N p

6.023 × 10 23 mol -1 × 0.0100 × 1 × 10 5 Pa
~
N Ar = A Ar =
= 2.41× 10 23 m -3
−1
−1
RT
8.314472 J K mol × (300 K )

(

)

The total number at Ar atoms in the atmosphere is:

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N Ar

∞

∞

0

0

−
~
~
=  N Ar A dz = N Ar  e

(2.41×10
=

23

)

M Ar g z
RT

RT
~
A dz = N Ar A
M Ar g

(
(

) (
) (

)

m -3 × 4π × 6.37 × 10 6 m × 8.314472 J K −1 mol −1 × (300 K )
= 7.85 × 10 41
39.9 × 10 −3 kg × 9.81 m s -1
2

)

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

(

) (

)

~
N Ar V 2.41×10 23 m -3 × 0.500 ×10 −3 m 3
=
= 1.53 ×10 − 22
41
N Ar
7.85 ×10

(

)

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

) ( (

) (

)

pV
0.0100 × 1× 10 5 Pa × 0.500 × 10 −3 m 3
23
-1
N = NA
= 1.21× 10 20
= 6.023 × 10 mol ×
−1
−1
RT
8.314472 J K mol × (300 K )

(

)

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
− M i g RT

as Pi = Pi0 e

where Pi is the partial pressure at the height z, Pi 0 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 x N = 0.600 and xCO = 0.400 and that T = 300. K. Near sea level, the total pressure is
2

2

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?

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N2

pN2 = p e

−

Chapter 1/Fundemental Concepts of Thermodynamics

M N2 g z
RT

 ( 28.04 × 10−3 kg mol -1 ) × ( 9.81 m s-1 ) × ( 50 × 103 m ) 

= ( 0.600 ) × (1.01325 × 105 Pa ) × Exp  −
 = 248 Pa
(8.314472 J K −1 mol −1 ) × ( 300 K )



pCO2 = p


CO2

e

−

M CO2 g z
RT

 ( 44.04 × 10−3 kg mol -1 ) × ( 9.81 m s-1 ) × ( 50 × 103 m ) 
= ( 0.400 ) × (1.01325 × 105 Pa ) × Exp  −
 = 7.02 Pa
(8.314472 J K −1 mol −1 ) × ( 300 K )



x CO2 =

(p

pCO2

CO2

+ p N2

( 7.02 Pa )

=

) ( 7.02 Pa + 248 Pa )

= 0.028

x N 2 = 1 − x CO 2 = 0.972
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.
− Mi g z 
 the barometric pressure at 1600 m is:
 RT 

Using Pi = Pi0 × Exp 

(

) (

)

 − 28.9 × 10 −3 kg mol -1 × 9.80665 m s -2 × (1600 m ) 

4
Pi = 1 atm × Exp 
 = 0.833 atm = 8.44 × 10 Pa
−1
−1
(
)
8.314472
J
K
mol
298.15
K
×



(

)

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?
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We need to consider two equilibria:
CuO (s) + H2 (g) → H2O (l) + Cu (s)
at equilibrium

α −β

n H2 − α

Cu (s) + ½ O2 (g) → CuO (s)
1
n O2 − β
2

α −β

at equilibrium

In the final state, only O2 is present. Therefore, n O − α . In an excess of O2, all the copper is oxi2

dized, and α − β = 0 and n O = β . That means that n O = n O − 1 n H .
2

2

2

2

2

Let V1 and V2 be the initial and final volumes:

(

V1 = n H2 + n O2

) RpT

and V2 =  n O + 1 n H  R T


2

2

2

 p

Dividing the second equation by the first yields:
n O
n H2
V2
1
1

1
3
=  2  −
= x O2 − x H2 = 1 − x H2 = 1 − x H2


V1 n H2 + n O2 2 n H2 + n O2
2
2
2

(

)

(

)

(
(

) = 0.103, and x
) 

2 V  2
84.5 cm3
x H2 = 1 − 2  = 1 −
3  V1  3 
100 cm3


O2

= 1 − x H2 = 0.897

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

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Chapter 1/Fundemental Concepts of Thermodynamics

oxygen behaves ideally at STP.
The number of moles of O2 according to the stoichiometry of the equation is:
n (O 2 ) = 6 × n (glucose) = 6 ×

(10 g )

(180.18 g mol )
−1

= 0.0555 mol

Therefore, the volume of O2 required is:
VO2 =

n O2 R T
p

=

(0.0555 mol) × (8.314472 J K −1 mol−1 ) × (273 K ) = 7.47 × 10-3 m 3 = 7.47 L
(101325 Pa )

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 oxidation of 0.0100 g of glycine.
The number of moles of CO2 according to the stoichiometry of the equation is:
n (CO 2 ) = 3 × n (glycine) = 3 ×

(0.0100 g )

(75.05 g mol ) = 3.997 × 10
−1

−4

mol

Therefore, the volume of CO2 evolved is:
VCO2 =

n CO2 R T
p

=

(3.997 × 10

−4

) (

)

mol × 8.314472 J K −1 mol −1 × (310 K )
= 1.0168 × 10 -5 m 3 = 1.02 × 10 -2 L
(101325 Pa )

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:

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