Chapter 2: Heat, Work, Internal Energy, Enthalpy, and the
First Law of Thermodynamics
Problem numbers in italics indicate that the solution is included in the
Student’s Solutions Manual.
Questions on Concepts
Q2.1) Electrical current is passed through a resistor immersed in a liquid in an adiabatic
container. The temperature of the liquid is varied by 1ºC. The system consists solely of
the liquid. Does heat or work flow across the boundary between the system and
surroundings? Justify your answer.
Although work is done on the resistor, this work is done in the surroundings. Heat
flows across the boundary between the surroundings and the system because of
the temperature difference between them.
Q2.2) Explain how a mass of water in the surroundings can be used to determine q for a
process. Calculate q if the temperature of 1.00-kg water bath in the surroundings
increases by 1.25ºC. Assume that the surroundings are at a constant pressure.
If heat flows across the boundary between the system and the surroundings, it will
q
lead to a temperature change in the surroundings given by T 
. For the
CP
case of interest,
q  qsurroundings  mCP T  1000 g  4.19 J g1K1  1.25 K  5.24 103 J.
Q2.3) Explain the relationship between the terms exact differential and state function.
In order for a function f(x,y) to be a state function, it must be possible to write the
 f 
 f 
total differential df in the form df    dx    dy. If the form df as
 x  y
 y  x
written exists, it is an exact differential.
Q2.4) Why is it incorrect to speak of the heat or work associated with a system?
Heat and work are transients that exist only in the transition between equilibrium
states. Therefore, a state at equilibrium is not associated with values of heat or
work.
Q2.5) Two ideal gas systems undergo reversible expansion starting from the same P and
V. At the end of the expansion, the two systems have the same volume. The pressure in
2-1
Chapter 2/ Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
the system that has undergone adiabatic expansion is lower than in the system that has
undergone isothermal expansion. Explain this result without using equations.
In the system undergoing adiabatic expansion, all the work done must come
through the lowering of U, and therefore of the temperature. By contrast, some
of the work done in the isothermal expansion can come at the expense of the heat
that has flowed across the boundary between the system and surroundings.
Q2.6) A cup of water at 278 K (the system) is placed in a microwave oven and the oven
is turned on for 1 minute during which it begins to boil. Which of q, w, and U are
positive, negative, or zero?
The heat q is positive because heat flows across the system-surrounding boundary
into the system. The work w is negative because the vaporizing water does work
on the surroundings. U is positive because the temperature increases and some
of the liquid is vaporized.
Q2.7) What is wrong with the following statement?: Because the well-insulated house
stored a lot of heat, the temperature didn't fall much when the furnace failed. Rewrite the
sentence to convey the same information in a correct way.
Heat can’t be stored because it exists only as a transient. A possible rephrasing
follows. Because the house was well insulated, the walls were nearly adiabatic.
Therefore, the temperature of the house did not fall as rapidly when in contact
with the surroundings at a lower temperature as would have been the case if the
walls were diathermal.
Q2.8) What is wrong with the following statement?: Burns caused by steam at 100ºC
can be more severe than those caused by water at 100ºC because steam contains more
heat than water. Rewrite the sentence to convey the same information in a correct way.
Heat is not a substance that can be stored. When steam is in contact with your
skin, it condenses to the liquid phase. In doing so, energy is released that is
absorbed by the skin. Hot water does not release as much energy in the same
situation, because no phase change occurs.
Q2.9) Describe how reversible and irreversible expansions differ by discussing the
degree to which equilibrium is maintained between the system and the surroundings.
In a reversible expansion, the system and surroundings are always in equilibrium
with one another. In an irreversible expansion, they are not in equilibrium with
one another.
2-2
Chapter 2/ Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
Q2.10) A chemical reaction occurs in a constant volume enclosure separated from the
surroundings by diathermal walls. Can you say whether the temperature of the
surroundings increases, decreases, or remains the same in this process? Explain.
No. The temperature will increase if the reaction is exothermic, decrease if the
reaction is endothermic, and not change if no energy is evolved in the reaction.
Problems
P2.1) 3.00 moles of an ideal gas at 27.0ºC expands isothermally from an initial volume
of 20.0 dm3 to a final volume of 60.0 dm3. Calculate w for this process a) for expansion
against a constant external pressure of 1.00 × 105 Pa and b) for a reversible expansion.
a) w   Pexternal V  1.00×105 Pa×  60.0-20.0 ×103 m3  4.00×103 J
b)
Vf
60.0 dm 3
1 1
wreversible   nRT ln
 3.00mol×8.314 J mol K ×300 K×ln
 8.22×10 3J
3
Vi
20.0 dm
P2.2) 3.00 moles of an ideal gas are compressed isothermally from 60.0 to 20.0 L using
a constant external pressure of 5.00 atm. Calculate q, w, U, and H.
w  2.03  104 J; U  0 and H  0
q  2.03  104 J
P2.3) A system consisting of 57.5 g of liquid water at 298 K is heated using an
immersion heater at a constant pressure of 1.00 bar. If a current of 1.50 A passes through
the 10.0-ohm resistor for 150 s, what is the final temperature of the water? The heat
capacity for water can be found in Appendix A.
I 2 Rt  nCP ,mTi
q  I 2 Rt  nCP ,m T f  Ti ; T f 
nCP ,m


1.50A 2 ×10.0 ohm  150 s +
Tf 
57.5g
×75.291 J mol 1K 1×298 K
18.02 g mol 1
57.5 g
×75.291 J mol 1K 1
18.02 g mol 1
 312 K
P2.4) For 1 mol of an ideal gas, Pexternal = P = 200 × 103 Pa. The temperature is changed
from 100ºC to 25.0ºC, and CV,m = 3/2R. Calculate q, w, U, and H.
3
U  nCV ,m T   8.314 J mol1K 1   298 K  373 K   935 J
2
2-3
Chapter 2/ Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics


H  nCP ,m T  n CV ,m  R T
5
 8.314 J mol 1K 1   298 K  373 K 
2
 1.56  103 J

= qP
w  U  qP  935 J  1.56  103 J  624 J
P2.5) Consider the isothermal expansion of 5.25 mol of an ideal gas at 450 K from an
initial pressure of 15.0 bar to a final pressure of 3.50 bar. Describe the process that will
result in the greatest amount of work being done by the system with
Pexternal  3.50 bar and calculate w. Describe the process that will result in the least amount
of work being done by the system with Pexternal  3.50 bar and calculate w. What is the
least amount of work done without restrictions on the external pressure?
The greatest amount of work is done in a reversible expansion. The work is given by
Vf
P
15.0 bar
wreversible   nRT ln
 nRT ln i  5.25 mol×8.314 J mol 1K 1 ×450 K ×ln
Vi
Pf
3.50 bar
 28.6×103J
The least amount of work is done in a single stage expansion at constant pressure with the
external pressure equal to the final pressure. The work is given by
 1 1
w   Pexternal V f  Vi   nRTPexternal   
P P
i 
 f
1
1


3
 5.25 mol×8.314 J mol 1K 1 ×450 K×3.50 bar× 

  15.1×10 J
 3.50 bar 15.0 bar 
The least amount of work done without restrictions on the pressure is zero, which occurs
when Pexternal = 0.
P2.6) Calculate H and U for the transformation of 1 mol of an ideal gas from 27.0ºC
and 1.00 atm to 327ºC and 17.0 atm if CP ,m  20.9  0.042
2-4
T
in units of J K 1mol 1.
K
Chapter 2/ Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
Tf
H  n  CP ,m dT
Ti
600
 1 mol 
T T

  20.9  0.042 K d K
300
600
2

T 
 20.9   600  300  J  0.021    J
 K   300

 6.27  103 J  5.67  103 J
 1.19  104 J
U  H    PV   H  nRT
 1.19  104 J  8.314 J K 1mol 1  300 K
 9.41  103 J
P2.7) Calculate w for the adiabatic expansion of 1 mol of an ideal gas at an initial
pressure of 2.00 bar from an initial temperature of 450 K to a final temperature of 300 K.
Write an expression for the work done in the isothermal reversible expansion of the gas at
300 K from an initial pressure of 2.00 bar. What value of the final pressure would give
the same value of w as the first part of this problem? Assume that CP,m = 5/2R.
3
wad  U  n CP ,m  R T   ×8.314 J mol1K 1  150 K  1.87×103 J
2
Pi
Pi  wreversible
wreversible   nRT ln ;ln

Pf
Pf
nRT


Pi
nRT
1.87×103 J
ln


 0.7497
Pf wreversible 1 mol×8.314 J mol1K 1 ×300 K
Pf  0.472 Pi  0.944 bar
P2.8) In the adiabatic expansion of 1 mol of an ideal gas from an initial temperature of
25ºC, the work done on the surroundings is 1200 J. If CV,m = 3/2R, calculate q, w, U,
and H.
q  0 because the process is adiabatic
U  w  1200 J

U  nCV ,m T f  Ti
Tf 

U  nCV ,mTi
nCV ,m
1200 J  7.5  8.314 J mol 1K 1  298 K
1.5  8.314 J mol1K 1
 202 K

2-5
Chapter 2/ Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics


 
H  nCP ,m T f  Ti  n CV ,m  R T f  Ti
1
 2.5  8.314 J mol K
1

 202 K  298 K 
 2.00  103 J
P2.9) An ideal gas undergoes an expansion from the initial state described by Pi, Vi, T to
a final state described by Pf, Vf, T in a) a process at the constant external pressure Pf and
b) in a reversible process. Derive expressions for the largest mass that can be lifted
through a height h in the surroundings in these processes.
a)
w  mgh   Pf V f  Vi  ; m  
b)
w  mgh  nRT ln
Vf
Vi
; m 
Pf V f  Vi 
gh
nRT V f
ln
gh
Vi
P2.10) An automobile tire contains air at 320  103 Pa at 20ºC. 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 = 5/2R. Calculate the final
temperature. Assume ideal gas behavior.
because q  0, U  w
nCV ,m T f  Ti    Pext V f  Vi 
 nRT f nRTi 
nCV ,m T f  Ti    Pext 


 P
Pi 
 f
The factor n cancels out. Rearranging the equation


RP 
RP 
 CV ,m  ext  T f   CV ,m  ext  Ti
Pf 
Pi 


RP
CV ,m  ext
Tf
Pi

Ti C  RPext
V ,m
Pf
8.314 J mol1K 1 105 Pa
3.20 105 Pa

8.314 J mol1K 1 105 Pa
2.5  8.314 J mol1K 1 
105 Pa
T f  0.804Ti T f  235 K
2.5  8.314 J mol1K 1 
P2.11) 3.50 moles of an ideal gas is expanded from 450 K and an initial pressure of
2-6
Chapter 2/ Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
5.00 bar to a final pressure of 1.00 bar, and CP,m = 5/2R. Calculate w for the following
two cases.
a) The expansion is isothermal and reversible.
b) The expansion is adiabatic and reversible.
Without resorting to equations, explain why the result to part (b) is greater than or less
than the result to part (a).
a)
w   nRT ln
Vf
Vi
  nRT ln
Pi
Pf
 3.50 mol×8.314J mol 1K 1 ×450 K×ln
5.00 bar
 21.1 103J
1.00 bar
b) Because q = 0, w = U. In order to calculate U, we first calculate Tf.
1
Vf 
 
Ti  Vi 
Tf
1
 Tf 
 
 Ti 
 Pi

 Pf
1



1

P 
 Tf 
;    i 
P 
 Ti 
 f 
;
P
 i
Ti  Pf
Tf



1

5
3
5
3
1
 5.00 bar 

  0.525
Ti  1.00 bar 
T f  0.525×450 K  236 K
Tf
3  8.314 J mol 1K 1
×  236 K  450 K   9.34×103J
2
Less work is done on the surroundings in part (b) because in the adiabatic expansion, the
temperature falls and therefore the final volume is less than that in part (a).
w  U  nCV ,m T  3.50 mol×
P2.12) 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?
2-7
Chapter 2/ Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
n
PV
1.00 bar×10.0 L
i i

 0.401 mol
RTi 8.3145×102 L bar mol1K 1 ×300 K
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.
P
10.0 bar
T1  Ti 1  300 K ×
 3000 K
Pi
1.00 bar
Vf
P
w   nRT1 ln
  nRT1 ln i
Vi
Pf
 0.401 mol× 8.314 J mol1K 1 × 3000 K×ln
10.0 bar
 23.0×103J
1.00 bar
In step 3,
PV
1 i  PV
i 2 ; V2 
PV
1 i
 10Vi  100 L
Pi
105 Pa
10 3m 3
× 10 L  100 L  ×
 9.00×10 3J
bar
L
3
3
3
 0  23.0 10 J  9.00 10 J  14.0 10 J
w   Pexternal V  1.00 bar×
wcycle
If the cycle were traversed in the opposite direction, the magnitude of each work term
would be unchanged, but all signs would change.
2-8
Chapter 2/ Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
P2.13) 3.00 moles of an ideal gas with CV,m = 3/2R initially at a temperature Ti = 298 K
and Pi = 1.00 bar is enclosed in an adiabatic piston and cylinder assembly. The gas is
compressed by placing a 625-kg mass on the piston of diameter 20.0 cm. Calculate the
work done in this process and the distance that the piston travels. Assume that the mass
of the piston is negligible.
F mg 625 kg×9.81ms 2
Pexternal   2 
 1.95×105 Pa
2
A r
π× 0.100 m
Vi 
nRT 3.00 mol×8.314 J mol1K 1 ×298 K

 7.43  102 m3  74.3 L
Pi
105 Pa
Following Example Problem 2.6,


RPexternal 
8.314 J mol 1K 1×1.95 105 Pa
12.47 J mol 1K 1 +
 CV ,m 


Pi
1.00 105 Pa
  298 K  
T f  Ti 
1 1
5
 C  RPexternal 

1 1 8.314 J mol K ×1.95  10 Pa
12.47
J
mol
K
+
V
,
m



Pf
1.95 105 Pa



= 411 K
nRT 3.00 mol×8.314 J mol1K 1 ×411 K
Vf 

 5.25  102 m3
Pf
1.95×105 Pa










w   Pexternal V f  Vi  1.95×105 Pa× 5.25×102 m3  7.43  102 m3  4.25  103J
h
3
w
4.25×10 J

 0.69 m
mg 625 kg×9.81 m s-2
P2.14) A bottle at 21.0ºC contains an ideal gas at a pressure of 126.4 × 103 Pa. The
rubber stopper closing the bottle is removed. The gas expands adiabatically against
Pexternal = 101.9 × 103 Pa, and some gas is expelled from the bottle in the process. When
P = Pexternal, the stopper is quickly replaced. The gas remaining in the bottle slowly warms
up to 21.0ºC. What is the final pressure in the bottle for a monatomic gas, for which
CV,m = 3/2R, and a diatomic gas, for which CV,m = 5/2R?
In this adiabatic expansion, U  w
2-9
Chapter 2/ Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
nCV ,m T f  Ti    Pext V f  Vi 
 nRT nRT 
nCV ,m T f  Ti    Pext 


 V
V
f
i




RP 
RP 
 CV ,m  ext  T f   CV ,m  ext  Ti
Pf 
Pi 


RP
CV ,m  ext
Tf
Pi

Ti C  RPext
V ,m
Pf
8.314 J mol 1K 1 101.9 103 Pa
126.4 103 Pa

8.314 J mol 1K 1 101.9 103 Pa
1  1
1.5  8.314 J mol K 
101.9 103 Pa
1.5  8.314 J mol 1K 1 
Tf
Ti
 0.923 , T f  271 K
Once the stopper is put in place, the gas makes a transformation from
Ti  214 K, Pi  101.9 103 Pa to T f  294 K and Pf
PV
PV
i i
 f f , but Vi  V f
Ti
Tf
Tf
294 K
 101.9  103 Pa  110.5  103 Pa
Ti
271 K
5
The same calculation carried out for CV ,m  R gives
2
Tf
 0.945, T f  278 K
Ti
Pf 
Pi 
Pf  107.8  103 Pa
P2.15) A pellet of Zn of mass 10.0 g is dropped into a flask containing dilute H2SO4 at a
pressure of 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
2-10
Chapter 2/ Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
1mol H 2 8.314 J mol1K 1  298 K
V
×
×
 3.79  103m3
1
5
1×10 Pa
65.39 g  mol Zn  1mol Zn
10.0 g
w   Pexternal V
V  volume of H 2 produced.
w  1×105 Pa ×3.79×103m3  379 J
P2.16) 1 mol of an ideal gas for which CV,m = 20.8 J K–1 mol–1 is heated from an initial
temperature of 0ºC to a final temperature of 275ºC at constant volume. Calculate q, w,
U, and H for this process.
w = 0 because V = 0
U  q  CV T  20.8 J mol1K 1 ×275 K  5.72  103J
H  U    PV   U  RT  5.72×103J  8.314 J mol1K 1  275 K  8.01 103J
P2.17) 1 mol of an ideal gas, for which CV,m = 3/2R, initially at 20.0ºC and 1.00 × 106 Pa
undergoes a two-stage transformation. For each of the stages described in the following
list, calculate the final pressure, as well as q, w, U, and H. Also calculate q, w, U,
and H for the complete process.
a) The gas is expanded isothermally and reversibly until the volume doubles.
b) Beginning at the end of the first stage, the temperature is raised to 80.0ºC at constant
volume.
a) P2 
PV
P
1 1
 1  0.500 106 Pa
V2
2
w  nRT ln
V2
 8.314 J mol1K 1  ln 2  1.69 103 J
V1
U  0 and H  0 because T  0
q   w  1.69 103 J
T1 T2
T2 P1 353 K × 0.500 × 106 Pa
b)
 ; P2 

 6.02 × 105 Pa
P1 P2
T1
293 K
U  nCV ,m T  1.5  8.314 J mol 1K 1   353 K  293 K   748 J
w  0 because V  0
q  U  748 J


H  nCP ,m T  n CV ,m  R T 
3
 8.314 J mol 1K 1   353 K  293 K 
2
 1.25  103 J
For the overall process,
2-11
Chapter 2/ Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
q  1.69  103 J  748 J  2.44  103 J
w  1.69  103 J  0  1.69  103 J
U  0  748 J  748 J
H  0  1.25  103 J  1.25  103 J
P2.18) 1 mol 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.
1
Vf 
 
Ti  Vi 
Tf
1
 Tf 
 
 Ti 
 Pi

 Pf
1




P
 Tf 
;     i
P
 Ti 
 f
1



;
P
 i
Ti  Pf
Tf



1

5
3
5
3
1
 1.00×105 Pa 
0.4

   0.100   2.51
6
Ti  1.00×10 Pa 
T f  2.51×298 K  749 K
q = 0 because the process is adiabatic.
3×8.314 J mol 1K 1
w  U  nCV ,m T  1 mol×
  749 K  298 K   5.62  103J
2
H  U    PV   U  RT  5.62  103J  8.314 J mol 1K 1   749 K  298 K 
Tf
H  9.37  103 J
P2.19) 1 mol of an ideal gas, for which CV,m = 3/2R, is subjected to two successive
changes in state:
a) From 25.0ºC and 100  103 Pa, the gas is expanded isothermally against a constant
pressure of 20.0  103 Pa to twice the initial volume.
b) At the end of the previous process, the gas is cooled at constant volume from 25.0ºC to
–25.0ºC. Calculate q, w, U, and H for each of the stages. Also calculate q, w, U, and
H for the complete process.
a) Vi 
nRT 8.314 J mol 1K 1  298 K

 2.48  10 2 m 3
Pi
100 R  103 Pa
V f  2Vi  4.96  10 2 m 3




w   Pext V f  Vi  20.0  103 Pa  4.96  10 2 m 3  2.48  10 2 m 3  496 J
U and H  0 because T  0
q   w  496 J
2-12
Chapter 2/ Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics


b) U  nCV ,m T f  Ti  1.5  8.314 J mol1K 1   248 K  298 K   623 J
w  0 because V  0
q  U  623 J

 

H  nCP ,m T f  Ti  n CV ,m  R T f  Ti

 2.5  8.314 J mol 1K 1   248 K  298 K 
 1.04  103J
U total  0  623 J = 623 J
wtotal  0  496 J =  496 J
qtotal  496 J  623 J =  127 J
H total  0  1.04  103J =  1.04  103J
P2.20) The temperature of 1 mol of an ideal gas increases from 18.0º to 55.1ºC as the
gas is compressed adiabatically. Calculate q, w, U, and H for this process assuming
that CV,m = 3/2R.
q = 0 because the process is adiabatic.
3×8.314 J mol 1K 1
w  U  nCV ,m T  1 mol×
× 55.1 C  18.0 C  463 J
2
H  U    PV   U  RT  463 J  8.314 J mol 1K 1  55.1 C  18.0 C




H  771 J
P2.21) A 1-mol sample of an ideal gas for which CV,m = 3/2R undergoes the following
two-step process.
a) From an initial state of the gas described by T = 28.0ºC and P = 2.00  104 Pa, the gas
undergoes an isothermal expansion against a constant external pressure of 1.00  104 Pa
until the volume has doubled.
b) Subsequently, the gas is cooled at constant volume. The temperature falls to –40.5ºC.
Calculate q, w, U, and H for each step and for the overall process.
a) For the first step, U = H = 0 because the process is isothermal.
1 1
nRTi 1 mol×8.314 J mol K ×  273.15 + 28.0 K
Vi 

 1.25  102 m3
4
Pi
2.00×10 Pa
w   q   Pexternal V  1.00×104 Pa×0.125×102 m3  1.25×103J
b) For the second step, w = 0 because V = 0.
2-13
Chapter 2/ Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
3×8.314 J mol1K 1
× 28.0o C + 40.5o C  854 J
2
H  U    PV   U  RT  854 J +8.314 J mol1K 1 × 28.0o C + 40.5o C

q  U  CV T  1 mol×



H  1.42  103J
For the overall process, w = – 1.25×103 J, q = 854 + 1.25×103 J = 2.02×103 J, U = 854 J,
and H = 1.42  103 J.
P2.22) A cylindrical vessel with rigid adiabatic walls is separated into two parts by a
frictionless adiabatic piston. Each part contains 50.0 L of an ideal monatomic gas with
CV,m = 3/2R. Initially, Ti = 298 K and Pi = 1.00 bar in each part. Heat is slowly
introduced into the left part using an electrical heater until the piston has moved
sufficiently to the right to result in a final pressure Pf = 7.50 bar in the right part.
Consider the compression of the gas in the right part to be a reversible process.
a) Calculate the work done on the right part in this process and the final temperature in
the right part.
b) Calculate the final temperature in the left part and the amount of heat that flowed into
this part.
The number of moles of gas in each part is given by
PV
1.00 bar×50.0 L
n i i 
 2.02 mol
RTi 8.3145×102 L bar mol1K 1 ×298 K
a) We first calculate the final temperature in the right side.
1
Vf 
 
Ti  Vi 
Tf
1
 Tf 
 
 Ti 
 Pi

 Pf
1



1

P 
 Tf 
;     i 
P 
 Ti 
 f 
P
;
 i
Ti  Pf
Tf



1

5
3
5
3
1
 1.00 bar 

  2.24
Ti  7.50 bar 
T f  2.24×298 K  667 K
Tf
3×8.314 J mol1K 1
U  w  nCV ,m T  2.02 mol×
  667 K  298 K   9.30  103 J
2
b) We first calculate the final volume of the right part.
RTrf 2.02 mol×8.314×102 L bar mol 1K 1 ×667 K
Vrf 

 14.9 L
Prf
7.50 bar
Therefore, Vlf = 100.0 L – 14.9 L = 85.1 L.
2-14
Chapter 2/ Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
Tlf 
Plf Vlf
nR

7.50 bar×85.1 L
 3.80  103K
2
1 1
2.02 mol×8.314×10 L bar mol K


U  nCV ,m T  2.02 mol×8.314 J mol 1K 1 × 3.80  103K  298 K  58.8 103J
From part (a), w = –9.30  103J
q = U – w = 58.3  103 J + 9.30  103 J = 67.3  103 J
P2.23) A vessel containing 1 mol of an ideal gas with Pi = 1.00 bar and CP,m = 5/2R is in
thermal contact with a water bath. Treat the vessel, gas, and water bath as being in
thermal equilibrium, initially at 298 K, and as separated by adiabatic walls from the rest
of the universe. The vessel, gas, and water bath have an average heat capacity of
CP = 7500 J K–1. The gas is compressed reversibly to Pf = 10.5 bar. What is the
temperature of the system after thermal equilibrium has been established?
Assume initially that the temperature rise is so small that the reversible compression can
be thought of as an isothermal reversible process. If the answer substantiates this
assumption, it is valid.
Vf
P
w   nRT1 ln
  nRT1 ln i
Vi
Pf
 1 mol× 8.314 J mol 1K 1 × 298 K×ln
1.00 bar
 5.83 103J
10.5 bar
U combined system  CP T
T 
U combined system
CP

5.83×103J
 0.777 K
7500 J K 1
T f  299 K
The result justifies the assumption.
P2.24) The heat capacity of solid lead oxide is given by
T
in units of J K 1mol 1. Calculate the change in enthalpy of
K
1 mol of PbO(s) if it is cooled from 500 K to 300 K at constant pressure.
CP ,m  44.35  1.47  103
2-15
Chapter 2/ Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
Tf
H  n  C p ,m dT
Ti
300


  44.35  1.47 10
3
500
T  T 
d  
K K
= 44.35   300 K  500 K 
300 K
1.47 103  T 2 

  
2
 K   500 K

 8870 J  117 J
 8.99 103 J
P2.25) Consider the adiabatic expansion of 0.500 mol of an ideal monatomic gas with
CV,m = 3/2R. The initial state is described by P = 3.25 bar and T = 300 K.
a) Calculate the final temperature if the gas undergoes a reversible adiabatic expansion to
a final pressure P = 1.00 bar.
b) Calculate the final temperature if the same gas undergoes an adiabatic expansion
against an external pressure of P = 1.00 bar to a final pressure P = 1.00 bar.
Explain the difference in your results for parts (a) and (b).
a)
1
V 
 f 
Ti  Vi 
Tf
1
T 
 f 
 Ti 
 Pi

 Pf
1



1

T   P 
;  f    i 
 
 Ti   Pf 
5
3
5
3
1
 3.25bar 

  0.626
Ti  1.00 bar 
T f  0.626  300 K  188 K
Tf
b)
U  nCV ,m T f  Ti    Pexternal V f  Vi 
2-16
P
;  i
Ti  Pf
Tf



1

Chapter 2/ Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
T T 
nCV ,m T f  Ti   nRPexternal  f  i 
P P
i 
 f

nRPexternal
T f  nCV ,m 

Pf



nRPexternal 
  Ti  nCV ,m 

Pi




RPexternal
 CV ,m 
Pi
T f  Ti 
 C  RPexternal
 V ,m
Pf

T f  217 K
1 1


1 1 8.314 J mol K ×1.00 bar
1.5×8.314
J
mol
K
+


3.25bar
  300 K  
1 1


1 1 8.314 J mol K ×1.00 bar
1.5×8.314
J
mol
K
+


1.00 bar








More work is done on the surroundings in the reversible expansion, and therefore U and
the temperature decrease more than for the irreversible expansion.
P2.26) An ideal gas undergoes a single-stage expansion against a constant external
pressure Pexternal at constant temperature from T, Pi, Vi, to T, Pf, Vf.
a) What is the largest mass m that can be lifted through the height h in this expansion?
b) The system is restored to its initial state in a single-state compression. What is the
smallest mass m' that must fall through the height h to restore the system to its initial
state?
c) If h = 10.0 cm, Pi = 1.00  106 Pa, Pf = 0.500  106 Pa , T = 300 K, and n = 1.00 mol,
calculate the values of the masses in parts (a) and (b).
Consider the expansion
a) w  mgh   Pext V f  Vi 
m
 Pext V f  Vi 
gh
for the final volume to be V f , the external pressure can be no bigger than Pf
mmax 
 Pf V f  Vi 
gh
b) Consider the compression
m 
 Pext Vi  V f 
gh
for the final volume to be Vi , the pressure can be no smaller than Pi
'
mmin

 Pi Vi  V f 
gh
2-17
Chapter 2/ Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
c)
Vi 
nRT 1.00 mol× 8.314 J mol1K 1× 300 K

 2.49  103 m3
Pi
1.00×106 Pa
Pf V f  PV
i i
Vf 
PV
1.00 × 106 Pa × 2.49 × 103m3
i i

 4.98  103 m3
6
Pf
0.500 × 10 Pa

0.500  106 Pa  4.98  103 m3  2.49  103 m3
mmax 
2
9.81 m s  0.100 m
1.00  106 Pa   2.49  103 m3  4.98  103 m3 
 
mmin
9.81 m s2  0.100 m
  1.27  10 kg
3
 2.54  103 kg
P2.27) Calculate q, w, U, and H if 1.00 mol of an ideal gas with CV,m = 3/2R
undergoes a reversible adiabatic expansion from an initial volume Vi = 5.25 m3 to a final
volume Vf = 25.5 m3. The initial temperature is 300 K.
q = 0 because the process is adiabatic.
1
Vf 
 
Ti  Vi 
Tf
1
5
 25.5 L  3

  0.349
Ti  5.25L 
T f  0.349×300 K  105 K
Tf
3  8.314 J mol1K 1
× 105 K  300 K   2.43  103J
2
H  U  nRT  2.43×103J + 1.00 mol×8.314 J mol 1K 1 × 105 K  300 K 
U  w  nCV ,n T  1.00 mol×
H  4.05  103 J
P2.28) A nearly flat bicycle tire becomes noticeably warmer after it has been pumped
up. Approximate this process as a reversible adiabatic compression. Take the initial
pressure and temperature of the air before it is put in the tire to be Pi = 1.00 bar and
Ti = 298 K. The final volume of the air in the tire is Vf = 1.00 L and the final pressure is
Pf = 5.00 bar. Calculate the final temperature of the air in the tire. Assume that CV,m =
5/2R.
2-18
Chapter 2/ Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
1
Vf 
 
Ti  Vi 
Tf
1
 Tf 
 
 Ti 
 Pi

 Pf
1



1

P 
 Tf 
;     i 
P 
 Ti 
 f 
P
;
 i
Ti  Pf
Tf



1

7
5
7
5
1
0.285
 1.00 bar 

 1.58
   0.200 
Ti  5.00 bar 
T f  1.90×300 K  475 K
Tf
P2.29) 1 mol of an ideal gas with CV,m = 3/2R is expanded adiabatically against a
constant external pressure of 1.00 bar. The initial temperature and pressure are Ti = 300
K and Pi = 25.0 bar. The final pressure is Pf = 1.00 bar. Calculate q, w, U, and H for
the process.
U  nCV ,m T f  Ti    Pexternal V f  Vi   w
q = 0 because the process is adiabatic.
T T 
nCV ,m T f  Ti   nRPexternal  f  i 
P P
i 
 f

nRPexternal
T f  nCV ,m 

Pf



nRPexternal 
  Ti  nCV ,m 

Pi




RPexternal
 CV ,m 
Pi
T f  Ti 
 C  RPexternal
 V ,m
Pf

T f  185 K
1 1


1 1 8.314 J mol K ×1.00 bar

 1.5×8.314 J mol K +
25.0 bar
  300 K  
1 1


1 1 8.314 J mol K ×1.00 bar
1.5×8.314
J
mol
K
+


1.00 bar








3×8.314 J mol 1K 1
 185 K  300 K   1.43  103J
2
H  U  nRT  1.43×103J + 1.00 mol×8.314J mol 1K 1  185 K  300 K 
U  w  nCV ,n T  1.00 mol×
H  2.39×103J
P2.30) One mole of N2 in a state defined by Ti = 300 K and Vi = 2.50 L undergoes an
isothermal reversible expansion until Vf = 23.0 L. Calculate w assuming a) that the gas is
described by the ideal gas law and b) that the gas is described by the van der Waals
equation of state. What is the percent error in using the ideal gas law instead of the van
der Waals equation? The van der Waals parameters for N2 are tabulated in Table 7.4.
a) for the ideal gas
wreversible   nRT ln
Vf
Vi
 1mol×8.314 J mol 1K 1 ×300 K ×ln
2-19
23.0 L
 5.54 103J
2.50 L
Chapter 2/ Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
b) for the van der Waals gas
Vf
Vf
 RT
a 
w    Pexternal dV    
 2  dV
V  b Vm 
Vi
Vi  m
f
 RT 
 a 
  
dV


 2  dV

V b 
V
Vi  m
Vi  m 
Vf
V
The first integral is solved by making the substitution y = Vm – b.
Vf
yf
 RT 
 RT 
 
dy   RT ln V f  b   ln Vi  b  
 dV    
V b
y 
Vi  m
yi 
Therefore, the work is given by
V f  b   a  1  1
w  nRT ln

Vi  b   Vi V f



23.0 L  0.0380 L
2.50 L  0.0380 L
5
6 6

10 Pa 10 m 
1
1
 1.366 L2 bar×
×


2
3 3
3 3 
bar
L  2.50×10 m 23.0×10 m 
 1 mol×8.314 J mol1K 1 ×300 K ×ln
w  5.56 103 J  48.7 J  5.52 103J
5.52×103J + 5.54×103J
Percent error = 100 
 0.4%
5.52 103J
2-20