Chem. 444
Instructor: Nancy Makri
PROBLEM McQ 20-6
Calculate qrev and S for a reversible cooling of one mole of an ideal gas at a
constant volume V1 from P1, V1, T1 to P2, V1, T4 followed by a reversible expansion at constant pressure P2 from P2 , V1, T4 to P2, V2, T1 (the final state for all the processes shown in
Figure 20.3). Compare your result for S with those for paths A, B+C, and D+E in Figure
20.3.
Beginning with the first law of thermodynamics:
dU   q   w
We recall that for an ideal gas U α T and that for the overall process Ti = Tf so,
dU  0   q   w
And with a little more work,
V1
V2
V2
V1
V1
V1
qrev   w   PdV   PdV  0  P2  dV  P2 (V2  V1 )
It then follows for the entropy that
dS 
dqrev PdV RdV


T
T
V
where n = 1 mole as suggested by the problem statement. It then follows that
S  
V1
V1
V 
RdV 2 RdV

R ln  2 
V
V
 V1 
V1
V
The process described in this problem and in Figure 20.3 trace out different paths in
PV -space but have the same final and initial points. Hence the heat evolved, being described
by an inexact differential, has different values depending on the exact process taken. [1 point]
However, the entropy, being a state function, depends only on the final and initial points and
therefore has the same value for all these four processes.
PROBLEM McQ 20-8
Calculate the value of S if one mole of an ideal gas is expanded reversibly and isothermally from 10.0 dm3 to 20.0 dm3. Explain the sign of S .
V 
 20.0dm3 
S  nR ln  2   8.31JK 1 ln 
 5.76 JK 1
3 
 10.0dm 
 V1 
The change in entropy is positive because the gas is expanding. [1 point] An expansion of the
gas increases its disorder because the number of available microstates in the larger volume is
greater.
PROBLEM McQ 20-9
Calculate the value of S if one mole of an ideal gas is expanded reversibly and isothermally from 1.00 bar to 0.100 bar. Explain the sign of S .
V 
S  nR ln  2 
 V1 
PV
1 1  PV
2 2 
P
V2 P1
  S  nR ln  1 
V1 P2
 P2 
 1.00bar 
1
S  8.31JK 1 ln 
  19.1JK
 0.10bar 
The change in entropy is positive because the gas is expanding. This can be deduced from the
ideal gas equation, which implies that a decrease in the pressure is concomitant with an increase in volume. An expansion of the gas increases its disorder because the number of available microstates in the larger volume is greater.
PROBLEM McQ 20-13
Show that
T2
T1
for a constant-pressure process if C p is independent of temperature. Calculate the change in
S  C p ln
entropy of 2.00 moles of H2O(l) ( C p = 75.2 J·K-1·mol-1) if it is heated from 10 ºC to 90 ºC.
Applying a change of variables to the definition of the entropy,
dS 
dqrev  q  dT  U  dT
dT


 CP


T
T
 T  P T  T  P T
We then integrate to get
T2
S   CP
T1
T 
dT
 363K 
1
 CP ln  2   (2.00mol )(75.2mol 1K 1 ) ln 
  37.4 JK
T
 283K 
 T1 