CHE 303 (Answers for Homework)
Answers for problem set #5, Winter 2009
1. (5.22) Consider filling a gas cylinder with ethane from a high-pressure supply line. Before
filling, the cylinder is empty (vacuum). The valve is then opened, exposing the tank to a
3-MPa line at 500oK until the pressure of the cylinder reaches 3 MPa. The valve is then
closed. The volume of the cylinder is 50 L. For ethane, use the truncated virial equation
of state, in pressure:
z=
Pv
= 1 + B’P, where B’ =  2.810-8 [m3/J]
RT
(a) What is the temperature immediately after the valve is closed?
(b) If the cylinder then sits in storage at 293oK for a long time, what is the entropy change of the
universe (from the original unfilled, state)?
Ans:
T2  552 K
(b)

 J 
 J 
S univ  75.7 mol  54.08 
 46.9 


 mol  K 
 mol  K  

J
S univ  544  
K 
2. (5.35) The speed of sound, Vsound [m/s], is formally equal to the partial derivative of
pressure with respect to density at constant entropy:
 P 
2
Vsound
  
   s
 P 
v2
Show that    
MW
   s
 P 
  where MW is the molecular weight.
 v  s
Ans:
 P 
v2
2
Vsound
    
MW
   s
 P 
 
 v  s
3.
 P 
2
   , use the thermodynamic web to
(5.36) Based on the definition Vsound
   s
come up with an expression for [Vsound] in air. What is the value of [Vsound] in air at 20oC?
You may consider air to be an ideal gas with cp = (7/2)R. Based on this result, how far
away is a bolt of lightning if you hear the thunder four seconds after you see the lightning.
Ans:
Vsound  343 m/s  The lightening bolt is 1360 m away.
4. (5.401) Gas A expands through an adiabatic turbine. The inlet stream flows in at 100 bar
and 600oK while the outlet is at 20 bar and 445oK. Calculate the work produced by the
turbine. The following data are available for gas A. The ideal gas heat capacity for this
process is
cp = 30.0 + 0.02T, where cp is in [J/moloK] and T is in [oK].
PvT data has been fit to the following equation
p(v  b) = RT +
where
aP 2
T
a = 0.001 [Km3/(barmol)] and b = 810-5 [m3/mol].
Solution
 J 
h  h1  h2  h3  8487 
 mol 
 J 
ws  8487 
 mol 
In other words, for every mole of gas that flows through the turbine, 8487 joules of work are
produced.
5. (6.121) Pure ethanol boils at a temperature of 63.5oC at a pressure of 400 torr. It also boils
at 78.4oC and 760 torr. Estimate the saturation pressure for ethanol at 100oC.
Solution
  h vap
P3sat  P2sat exp 

R



 kJ 
  42.39 



1
1 
1
1
mol  




     760 torr  exp 
 kJ   373.15 K 351.55 K  
 T3 T2  
 0.008314 

 mol  K 


P3sat  1760 torr  2.32 atm
In comparison, ThermoSolver gives a value of 2.23 atm, using the Antoine equation.
6.
(6.141) At 922oK, the enthalpy of liquid Mg is 26.780 [kJ/mol] and the entropy is
73.888 [kJ/mol]. Determine the Gibbs energy of liquid Mg at 1300oK. The heat capacity
of the liquid is constant over this temperature range and has a value of 32.635 [J/moloK].
Solution
 J 
g 2  h2  T2 s2  71,500 
 mol 
7. (6.151) Solid sulfur undergoes a phase transition between the monoclinic (m) and
orthorhombic (o) phases at a temperature of 368.3oK and pressure of 1 bar. Calculate the
difference in Gibbs energy between monoclinic sulfur and orthorhombic sulfur at 298oK
and a pressure of 1 bar. Which phase is more stable at 298oK? Take the entropy in each
phase to be given by the following expression:
Monoclinic phase:
sm = 13.8 + 0.066T [J/moloK]
Monoclinic phase:
so = 11.0 + 0.071T [J/moloK]
Solution
monoclinic
orthorhombic
T = 298 [K]
g1
g3
T = 368.3 [K]
monoclinic
g2 = 0
orthorhombic
 J 
m o
g 298
K  79.5 
 mol 
Therefore, the transition from the monoclinic to orthorhombic state occurs spontaneously. The
orthorhombic state is more stable.