PHYS-4420 THERMODYNAMICS & STATISTICAL MECHANICS
SPRING 2006
Homework Problems
Assignment 6. Due Friday 3/10/06: 9-8, Supplemental
9-8. The sketch below shows the situation before and after mixing. The gas on the left is the
neon, at 4 atmospheres.
Before mixing
After mixing
Eventually, n2 will be needed, so it will be calculated. Since the two ideal gases are at the
same temperature and volume, their pressures are proportional to the number of moles
P
1 atm
1
P
n
1 mole  mole .
present, since PV = nRT. Then, 2i  2 , so n2  2i n1 
P1i
4 atm
4
P1i n1
a) Since both gases are initially at the same temperature, and there is no chemical
interaction, the temperature does not change upon mixing.
T = 300 K
The final pressure is the sum of the partial pressures of the two gases, P  P1 f  P2 f .
Since each gas doubles its volume at constant temperature, its pressure is halved. Then
P  P2i 4 atm  1 atm
P1 f  12 P1i and P2 f  12 P2i , so P  P1 f  P2 f  1i

P = 2.5 atm
2
2
This could also be calculated from P1 f  x1P , so P 
x1 
n1
1
4

 .
n1  n2 1  1 / 4 5
Then,
P
P1 f
x1

P1 f
x1
, where
P1i / 2 5
 P1i  2.5 atm
4/5 8

P 
 P 
b) From the class notes, G  RT n1 ln  1 f   n2 ln  2 f , T is constant, so P  V 1 .
 P1i 
 P2i 


V 
 V 
1
G  RT n1 ln  0   n2 ln  0   RT (n1  n2 ) ln     RT (n1  n2 ) ln 2
2
 2V0 
 2V0 

G   RT (n1  n2 ) ln 2  (8.31  103 J  kmole -1K -1 )(300 K)1  14 kmole ln 2
G = – 2.16 ×106 J
c) From class notes, S  R(n1  n2 ) ln 2  
G
 2.16  106 J

T
300 K
S = 7.20 ×103 J·K-1
Supplemental
1. When a certain acid condenses from the vapor phase at room temperature, it releases latent
heat of 4600 kcal/kmole.
a) Assuming the self-interactions (hence the chemical potential) negligible in the vapor
state, use the information to calculate the chemical potential of this acid in the liquid form
at room temperature. Express your answer in electron volts per molecule.
Q
= 4600 kcal/kmole = – (4600 kcal/kmole)(4184 J/kcal) = – 1.925 ×107 J/kmole
n
1
  (1.925  107 J/kmole)
19
(1.6  10 J/eV) (6.02  1026 molecules/ kmole)

 = – 0.200 eV/molecule
b) When 10-5 kmole of this acid is added to one liter of water at room temperature, the
temperature of the water rises 0.1ºC. What is the chemical potential for this acid in water?
Express your answer in electron volts per molecule.
Q  mcP T
Q  (1.0 kg)(1.0 kcal  kg -1  K -1 )(0.10 K)  (0.10 kcal)(4184 J/kcal)  418.4 J
Q
418.4 J

 4.184  107 J/kmole
-5
n
1.0  10 kmoles
 4.184  107  107 J/kmole
 6.950  10 20 J/molecule


6.02  1026 molecules/ kmole
1.6  1019 J/eV

 = – 0.434 eV/molecule
2. A certain sugar has a chemical potential of – 1.6 eV in water and – 0.9 eV in oil. Suppose
some of this sugar was dissolved in oil and you wished to remove it. Can you suggest a way
to do it, taking advantage of the difference in the two chemical potentials? If you can, explain
how you would do it. (You should be aware that if oil and water are mixed, and then set
aside, they will separate. The end result will be a layer of oil floating on the water.)
Mix the oil and water. The sugar will move from the oil to the water because water has a
lower chemical potential. Then wait until the oil and water separate. The sugar-free oil can
be skimmed from the top of the water. That is all that is required to answer the question.
If you wish to get the sugar back, you could boil away the water, and leave the sugar.