非習題 (55% total, 5% each)

1. Briefly describe the definition of chemical potential, and why it has a simple correlation to Gibbs free
energy: G = N  for a system of N molecules, but not for other energy state functions such as internal
energy U.
 G 
Ans: dG = −SdT + VdP +  dN ; or  =   . Thus, chemical potential is defined as the Gibbs free
  N T , P
energy change at const. temperature and pressure when inserting or removing a particle from the system.
Since only intensive properties are controlled, there is this simple correlation G = N  .

aN 2
2. A van der Waals model is written as ( P + ) (V − Nb ) = NkT . Briefly explain the corrections made in it
V2
in comparison to the ordinary ideal gas model.

Ans: There are two corrections: (1) a minimum volume existed; (2) a pressure correction due to
particle-particle interaction, which in turn leads to lower effective pressure.

3. Boiling and freezing temperature of dilute solution are known as its colligative property. Explain why it gets
the name.:

Ans: The chemical potential of a dilute solution will only be influenced by the amount of solute, rather than
the characteristics/properties of the solute. Since boiling and freezing temperatures are dependent upon the
chemical potential of the dilute solution, they are solely affected by the amount of solute. Therefore, they
are called “colligative properties.
”
4. Explain the phenomenon of osmosis, and how is the reverse osmosis (RO) process being conducted to
acquire “purified” water?

Ans. Osmosis is when there is a membrane (which allows only solvent to penetrate) existed between two
solutions, one is with more “solute” than the other. The chemical potential will then drive the solvent to
transport from the one with less solute to the one with more solute before reaching equilibrium. This
chemical potential imbalance can be reversed by applying pressure. When the applied pressure is larger than
the solute induced chemical potential difference, you can reverse the solvent flow direction and get
“purified”.

5. Prove that, for any system in equilibrium with a reservoir at temperature T, the average value of the energy
is
1 Z 
E=− =− ln Z , where  = 1 kT .
Z  
6. The speed of molecules of an ideal gas is
known to follow the Maxwell speed
distribution as shown to the right. Please
explain why the curve is skewed to the right
in comparison to Boltzmann distribution.

Ans: The distribution function D (v ) is

proportional to (1) probability of a molecule
having velocity v ; (2) number of vectors corresponding to speed v . Of the two, part (1) obeys
Boltzmann distribution while part (2) is proportional to 4 v 2 . Therefore, the curve shown is shifted to the
right in comparison to Boltzmann distribution.
7. Consider N 2 (species A) and O 2 (species B) form nearly ideal mixtures in both liquid and gas phases, and the
phase diagram at 1 atm is shown to the right. Please answer the following questions.

(a) The formula for the entropy of a monatomic ideal gas is given by Sackur-Tetrode equation:
  V  4 mU 3/ 2  5 
S = Nk  ln   2  + 
  N  3 Nh   2 

Consider the two gases are enclosed in a container and

separated by a partition according to the molar ratio
initially. Show that the mixing entropy is given by
expression

S mixing = − Nk  x ln x + (1 − x ) ln (1 − x )  .

Ans. Assuming the number of B molecules is xN . When

the partition is removed, the molecules expand to fill a volume that is greater by a factor of 1 / x , so
1
their entropy increases by S B = xNk ln = − Nkx ln x according to Sackur-Tetrode equation. Similarly,
x

the number of A molecules is (1 − x) N and expand to the volume of 1/ (1 − x ) , so

1
S A = (1 − x ) N  k ln = − Nk (1 − x ) ln(1 − x ) .
1− x
The total entropy increases upon mixing is

Smix = S A + S B = − Nk  x ln x + (1 − x) ln(1 − x) .

(b) Write down the equation of Gibbs free energy (G), and schematically draw the Gibbs free energy versus
molar percentage (x) diagram of the two comprising phases at oxygen boiling temperature, T1, T2 = 84
K, and nitrogen boiling temperature, T3. Explain how the Gibbs free energy of either phase change with
decreasing temperature from T1 to T2 to T3.

G = xN B0 + (1 − x) N  A0 − NkT  x ln x + (1 − x) ln(1 − x)

 (c) Explain why there is a miscibility gap at 84 K.
Ans. Based on curves drawn in Pb. (b), when T=84 K the Gibbs free energy curve becomes a “double
well”. This indicates that the system is more stable when it separates into 2 phases (liquid and gas)
rather than a uniform mixture. Thus, there is miscibility gap at 84 K.

8. In a soluble lead flow battery the electrochemical reaction is as the following:

2+ +
2Pb(aq) + 2H 2O(l ) Pb(s) + PbO2(s) + 4H (aq)

(a) Write down the equilibrium condition for this reaction.

(b) Use the attached property table to find the equilibrium constant of the aforementioned reaction.

Table 1. Thermodynamic
properties of Selected
Substances (all values are
for one mole of material at
298 K and 1 bar)

Pb 2 + (aq) −1.7 −24.43 10.5

Ans.
(a)
2  Pb 2+ + 2  H 2O =  Pb +  PbO2 + 4  H +

(
2  Pb
0
2+ + RT ln C
Pb 2+ ) ( )
+ 2  H0 2O + RT ln C H 2O =  Pb
0
+ RT ln C Pb +  PbO
0
2
+ RT ln C PbO2 + 4 (  H0 + + RT ln C H + )

(b) Rearrange the aforementioned equation:

(
2  Pb
0
2+) ( )
+ 2  H0 2O −  Pb
0
−  PbO
0
2
− 4  H0 + = RT ln C Pb + RT ln C PbO2 + 4 ( RT ln C H + ) − 2 RT ln C Pb2+ − 2 RT ln C H 2O

−Grxn
0
= 2  f GPb 2+ + 2  f G H 2O −  f GPb −  f GPbO2 − 4 f G H +
  2  −24.43 − 237.13 + 217.33  kJ 
 ( )
Thus,  − Grxn
0
   = exp  −30579  = 4.337  10 −6
K = exp   = exp  
 RT  

 K (
8.31 J ( 298K ) ) 

 2476.38 