Solutions for Statistical Mechanics for Chemistry and Materials Science, 1st Edition by Bagchi

Solution Manual
Statistical Mechanics for
Chemistry and Materials
Science


Biman Bagchi




2

, Chapter 1: Preliminaries

Problem 1
Start from dE = TdS − PdV and show that the equation of state for ideal gases, that is, PV = n
k B T, actually implies that the energy depends only on temperature. Also show that for a van
der Waals gas, Cv is a function of temperature only.


Solution:
We know, dE  TdS  PdV

So constant temperature, it can be written as

 E   S 
  T  P
 V T  V T

 E   p 
  T   P (From Maxwell’s Relations)
 V T  T V

 p  nk B
Now for ideal gas,   
 T V T

 E  nkB
  T P0
 V T V

Thus energy is not dependent on volume of the system.

For ideal gas, atoms do not interact with each other, thus only thermal motion will contribute
in energy of the system.

3
E Nk BT
2

i.e. for ideal gas energy depends only on temperature.

Equation of state for van der Waal gas

 a 
 p  2  V  b   RT
 V 

 E 
Cv   
 T v

4

, Cv     E      E  
         
 V T  V  T V T  T  V T V

 E   P 
  T  P (1)
 V T  T V

 P     RT a  R
Now, for van der Waals gas      2   (2)
 T V  T  V  b V   V  b

 E  a
   2
 V T V

   E  
    =0
 T  V T V

 Cv 
  0
 V T

Cv is function of temperature only.

Problem 2
The Clausius-Clapeyron equation describes the change in boiling point with pressure. It is
derived from the fact that the chemical potential of liquid and gas phase are the same at the
point of co-existence. For an alternate derivation, consider a Carnot engine using one mole of
water. At the source (P, T) the latent heat L is supplied, converting water to steam. There
occurs an increase in volume. The pressure is adiabatically decreased to -dP. At the sink (P-
dp, T-dT) the water was condensed back.
Show the work output of the engine is W  VdP  O(dP2 ) . Hence obtain the Clausius-
Clapeyron equation from there.


Solution:
Let us consider process is taking place at constant Volume.

W  PdV  PV  (P  dP)V  VdP
We have considered volume of gas only, neglected volume of liquid .Thus error due to this
assumption is



5

,  V 
 dPdP  O(dP )
2

 P  s

W  VdP  O(dP2 )

Efficiency of Carnot cycle

W Q  QH TC  TH
  C 
QH QH TC

Latent heat L was supplied

QH  L , TH  T ,W  Vdp,TC  T  dT

Substituting these values

VdP T  dT dT
1 =
L T T

dP L

dT T .V

Problem 3
Thermodynamic quantities that acquire unique values for a given state of a system are known
as state functions (for example Internal energy [E], Enthalpy [H], Entropy [S], Free energies
2Z 2Z
[F or G], etc.). Mathematically if Z = Z(x,y) and Z is a state function, then  . Use
xy yx
this definition to prove the following thermodynamic relations.

 T   P 
a.   =  
 V  S  S V

 T   V 
b.    
 P  S  S  P

 S   P 
c.   =  
 V T  T V

 S   V 
d.     
 P   T  P




6