Physical Chemistry II
Chem 402
Spring 2011
Chapter 3 (1, 7, 10, 13, 14, 15, 21, 25, 26, 27)
P3.1) Obtain an expression for the isothermal compressibility  T  1 v (v P)T for a van der Waals
gas.
1  v 
1
1


 v T
a 
v 
 2 
 v  v  b v  
 T


T      
v  P T
 P 
    RT
v
T  
1
 2a
RT 
v 3 
2
 v  b  
 v
P3.7) Integrate the expression   1 v (v T ) P assuming that  is independent of pressure. By doing
so, obtain an expression for V as a function of T and  at constant P.

1  V 


V  T  P
dV
  dT
V
Vf
dV

dT

or
ln
  T f  Ti 
V 
Vi
if can be assumed constant in the temperature interval of interest.
P3.10) Derive the following expression for calculating the isothermal change in the constant volume heat
capacity: (CV V )T  T ( 2 P T 2 )V .
 
 CV 
 V    V

T 
 
 U  

  
 T V T  T
 U  

 
 V T T
The order of differentiation can be reversed because U is a state function.
 U 
 P 
Using the equation 
 T
 P
 V T
 T V
    P 

 CV 
T

 
  P

 V T  T   T V
 V
  2 P   P 
 2P 
 P 
=
T
T




 T 2   T 
 T 2 
 T V
V

V 

V
P3.13) Equation (3.38), CP  CV  TV ( 2  T ) , links CP and CV with  and  T . Use this equation to
evaluate CP – CV for an ideal gas.
1  v 
1  v 
1R
RT
1
   
; T   

 
v  T  P v P
v  P T vP 2 P
2
R2
1 R
 Tv 
R
CP  CV  TV
 P T
T
vP
v P
2
P3.14) Use the result of Problem P3.26 to derive a formula for (CV V )T for a gas that obeys the
Redlich–Kwong equation of state, P 
We use the relationship
 2 P 
 CV 
T

 2
 V 
  T
 T V
RT
a
P
v  b T v v  b
R
a
 P 
 3/2

 
 T V v  b 2T v  v  b 
 2 P 
3a
 2   5/2
 T V 4T v  v  b 
3a
 CV 
 V    4T 3/ 2 v v b
  
  T
RT
a
1

v b
T v (v  b)
P3.15) The function f(x,y) is given by f ( x, y )  xy sin 5 x  x 2 y ln y  3e2 x cos y . Determine
2
2
2
 f   f    f    f     f  
,
,
,
     2   2 ,    
 x  y  y  x  x  y  y  x  y  x  y  x
   f  
and     Is
 x  y  
x y

   f      f  
         ?
 y  x  y  x  x  y  x  y
Obtain an expression for the total differential df.
 f 
2 x 2
cos y  2 x y ln y
   5 xy cos 5 x  y sin 5 x  12 xe
 x  y
2
 f 
x2
x 2 ln y
x
x
sin
5



 3e 2 x sin y
 y 
y
2 y
 x
2
 2 f 
2 x 2
cos y  48e 2 x x 2 cos y  2 y ln y
 x 2   10 y cos 5 x  25 xy sin 5 x  12e

y
 2 f 
x 2 ln y
2 x2
 2   3e cos y 
3
 y  x
4y2
   f  
2
2 x x ln y

 sin 5 x  5 x cos 5 x  12e 2 x x sin y
    
y
y
 x  y  x  y
   f  
   f  
2 x x ln y
2 x 2

 sin 5 x     
     5 x cos 5 x  12e x sin y 
y
y
 x  y  x  y
 y  x  y  x


2
 f 
 f 
df    dx    dy  5 xy cos 5 x  y sin 5 x  12 xe 2 x cos y  2 x y ln y dx
 x  y
 y  x


2
x2
x 2 ln y
  x sin 5 x 

 3e 2 x sin y  dy


y
2 y


P3.21) Starting with the van der Waals equation of state, find an expression for the total differential dP in
terms of dV and dT. By calculating the mixed partial
derivatives ( (P V )T T )V and ( (P T )V V )T , determine if dP is an exact differential.
 

 V
 RT
a 
2a
RT

 3 

2
2 

v  b
 v  b v  T v
 

 T
 RT
a 
R
 2   

 v  b v  V  v  b 
 2a
RT 
R
dP   3 
dT
 dV 
2
v


v
b



v
b




   P    


  =
 T  V T  V  T
 2a
RT  
R
 
 3 
2
2
 v  b   V  v  b 
 v
   P      R  
R

   


   
2
v  b
 V  T V T  V   v  b   T
Therefore, dP is an exact differential.
P3.25) Show that d  m  m   dT  T dP where m is the mass density  m  m V . Assume that the
mass, m, is constant.
d m
m

d m V 
mV
Vd 1 V   
d m
m
1
V
 Vd 1 V  . Because
d 1 V 
dV

1
dV
, Vd 1 V   
2
V
V
 V 

1
 V 
 dT  
 dP    V  dT  V  T dP 

V
 P T 
 T  P
  dT   T dP
P3.26) For a gas that obeys the equation of state v 
RT
 B(T )
P
dB(T )
H 
derive the result 
 B(T )  T

dT
 P T
 H 
 V 

 V T 

 P T
 T  P
For v 
RT
 B T 
P
R dB
 V 

  
 T  P P dT
dB  T 
RT
RT
 H 
 R dB  RT
 B T   T  
 B T  
T

 

P
P
P
dT
 P T
 P dT 
dB T 
 B T   T
dT
P3.27) Because V is a state function, ( (V T ) P P )T  ( (V P)T T ) P . Using this relationship,
show that the isothermal compressibility and isobaric expansion coefficient are related by
( P)T  ( T ) P .
   V      V  
 
  

 
 P  T  P T  T  P T  P
1  V 
1  V 
Because   
 and  T   

V  T  P
V  P T
  V   
  V  T  
  
  T 

  
 or 
  

 P T
 T  P
 P T
 T  P