PHYS-4420 THERMODYNAMICS & STATISTICAL MECHANICS
SPRING 2006
Class Activity - Class 2 January 20, 2006
Name______Solution______________________
Do problem 2-5 from the textbook. Since many of you do not have the textbook with you, here it
is:
Noting that at the critical point, the three roots of the van der Waals equation are equal
(i. e., (v – vC )3 = 0), show that the critical values of the specific volume, temperature, and
v
T
P
pressure are given by Equation (2.10). Show that in terms of v'  , T '  , and P' 
, the
vC
TC
PC
3  1  8

van der Waals equation becomes:  P' 2  v'   T '
v'  3  3

Solution:
P
2P
 0 and
The critical point occurs where both
 0 . Then,
v
v 2
RT
a
P
 RT
2a
2P
2 RT
6a
P
 2 , so
and



 4 . From the last two
2
3
2
3
v b v
v (v  b)
v
v
(v  b) v
2 RTC
6a
RTC
2a
 4 . Divide the first
 3 , and
equations, at the critical point we have,
2
3
(vC  b) vC
(vC  b)
vC
v  b vC
2v
v

equation by the second to get, C
, and vC  b  C . Then, C  b .
vC = 3b
2
3
3
3
This result can be put into the equation above from the first derivative. Then,
RTC
2a
RTC
2a
8a

, and
, so

TC 
2
3
2
3
4b
27b
(3b  b)
(3b)
27bR
From the starting equation,
 8a 
R

RTC
a
a
8a
a
4a
3a
a
27bR 

PC 
 2 


 2 

PC 
2
2
2
vC  b vC
3b  b
(3b)
27b(2b) 9b
27b 27b
27b 2
a
a
a 
8a


 2 2 (v'3b  b)  RT '
Finally,  P  2 (v  b)  RT becomes,  P'
.
2
v 
v' 9b 
27bR

 27b
1 
8
 1
After careful cancellation of the constants, this becomes,  P ' 2 (3v'1)  T ' .
9v ' 
27
 27
Then after handling the numbers this becomes,
3  1  8

 P' 2  v'   T '
v'  3  3
