# PHYS-4420 THERMODYNAMICS & STATISTICAL MECHANICS ```PHYS-4420 THERMODYNAMICS &amp; STATISTICAL MECHANICS
SPRING 2006
Class Activity - Class 9 February 14, 2006
Do problem 8-13 from the textbook. Since many of you do not have the textbook with you, it is
given here.
The equations of the sublimation and vaporization curves of a particular material are given by:
6
ln P  0.04 
(sublimati on)
T
4
ln P  0.03 
(vaporizat ion)
T
where P is in atmospheres.
(a) Find the temperature of the triple point.
At the triple point the two curves will cross, so the values of P and T will be equal.
6
4
0.04   0.03  , so 0.04 T – 0.03 T = 6 – 4,
0.01 T = 2
T
T
TTP = __200 K______
(b) Show that the specific latent heats of vaporization and sublimation are 4R and 6R
respectively. (You may assume that the specific volume in the vapor phase is much larger
than the specific volume in the liquid and solid phases.)
13
 dP 
 dP 
. We can get 

 
 from the equations given.
 dT 13 T (v  v)
 dT 13
d
ln P   d  0.04  6 , so 1 dP  62 , and dP  6 P2
dT
dT 
T
P dT T
dT T
6 Pv
6P
13

 13 
 6R

 13
since v  v . Then,
2
T
T
T (v  v) Tv
4 Pv
 23 
 4R
The same procedure with the second equation yields
T
(c) Find the latent heat of fusion.
12   23   31  0
or
12   23  13  0 .
Then 12  13   23  6R  4R
 12 = __2R_________
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