Problems and Solutions on Thermodynamics and Statistical Mechanics

Statistical Physics 275

Solution:

p = - (g) , where F is the free energy, F = E - TS. When
T
T = 0 K, F = E, and

p=-(g) T.

2
Using pV = -El we have
3

or

p=-(g)T=-[-.-(fpV)]T=-;[V(g)T+P] a ,

5

1 av 3

Hence K. = --

(-)

= - (T = 0 K).

ap T 5P

At T = 0 K,

h2k2

d3k-

2E 2 V

3V 3V (2~)~ /,,,, 2m

p= - = -.2.-

we obtain

For an ideal gas, the energy of a particle is

h2 k2

2m

&(k) = -.

Thus

h2 k;

2m

&F=-.

Therefore,

2

5

p= -n.&F, (T= 0 K) ,

and

3

K.= -. (T=OK)

2n&F