Problems and Solutions on Thermodynamics and Statistical Mechanics

349

This can be written as

with

Tt

((Az)~) = -

x

X = 7/2k cx 7

2161
A box of volume 2V is divided into halves by a thin partition. The
left side contains a perfect gas at pressure po and the right side is initially
vacuum. A small hole of area A is punched in the partition. What is
the pressure pl in the left hand side as a function of time? Assume the
temperature is constant on both sides. Express your answer in terms of the
average velocity v.
( wzs c 0 ns in)
Solution:
Because the hole is small, we can assume the gases of the two sides
are at thermal equilibrium at any moment. If the number of particles of
the left side per unit volume at t =^0 is no, the numbers of particles of the
left and right sides per unit volume at the time t are nl(t) and no - nl(t)
respectively. We have

dni(t) A A

V- = --n1v + -(no - nl)v ,

dt 4 4

where w = - is the average velocity of the particles. The first term is

the rate of ecrease of particles of the left side due to the particles moving

to the right side, the second term is the rate of increase of particles of

the left side due to the particles moving to the left side. The equation is

simplified to

d’::

dn1(t) A A

+ -n1v = -now.

dt 2V 4v

With the initial condition nl(0) = no, we have

and