Question 2.161: A box of volume 2V is divided into halves by a thin partitio...
A box of volume 2V is divided into halves by a thin partition. The left side contains a perfect gas at pressure p_{0} and the right side is initially vacuum. A small hole of area A is punched in the partition. What is the pressure p_{1} 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.
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 n_{0}, the numbers of particles of the left and right sides per unit volume at the time t are n_{1}(t) \text { and } n_{0}-n_{1}(t) respectively. We have
V \frac{d n_{1}(t)}{d t}=-\frac{A}{4} n_{1} v+\frac{A}{4}\left(n_{0}-n_{1}\right) v.
where v=\sqrt{\frac{8 k T}{\pi m}} is the average velocity of the particles. The first term is the rate of decrease 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
\frac{d n_{1}(t)}{d t}+\frac{A}{2 V} n_{1} v=\frac{A}{4 V} n_{0} v.
\text { With the initial condition } n_{1}(0)=n_{0} \text {, we have }n_{1}(t)=\frac{n_{0}}{2}\left(1+e^{\frac{-A v t}{2 V}}\right).
and
p_{1}(t)=\frac{p_{0}}{2}\left(1+e^{-\frac{A v t}{2 v}}\right).