Diffusion Problem
by yayups, Aug 7, 2019, 8:52 PM
In this post, we'll be solving the following very instructive physics problem.
A vessel with a small hole of diameter 
is place inside a chamber much larger than it that is filled with a gas of temperature 
and pressure 
. The pressure is very low, so in particular, the mean free path 
of the gas is much larger than . The vessel is kept a constant temperature 
. Once steady state is reached, what is the pressure in the vessel?
Since the chamber is much larger than the vessel, gas entering the vessel and leaving from the vessel won't affect the pressure or temperature of the chamber.
At this point, it is very easy to fall into the following trap. One may argue that steady state is reached when the pressures in the vessel and chamber are the same. After all, isn't that what happens when two gases come in equilibrium?
The issue with this reasoning is due to the fact that we are in the regime where
. What this means is that for particles near the hole that have a chance of passing through it, there are unlikely to be any collisions. The effect of pressure arises through particle collisions, so in this case, particles will pass through the hole not due to a pressure difference, but rather through the random chance that their velocity has an appropriate magnitude and direction to go through the hole (this is diffusion). Steady state is reached when the rate of particles going from the chamber to the vessel is the same as the rate of particles going from the vessel to the chamber.
Thus, to solve the problem, we need to calculate the rate that particles leave through a hole in a box of gas, given that the size of the hole is much smaller than the mean free path.
The finish from here is easy. Suppose the particle number density in the chamber is
and in the vessel is 
. Steady state implies the rate of particles flowing through the hole is the same from both sides, or that
![\[\eta_0\langle v_0\rangle = \eta_1\langle v_1\rangle\implies \eta_0\sqrt{T_0}=\eta_1\sqrt{T_1}\implies \eta_0=2\eta_1.\]](//latex.artofproblemsolving.com/5/5/3/553ae696d12775b53e1e0baa37fd4f7889e4f04e.png)
The ideal gas law tells us that 
, so 
and 
.
Some Graduate Entrance Exam Problem wrote:
is place inside a chamber much larger than it that is filled with a gas of temperature 
and pressure 
. The pressure is very low, so in particular, the mean free path 
of the gas is much larger than . The vessel is kept a constant temperature 
. Once steady state is reached, what is the pressure in the vessel?Since the chamber is much larger than the vessel, gas entering the vessel and leaving from the vessel won't affect the pressure or temperature of the chamber.
At this point, it is very easy to fall into the following trap. One may argue that steady state is reached when the pressures in the vessel and chamber are the same. After all, isn't that what happens when two gases come in equilibrium?
The issue with this reasoning is due to the fact that we are in the regime where
. What this means is that for particles near the hole that have a chance of passing through it, there are unlikely to be any collisions. The effect of pressure arises through particle collisions, so in this case, particles will pass through the hole not due to a pressure difference, but rather through the random chance that their velocity has an appropriate magnitude and direction to go through the hole (this is diffusion). Steady state is reached when the rate of particles going from the chamber to the vessel is the same as the rate of particles going from the vessel to the chamber.Thus, to solve the problem, we need to calculate the rate that particles leave through a hole in a box of gas, given that the size of the hole is much smaller than the mean free path.
Lemma wrote:
Given a box with an (ideal) gas of particle mass 
, temperature 
, and number density 
(number of particles per unit volume), the number of particles leaving a small hole of area 
per unit time is
![\[\frac{\eta A}{4}\langle v\rangle=\frac{\eta A}{4}\sqrt{\frac{8k_BT}{\pi m}}\]](//latex.artofproblemsolving.com/8/8/6/886cf171b31e20a20b310f5185d3b7eaab1795dd.png)
assuming that 
.
, temperature 
, and number density 
(number of particles per unit volume), the number of particles leaving a small hole of area 
per unit time is![\[\frac{\eta A}{4}\langle v\rangle=\frac{\eta A}{4}\sqrt{\frac{8k_BT}{\pi m}}\]](http://latex.artofproblemsolving.com/8/8/6/886cf171b31e20a20b310f5185d3b7eaab1795dd.png)
assuming that 
.
be the Maxwell-Boltzmann distribution of velocities of the particles. What this means is that the probability that a particle has velocity in ![$[v_x,v_x+dv_x]\times[v_y,v_y+dv_y]\times[v_z,v_z+dv_z]$](http://latex.artofproblemsolving.com/2/3/4/2347751741e1dbe9846903d39eae4b040d7275a4.png)
is![\[f\left(\sqrt{v_x^2+v_y^2+v_z^2}\right)dv_xdv_ydv_z.\]](http://latex.artofproblemsolving.com/e/d/9/ed939a866b843f7a218b9fc1c8ff467f933d8332.png)
Set up spherical coordinates with origin at the hole. We will now count the number of particles that hit the hole in a time 
using a funny double counting argument, where we start by counting the number of particles that hit the hole with a certain velocity and then integrate over all velocities.
(technically speed in ![$[v,v+dv]$](http://latex.artofproblemsolving.com/e/3/6/e36557a3f0a889236a1e6388bc09705240ccb880.png)
,
.
means pointing toward the hole, and 
is parallel to the plane of the hole (the spherical coordinates for the velocity are flipped compared to those for space, since the 
,![\[(\eta dV)\cdot f(v)\cdot v^2\sin\theta \,dv \,d\theta\, d\phi.\]](http://latex.artofproblemsolving.com/8/9/9/89937eabed07f31a25ce99d3d8736a99df1e7688.png)
For this given velocity, the volume in space that will allow such particles to hit the hole is a tilted cone object with base 
,
,
direction. In particular, its volume is 
,
hitting the hole in time ![\[(\eta A dt)\cdot f(v)\cdot v^3\sin\theta\cos\theta \,dv \,d\theta\, d\phi.\]](http://latex.artofproblemsolving.com/8/4/3/843724208a7990c3e3524499b8b444cf5a48aefc.png)
Thus, the rate of particles leaving is![\[\alpha=\eta A\int_0^\infty v^3f(v)\,dv\int_0^{\pi/2}\sin\theta\cos\theta\,d\theta\int_0^{2\pi}d\phi=\pi\eta A\int_0^\infty v^3f(v)\,dv.\]](http://latex.artofproblemsolving.com/2/c/6/2c62a0f3e36a3c0148d8dd4ba9dd2de33fa2026d.png)
Note that![\[\langle v\rangle=\int_0^\infty\int_0^\pi\int_0^{2\pi} v\cdot f(v)\cdot v^2\sin\theta\,dv\,d\theta\,d\phi=4\pi\int_0^\infty v^3f(v)\,dv,\]](http://latex.artofproblemsolving.com/c/8/c/c8cb9af6a1e632d0586359560e0b24ccaf0f11cb.png)
which tells us that![\[\alpha=\frac{\eta A}{4}\langle v\rangle,\]](http://latex.artofproblemsolving.com/f/e/2/fe26c6f89cd559d3461d875d9628e80638cdf505.png)
as desired. Note that proof didn't depend on the particular form of the Maxwell-Boltzmann speed distribution.The finish from here is easy. Suppose the particle number density in the chamber is
and in the vessel is 
. Steady state implies the rate of particles flowing through the hole is the same from both sides, or that![\[\eta_0\langle v_0\rangle = \eta_1\langle v_1\rangle\implies \eta_0\sqrt{T_0}=\eta_1\sqrt{T_1}\implies \eta_0=2\eta_1.\]](http://latex.artofproblemsolving.com/5/5/3/553ae696d12775b53e1e0baa37fd4f7889e4f04e.png)
The ideal gas law tells us that 
, so 
and 
.
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