Kinetic Energy of Molecules Escaping Through Small Hole

[TheAmorphist]

1. Homework Statement
"A perfect gas containing a single species of molecular weight M is in a container at equilibrium. Gas escapes into a vacuum through a small circular hole of Area A in the wall of the container. Assume wall container is negligibly thick and planer in vicinity to hole. The diameter of the hale is appreciably smaller than the mean free path, but larger than molecular diameter.
a. Show that the number of molecules escaping from the hole per unit area per unit time is given by nC/4.
b.Obtain an expression for the rate of mass outflow.
(What I actually need help on)c. Show that the mean kinetic energy of the escaping molecules if greater than that of the molecules inside the container in the ratio of 4/3.


2. Homework Equations
The flux equation of F=(Int)nf(Ci)QCnDVc, where f(Ci) is the Maxwellian Velocity Distribution, Q is some quantity (energy here) and integration is performed over the range of velocity space of interest.


3. The Attempt at a Solution

I have completed the nitty gritty of parts a and b, but simply cannot make any progress on part C. I'm assuming that the gas "within the container" have a mean kinetic energy of (3/2)KT, as they have 3 degrees of freedom. My attempt at finding the kinetic energy of those escaping the hole has consisted of integrating the flux equation a number of times to find a constant factor that resulted in a ratio of "4/3" but I feel like this is in inappropriate way to go about this. Any help or at least pointing me in the right direction would be really helpful.

Text is "Physical Gas Dynamics" -Vincenti and Krieger

 

[Jhero]

(1965)

Hello,

Thank you for bringing this problem to our attention. It seems that you have made good progress on parts a and b, but are stuck on part c. Let me try to guide you in the right direction.

First, let's consider the mean kinetic energy of the molecules inside the container. As you correctly stated, for a monatomic gas, the mean kinetic energy is given by (3/2)kT, where k is the Boltzmann constant and T is the temperature of the gas. This is because monatomic gases have three translational degrees of freedom.

Now, let's consider the mean kinetic energy of the molecules escaping through the hole. We can use the same approach as in parts a and b, by using the flux equation and integrating over the velocity space. However, instead of considering the entire range of velocities, we will only consider the velocities of the molecules that are escaping through the hole. This will give us a more accurate picture of the kinetic energy of these molecules.

To do this, we can use the fact that the diameter of the hole is smaller than the mean free path, but larger than the molecular diameter. This means that the molecules escaping through the hole will have a range of velocities that is close to the average speed of the gas, which is given by (8kT/πM)^1/2. You can use this information to set the limits of integration for the velocity space.

Once you have set the limits of integration, you can use the flux equation to calculate the mean kinetic energy of the escaping molecules. You should get a result that is greater than (3/2)kT, and the ratio of this value to (3/2)kT should be equal to 4/3.

I hope this helps. Good luck with your problem!