# Kinteic Energies of particles hitting the walls of a container

1. May 6, 2006

### mmh37

I was wondering whether anyone was able to give me a hint on the following problem:

Is the average kinetic energy of all particles in a gas hitting the sides of a container less, equal or more than 3/2kT? And if it is not equal how can you calculate the mean kinetic energy?

I was thinking that since the flux and the pressure is the same everywherein the container, the kinteic energy should be the same as well. Thus the average KE is 3/2kT. I'm not really convinced by my argument here. But I can't really find a counter-argument (except maybe that the wall has 2D instead of 3D, but I'm not sure how and whether that would influence the flux or KE of the particles).

Thanks a lot for any hints and suggestions!

2. May 12, 2006

### maverick280857

If the gas is ideal and the Kinetic Theory is applied, the answer is $\frac{3}{2}kT$

But how does that give you 1.5kT right away?? In fact, you need to calculate this from the standpoint of momentum transfer to the wall on collision. And the average kinetic energy is the mean kinetic energy. Average and mean imply the same thing.

Ideal gases under the Kinetic Theory model follow Maxwell-Boltzmann Statistics. If you know the distribution function of velocities, you can compute different parameters such as rms speed, most probable speed and average speed (its speed and not velocity). It is not correct that all molecules have the same velocity and the concentration of molecules is homogenous. In fact the number of molecules with a particular velocity is described by the distribution function.

The distribution function is given by

$$N(v)dv = 4 \Pi N_{0}(\frac{M}{2RT})^{\frac{3}{2}}e^{-\frac{Mv^2}{2RT}} dv$$

here N(v)dv is the number of molecules with velocity v and $N_{0}$ is the total number of molecules in the container (M is the molar mass. If you use molecular mass m, then replace R by k, the Boltzmann constant).

You can actually derive expressions for pressure and average kinetic energy by purely classical considerations. Consider a molecule traveling in the +x direction just about to hit a well. It undergoes a perfectly elastic collision with the wall which reverses its velocity. Using the (linear) distance $L$ traveled by the molecule between 2 successive collisions, you can find the collision time $t_{coll}$. It is simply

$$\frac{L}{v}$$

Now you can find the force exerted by the molecule on the wall (and hence the force exerted by the wall on the molecule). Repeating the procedure for motion of another molecule along the y axis and still another along the z-axis (or of an ensemble of particles with velocity components along all the 3 axes), you can get the expression for pressure (considering a cubical container helps). Most of this "derivation" can be found in any general physics textbook describing the Kinetic Theory of Gases.

Molecules hit the wall with random velocities. If the observation time is large then almost all kinds of molecules will hit the wall. By all kinds, I mean molecules with all possible velocities. In that case, the average kinetic energy of molecules striking the wall will be a good measure of the average kinetic energy of the gas itself.

Last edited: May 12, 2006
3. Oct 22, 2008

### Solet

Hi,

I was searching for this question, and would just like to point out that the given answer is wrong - the probability distribution of molecules actually hitting the side of a container is proportional to v * the Maxwellian Distribution given above (as faster molecules will impact more often than slower ones).

The actual distribution function (for the velocity of molecules hitting) is:

(m^2/(2*k^2*T^2)) * v^3 * exp(-(m*v^2/(2*k*T)))

working out, this gives an actual average energy of 2kT, greater than the 1.5kT average of the entire gas.