# Hit rate of molecules on a wall

Reviewing my final from last semester to prep for comps:

Question:

A piston of mass M can move freely in a tube with cross-section area A filled with ideal monoatomic gas with molecular mass m ≪ M and density n at temperature T.

The first part of the question asks:

Calculate the rate of molecular collisions with the piston (both sides).

I found an equation in one of my fav SM books (Blundell and Blundell) that I think would help here:

However, my professor's solution works entirely in 1-dimension and so using the 1-d maxwell-boltzmann distrubiton. And so I can justify to myself taking the $$1/2 cos(\theta) sin(\theta) d\theta$$ out of equation 6.12 to match what my professor has in his solution.

Thus

$$N= A \cdot v \cdot dt \cdot n \cdot f(v) \cdot dv$$

where n = number density (N/V), A is the area of the wall/piston, v is velocity, dt is some time interval, and f(v) is the 1-d maxwell boltzmann distribution.

My question is where my professors integration comes from in his provided solution:

$$\frac{d N}{dt} = 2 \cdot \bigg(\frac{m}{2 \pi T}\bigg)^{1/2}\cdot n \cdot A \cdot \int_0^\infty v e^{\frac{-mv^2}{2 T}} dv = n A \sqrt{\frac{2T}{\pi m}}$$

I have almost all these ingredients from the Blundell and Blundell equation except the integration on the right hand side, and the dN as opposed to just the N on the left hand side.

The only integration of the distribution I am familiar with is for the average velocity,

$$\bar{v} = \int_0^\infty v f(v) dv$$

What am I reading wrong regarding the missing integration in that Blundell and Blundell equation?

An instance of integrating to get N I do know about is in the case of the degenerate fermi gas at a small but non-zero temperature from equation 7.53 of Schroeder

$$N= \int_0^\infty g(\epsilon) \bar{n}_{FD}(\epsilon) d\epsilon$$

The 1 dimensional distribution you are referring to is speed (in contrast to velocity).

$$p(\vec v)d^3v$$ is really

$$p(v_x, v_y, v_z)dv_xdv_ydv_z\rightarrow p(v)v^2dv$$

since the volume element is a shell in velocity space.

For this problem, you're considering the flux across an area:

$$p(\vec v)\vec v \cdot \hat A = p(v_x)v_xA_x + p(v_y)v_yA_y + p(v_z)v_zA_z$$

where $$\hat A$$ is the normal to the surface. Since $$A_y=A_z=0$$, and $$A_x=1$$, you're just using:

$$p(v_x)v_x$$

• Im totally comfortable with this, my question is regarding the discrepancy in integration versus no integration Jan 20 at 14:03