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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.

enter image description here

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:

enter image description 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

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

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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:

$$p(v_x)v_x $$

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  • $\begingroup$ Im totally comfortable with this, my question is regarding the discrepancy in integration versus no integration $\endgroup$
    – Lopey Tall
    Jan 20 at 14:03

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