how is momentum being transported throughout the gas - if this were the case, surely layers would be changing their horizontal speeds and so we wouldn't have this steady state.
The gas is in a steady state, and momentum is being transferred. This means that a steady horizontal force on the top plate is required, and an equal and opposite force on the bottom plate. Each layer of gas receives momentum from above, and transfers the same momentum to the layer below it, so it undergoes no acceleration.
$F/A$ represents the change in momentum in unit time of unit area of this layer
The layer has a steady speed, so it has no change in momentum per unit time. $F/A$ is the momentum transferred from one layer to the one below it.
However how does this equal the x direction momentum of the molecules that passes through unit area in unit time of this layer? They seem to be unrelated.
At the microscopic level, the only way one layer can impart momentum to the adjacent layer is by swapping molecules. $F/A$ is the net momentum carried across the imaginary boundary from the upper to the lower layer.
why don't we just find the total momentum rather than the amount by which it exceeds the momentum of that layer?
$F/A$ is the net momentum carried across the imaginary boundary. For every molecule passing downwards through the boundary, there is on average a molecule passing upwards, carrying on average slightly less $x$ momentum, characteristic of the region it came from.
As you can see (I hope), the derivation was correct - all I have done is tweak the words. There is often a tendency in textbooks to assume that if a quantity obviously has the correct dimensions, then it must be what is wanted. This can lead to a casual use of words when describing the logic. hth
I notice that you also have solid plates at the top and bottom of your diagram, which require that you use the no-slip boundary condition. This is always good when the mean free path is small, but for an accurate result for arbitrary mean free path you would need to know something about the surface, and how it interacts with incident molecules.
The following is a self answer concerning the rigorous derivation of the relative velocity in a Maxwellian gas. It is here for comparison against other answers.
Assumptions are
- Two particle collisions are the most probable ones, and collisions involving more than this, contribute to the mean relative velocity by a negligible amount.
- The mean relative velocity is strictly positive.
I think that this derivation differs from the one by @Thorondor because within it, $\mathbf{v_1}^2+\mathbf{v_2}^2\neq\mathbf{v_r}^2$. Which is to say that the dot product of $\mathbf{v_1}$, and $\mathbf{v_2}$ is not assumed to be zero.
This forces us to integrate over all possible angles between the velocities, which may account for the ~0.07 difference. It is also why the derivation is significantly longer.
The mean free path of an atom/molecule in a Maxwellian gas, depends upon the average relative velocity of each particle to one another. In order to obtain it, we first find the magnitude of velocity $\mathbf{v}_1$ relative to all other particles moving with $\mathbf{v}_2$. This expression is then averaged for all values of $\mathbf{v}_2$, from zero to infinity, producing a mean relative speed given $\mathbf{v}_1$ exists.
Following modus ponens, the mean is multiplied by the probability that an atom/molecule has velocity $\mathbf{v}_1$, and then averaged for all $\mathbf{v}_1$ from zero to infinity.
Consider two velocities $\mathbf{v}_1$ and $\mathbf{v}_2$ inclined at an angle $\theta$, there relative velocity will be described by
\begin{align}
(v_r)_{12}&=\sqrt{v_1^2+v_2^2-2v_1v_2cos(\theta)}.
\end{align}
All directions are equally probable for $\mathbf{v}_2$. To calculate the average of $(v_r)_{12}$, we multiply it by the probability that it lies within some solid angle $d\Omega$.
\begin{align}
\langle(v_r)_{12}\rangle&=\int_{\Omega}\frac{d\Omega}{4\pi}(v_r)_{12}\\
&=\int_{\Omega}\frac{2\pi sin(\theta)d\theta}{4\pi}(v_r)_{12}\\
&=\frac{1}{2}\int_{0}^{\pi}sin(\theta)d\theta(v_1^2+v_2^2-2v_1v_2cos(\theta))^\frac{1}{2}
\end{align}
Consider a change of basis, where $cos(\theta)=x$, so $sin(\theta)d\theta=-dx$. The limits of integration will change to $cos(0)=1$ and $cos(\pi)=-1$.
\begin{align}
\therefore~\langle(v_r)_{12}\rangle&=\frac{1}{2}\int_{1}^{-1}dx(v_1^2+v_2^2-2v_1v_2x)^\frac{1}{2}\\
&=\frac{1}{3}\frac{(v_1^2+v_2^2-2v_1v_2x)^\frac{3}{2}}{2v_1v_2}\Biggr|_{1}^{-1}\\
&=\frac{1}{6v_1v_2}\left((v_1^2+v_2^2+2v_1v_2)^\frac{3}{2}-(v_1^2+v_2^2-2v_1v_2)^\frac{3}{2}\right)\\
&=\frac{1}{6v_1v_2}\left((v_1+v_2)^\frac{3}{2}(v_1+v_2)^\frac{3}{2}+(v_1-v_2)^\frac{3}{2}(v_1-v_2)^\frac{3}{2}\right)\\
&=\frac{1}{6v_1v_2}\left((v_1+v_2)^3+|v_1-v_2|^3\right)\\
\end{align}
We take the magnitude of $v_1-v_2$ because $\langle(v_r)_{12}\rangle$ should always be positive, and therefore $\left((v_1-v_2)^2\right)^\frac{3}{2}$ must also be. This splits the average into two parts.
\begin{align}
\langle(v_r)_{12}\rangle&=\frac{1}{6v_1v_2}\left((v_1+v_2)^3-(v_1-v_2)^3\right),~when~v_1\geq v_2,\\
&=\frac{1}{6v_1v_2}\left[2v_2^3+6v_1^2v_2\right]\\
&=\frac{3v_1^2+v_2^2}{3v_1}.\\
\langle(v_r)_{12}\rangle&=\frac{1}{6v_1v_2}\left((v_1+v_2)^3+(v_1-v_2)^3\right),~when~v_2>v_1,\\
&=\frac{3v_2^2+v_1^2}{3v_2}.
\end{align}
The average relative velocity of an atom/molecule possessing magnitude and direction $\mathbf{v_1}$, with respect to all other particles moving with $\mathbf{v_2}$, lying within speeds of zero to infinity is then
\begin{align}
\langle(v_r)_1\rangle&=\int_{0}^{\infty}P(v_2)\langle(v_r)_{12}\rangle dv_2,\\
P(v_2)dv_2&=4\pi\left(\frac{m}{2\pi k_BT}\right)^\frac{3}{2}e^{-\frac{mv_2^2}{2k_BT}}v_2^2dv_2.
\end{align}
Note that $\langle(v_r)_1\rangle$ has two distinct forms depending on the difference between speeds. Because of this we break the integral into two parts.
\begin{align}
\langle(v_r)_1\rangle=4\pi\left(\frac{m}{2\pi k_BT}\right)^\frac{3}{2}&\left[\int_{0}^{v_1}dv_2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_1^2+v_2^2}{3v_1}\right)v_2^2\right.\\
&+\left.\int_{v_1}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_2^2+v_1^2}{3v_2}\right)v_2^2\right]
\end{align}
To arrive at the relative velocity of any atom with another, we form a product of the above, and the probability a particle has velocity $v_1+dv_1$.
\begin{align}
\langle v_r \rangle&=\int_{0}^{\infty}P(v_1)\langle(v_r)_1\rangle dv_1
\end{align}
\begin{align}
\langle v_r \rangle=\left(4\pi\left[\frac{m}{2\pi k_BT}\right]^\frac{3}{2}\right)^2\int_{0}^{\infty}dv_1e^{-\frac{mv_1^2}{2k_BT}}v_1^2&\left[\int_{0}^{v_1}dv_2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_1^2+v_2^2}{3v_1}\right)v_2^2\right.\\
&+\left.\int_{v_1}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_2^2+v_1^2}{3v_2}\right)v_2^2\right]
\end{align}
which may be re-written as
\begin{align}
\langle v_r\rangle=\left(\frac{2m}{k_BT}\right)^3&\left[\int_{0}^{\infty}dv_1e^{-\frac{mv_1^2}{2k_BT}}v_1^2\int_{0}^{v_1}dv_2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_1^2+v_2^2}{3v_1}\right)v_2^2\right.\\
&+\left.\int_{0}^{\infty}dv_1e^{-\frac{mv_1^2}{2k_BT}}v_1^2\int_{v_1}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_2^2+v_1^2}{3v_2}\right)v_2^2\right],
\end{align}
or
\begin{align}
\langle v_r\rangle&=\left(\frac{2m}{k_BT}\right)^3\left[I_1+I_2\right],
\end{align}
where $I_1$ stands for the first integral in the square brackets of the above, and $I_2$ the second one.
\begin{align}
I_1&=\int_{0}^{\infty}dv_1e^{-\frac{mv_1^2}{2k_BT}}v_1^2\int_{0}^{v_1}dv_2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_1^2+v_2^2}{3v_1}\right)v_2^2\\
I_2&=\int_{0}^{\infty}dv_1e^{-\frac{mv_1^2}{2k_BT}}v_1^2\int_{v_1}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_2^2+v_1^2}{3v_2}\right)v_2^2
\end{align}
We wish to show that these two definite integrals are equivalent. To do so, a substitution, then change in the order of integration following Fubini's theorem is employed.
Consider the piecewise function,
\begin{align}
\mathbb{I}_{\{v_2\leq v_1\}}&=\begin{cases}
1&v_1\leq v_2\\
0&v_1>v_2\\
\end{cases}
\end{align}
which is zero outside the limit of the second integral of $I_1$. Using it, we can rewrite equation $I_1$ as
\begin{align}
I_1&=\int_{0}^{\infty}dv_1e^{-\frac{mv_1^2}{2k_BT}}v_1^2\int_{0}^{\infty}dv_2\mathbb{I}_{\{v_2\leq v_1\}}e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_1^2+v_2^2}{3v_1}\right)v_2^2.\\
\end{align}
We can combine these two integrals with a substitution $a=v_1$, allowing us to write the following
\begin{align}
I_1&=\int_{0}^{\infty}\int_{0}^{\infty}\mathbb{I}_{\{v_2\leq v_1\}}e^{-\frac{ma^2}{2k_BT}}a^2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_1^2+v_2^2}{3v_1}\right)v_2^2dv_2da.\\
\end{align}
Since the integrals limits are constant, we can apply Fubini's theorem exchange there order.
\begin{align}
I_1&=\int_{0}^{\infty}\int_{0}^{\infty}\mathbb{I}_{\{v_2\leq v_1\}}e^{-\frac{ma^2}{2k_BT}}a^2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_1^2+v_2^2}{3v_1}\right)v_2^2dadv_2\\
&=\int_{0}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}v_2^2\int_{0}^{\infty}dv_1\mathbb{I}_{\{v_2\leq v_1\}}e^{-\frac{mv_1^2}{2k_BT}}\left(\frac{3v_1^2+v_2^2}{3v_1}\right)v_1^2\\
&=\int_{0}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}v_2^2\int_{v_2}^{\infty}dv_1e^{-\frac{mv_1^2}{2k_BT}}\left(\frac{3v_1^2+v_2^2}{3v_1}\right)v_1^2\\
\end{align}
Then, because the integral is definite, we can interchange $v_1$ with $v_2$, and obtain an equivalent volume.
\begin{align}
I_1&=\int_{0}^{\infty}dv_1e^{-\frac{mv_1^2}{2k_BT}}v_1^2\int_{v_1}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_2^2+v_1^2}{3v_2}\right)v_2^2\\
&=I_2
\end{align}
Hence, the average relative velocity can be rewritten as
\begin{align}
\langle v_r\rangle&=2\left(\frac{2m}{k_BT}\right)^3I_2.
\end{align}
To evaluate $I_2$, first consider the integral
\begin{align}
I_0&=\int_{v_1}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}\left(\frac{3v_2^2+v_1^2}{3v_2}\right)v_2^2.
\end{align}
We can break this into parts,
\begin{align}
I_0&=\int_{v_1}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}v_2^3+\frac{v_1^2}{3}\int_{v_1}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}v_2,
\end{align}
then consider the substitution $\frac{m}{2k_BT}v_2^2=x$, so $dv_2=\frac{k_BT}{mv_2}$, and the lower limit of integration changes to $\frac{mv_1^2}{2k_BT}$.
\begin{align}
I_0&=\int_{\frac{mv_1^2}{2k_BT}}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}v_2^3+\frac{v_1^2}{3}\int_{\frac{mv_1^2}{2k_BT}}^{\infty}dv_2e^{-\frac{mv_2^2}{2k_BT}}v_2\\
&=2\left(\frac{k_BT}{m}\right)^2\int_{\frac{mv_1^2}{2k_BT}}^{\infty}e^{-x}xdx+\frac{v_1^2}{3}\frac{k_BT}{m}\int_{\frac{mv_1^2}{2k_BT}}^{\infty}e^{-x}dx
\end{align}
Since
\begin{align}
\int_{a}^{b}xe^{-x}dx&=-(x+1)e^{-x}\Biggr|_{a}^{b},
\end{align}
$I_0$ simplifies to
\begin{align}
I_0&=2\left(\frac{k_BT}{m}\right)^2\left[\frac{m}{2k_BT}v_1^2+1\right]e^{\frac{-mv_1^2}{2k_BT}}\\
&=e^{-\frac{mv_1^2}{2k_BT}}\left[\frac{4}{3}\left(\frac{k_BT}{m}\right)v_1^2+2\left(\frac{k_BT}{m}\right)^2\right]
\end{align}
Substituting this into $I_2$ yields
\begin{align}
I_2=&\int_{0}^{\infty}dv_1e^{-\frac{mv_1^2}{2k_BT}}v_1^2\left[\frac{4}{3}\left(\frac{k_BT}{m}\right)v_1^2+2\left(\frac{k_BT}{m}\right)^2\right]
\\=&2\left(\frac{k_BT}{m}\right)^2\int_{0}^{\infty}dv_1e^{-\frac{mv_1^2}{2k_BT}}v_1^2\\
&+\frac{4}{3}\frac{k_BT}{m}\int_{0}^{\infty}dv_1e^{-\frac{mv_1^2}{2k_BT}}v_1^4\\
\end{align}
$I_2$ is actually a standard integral of the form,
\begin{align}
\int_{0}^{\infty}e^{-ax^2}x^n&=\frac{1}{2a^{\frac{n+1}{2}}}\Gamma\left(\frac{n+1}{2}\right),
\end{align}
where $\Gamma$ is the gamma function, a generalization of the factorial. $I_2$ now simplifies to
\begin{align}
I_2&=\left(\frac{k_BT}{m}\right)^{\frac{7}{2}}\left[\Gamma\left(\frac{3}{2}\right)+\frac{2}{3}\Gamma\left(\frac{5}{2}\right)\right]\\
&=\left(\frac{k_BT}{m}\right)^{\frac{7}{2}}\left[\frac{\sqrt{\pi}}{2}+\frac{2}{3}\frac{3\sqrt{\pi}}{4}\right]\\
&=\left(\frac{k_BT}{m}\right)^{\frac{7}{2}}\sqrt{\pi}
\end{align}
Hence, the average relative speed of any Maxwellian gas particle with respect to any other is
\begin{align}
\langle v_r\rangle&=2\left[4\pi\left(\frac{m}{2\pi k_BT}\right)^\frac{3}{2}\right]^2\left(\frac{k_BT}{m}\right)^\frac{7}{2}\sqrt{\pi},\\
&=\frac{2\sqrt{\pi}\cdot16\pi^2}{8\pi^3}\left(\frac{k_BT}{m}\right)^{-3}\left(\frac{k_BT}{m}\right)^\frac{7}{2},\\
&=\frac{4}{\sqrt{\pi}}\sqrt{\frac{k_BT}{m}},\\
&=\sqrt{\frac{16}{\pi}\frac{k_BT}{m}}.
\end{align}
And the ratio of mean velocity $\langle v_r\rangle$ to mean relative velocity $\langle v\rangle$ is
\begin{align}
\frac{\langle v\rangle}{\langle v_r\rangle}&=\frac{\sqrt{\frac{8}{\pi}\frac{k_BT}{m}}}{\sqrt{\frac{16}{\pi}\frac{k_BT}{m}}}\\
&=\frac{1}{\sqrt{2}}
\end{align}
And thus the mean free path $\ell$ is
\begin{align}
\ell=\frac{1}{n\sigma\sqrt{2}}
\end{align}
Best Answer
You want the average energy of particles hitting the wall, so you need to normalise with respect to the number of particles hitting the wall (you need to divide your result by the number of particles hitting the wall). Using unnormalised distribution (the normalisation cancels), gives you $$\frac{\int_0^\infty m v^3 f(v) dv}{2\int_0^\infty v f(v) dv}\,,$$ where the integral over $\theta$ and all other constants from the flux (number of particles hitting a wall) cancelled. This gives you $2 k T$ when the integrals are evaluated.