= Solution
An elastic reflection transfers normal <momentum> $2mv\cos\theta$ to the wall. Multiplying this impulse by the collision flux and integrating gives
$$
p=2mnA I_4\int_0^{\pi/2}\sin\theta\cos^2\theta\,d\theta\int_0^{2\pi}d\phi=\frac{4\pi}{3}mnAI_4.
$$
Using the radial integral and $\alpha=m/(2k_BT)$ yields
$$
\boxed{p=\frac{nm}{2\alpha}=nk_BT.}
$$
This agrees with the ideal-gas <equation of state>. The calculation assumes a stationary wall and elastic normal momentum reversal, appropriate to a gas in equilibrium with it.
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