 Solution

ID: past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2016/paper-2/section-c/c12/b/solution

An elastic reflection transfers normal momentum

2 m v cos θ

to the wall. Multiplying this impulse by the collision flux and integrating gives

p = 2 mn A I_{4} \int_{0}^{π /2} sin θ cos^{2} θ d θ \int_{0}^{2 π} d ϕ = \frac{4 π}{3} mn A I_{4} .

Using the radial integral and

α = m / (2 k_{B} T)

yields

p = \frac{nm}{2 α} = n k_{B} T .

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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