Past exam of the physics B course of the University of Cambridge / 2014 / Paper 2 / Section C / C13

Molecules. In a dilute classical gas, collisions establish a Maxwell-Boltzmann distribution. The normalized speed distribution and its first two moments areMomentum transfer to a wall gives $p=nm\overline{v^2}/3=n{k}_{B}T$. Translational internal energy is $U=3N{k}_{B}T/2$; rotational and vibrational degrees of freedom add energy when thermally populated, without changing the dilute ideal-gas pressure law. Classical equipartition gives $k_{B}T/2$ per quadratic contribution, but quantum level spacing can freeze out modes.

Only the molecules moving towards a surface contribute to its one-sided number flux:For an effusive molecular energy flux, faster molecules are preferentially sampled, giving mean translational energy $2k_{B}T$ per crossing molecule and $F_E=2k_{B}TJ$. This differs from multiplying $J$ by the bulk mean $3k_{B}T/2$. Collisions and finite molecular interactions produce transport and departures from the ideal-gas equation of state.

Photons. All photons travel at $c$ in vacuum, so their thermal distribution is over frequency, not over a range of speeds. The two polarizations and Bose-Einstein occupation with zero chemical potential giveHere the spectral quantity is $u_{\nu }$ and the total energy density is $u$. Momentum flux gives $p=u/3$, while isotropic angular averaging gives the one-sided energy flux $F_E=cu/4={\sigma }_{\mathrm{SB}}{T}^4$ through a small aperture of a black-body cavity. The photon number flux is similarly $c{n}_{\gamma }/4$.

Both pressures arise from momentum transfer, but their equations of state differ: a classical molecular gas has $pV=N{k}_{B}T$ with fixed molecule number, whereas a photon gas has $pV=U/3$, $U=a_{R}V{T}^4$ and an equilibrium photon number that changes with temperature. Photon creation and absorption by the walls are essential to the zero-chemical-potential equilibrium description.