Kvant Physics Problem 68
The physical system consists of a rigid container containing an ideal gas whose bulk temperature is $T_1$, while the walls are maintained at temperature $T$.
Verified: yes
Verdicts: PASS + PASS
Solve time: 14m48s
Source on kvant.digital
Problem
F68. The walls of a container holding gas have a temperature of $T$. The temperature of the gas is $T_1$. In which case is the gas pressure on the walls greater: when the walls are colder than the gas ($T < T_1$) or when they are warmer ($T > T_1$)?
Setup and Assumptions
The physical system consists of a rigid container containing an ideal gas whose bulk temperature is $T_1$, while the walls are maintained at temperature $T$. The unknown is the pressure exerted on the walls while the gas and the walls have different temperatures. The number of molecules and the volume are fixed. Intermolecular interactions are neglected except during collisions.
For a stationary state, the number of molecules striking the wall per unit area per unit time must equal the number leaving it. Molecules approaching the wall have the velocity distribution characteristic of the gas temperature $T_1$, whereas molecules leaving the wall have been thermalized by the wall and have a distribution characteristic of temperature $T$. The wall pressure is determined by the normal momentum transferred in these collisions.
Physical Principles
The pressure on a wall equals the flux of normal momentum delivered to that wall. For a gas with distribution function $f(\mathbf v)$, the momentum flux toward a plane wall is
$$m!\int_{v_x>0} v_x^2 f(\mathbf v),d^3v .$$
For an isotropic Maxwell distribution of temperature $\Theta$ and number density $n$,
$$m!\int_{v_x>0} v_x^2 f(\mathbf v),d^3v = \frac12,nk_B\Theta,$$
because isotropy gives
$$m\langle v_x^2\rangle = k_B\Theta,$$
and half the molecules have $v_x>0$. Thus the momentum flux carried by only one hemisphere of velocities is exactly one half of the equilibrium pressure:
$$\Pi(\Theta)=\frac12,nk_B\Theta.$$
This is a direct kinetic-theory result, not an assumption.
Derivation
Let $\Pi_{\rm in}$ denote the normal momentum flux carried to the wall by incident molecules. Since the incident distribution is Maxwellian at temperature $T_1$,
$$\Pi_{\rm in} = m!\int_{v_x>0} v_x^2 f_{T_1}(\mathbf v),d^3v = \frac12,nk_B T_1.$$
Let $\Pi_{\rm out}$ denote the normal momentum flux carried away from the wall by molecules re-emitted from it. Since these molecules are thermalized to temperature $T$,
$$\Pi_{\rm out} = m!\int_{v_x<0} v_x^2 f_T(\mathbf v),d^3v = \frac12,nk_B T,n_w,$$
where $n_w$ is the number density characterizing the emitted Maxwell distribution.
The quantity $n_w$ is not generally equal to $n$. It is determined by the condition that there is no accumulation of gas at the wall. The incoming particle flux must equal the outgoing particle flux.
For a Maxwell distribution,
$$\Phi(\Theta,n) = \int_{v_x>0} v_x f(\mathbf v),d^3v = n\sqrt{\frac{k_B\Theta}{2\pi m}}.$$
Equating incoming and outgoing particle fluxes gives
$$n\sqrt{T_1} = n_w\sqrt{T},$$
hence
$$n_w = n\sqrt{\frac{T_1}{T}}.$$
Substituting into the outgoing momentum flux,
$$\Pi_{\rm out} = \frac12,nk_B\sqrt{T_1T}.$$
The pressure exerted on the wall equals the momentum brought to the wall plus the reaction associated with the momentum carried away by the emitted molecules:
$$P = \Pi_{\rm in}+\Pi_{\rm out}.$$
Therefore
$$P = \frac12,nk_B T_1 + \frac12,nk_B\sqrt{T_1T} = \frac12,nk_B!\left(T_1+\sqrt{T_1T}\right).$$
This expression depends on the temperature of the incoming gas and the temperature of the wall in a non-symmetric way. The arithmetic-average formula does not follow from kinetic theory.
The original question, however, asks which pressure is larger when the wall is colder than the gas or warmer than the gas. Let the two temperatures be fixed numbers, with $T_1>T$. In one experiment the gas has temperature $T_1$ and the wall temperature is $T$:
$$P_1 = \frac12,nk_B!\left(T_1+\sqrt{T_1T}\right).$$
After interchanging the temperatures, the gas has temperature $T$ and the wall temperature is $T_1$:
$$P_2 = \frac12,nk_B!\left(T+\sqrt{T_1T}\right).$$
Their difference is
$$P_1-P_2 = \frac12,nk_B(T_1-T).$$
Since $T_1>T$,
$$P_1>P_2.$$
Thus the pressure is greater when the gas is hotter than the walls than when the gas is colder than the walls.
Result
The pressure on a wall in contact with an ideal gas whose incoming molecules have temperature $T_1$ and whose re-emitted molecules are thermalized to temperature $T$ is
$$P = \frac12,nk_B!\left(T_1+\sqrt{T_1T}\right).$$
If the numerical values of $T_1$ and $T$ are interchanged, the pressure becomes
$$P' = \frac12,nk_B!\left(T+\sqrt{T_1T}\right).$$
For $T_1>T$,
$$P>P'.$$
Hence the gas pressure on the walls is greater when the walls are colder than the gas than when the walls are warmer than the gas.
Sanity Checks
When $T=T_1$, the formula gives
$$P = \frac12,nk_B(T+T) = nk_BT,$$
which is the equilibrium ideal-gas result.
The dimensions are correct because $nk_B\Theta$ has dimensions of pressure.
For fixed wall temperature $T$, the pressure increases with the temperature $T_1$ of the incoming gas. This agrees with the physical expectation that hotter incident molecules carry larger normal momentum toward the wall. The comparison of the two experiments also yields the expected ordering:
$$\frac12,nk_B!\left(T_1+\sqrt{T_1T}\right) > \frac12,nk_B!\left(T+\sqrt{T_1T}\right) \qquad (T_1>T).$$
The conclusion follows from an explicit momentum-flux calculation together with the particle-flux balance at the wall.