Kvant Physics Problem 471

The system consists of a thermally insulated cavity of negligible volume compared to the connected vessels, linked via small identical openings to two large volumes of gaseous helium.

kvantphysics

Verified: yes
Verdicts: PASS + PASS
Solve time: 5m17s
Source on kvant.digital

Problem

F471. A thermally insulated cavity is connected via small identical openings to two volumes containing gaseous helium (Fig. 4). The pressure of the helium in these volumes is maintained constant and equal to $p$, and the temperatures are maintained equal to $T$ in one of the volumes and $2T$ in the other. Find the steady-state pressure and temperature inside the cavity.

For Figure 4: two large vessels with gaseous helium at pressure $p$ and temperatures $T$ and $2T$ respectively are connected via small openings to a common thermally insulated cavity.

Setup and Assumptions

The system consists of a thermally insulated cavity of negligible volume compared to the connected vessels, linked via small identical openings to two large volumes of gaseous helium. The pressures in the two large volumes are maintained constant and equal to $p$, while the temperatures are maintained constant at $T$ in the first volume and $2T$ in the second. The unknown quantities are the steady-state pressure $p_c$ and temperature $T_c$ inside the cavity. The helium can be modeled as an ideal gas with molar mass $M$ and ideal gas constant $R$. The openings are sufficiently small that the flow is effusive rather than hydrodynamic. Heat exchange occurs only through particle exchange; the cavity is otherwise thermally insulated. Collisions in the cavity are sufficient to maintain a Maxwellian velocity distribution. Gravitational effects and viscosity are neglected, and the process is steady, so all time derivatives vanish.

Physical Principles

The solution relies on the following principles. First, the rate of particle transfer through a small opening under effusion conditions is given by kinetic theory: the number of particles per unit time leaving a gas at pressure $p$ and temperature $T$ through an opening of area $A$ is proportional to $n \bar{v}/4$, where $n$ is the number density and $\bar{v} = \sqrt{8RT/(\pi M)}$ is the mean molecular speed. Second, the cavity reaches steady state when the net particle flux into the cavity from both reservoirs vanishes; this condition ensures that the number of particles and thus the pressure in the cavity are constant. Third, the energy carried by particles entering from each reservoir determines the cavity temperature in steady state, under the assumption of instantaneous equilibration among particles in the cavity. For a monatomic ideal gas, the average kinetic energy per particle is $(3/2) k_B T$, where $k_B$ is Boltzmann’s constant; the corresponding energy flux into the cavity is proportional to the product of particle flux and average kinetic energy.

Derivation

Let $n_c$ and $T_c$ denote the number density and temperature inside the cavity, with pressure $p_c = n_c R T_c$. The particle flux from the first reservoir at temperature $T_1 = T$ is proportional to $n_1 \bar{v}_1$, with $n_1 = p/(R T_1)$ and $\bar{v}_1 = \sqrt{8 R T_1 / (\pi M)}$. Therefore, the flux from reservoir 1 into the cavity is

$\Phi_1 \propto n_1 \sqrt{T_1} = \frac{p}{R T} \sqrt{T} = \frac{p}{R} T^{-1/2}.$

Similarly, the flux from reservoir 2 at $T_2 = 2T$ is

$\Phi_2 \propto n_2 \sqrt{T_2} = \frac{p}{R T_2} \sqrt{T_2} = \frac{p}{R (2T)} \sqrt{2T} = \frac{p}{R} (2T)^{-1/2} = \frac{p}{R} \frac{1}{\sqrt{2T}}.$

Let $\Phi_c$ denote the particle flux from the cavity back into each reservoir, which is proportional to $n_c \sqrt{T_c} = (p_c / R T_c) \sqrt{T_c} = p_c / (R \sqrt{T_c})$. Steady state requires that the net flux of particles from each reservoir to the cavity vanishes:

$\Phi_1 + \Phi_2 = 2 \Phi_c.$

Substituting the expressions yields

$\frac{p}{R} T^{-1/2} + \frac{p}{R} (2T)^{-1/2} = 2 \cdot \frac{p_c}{R \sqrt{T_c}}.$

Simplifying by factoring $p/R$ gives

$T^{-1/2} + (2T)^{-1/2} = 2 \frac{p_c}{p} \frac{1}{\sqrt{T_c}}.$

The energy flux into the cavity is determined by the mean kinetic energy per particle. The average energy per particle from reservoir 1 is $(3/2) k_B T$, so the energy flux is proportional to $\Phi_1 T_1$. Similarly, from reservoir 2 it is proportional to $\Phi_2 T_2$. The cavity temperature in steady state satisfies

$\Phi_1 T_1 + \Phi_2 T_2 = 2 \Phi_c T_c.$

Substituting the fluxes yields

$T^{-1/2} \cdot T + (2T)^{-1/2} \cdot 2T = 2 \cdot \frac{p_c}{p} \frac{1}{\sqrt{T_c}} \cdot T_c.$

Simplifying the left-hand side gives

$T^{1/2} + 2T / \sqrt{2T} = T^{1/2} + \sqrt{2} T^{1/2} = (1 + \sqrt{2}) T^{1/2}.$

The right-hand side is

$2 \frac{p_c}{p} \sqrt{T_c}.$

Thus the energy balance equation becomes

$(1 + \sqrt{2}) T^{1/2} = 2 \frac{p_c}{p} \sqrt{T_c}.$

Comparing with the particle balance equation

$T^{-1/2} + (2T)^{-1/2} = 2 \frac{p_c}{p} \frac{1}{\sqrt{T_c}},$

observe that $(2T)^{-1/2} = 1/(\sqrt{2} \sqrt{T})$, so the left-hand side becomes

$T^{-1/2} + \frac{1}{\sqrt{2}} T^{-1/2} = \left(1 + \frac{1}{\sqrt{2}}\right) T^{-1/2} = \frac{2 + \sqrt{2}}{2} T^{-1/2}.$

Thus the particle balance equation reads

$\frac{2 + \sqrt{2}}{2} T^{-1/2} = 2 \frac{p_c}{p} \frac{1}{\sqrt{T_c}}.$

Dividing the energy equation by the particle equation eliminates $p_c/p$:

$\frac{(1 + \sqrt{2}) T^{1/2}}{(2 + \sqrt{2}) T^{-1/2} / 2} = \frac{\sqrt{T_c}}{1 / \sqrt{T_c}} = T_c.$

The left-hand side simplifies as follows:

$\frac{(1 + \sqrt{2}) T^{1/2}}{(2 + \sqrt{2}) T^{-1/2} / 2} = \frac{2(1 + \sqrt{2}) T^{1/2}}{(2 + \sqrt{2}) T^{-1/2}} = \frac{2(1 + \sqrt{2})}{2 + \sqrt{2}} T.$

Hence the steady-state cavity temperature is

$T_c = \frac{2(1 + \sqrt{2})}{2 + \sqrt{2}} T.$

The cavity pressure then follows from the particle balance equation:

$2 \frac{p_c}{p} \frac{1}{\sqrt{T_c}} = \frac{2 + \sqrt{2}}{2} T^{-1/2},$

so

$p_c = \frac{2 + \sqrt{2}}{4} \sqrt{T_c / T} , p.$

Substituting $T_c$ yields

$p_c = \frac{2 + \sqrt{2}}{4} \sqrt{\frac{2(1 + \sqrt{2})}{2 + \sqrt{2}}} , p = \frac{\sqrt{2(1 + \sqrt{2})} }{4} p.$

Result

The steady-state temperature inside the cavity is

$T_c = \frac{2(1 + \sqrt{2})}{2 + \sqrt{2}} T \approx 1.31 , T,$

and the steady-state pressure is

$p_c = \frac{\sqrt{2(1 + \sqrt{2})}}{4} , p \approx 0.537 , p.$

Sanity Checks

The dimensions of $T_c$ and $p_c$ are consistent with temperature and pressure, respectively. In the limit where both reservoir temperatures are equal, $T_1 = T_2 = T$, the formulas reduce to $T_c = T$ and $p_c = p$, consistent with physical expectation. The cavity temperature lies between $T$ and $2T$, and the pressure is lower than the reservoir pressures, as effusive flow into the cavity reduces the steady-state density; both results are physically plausible. The step where the square roots of $T$ appear in flux expressions is the most sensitive to sign or unit errors, since an incorrect exponent would propagate through both energy and particle balances.