IPhO 1981 · Theory problem 1

Container emptied by a released piston and escaping gas

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A static container of mass \(M\) and cylindrical shape is placed in vacuum. One of its ends is closed. A fixed piston of mass \(m\) and negligible width separates the volume of the container into two equal parts. The closed part contains \(n\) moles of monoatomic perfect gas with molar mass \(M_0\) and temperature \(T\). After releasing of the piston, it leaves the container without friction. After that the gas also leaves the container. What is the final velocity of the container?

The gas constant is \(R\). The momentum of the gas up to the leaving of the piston can be neglected. There is no heat exchange between the gas, container and the piston. The change of the temperature of the gas, when it leaves the container, can be neglected. Do not account for the gravitation of the Earth.

Official English text of the problem.

Answers

  • Problem 1 — Under the free-molecular escape approximation described below, the container’s final speed is \[ \boxed{ v_{\mathrm f} = \sqrt{\frac{3mnRT}{M(M+m)} \left(1-2^{-2/3}\right)} + \frac{n}{M} \sqrt{\frac{2M_0RT}{\pi\,2^{2/3}}} } \] directed toward its closed end. The statement supplies no numerical values. The second term holds for the free-molecular escape model stated in the solution.

Solution

Take the positive \(x\)-direction along the cylinder toward its open end.

Expansion while the piston remains inside.

Let the initial gas volume be \(V_0\). When the piston reaches the open end, the gas occupies \(2V_0\).

For a monoatomic ideal gas, \[ C_V=\frac32R,\qquad \gamma=\frac53. \] Using the uniform-pressure, quasistatic approximation associated with neglecting gas inertia, the expansion is adiabatic: \[ nC_V\,dT=-p\,dV=-\frac{nRT}{V}\,dV. \] Consequently, \[ TV^{2/3}=\text{constant}, \] so the temperature when the piston leaves is \[ \boxed{T_1=T\,2^{-2/3}}. \]

Write the container velocity at that instant as \(-v_1\), with \(v_1>0\), and the piston velocity as \(u_1>0\). Conservation of momentum, with the stipulated neglect of gas momentum, gives \[ -Mv_1+mu_1=0, \qquad u_1=\frac{M}{m}v_1. \]

The decrease in gas internal energy becomes kinetic energy of the piston and container: \[ \frac12Mv_1^2+\frac12mu_1^2 =\frac32nR(T-T_1). \] Substitution yields \[ \frac{M(M+m)}{2m}v_1^2 =\frac32nRT\left(1-2^{-2/3}\right), \] hence \[ \boxed{ v_1= \sqrt{\frac{3mnRT}{M(M+m)} \left(1-2^{-2/3}\right)} }. \] After leaving, the piston experiences no further force.

Recoil during molecular escape.

For this stage, use the following molecular idealization: molecules move freely during evacuation; reflections from the cylindrical side wall preserve their axial velocity component; reflection from the closed end reverses that component. Changes in molecular speeds caused by the container’s motion are neglected.

A molecule initially travelling toward the opening escapes directly. A molecule initially travelling toward the closed end reflects and subsequently escapes. Thus, if its initial axial velocity is \(v_x\), its final axial velocity is approximately \(|v_x|\).

Let \(\mu\) be the mass of one molecule and \(k_{\mathrm B}\) Boltzmann’s constant. The initial Maxwell distribution of axial velocities at temperature \(T_1\) is \[ f(v_x)= \sqrt{\frac{\mu}{2\pi k_{\mathrm B}T_1}}\, \exp\!\left(-\frac{\mu v_x^2}{2k_{\mathrm B}T_1}\right). \] The required average, counting every initially present molecule once, is \[ \begin{aligned} \langle |v_x|\rangle &=2\int_0^\infty v_x f(v_x)\,dv_x\\ &=\sqrt{\frac{2k_{\mathrm B}T_1}{\pi\mu}} =\sqrt{\frac{2RT_1}{\pi M_0}}. \end{aligned} \]

The gas has total mass \(nM_0\). Its final axial momentum is therefore \[ P_{\mathrm g} =nM_0\langle |v_x|\rangle =n\sqrt{\frac{2M_0RT_1}{\pi}}, \] directed toward the opening.

Conservation of momentum for the complete system now gives \[ -Mv_{\mathrm f}+mu_1+P_{\mathrm g}=0. \] Since \(mu_1=Mv_1\), \[ v_{\mathrm f}=v_1+\frac{P_{\mathrm g}}{M}. \] Inserting \(T_1=T2^{-2/3}\) gives \[ \boxed{ v_{\mathrm f} = \sqrt{\frac{3mnRT}{M(M+m)} \left(1-2^{-2/3}\right)} + \frac{n}{M} \sqrt{\frac{2M_0RT}{\pi\,2^{2/3}}} }. \] Both contributions propel the container toward its closed end.

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.