IPhO 1974 · Theory problem 3

Engine using a gas source and semipermeable membranes

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A scientific expedition stayed on an uninhabited island. The members of the expedition had had some sources of energy, but after some time these sources exhausted. Then they decided to construct an alternative energy source. Unfortunately, the island was very quiet: there were no winds, clouds uniformly covered the sky, the air pressure was constant and the temperatures of air and water in the sea were the same during day and night. Fortunately, they found a source of chemically neutral gas outgoing very slowly from a cavity. The pressure and temperature of the gas are exactly the same as the pressure and temperature of the atmosphere.

The expedition had, however, certain membranes in its equipment. One of them was ideally transparent for gas and ideally non-transparent for air. Another one had an opposite property: it was ideally transparent for air and ideally non-transparent for gas. The members of the expedition had materials and tools that allowed them to make different mechanical devices such as cylinders with pistons, valves etc. They decided to construct an engine by using the gas from the cavity.

Show that there is no theoretical limit on the power of an ideal engine that uses the gas and the membranes considered above.

Official English text of the problem.

Answers

  • Problem 3: Let \(\dot n_g>0\) be the molar flow of source gas, \(T_0\) the ambient temperature, and \(r\) the number of moles of air used per mole of gas. An ideal membrane engine can deliver \[ \boxed{\mathcal P(r)=\dot n_gRT_0 \left[\ln(1+r)+r\ln\!\left(1+\frac1r\right)\right].} \] Since \(\mathcal P(r)\to\infty\) as \(r\to\infty\), there is no finite theoretical upper bound on its power, even at a fixed, small gas flow.

Solution

Denote the common atmospheric and source pressure by \(p_0\), and their temperature by \(T_0\). Use the ideal-gas approximation, neglect friction, and allow isothermal operation through heat exchange with the surroundings. The available resource is the initially separate state of the source gas and atmospheric air: the membranes allow their mixing to produce mechanical work.

Consider a charge consisting of \(n\) moles of source gas and \(rn\) moles of air, initially at \(p_0,T_0\). Their volumes are \[ V_g=\frac{nRT_0}{p_0},\qquad V_a=\frac{rnRT_0}{p_0}=rV_g, \] where \(R\) is the universal gas constant.

Place these gases in a cylinder of fixed total volume \[ V=V_g+V_a, \] with source gas on the left and air on the right. Initially, two adjacent membrane pistons separate them:

  • The left membrane passes source gas but blocks air.
  • The right membrane passes air but blocks source gas.

Allow the left membrane to move toward the left end and the right membrane toward the right end, each driving a mechanical load. Between the membranes there is a mixture; outside them there is pure source gas on the left and pure air on the right. Ultimately, both components occupy the entire cylinder.

The essential property is that a permeating component has equal partial pressures on the two sides of its membrane in reversible operation. Thus, across the air-permeable piston, the forces due to air cancel. Only the source-gas partial pressure drives that piston. If the volume accessible to the source gas is \(v_g\), its work is \[ dW_g=p_g\,dv_g=\frac{nRT_0}{v_g}\,dv_g. \] Similarly, the gas-permeable piston extracts work from the expanding air: \[ dW_a=p_a\,dv_a=\frac{rnRT_0}{v_a}\,dv_a. \]

Consequently, the total work obtained in reversible isothermal mixing is \[ \begin{aligned} W &=\int_{V_g}^{V}\frac{nRT_0}{v_g}\,dv_g +\int_{V_a}^{V}\frac{rnRT_0}{v_a}\,dv_a\\ &=nRT_0\ln\frac{V}{V_g} +rnRT_0\ln\frac{V}{V_a}\\ &=\boxed{nRT_0\left[ \ln(1+r)+r\ln\!\left(1+\frac1r\right) \right]}. \end{aligned} \]

At the end, the mixture has total pressure \[ p_f=\frac{(1+r)nRT_0}{V}=p_0. \] It can therefore be exhausted and replaced with fresh gas and air at the common pressure \(p_0\). Ordinary pistons and bypass valves can perform this exchange and reset the membranes. Since the incoming and outgoing total volumes are equal, their pressure-volume work cancels: \[ W_{\text{intake/exhaust}}=p_0(V_g+V_a)-p_0V=0. \] The spent mixture is discharged, so the engine does not need to separate it again. Thus \(W\) is useful work per charge of \(n\) moles of source gas.

For a steady gas supply \(\dot n_g\), operating successive or parallel cylinders gives the ideal mean power \[ \mathcal P(r)=\frac{\dot n_g}{n}W =\dot n_gRT_0 \left[\ln(1+r)+r\ln\!\left(1+\frac1r\right)\right]. \] Equivalently, if \(q_g\) is the source’s volume flow measured at \(p_0,T_0\), then \[ \mathcal P(r)=p_0q_g \left[\ln(1+r)+r\ln\!\left(1+\frac1r\right)\right], \] with \(p_0q_g\) measured in watts.

Now increase the amount of air used per mole of source gas. Since \[ \ln(1+r)\longrightarrow\infty, \qquad r\ln\!\left(1+\frac1r\right)\longrightarrow1, \] we obtain \[ \mathcal P(r)=\dot n_gRT_0\,[\ln r+1+o(1)] \longrightarrow\infty. \] Hence even an arbitrarily small, fixed, positive gas flow imposes no finite theoretical power ceiling.

This conclusion allows unrestricted apparatus size, air throughput, membrane area and heat exchange. Slow, nearly reversible strokes can be operated in sufficiently many parallel cylinders to process the prescribed gas flow. Increasing the power requires an increasingly large amount of fresh air per mole of gas.

Finally, this respects the second law. For ideal gases at unchanged temperature, the mixing process leaves internal energy unchanged, so the absorbed heat equals the work: \[ Q=W. \] The entropy increase of the gases is \[ \Delta S_{\mathrm{mix}} =nR\ln\frac{V}{V_g}+rnR\ln\frac{V}{V_a} =\frac{W}{T_0}, \] which balances the heat reservoir’s entropy decrease in reversible operation. The engine consumes the initial separation of gas and air. The divergence assumes access to arbitrarily much fresh air; finite reservoirs and finite equipment impose practical limits.

The solution was checked against the official solution.