IPhO 1976 · Theory problem 2

Helium in a cylinder with a piston and a valved dividing wall

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The walls of a cylinder of base \(1\,\mathrm{dm}^2\), the piston and the inner dividing wall are perfect heat insulators (Fig. 10). The valve in the dividing wall opens if the pressure on the right side is greater than on the left side. Initially there is \(12\,\mathrm{g}\) helium in the left side and \(2\,\mathrm{g}\) helium in the right side. The lengths of both sides are \(11.2\,\mathrm{dm}\) each and the temperature is \(0^\circ\mathrm{C}\). Outside we have a pressure of \(100\,\mathrm{kPa}\).

The specific heat at constant volume is \(c_v=3.15\,\mathrm{J/gK}\), at constant pressure it is \(c_p=5.25\,\mathrm{J/gK}\). The piston is pushed slowly towards the dividing wall. When the valve opens we stop then continue pushing slowly until the wall is reached. Find the work done on the piston by us.

Figure 1

Figure 10

Official English text of the problem.

Answers

  • Problem 2: The work done on the piston by us is \[ \boxed{W_{\mathrm{us}}\approx 3.68\times10^3\,\mathrm{J}=3.68\,\mathrm{kJ}.} \]

Solution

Treat helium as an ideal gas with the given constant specific heats. Neglect friction and kinetic energy; the slow piston motion makes the compression quasistatic.

The initial volume of each compartment is \[ V_0=A\ell_0=(1\,\mathrm{dm}^2)(11.2\,\mathrm{dm}) =1.12\times10^{-2}\,\mathrm{m}^3. \] Also, \[ T_0=273.15\,\mathrm{K},\qquad R_s=c_p-c_v=2.10\,\mathrm{J\,g^{-1}K^{-1}},\qquad \gamma=\frac{c_p}{c_v}=\frac53. \]

Before the valve opens. The left-hand gas remains at fixed volume and temperature, so its pressure stays at \[ p_{L0}=\frac{m_LR_sT_0}{V_0}=614.6\,\mathrm{kPa}. \] The initial right-hand pressure is \[ p_{R0}=\frac{m_RR_sT_0}{V_0}=102.4\,\mathrm{kPa}, \qquad \frac{p_{L0}}{p_{R0}}=\frac{12}{2}=6. \]

The right-hand gas undergoes quasistatic adiabatic compression: \[ p_Rv^\gamma=p_{R0}V_0^\gamma, \] where \(v\) is its instantaneous volume. In the quasistatic limit, the valve starts opening when \(p_R=p_{L0}\). Hence the right-hand volume at opening is \[ v_*= V_0\left(\frac{p_{R0}}{p_{L0}}\right)^{1/\gamma} =V_0\,6^{-3/5} =3.822\times10^{-3}\,\mathrm{m}^3. \]

The pause contributes no work because the piston does not move. Any internal equilibration during the pause also leaves the total energy unchanged.

After the valve opens. Continued slow compression transfers helium into the left compartment. The pressure difference needed to open the ideal valve is infinitesimal, so the two compartments have a common pressure \(p\).

Their temperatures need not be equal. For either compartment, the ideal-gas equation gives \[ U_i=m_ic_vT_i=\frac{c_v}{R_s}pV_i =\frac{pV_i}{\gamma-1}. \] Thus, writing the total gas volume as \(\mathcal V=V_0+v\), \[ U=\frac{p\mathcal V}{\gamma-1}. \]

The whole gas system is thermally insulated, and only the piston does boundary work. Therefore, \[ dU=-p\,d\mathcal V. \] Substitution gives \[ \frac{\mathcal V\,dp+p\,d\mathcal V}{\gamma-1} =-p\,d\mathcal V, \] or \[ \mathcal V\,dp+\gamma p\,d\mathcal V=0. \] Consequently, \[ p\mathcal V^\gamma=\text{constant}. \] This relation follows from energy conservation and equal pressures; it does not require equal compartment temperatures.

At the start of this stage, \(\mathcal V=V_0+v_*\) and \(p=p_{L0}\). When the piston reaches the dividing wall, all the helium occupies \(V_0\). Its final pressure is therefore \[ \begin{aligned} p_f &=p_{L0}\left(\frac{V_0+v_*}{V_0}\right)^\gamma\\ &=p_{L0}\left(1+6^{-3/5}\right)^{5/3}\\ &\approx1.00257\times10^6\,\mathrm{Pa}. \end{aligned} \]

Work supplied by us. Since no heat enters the gas, the total work done on it equals its internal-energy increase: \[ \begin{aligned} W_{\mathrm{on\,gas}} &=U_f-U_i\\ &=\frac{p_fV_0-(p_{L0}+p_{R0})V_0}{\gamma-1}\\ &\approx4797.3\,\mathrm{J}. \end{aligned} \]

The atmosphere supplies part of this work. The piston sweeps out the entire initial right-hand volume, so \[ W_{\mathrm{atmosphere}} =p_{\mathrm{ext}}V_0 =(100000)(0.0112) =1120\,\mathrm{J}. \] With negligible piston kinetic energy and friction, \[ W_{\mathrm{us}}+W_{\mathrm{atmosphere}} =W_{\mathrm{on\,gas}}. \] Hence \[ \boxed{ W_{\mathrm{us}}=4797.3-1120 \approx3.68\times10^3\,\mathrm{J}. } \]

The solution was checked against the official solution.