IPhO 1972 · Theory problem 2

Argon and oxygen in two cylinders with connected pistons

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Two cylinders A and B, with equal diameters have inside two pistons with negligible mass connected by a rigid rod. The pistons can move freely. The rod is a short tube with a valve. The valve is initially closed (fig. 2.1).

Figure 1

Fig. 2.1

The cylinder A and his piston is adiabatically insulated and the cylinder B is in thermal contact with a thermostat which has the temperature \(\theta = 27^\circ\mathrm{C}\).

Initially the piston of the cylinder A is fixed and inside there is a mass \(m=32\ \mathrm{kg}\) of argon at a pressure higher than the atmospheric pressure. Inside the cylinder B there is a mass of oxygen at the normal atmospheric pressure.

Liberating the piston of the cylinder A, it moves slowly enough (quasi-static) and at equilibrium the volume of the gas is eight times higher, and in the cylinder B de oxygen’s density increased two times. Knowing that the thermostat received the heat \(Q'=747{,}9\cdot10^4\mathrm{J}\), determine:

a) Establish on the base of the kinetic theory of the gases, studying the elastic collisions of the molecules with the piston, that the thermal equation of the process taking place in the cylinder A is \(TV^{2/3}=\text{constant}\).

b) Calculate the parameters \(p\), \(V\), and \(T\) of argon in the initial and final states.

c) Opening the valve which separates the two cylinders, calculate the final pressure of the mixture of the gases.

The kilo-molar mass of argon is \(\mu=40\ \mathrm{kg/kmol}\).

Official English text of the problem.

Answers

The following are the formal results from the requested adiabatic relation, final mechanical equilibrium, and overall energy conservation. The quasistatic wording has a physical consistency issue: on release the argon is at \(64p_0\) but the oxygen only at \(p_0\), so the massless piston assembly cannot move quasistatically.

  • a) \(\boxed{TV^{2/3}=\text{constant}}\), equivalently \(pV^{5/3}=\text{constant}\).
  • b) The argon states, before opening the valve, are:
State Pressure \(p\) Volume \(V\) Temperature \(T\)
Initial \(64p_0 \simeq 6.48\times10^6\ \mathrm{Pa}\) \(1.03\ \mathrm{m^3}\) \(1.00\times10^3\ \mathrm{K}\)
Final \(2p_0 \simeq 2.03\times10^5\ \mathrm{Pa}\) \(8.20\ \mathrm{m^3}\) \(250\ \mathrm{K}\)
  • c) After opening the valve and reaching thermal equilibrium, \[ \boxed{p_{\mathrm{mix}}=\frac{166}{75}p_0 \simeq 2.24\times10^5\ \mathrm{Pa}}. \]

Here \(p_0\) is normal atmospheric pressure.

Solution

Treat both gases as ideal and neglect the volume of the short connecting tube. Use \[ p_0=1.01325\times10^5\ \mathrm{Pa},\qquad T_0\simeq300\ \mathrm{K},\qquad R=8.31\times10^3\ \mathrm{J\,kmol^{-1}\,K^{-1}}. \] The amount of argon is \[ n_A=\frac{m}{\mu}=\frac{32}{40}=0.800\ \mathrm{kmol}. \]

Subscripts \(1\) and \(2\) denote the initial state and the equilibrium state reached with the valve still closed.

a) Derivation from molecular collisions

Let the argon piston have area \(S\) and move outward with speed \(u\), small compared with molecular thermal speeds. Consider an argon atom of mass \(m_0\) approaching it with velocity component \(v_x\) normal to the piston.

In the instantaneous rest frame of the piston, an elastic collision reverses the normal velocity. Thus, in the laboratory frame, \[ v_x'=2u-v_x. \] The other velocity components remain unchanged. The atom’s energy change is therefore \[ \Delta\varepsilon =\frac{m_0}{2}\left[(2u-v_x)^2-v_x^2\right] =-2m_0u(v_x-u). \] The impulse delivered to the piston is \[ I=2m_0(v_x-u), \] so \[ \Delta\varepsilon=-uI. \]

Summing over molecular collisions, the impulse delivered per unit time is the force \(p_AS\). Consequently, \[ \frac{dU_A}{dt}=-p_ASu=-p_A\frac{dV_A}{dt}, \] or \[ dU_A=-p_A\,dV_A. \]

For an equilibrium monatomic ideal gas, kinetic theory gives \[ p_A=\frac{Nm_0}{V_A}\langle v_x^2\rangle =\frac{Nm_0}{3V_A}\langle v^2\rangle, \qquad U_A=\frac12Nm_0\langle v^2\rangle. \] Hence \[ p_AV_A=\frac23U_A, \qquad U_A=\frac32Nk_BT_A. \] The slow-motion assumption allows intermolecular collisions to maintain an isotropic equilibrium velocity distribution.

Substitution into the energy equation yields \[ \frac32Nk_B\,dT_A =-\frac{Nk_BT_A}{V_A}\,dV_A. \] Therefore \[ \frac{dT_A}{T_A}=-\frac23\frac{dV_A}{V_A}. \] Integrating, \[ \boxed{T_AV_A^{2/3}=\text{constant}}. \] Combining this with \(p_AV_A=n_ART_A\) also gives \[ \boxed{p_AV_A^{5/3}=\text{constant}}. \]

b) Initial and final argon states

The specified argon expansion gives \[ V_{A2}=8V_{A1}. \] Using part a), \[ \frac{T_{A2}}{T_{A1}} =\left(\frac{V_{A1}}{V_{A2}}\right)^{2/3} =8^{-2/3}=\frac14. \] Thus \[ T_{A2}=\frac14T_{A1}. \]

The oxygen mass is unchanged before the valve opens. Its doubled density therefore means \[ V_{B2}=\frac12V_{B1}. \] At both equilibrium endpoints, oxygen is at the thermostat temperature \(T_0\). Its ideal-gas equation then gives \[ p_{B2}=p_{B1}\frac{V_{B1}}{V_{B2}}=2p_0. \]

At final mechanical equilibrium, the forces on the connected piston assembly balance. The atmospheric forces cancel because the piston areas are equal, leaving \[ p_{A2}=p_{B2}=2p_0. \] The argon adiabatic relation now gives \[ p_{A1} =p_{A2}\left(\frac{V_{A2}}{V_{A1}}\right)^{5/3} =2p_0\,8^{5/3} =\boxed{64p_0}. \]

To determine the temperatures, consider both gases together. Equal piston displacements and equal cross-sectional areas imply \[ dV_A=-dV_B. \] Thus the combined gas volume is constant, and the net work against the atmosphere is zero. Work transmitted through the rod is internal to the combined system. Since the pistons are at rest at both endpoints, overall energy conservation gives \[ \Delta U_A+\Delta U_B=-Q'. \] Oxygen has the same initial and final temperature, so, for an ideal gas, \[ \Delta U_B=0. \] Therefore \[ Q'=\frac32n_AR(T_{A1}-T_{A2}) =\frac98n_ART_{A1}. \] With \(Q'=7.479\times10^6\ \mathrm{J}\), \[ T_{A1} =\frac{8Q'}{9n_AR} =\frac{8(7.479\times10^6)} {9(0.800)(8.31\times10^3)} =\boxed{1000\ \mathrm{K}}, \] and \[ \boxed{T_{A2}=250\ \mathrm{K}}. \]

Finally, \[ V_{A1}=\frac{n_ART_{A1}}{p_{A1}} =\frac{(0.800)(8.31\times10^3)(1000)} {64(1.01325\times10^5)} =1.025\ \mathrm{m^3}, \] and \[ V_{A2}=8V_{A1}=8.201\ \mathrm{m^3}. \]

To three significant figures, \[ \boxed{ (p_{A1},V_{A1},T_{A1}) =(6.48\times10^6\ \mathrm{Pa},\ 1.03\ \mathrm{m^3},\ 1.00\times10^3\ \mathrm{K}) } \] and \[ \boxed{ (p_{A2},V_{A2},T_{A2}) =(2.03\times10^5\ \mathrm{Pa},\ 8.20\ \mathrm{m^3},\ 250\ \mathrm{K}) }. \]

c) Final pressure after opening the valve

The piston geometry gives \[ V_{A2}-V_{A1}=V_{B1}-V_{B2}. \] Using \(V_{A2}=8V_{A1}\) and \(V_{B2}=V_{B1}/2\), \[ 7V_{A1}=\frac12V_{B1}. \] Hence \[ V_{B1}=14V_{A1},\qquad V_{B2}=7V_{A1}, \] and the constant total volume is \[ V_{\mathrm{tot}}=15V_{A1}. \]

After the valve opens, the gases mix and eventually reach the thermostat temperature \(T_0\). Although cylinder A is insulated, its gas can exchange energy through the open connection with cylinder B.

The conserved amount of oxygen satisfies \[ n_BRT_0=p_0V_{B1}=14p_0V_{A1}. \] For argon, \[ n_ART_0 =p_{A1}V_{A1}\frac{T_0}{T_{A1}} =64p_0V_{A1}\frac{300}{1000} =19.2p_0V_{A1}. \] Applying the ideal-gas equation to the final mixture, \[ p_{\mathrm{mix}} =\frac{(n_A+n_B)RT_0}{V_{\mathrm{tot}}} =\frac{(19.2+14)p_0V_{A1}}{15V_{A1}}. \] Therefore \[ \boxed{ p_{\mathrm{mix}} =\frac{166}{75}p_0 =2.2133p_0 \simeq2.24\times10^5\ \mathrm{Pa} }. \]

The solution was checked against the official solution.