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For a reversible quasi-static process, the compressive normal stress exerted by the gas on the piston face is given by the ideal gas law: $$\sigma=\frac{RT}{v}$$where v is the molar volume of the gas. For an irreversible non-quasistatic process, Newton's law of viscosity adds the basic viscous contribution to the normal stress (see Transport Phenomena by Bird, Stewart, and Lightfoot) such that, at the piston face, $$\sigma=\frac{RT}{v}-\frac{4}{3}\frac{\mu}{v}\frac{d v}{d t}$$where ##\mu## is the gas viscosity, and where all the properties are evaluated at the piston face.Andrew Mason said:Sure. It is difficult to analyse a dynamic process. You might be interested in this paper I found which is similar to an analysis of the left side of the present problem.

AM

If the piston velocity is small compared to the speed of sound in the gas, the molar volume of the gas will be essentially uniform, and v in these equations can be replaced by the total gas volume V divided by the number of moles of gas n. In addition, if, as expected, the rapid irreversible deformation of the gas is turbulent, the gas viscosity in this equation can be replaced by the turbulent viscosity ##\mu_T## which is relatively independent of temperature. Under these circumstances, the above equation reduces to:

$$\sigma=\frac{nRT}{V}-\frac{4}{3}\frac{\mu_T}{V}\frac{d V}{d t}$$

So, as a result of these considerations, the compressive normal stresses exerted by the gases on the left and right faces of the partition in the present problem are given by:$$\sigma_L=\frac{n_LRT_L}{V_L}-\frac{4}{3}\frac{\mu_T}{V_L}\frac{d V_L}{d t}$$and $$\sigma_R=\frac{n_RRT_R}{V_R}-\frac{4}{3}\frac{\mu_T}{V_R}\frac{d V_R}{d t}=\frac{n_RRT_R}{V_R}+\frac{4}{3}\frac{\mu_T}{V_R}\frac{d V_L}{d t}$$where use has been made here of the fact that the sum of the two gas volumes is constant.

Since the piston is massless and frictionless, we have that $$\sigma_L=\sigma_R=\sigma$$and thus

$$\sigma=\frac{n_LRT_L}{V_L}-\frac{4}{3}\frac{\mu_T}{V_L}\frac{d V_L}{d t}=\frac{n_RRT_R}{V_R}+\frac{4}{3}\frac{\mu_T}{V_R}\frac{d V_L}{d t}$$Eliminating ##dV_L/dt##from these relationshiips yields $$\sigma=\frac{n_LRT_L+n_RRT_R}{(V_L+V_R)}\tag{1}$$ But, since the internal energy of the combined system is constant, we have $$n_LT_L+n_RT_R=n_LT_{Li}+n_RT_{Ri}$$and Eqn. 1 becomes:$$\sigma=\frac{n_LRT_{Li}+n_RRT_{R_i}}{(V_{Li}+V_{Ri})}=\frac{(P_{Li}V_{Li}+P_{Ri}V_{Ri})}{(V_{Li}+V_{Ri})}=P_f\tag{2}$$where ##P_f## is the final pressure. Eqn. 2 indicates that, at all times during the irreversible process, the normal stress on both sides of the partition is approximately equal to the final pressure.