This is a continuation of the analysis begun in the previous post.
We can solve Eqns. 3 (representing the non-quasistatic Mungan gas model recommended by
@Andrew Mason) to obtain the time derivative of the left compartment volume ##dV_L/dt## and the "dynamic pressure" ##\hat{P}## solely in terms of the compartment temperatures, volumes, and gas viscosities:
$$\frac{dV_L}{dt}=nR\frac{(T_LV_R-T_RV_L)}
{(\mu_LV_R+\mu_RV_L)}\tag{4}$$and$$\hat{P}=nR\frac{(T_L\mu_R+T_R\mu_L)}{(\mu_LV_R+\mu_RV_L)}\tag{5}$$where ##\mu_L=\mu(T_L)## and ##\mu_R=\mu(T_R)##.
Next, if we add Eqns. 1 and 2, we obtain:$$nC_v\frac{d(T_L+T_R)}{dt}=0$$The solution to this equation, subject to the prescribed initial conditions is $$\frac{T_L+T_R}{2}=\bar{T}=\frac{3}{2}T_B\tag{6}$$where ##\bar{T}## is the (constant) average of the two compartment temperatures throughout the deformation.
Next, if we subtract Eqn. 2 from Eqn. 1, we obtain:$$nC_v\frac{d(T_L-T_R)}{dt}=2\hat{P}\frac{dV}{dt}$$or, equivalently, $$nC_v\frac{d(T_L-T_R)}{dV_L}=-2\hat{P}\tag{7}$$At this juncture, we define the following substitutions: $$V_L=V(1+\xi_V)\tag{8a}$$
$$V_R=V(1-\xi_V)\tag{8b}$$
$$T_L=\bar{T}(1+\xi_T)\tag{8c}$$
$$T_R=\bar{T}(1-\xi_T)\tag{8d}$$
$$\mu_L=\mu(\bar{T})\sqrt{1+\xi_T}\tag{8e}$$
$$\mu_R=\mu(\bar{T})\sqrt{1-\xi_T}\tag{8e}$$where, in writing Eqns. 8e and 8f, we have made use of the fact that, in the Mungan model, ##\mu## is proportional to ##\sqrt{T}##. If we substitute Eqns. 8 into Eqns. 4, 5, and 6, we obtain:$$\frac{d\xi_V}{d\tau}=\frac{2(\xi_T-\xi_V)}{\sqrt{1+\xi_T}(1-\xi_V)+\sqrt{1-\xi_T}(1+\xi_V)}\tag{9}$$
$$\hat{P}=\frac{nR\bar{T}}{V}\left[\frac{\sqrt{1+\xi_T}(1-\xi_T)+\sqrt{1-\xi_T}(1+\xi_T)}{\sqrt{1+\xi_T}(1-\xi_V)+\sqrt{1-\xi_T}(1+\xi_V)}\right]\tag{10}$$
$$\frac{d\xi_T}{d\xi_V}=-(\gamma - 1)\left[\frac{\sqrt{1+\xi_T}(1-\xi_T)+\sqrt{1-\xi_T}(1+\xi_T)}{\sqrt{1+\xi_T}(1-\xi_V)+\sqrt{1-\xi_T}(1+\xi_V)}\right]\tag{11}$$
where the dimensionless time ##\tau## is given by ##\tau=\frac{nR\bar{T}}{V\mu(\bar{T})}t##. For the present problem, initial conditions of these ordinary differential equations are: $$\xi_V(0)=0$$
and $$\xi_T(0)=\frac{1}{3}$$
One will note from Eqn. 11 that, even though this is a time-dependent problem, the trajectory of the dimensionless temperature variation ##\xi_T## as a function of the dimensionless volume variation ##\xi_V## is unique, and independent of the time history. Therefore, Eqn. 11 can be solved once-and-for-all for ##\xi_T## (and the dynamic pressure ##\hat{P}##) vs ##\xi_V##. In the present analysis this has been done by integrating Eqn. 11 numerically from ##\xi_V=0## to the final equilibrium point where, according to Eqn. 9, ##\xi_T=\xi_V##. The results of these numerical calculations using the recommended Mungan model will be presented in the next post, together with corresponding predictions using Chet Miller's approximation and using Andrew Mason's approximation (to determine which approximation provides a better match to the Mungan model).