Let's consider an isolated container containing a mole of an ideal monoatomic gas $A$ and a mole of a biatomic gas $B$, separated by an ideal piston (hermetic). The gases slowly exchanges heat and the piston slides (no friction). At the very beginning it is given the volume of $A$ ($V_{A,0}=V_0$), the temperature of $A$ ($T_{A,0}=T_0$) and the volume of $B$ ($V_{B,0}=3V_0$).
Three questions:
- Which are the final states of the gases ($(p_A, V_A, T_A)$ and $(p_B, V_B, T_B)$)?
- Which are the entropy variations of the gases ($\Delta S_A$ and $\Delta S_B$)?
- What can you say about the sum of the two entropy my variation?
My approach
At final state, the gases will reach both thermal and dynamical equilibria. This means that:
$$T_A = T_B = T ~\text{and}~p_A = p_B = p.$$
As a consequence:
$$V_A = V_B = V.$$
Since the system is isolated, then the total volume is constant over time. This implies that:
$$V_A + V_B = V_{A,0} + V_{B,0} = 4V_0 \Rightarrow V = 2V_0.$$
The entropy variations are:
$$\Delta S_A = R \log \frac{V_A}{V_{A,0}} + \frac{3}{2}R\log \frac{T_A}{T_{A,0}} = \\ =R \log 2 + \frac{3}{2}R\log \frac{T}{T_{0}} $$
and
$$\Delta S_B = R \log \frac{V_B}{V_{B,0}} + \frac{5}{2}R\log \frac{T_B}{T_{B,0}} = \\ =-R \log 2 + \frac{5}{2}R\log \frac{T}{T_{B,0}}.$$
The sum of the entropy variations is $0$ since the system is isolated (is this true???).
Then $\Delta S_A + \Delta S_B = 0$ implies that:
$$3\log \frac{T}{T_{0}} + 5\log \frac{T}{T_{B,0}} = 0 \Rightarrow \\ \log \frac{T^3}{T^3_{0}} + \log \frac{T^5}{T^5_{B,0}} = 0 \Rightarrow \\ \log \frac{T^8}{T^3_{0}T^5_{B,0}} = 0 \Rightarrow \\ \frac{T^8}{T^3_{0}T^5_{B,0}} = 1,$$
yielding to
$$T = T_0^{\frac{3}{8}} T_{B,0}^{\frac{5}{8}}.$$
Are all the reasonings I wrote here right?
