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One mole of diatomic ideal gas ($c_V= {5\over 2}R$) is at an initial state $A$ at a volume $V_A$ and a temperature $T_1$. From $A$, the gas undergoes an isothermal process and expands to state $B$, with volume $V_B=2V_A$. Then, during an isochoric process to state $C$, the temperature of the gas is brought to $T_2=T_1/2$. Next, the gas undergoes another isothermal process to state $D$ where it has a volume $V_D=V_A$, and lastly it is brought back to state $A$ through an isochoric process. All processes are reversible.

Calculate the entropy change $\Delta S_1$ and $\Delta S_2$ of the reservoirs of temperatures $T_1$ and $T_2$ respectively.

Sorry for my English. Here is what I tried to do.

So apparently the cycle is isothermal at $T_1 \to$ isochoric at $V_1 \to$ isothermal at $T_2 \to$ isochoric at $V_2$. All processes are reversible, so the total entropy $\Delta S=\Delta S_1+\Delta S_2$ must be $0$.

I tried to calculate the heats of the various processes ($n=1$ because is one mole): $$Q_1=nRT_1 \ln{V_B\over V_A} = RT_1 \ln{2V_A\over V_A} = RT_1 \ln{2}$$ $$Q_2=\Delta U=nc_V(T_2-T_1)=c_V(T_2-2T_2)=-{5\over 2} RT_2$$ $$Q_3 =nRT_2 \ln{V_A\over V_B} = -nRT_2 \ln{V_B\over V_A} = -RT_2 \ln{2}$$ $$Q_4=\Delta U=nc_V(T_1-T_2)=c_V(2T_2-T_2)={5\over 2} RT_2=-Q_2$$

The reservoir at temperature $T_1$ comes into play in processes $1$ (AB) and $4$ (DA), and it releases to the system a total heat of $-Q_1-Q_4<0$, and reservoir at temperature $T_2$ absorbs from the system a total heat of $-Q_2-Q_4>0$. But the relative entropies don't sum up to zero:

$$\Delta S_1 = {-Q_1-Q_4\over T_1} = {-RT_1 \ln{2}-{5\over 2} RT_2 \over T_1}={-2RT_2 \ln{2}-{5\over 2} RT_2 \over 2T_2}=-R\left(\ln{2}+{\frac54}\right)\simeq-16.16 \text{ J/K}$$

$$\Delta S_2 = {-Q_2-Q_3\over T_2} = {{5\over 2} RT_2 + RT_2 \ln{2} \over T_2} = R\left(\ln{2}+{\frac52}\right)\simeq +26.55 \text{ J/K}$$

So $\Delta S=\Delta S_1 + \Delta S_2\simeq 10.40 \neq 0$. What did I do wrong?

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  • $\begingroup$ You acknowledged that all processes are reversible but, as Chet Miller pointed out, you calculated the entropy changes for the isochoric processes based on them being irreversible. That's what you did wrong. $\endgroup$
    – Bob D
    May 20 at 15:11
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Hint: Only the isothermal processes are involved with the entropy changes of thermal reservoirs $T_1$ and $T_2$. The isochoric processes involve entropy changes for an infinite series of thermal reservoirs between $T_1$ and $T_2$.

Hope this helps.

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  • $\begingroup$ Thank you, actually considering only the isothermal processes, the entropy change is zero as expected. $\endgroup$ May 21 at 13:26
  • $\begingroup$ @EdoardoFiocchi Yes, but it is also zero if you calculated the entropy change for reversible isochoric processes, which you didn't do. For those, the changes should be $$\Delta S=c_{v}\ln\frac{T_{final}}{T_{initial}}$$ $\endgroup$
    – Bob D
    May 21 at 13:50
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You didn't do anything wrong. The process, as you describe it, is irreversible, specifically the isochoric steps in the process. When you put the working fluid in contact with a constant temperature reservoir at a different temperature than the working fluid, the subsequent isochoric equilibration of temperature is irreversible. So entropy was generated within the working fluid during these steps, and this entropy was transferred to the reservoirs. So, even though the working fluid went through a cycle and its entropy did not change, the irreversibility resulted in an increase in entropy of the reservoirs.

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  • $\begingroup$ You said he did nothing wrong, but the problem stated "all processes are reversible" $\endgroup$
    – Bob D
    May 20 at 12:04
  • $\begingroup$ @BobD Oops. Missed that. $\endgroup$ May 20 at 12:08
  • $\begingroup$ I initially missed it too! $\endgroup$
    – Bob D
    May 20 at 12:37

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