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This question is one of three follow-ups to this other question, which I plan to post in sequence.

Why is the entropy change for a system in an irreversible isobaric transformation the same as in a reversible isobaric transformation?

This question wants to focus on the specific case of the isobaric process, both reversible and irreversible. There is evidence at the end of the page that might suggest a contradiction to the question. Therefore, detailed answers are welcome.

Generally, the entropy change for a closed system for an irreversible transformation is not the same in a closed system for a reversible transformation.

  1. Change in entropy in isobaric reversible transformation for a closed system.

Since $\delta Q_{\mathrm{reversible}} = n \, c_P \, \mathrm{d}T$ in a reversible isobaric process, we have $$\begin{align*} \Delta S_{\mathrm{reversible}}^{\mathrm{system, \ isobaric}} &= \displaystyle \int\limits_{T_I}^{T_F} \frac{n \, c_P \, \mathrm{d}T}{T} \\ &= n \, c_P \ln \left(\frac{T_F}{T_I}\right) \\ &= n \, (c_V + R) \ln \left(\frac{T_F}{T_I}\right) \qquad &\text{(since $c_P = c_V + R$ by Mayer's relation)} \\ &= n \, c_V \ln \left(\frac{T_F}{T_I}\right) + n R \ln \left(\frac{T_F}{T_I}\right) \\ &= n \, c_V \ln \left(\frac{T_F}{T_I}\right) + n R \ln \left(\frac{p_F V_F}{P_I V_I}\right) \qquad &\text{(from $pV = nRT$ by Ideal Gas Law)} \\ &= n \, c_V \ln \left(\frac{T_F}{T_I}\right) + n R \ln \left(\frac{V_F}{V_I}\right), \\ \end{align*}$$ so $$\color{red}{\Delta S_{\mathrm{reversible}}^{\mathrm{system, \ isobaric}}} = n \, c_V \ln \left(\frac{T_F}{T_I}\right) + n R \ln \left(\frac{V_F}{V_I}\right).$$

  1. Change in entropy in isobaric irreversible transformation for a closed system.

We can't take a reversible isobaric process and evaluate the change in entropy. We have to devise a reversible path between the same two thermodynamic equilibrium initial $(I)$ and final $(F)$ states which need not be the same path as the reversible one.

We can take this reversible path as the combination of an isochoric reversible process and an isothermal reversible process.

  • Isochoric reversible process between initial state $I$ and transition equilibrium midpoint state $M$.

From the $1^{\text{st}}$ Law of Thermodynamics $\delta Q = \mathrm{d}U + \delta W$, and since $\delta W = 0$ for an isochoric reversible process, we have $\delta Q_{\mathrm{reversible}} = \mathrm{d}U = n \, c_V \mathrm{d}T$.
So: $$\Delta S_{\mathrm{reversible}}^{\mathrm{system, \ isochoric}} = \displaystyle \int \frac{\delta Q_{\mathrm{reversible}}}{T} = \displaystyle \int\limits_{T_I}^{T_M} \frac{n \, c_V \mathrm{d}T}{T} = n \, c_V \ln \left(\frac{T_M}{T_I}\right).$$

  • Isothermal reversible process between transition equilibrium midpoint state $M$ and final state $F$.

From the $1^{\text{st}}$ Law of Thermodynamics $\delta Q = \mathrm{d}U + \delta W$, and since $\delta U = 0$ for an isothermal reversible process, we have $\delta Q_{\mathrm{reversible}} = \delta W = p \, \mathrm{d}V = \dfrac{nRT}{V}\mathrm{d}V$.
So: $$\Delta S_{\mathrm{reversible}}^{\mathrm{system, \ isothermal}} = \displaystyle \int \frac{\delta Q_{\mathrm{reversible}}}{T} = \displaystyle \int\limits_{V_M}^{V_F} \frac{\frac{nRT}{V}\mathrm{d}V}{T} = n R \ln \left(\frac{V_F}{V_M}\right).$$

Finally:

$$\Delta S_{\mathrm{irreversible}}^{\mathrm{system, \ isobaric}} = \Delta S_{\mathrm{reversible}}^{\mathrm{system, \ isochoric}} + \Delta S_{\mathrm{reversible}}^{\mathrm{system, \ isothermal}} = n \, c_V \ln \left(\frac{T_M}{T_I}\right) + n R \ln \left(\frac{V_F}{V_M}\right).$$

Since $T_M = T_F$ and $V_M = V_I$:

$$\color{blue}{\Delta S_{\mathrm{irreversible}}^{\mathrm{system, \ isobaric}}} = n \, c_V \ln \left(\frac{T_F}{T_I}\right) + n R \ln \left(\frac{V_F}{V_I}\right).$$

So: $$\boxed{\color{red}{\Delta S_{\mathrm{reversible}}^{\mathrm{system, \ isobaric}}} = n \, c_V \ln \left(\frac{T_F}{T_I}\right) + n R \ln \left(\frac{V_F}{V_I}\right) = \color{blue}{\Delta S_{\mathrm{irreversible}}^{\mathrm{system, \ isobaric}}}}.$$

So, the entropy change for a closed system in the reversible isobaric case turns out to be exactly the same as the entropy change for a closed system in the irreversible isobaric case; thus, we could have chosen the same reversible pathway to evaluate the entropy change in isobaric transformation.

However, this insight suggests that, for the same change in temperature, the work done in an irreversible isobaric process is higher than the work done in a reversible isobaric process, and the heat flux exchanged in an irreversible isobaric process is also higher than the heat flux exchanged in a reversible isobaric process.
So, intuitively, the entropy change for a closed system in an irreversible isobaric process is higher than the entropy change for a closed system in a reversible isobaric process.

With the main question taken into account, what is going on here?

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    $\begingroup$ @BySymmetry I know the question you linked is similar, because I asked it a while ago. I explicitly wrote at the beginning of this question that this is a follow-up to that question, wanting to focus on a specific case. $\endgroup$ Commented Dec 26, 2024 at 23:47
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    $\begingroup$ What is your definition of an irreversible isobaric process? (a) you control the external pressure so that at all times through the process, the external pressure is equal to the initial pressure/, or (b) at time zero, you suddenly drop the external pressure to a new value, and then hold the external pressure at this new value until the system re-equilibrates? $\endgroup$ Commented Dec 27, 2024 at 14:00
  • $\begingroup$ @ChetMiller The same as in this answer, so I think it is (b). $\endgroup$ Commented Dec 27, 2024 at 14:02
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    $\begingroup$ In that case, the final pressure you get for the irreversible process is not the same as the initial pressure before the change, and that means that your first analysis taking into account only the temperature change does not account for the pressure change. The entropy change calculation for the irreversible process must involve the temperature change at constant initial pressure plus the pressure change at constant final temperature. $\endgroup$ Commented Dec 27, 2024 at 14:43
  • $\begingroup$ @ChetMiller Could you write an answer in which you show the calculations and the result of the entropy change (according to what you just said) so that me and the whole site can benefit from it? I look forward to seeing it, thank you. $\endgroup$ Commented Dec 27, 2024 at 14:56

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Why is the entropy change for a system in an irreversible isobaric transformation the same as in a reversible isobaric transformation?

What you "discovered" applies to any transformation between the same two equilibrium states, not just an isobaric transformation.

That's because the entropy change of the system between the same two equilibrium states is the same regardless of the type of transformation (isobaric, isothermal, isochoric, etc.) and regardless of whether the transformation is reversible or irreversible. Entropy, like internal energy, enthalpy, temperature, pressure, etc., is a system property which makes its change independent of the transformation.

It is the total entropy change of the system plus the surroundings that is different for a reversible and irreversible process. For a reversible process $\Delta S_{total}=0$. For an irreversible process $\Delta S_{total}\gt 0$.

Hope this helps.

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  • $\begingroup$ Thank you for the answer. I already knew that entropy is a system property which makes its change independent of the transformation. But, in the previous question I received a comment by @ChetMiller: "Irreversible and reversible isobaric processes do not go between the same initial and final states for adiabatic isobaric processes". So, how can we justify the fact that the entropy change for a system in an irreversible isobaric transformation is the same as in a reversible isobaric transformation? $\endgroup$ Commented Dec 27, 2024 at 8:34
  • $\begingroup$ @Bml Any type of irreversible adiabatic transformation can’t connect the same two equilibrium states as a reversible adiabatic transformation because the entropy generated in the irreversible transformation becomes “trapped” in the system. The only way the system can get rid of the generated entropy is by transferring it to the surroundings in the form of heat, which by definition of adiabatic is impossible. The second law is satisfied since the total entropy change is greater than zero $\endgroup$ Commented Dec 27, 2024 at 10:24
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Consider two different examples of irreversible processes that are regarded as isobaric according to criterion $\text{(b)}$ in your comment above. In each case you have $n$ moles of an ideal gas in a cylinder with a massless frictionless piston. The initial temperatures and pressures of the gas are $T_1$ and $P_1$, respectively.

Process 1: The cylinder is suddenly placed in contact with an ideal isothermal reservoir at $T_2$ and an external pressure of $P_2$, and allowed to re-equilibrate to the final state of $T_2$ and $P_2$. In this case, the entropy change is $$\Delta S = nC_P\ln{(T_2/T_1)}-nR\ln{(P_2/P_1)}.$$

Process 2: The cylinder is thermally insulated, and is suddenly placed in contact with an external pressure of $P_2$, and allowed to re-equilibrate to the final state of $T_2$ and $P_2$. In this case of an insulated cylinder, the first law of thermodynamics tells us that $$nC_v(T_2-T_1)=-P_2(V_2-V_1)=-nR\left(T_2-T_1\frac{P_2}{P_1}\right).$$Solving for $T_2$ in this insulated cylinder case gives: $$T_2=T_1\left(\frac{C_v}{C_p}+\frac{R}{C_p}\frac{P_2}{P_1}\right).$$For Process 2, the entropy change is given by the same formula as Process 1, except that, in this case of an insulated cylinder, the final temperature $T_2$ is different:$$\Delta S = nC_P\ln{(T_2/T_1)}-nR\ln{(P_2/P_1)}.$$

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  • $\begingroup$ Thank you. Could you edit your answer to include all the steps that lead to your result for $\Delta S$? I don't quite understand why you come up with a term $nR \ln (P_2/P_1)$, since I get $nR \ln (V_2/V_1)$. $\endgroup$ Commented Dec 27, 2024 at 20:08
  • $\begingroup$ You go reversibly from the initial state to the intermediate state of T1 ,P2 at constant temperature, and then from this intermediate state to the final temperature at constant pressure . Or you go reversible from the initial state to the intermediate state T2, P1 at constant pressure, and then from this intermediate state to the final pressure at constant temperatures. Both reversible paths give the exact same value for the change in entropy (as do all other conceivable reversible paths). $\endgroup$ Commented Dec 27, 2024 at 20:21
  • $\begingroup$ Ah, OK. I got it. Thank you. $\endgroup$ Commented Dec 27, 2024 at 20:26

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