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What I understand (Isothermal process):

In an isothermal process, the net entropy stays constant.

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Suppose, I have a cylinder filled with a monoatomic ideal gas. Now, I place it on an infinitely large reservoir.

Suppose both the cylinder and the infinitely large reservoir are at the same temperature $T$. Now, if I remove one tiny pebble from the top of the frictionless piston, then $dQ$ amount of heat will flow from the reservoir to the cylinder (I'm removing one tiny pebble because if I remove all of the pebbles, then the temperature of the cylinder won't even be defined as some parts of the cylinder will have a higher temperature than others, and we won't be able to do the math for entropy: please correct me if my explanation is wrong). So, the entropy change of the cylinder, $\Delta S_c=\frac{dQ}{T}$ and the entropy change of the reservoir, $\Delta S_r=\frac{-dQ}{T}$. Now, net change in entropy, $\Delta S=\Delta S_c+\Delta S_r=\frac{dQ}{T}+\frac{-dQ}{T}=0$. So, the entropy of the system stays constant.

What I don't understand (Isobaric process):

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Suppose, I have a cylinder filled with a monoatomic ideal gas with a frictionless piston. Now, if I add heat to the system very slowly, so that the macrostates are defined the whole time, and allow the piston to move freely, the gas will expand at constant pressure. Some of the added heat will be used to change the internal energy of the cylinder and some of the heat will be used to do work by the cylinder.

Now, how is heat added very slowly in case of an isobaric process? In the case of an isothermal process, there is a reservoir present: when a tiny pebble is removed, heat flows from the reservoir to the cylinder, and the temperature of the cylinder and the reservoir stay the same, and we are able to calculate the entropy change easily. But what happens in the case of an isobaric process? I think that an isobaric process is an irreversible process and the entropy of the system increases, unlike that of an isothermal process. Am I correct?

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2 Answers 2

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Now, for heat to be added reversibly, it has to be an isothermal process as well, right?

No.

For heat to be added reversibly, the system (gas in cylinder) has to constantly be in thermal as well as mechanical equilibrium with the surroundings. This can't be accomplished with a single reservoir as shown in your diagram.

The gas need to be exposed to an infinite series of thermal reservoirs, ranging from the initial gas temperature $T_i$ to the final gas temperature $T_f$, where each reservoir is infinitesimally greater in temperature than the previous reservoir and infinitesimally greater than the temperature of the gas. That keeps the temperature of the gas constantly in thermal equilibrium with its surroundings, as well as in mechanical equilibrium (constant pressure).

Then the change in entropy of the system becomes

$$\Delta S= mC_{p}\ln\frac{T_f}{T_i}$$

Hope this helps.

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    $\begingroup$ Oh I see. Similarly, if I want to cool the cylinder to its original state, I would have to expose it to an infinite number of thermal sinks, ranging from the final gas temperature $T_f$ to the initial gas temperature $T_i$. In that way, the entropy change of the system, consisting of the infinite number of thermal sinks and the cylinder, is 0. Is my understanding correct kind sir? $\endgroup$ Commented Dec 3, 2021 at 15:33
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    $\begingroup$ Yes a reversible isobaric compression back to the original state will result in zero change in entropy for both the system and surroundings. But just to make sure you understand, if your return the system to its original state the change in entropy of the SYSTEM will be zero even if the process is not reversible, because entropy is a state function. But if the process is irreversible then there will be a positive change in entropy of the surroundings; $\endgroup$
    – Bob D
    Commented Dec 3, 2021 at 15:43
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    $\begingroup$ Oh, I see. So, it doesn't matter whether the process is reversible or irreversible: if I heat the cylinder up and bring it back to its original condition, the entropy change of the CYLINDER will be zero. The difference is that in the irreversible process, the entropy change of the CYLINDER AND THE INFINITE NUMBER OF RESERVOIRS is not zero; it is positive, but in the case of the reversible process the entropy change of the CYLINDER AND THE INFINITE NUMBER OF RESERVOIRS is zero as well. I had another question sir... $\endgroup$ Commented Dec 3, 2021 at 15:50
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    $\begingroup$ If I just heat up the cylinder reversibly from $T_i$ to $T_f$ at constant pressure, will the entropy change of the CYLINDER AND THE INFINITE NUMBER OF RESERVOIRS be zero? If not, why does everyone say that in reversible processes, the entropy stays constant? $\endgroup$ Commented Dec 3, 2021 at 15:52
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    $\begingroup$ Regarding the first comment, for the process to be irreversible you would expose it to a SINGLE thermal reservoir of temperature equal to the final gas temperature, not an infinite series of reservoirs. Regarding the second comment, if the process is reversible then the entropy change of the system plus surroundings will always be zero. I don't understand your last statement. What entropy stays constant? $\endgroup$
    – Bob D
    Commented Dec 3, 2021 at 16:03
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In an isothrrmal process, entropy is change that is work done divided by temperature, $dS=nR\ln \frac{V_f}{V_i}$. Entropy change of cylinder or system is not equal to entropy change of reservoir or surrounding, $dS_c\ne dS_r$ but heat exchange is same, $dQ_r=dQ_c$ because if temperature of system and surrounding is same then transferrence of heat is not possible. Also per cycle, heat of surrounding during isothermal compression increases which decreases heat expel from system thus increases lower temperature and worsen efficiency of an engine.

In an isobaric process $dQ=dH$, so $dH=dW=pdV=mC_pdt$ and change in entropy is, $dS=mC_p\ln \frac{T_f}{T_i}=mC_p\ln \frac{V_f}{V_i}$. There is no need of slowly adding heat in isobaric process, but pressure of system is first increases through high compression because more the pressure more the work. The reason for isothermal to be slow because that allows to extract more work from system. So entropy not remain constant in an isobaric process, both its temperature and entropy changes.

Comparing change in entropy of an isothermal and an isobaric process, entropy of isothermal is lesser by $dS=nC_v\ln \frac{V_f}{V_i}$ than isobar. That is why efficiency of carnot engine is high.

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