Work done in reversible and irreversible processes

[Decimal]

Hello,

I am encountering some confusion understanding the difference in working with reversible and irreversible processes in thermodynamics. Let's say I have a process where an ideal gas at a certain starting temperature $T_i$ expands from volume $V_i$ to $V_f$. The temperature of the surroundings stays constant at $T_i$. There are two ways in which this process can happen, namely as an isothermal reversible process, and an isothermal process with a constant external pressure $P_{ext}$. I would like to calculate the entropy change of the gas in both of these processes.

I will use the following equation, which according to my book holds for both reversible and irreversible processes: $$ dU = TdS - PdV $$ Both processes are isothermal so $dU = 0$. This gives the following relation: $$dS = \frac {P} {T} dV$$ Working this out for the first process gives me: $$\Delta S = n*R *\int{\frac {dV} {V}}$$ Here I used the ideal gas equation. Now for the second process I again do the same calculation, with a constant pressure, I get: $$\Delta S = \frac {P_{ext}} {T} \int dV$$ So $$\Delta S = P_{ext}*(V_f - V_i)$$

However this answer is much different from the first answer, whilst the entropy should be a state variable for the gas. My book says that since the state is the same at the end of the process, the entropy should also be the same at the end for both processes. In my calculation this isn't a case. I feel like I am either missing a difference between the work done in reversible and irreversible processes, or I am not using the constant pressure correctly. Any help would be much appreciated!

 

[Chestermiller]

Hello,

I am encountering some confusion understanding the difference in working with reversible and irreversible processes in thermodynamics. Let's say I have a process where an ideal gas at a certain starting temperature $T_i$ expands from volume $V_i$ to $V_f$. The temperature of the surroundings stays constant at $T_i$. There are two ways in which this process can happen, namely as an isothermal reversible process, and an isothermal process with a constant external pressure $P_{ext}$. I would like to calculate the entropy change of the gas in both of these processes.

I will use the following equation, which according to my book holds for both reversible and irreversible processes: $$ dU = TdS - PdV $$

It really gives me a pain when books say that this relationship holds for both reversible and irreversible process. This equation actually expresses the relationship between dU, dS, and dV between two closely neighboring (differentially separated) thermodynamic equilibrium states of a material, so it is independent of any specific process. Since a reversible process involves differential changes along a continuous sequence of thermodynamic equilibrium states, it can also be used for reversible process between thermodynamic equilibrium states that are separated by large changes in U, S, and V at the end states. However, since an irreversible process does not involve differential changes along a continuous sequence of thermodynamic equilibrium states, the equation cannot be applied directly to relate the changes between end states the are separated by large changes. However, if you can identify a reversible path between the same two end states as the irreversible process, the equation can be applied to that.

Both processes are isothermal so $dU = 0$. This gives the following relation: $$dS = \frac {P} {T} dV$$ Working this out for the first process gives me: $$\Delta S = n*R *\int{\frac {dV} {V}}$$ Here I used the ideal gas equation. Now for the second process I again do the same calculation, with a constant pressure, I get: $$\Delta S = \frac {P_{ext}} {T} \int dV$$ So $$\Delta S = P_{ext}*(V_f - V_i)$$

Like I said, you can't apply the equation directly to the irreversible process. However, since the two end states for the reversible and irreversible paths are the same for this process, your result for the entropy change for the reversible path also applies to the irreversible path; they are equal.

If you want to get an understanding of how to determine the entropy change for an irreversible process process, there is a cookbook recipe for doing this. I've discussed this in a Physics Forums Insights article I wrote a while ago (which also discusses the rationale for the methodology): https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

Enjoy!

 

[Decimal]

Thanks a lot for the elaborate answer! I will definitely read your article. Looking back I also notice that in my calculation I assumed the work done by the gas, dependent on the external force, to be equal to $P dV$, which of course only holds for reversible processes.

 

[Chestermiller]

Thanks a lot for the elaborate answer! I will definitely read your article. Looking back I also notice that in my calculation I assumed the work done by the gas, dependent on the external force, to be equal to $P dV$, which of course only holds for reversible processes.

But, remember also that, irrespective of whether a process is reversible or irreversible, the volumetric work done by a system on its surroundings is always $W=\int{P_{ext}dV}$, where $P_{ext}$ is the force per unit area at the interface between the system and its surroundings.

 

[Decimal]

Yeah I understand. Where I went wrong is equating $P_{ext}dV$ to $P_{gas}dV$which is only true in a reversible process. Thanks for the help!