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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!

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!

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