# Entropy change of isothermal irreversible expansion of ideal gas [duplicate]

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Two mole of an ideal gas is subjected to isothermal expansion from $$\pu{2 atm}$$ to $$\pu{1 atm}$$ at $$\pu{300 K}$$. Calculate entropy change of the system, surrounding and total, if the process is irreversible.

Since

$$ΔS_\mathrm{system} = nR\ln\left(\frac{P_1}{P_2}\right),$$

I got the $$ΔS$$ for the system as $$1.4R$$. I know that for reversible process the entropy should not change, hence the entropy of surroundings decreases by $$1.4R$$. But how do we calculate the same for an irreversible one?

## marked as duplicate by Curt F., Jon Custer, tschoppi, Mithoron, airhuffFeb 19 at 21:00

Here is a cookbook recipe for determining the change in entropy for a system that has suffered an irreversible process:

THE RECIPE

1. Apply the First Law of Thermodynamics to the irreversible process to determine the final thermodynamic equilibrium state of the system

2. Totally forget about the actual irreversible process (entirely), and focus instead exclusively on the initial and final thermodynamic equilibrium states. This is the most important step.

3. Devise a reversible (alternative) path between the same two thermodynamic equilibrium states (end points). This reversible path does not have to bear any resemblance whatsoever to the actual irreversible process path. For example, even if the actual irreversible process is adiabatic, the reversible path you devise does not have to be adiabatic. You can even separate various parts of the system from one another, and subject each of them to a different reversible path, as long as they all end up in their correct final states. Plus, there are an infinite number of reversible process paths that can take you from the initial state to the final state, and they will all give exactly the same value for the change in entropy. So try to devise a path that is simple to work with (i.e., for which it is easy to apply step 4).

4. For the selected reversible path, evaluate the integral of dq/T from the initial state to the final state, where dq is the incremental amount of heat added to the system along the sequence of changes comprising the reversible path. This will be your change of entropy S. That is, $$\Delta S=\int{\frac{dq_{rev}}{T}}$$ , where the subscript rev refers to the reversible path.

For further details and worked examples, see: Reference https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

One way to proceed is as follows. Start from two things you know from the problem statement:

1. You have an ideal gas
2. The process is isothermal

Those two conditions mean the energy of the system does not change ($$\Delta U=0$$) during the process. It follows, from the first law, that

$$q_\mathrm{sys} = \Delta U -w_\mathrm{sys} = -w_\mathrm{sys}$$

You are only told the process is spontaneous. But you can assume that the expansion was against a constant pressure $$P_\mathrm{ext}=P_\mathrm{fin}=\pu{1atm}$$, so that

$$w_\mathrm{sys}=-P_\mathrm{ext}\Delta V = -nRT(1-\frac{P_\mathrm{fin}}{P_\mathrm{ini}})$$

Finally, the entropy change of the surroundings is

$$\Delta S_\mathrm{surr} = \frac{-q_\mathrm{sys}}{T_\mathrm{surr}}= \frac{w_\mathrm{sys}}{T_\mathrm{surr}}$$

Since the process is isothermal $$T_\mathrm{surr}=T_\mathrm{sys}=T$$ and

$$\Delta S_\mathrm{surr} = -nR(1-\frac{P_\mathrm{fin}}{P_\mathrm{ini}})$$

Plugging in values into this equation should give the desired answer ($$-R$$).

• The problem statement seems to imply that $P_{ext}=P_{fin}$ – Chet Miller Feb 17 at 14:48
• @ChesterMiller Thanks, I know, I didn't write that explicitly, but now made it a little tidier. – Night Writer Feb 17 at 14:58