## 9.47 [ENDORSED]

$\Delta S = \frac{q_{rev}}{T}$

Evelyn L 1H
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Joined: Fri Sep 29, 2017 7:05 am

### 9.47

This question:
Initially, a sample of ideal gas at 323 K occupies 1.67 L at 4.95 atm. The gas is allowed to expand to 7.33 L by two pathways: (a) isothermal, reversible expansion; (b) isothermal, irreversible free expansion. Calculate delta Stot, delta S, and delta Ssurr for each pathway.

Why for the isothermal, irreversible free expansion, why would delta S surrounding equal zero and delta S equal 3.84 because there is free expansion and there's no work or heat. I thought that delta S surrounding would equal 3.84 and delta S be 0.

Chem_Mod
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### Re: 9.47  [ENDORSED]

We use the fact that entropy is a state function (we are thinking of the states of the system here) to deduce the values of ΔS for the isothermal irreversible free expansion based on answers for the isothermal reversible expansion, which we know how to calculate directly.

Since it does not matter if the system went from state 1 to state 2 as a reversible or irreversible process, we know the ΔS must be the same for the system. We can't make similar conclusions for ΔSsurr or ΔStotal. We know that no heat is transferred to the surroundings during a free expansion, but that does not mean that the entropy of the system remains unchanged. Imagine a gas under compression suddenly and irreversibly expanding into a vacuum. There are obviously more microstates available for the the system now that it occupies a much larger volume so it couldn't possibly have a 0 entropy change. However, because of no heat transfer to the surroundings, the surroundings do have a 0 entropy change. Adding ΔSsys + ΔSsurr results in the positive value for ΔStotal. This implies the process is spontaneous, which we immediately understand it must be according to our intuition. A confined gas being suddenly allowed to enter a vacuum space will immediately begin expanding into that space spontaneously.