Challenge: the Gibbs paradox

Problem

Two distinct gases, each made up of $n$ moles and occupying a volume $V$ at the same pressure and temperature, are separated by a partition in an isolated container. The partition is removed and the gases mix. Calculate the entropy change $\Delta S$. What changes if the two gases are identical?

Solution

DISTINCT GASES case: once the partition is removed, each gas expands freely from its own volume $V$ to the total volume $2V$. The isothermal free expansion of $n$ moles increases the entropy by $nR\ln(2V/V)=nR\ln 2$. The two gases are independent, so the contributions add up: $\Delta S=nR\ln 2+nR\ln 2=2nR\ln 2>0$ This is the entropy of mixing, positive: the process is irreversible.

IDENTICAL GASES case: removing the partition produces no state physically distinguishable from the previous one (same gas, same density, same $T$ and $p$ everywhere). There is no real expansion nor observable mixing, so $\Delta S=0$

$\ev{\text{distinct: }\Delta S=2nR\ln 2;\quad \text{identical: }\Delta S=0}$ This is the Gibbs paradox: the “naive” formula would give $2nR\ln 2$ even for identical gases, generating an impossible discontinuity. It is resolved by the quantum indistinguishability of identical particles (Gibbs’s $1/N!$ factor), which correctly zeroes out the mixing entropy for equal gases.

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Topics: Entropia e secondo principio Concepts: Entropia Skills: Interpretazione micro-macro · Analisi di casi limite e fantafisica Objects: Gas ideale