Two Roads to the Same State, One Entropy

physicsthermodynamics/entropydifficulty 4/5#entropy#state-function#free-expansion#irreversibility#false-attractor

An ideal gas of $n$ moles starts in volume $V$ at temperature $T$ and ends in volume $2V$ at the
same temperature, by two different routes.

(a) Route 1 is a quasi-static isothermal expansion against a piston. Find $\Delta S_{\mathrm{gas}}$
from $\int \mathrm{d}Q/T$. (b) Route 2 is a free expansion into an evacuated chamber (no
piston, no heat, no work). A tempting claim: "no heat flows, so $\Delta S_{\mathrm{gas}} = 0$." Is
it? (c) Reconcile the two routes using (1), and say where the irreversibility of Route 2
actually lives.

Route 1: quasi-static slowslow Route 2: free expansion vacuumvacuum

The entropy change is fixed by the endpoints alone, because $S$ is a state function:

$$ \Delta S_{\mathrm{gas}} = \int_V^{2V}\frac{nR}{V'}\,\mathrm{d}V' = nR\ln 2 $$(1)

The isotherm the quasi-static route rides along:

0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 0.4 0.6 0.8 1 1.2 1.4 1.6 start end V P

Check your answer

Show answer & solution

Answer

(a) $ \Delta S_{\mathrm{gas}} = nR\ln 2 $

(b) $ \Delta S_{\mathrm{gas}} = nR\ln 2 \ \text{(same — } S \text{ is a state function; not } 0). $

(c)

$\Delta S_{\mathrm{gas}}=nR\ln2$ for both routes. Route 1 is reversible:
$\Delta S_{\mathrm{surr}} = -nR\ln2$, so $\Delta S_{\mathrm{univ}}=0$. Route 2 does no work on
anything, so $\Delta S_{\mathrm{surr}}=0$ and $\Delta S_{\mathrm{univ}} = nR\ln2 > 0$ — the
irreversibility lives in the universe, not in the gas.

Solution

Necessary insight. $S$ is a state function: $\Delta S_{\mathrm{gas}}$ depends only on the
endpoints, so it is identical for the two routes. To compute it, integrate along any reversible
path connecting them — the quasi-static isotherm — even for the free expansion.

(a) Isothermal, so $\mathrm{d}U=0$ and $\mathrm{d}Q = \mathrm{d}W = P\,\mathrm{d}V = nRT\, \mathrm{d}V/V$. Then $\Delta S_{\mathrm{gas}} = \int \mathrm{d}Q/T = nR\ln 2$ — equation (1).

(b) The false attractor is "$Q=0 \Rightarrow \Delta S=0$." That formula, $\Delta S=\int \mathrm{d}Q/T$, holds only for a reversible path; the free expansion is irreversible. Because $S$
is a state function and the endpoints are identical, $\Delta S_{\mathrm{gas}} = nR\ln 2$ here too.
(For an ideal gas $U=U(T)$, and $T$ is unchanged, consistent with this.)

(c) Route 1: the surroundings lose the same entropy the gas gains, $\Delta S_{\mathrm{univ}}=0$
(reversible). Route 2: nothing else changes, so $\Delta S_{\mathrm{univ}} = nR\ln 2 > 0$. Same gas,
same $\Delta S_{\mathrm{gas}}$ — the routes differ only in what happens to the rest of the world.

Grading

  • 3 pts (a) Quasi-static isothermal $\Delta S_{\mathrm{gas}} = nR\ln 2$ from $\int dQ/T$.
  • 4 pts (b) Free expansion: recognises $\Delta S_{\mathrm{gas}}$ is the same (state function), not zero.
  • 3 pts (c) Distinguishes $\Delta S_{\mathrm{gas}}$ from $\Delta S_{\mathrm{universe}}$; irreversibility located.

Total: 10 pts