Worked exercise — reversible vs irreversible isothermal expansion

Problem

A diatomic gas expands isothermally at $T = 300\;\text{K}$, doubling its volume ($V_B = 2V_A$). Calculate $Q$ and $L_\text{gas}$ in the two cases:

1. Reversible transformation (infinite thermostat at $300\;\text{K}$, pressure decreasing gradually).
2. Irreversible transformation (the gas expands suddenly into a vacuum, against a wall of zero pressure).

Discuss in each case $\Delta E_\text{int}$, $\Delta S_\text{gas}$ and $\Delta S_\text{tot}$. Use $n = 1\;\text{mol}$, $R = 8{,}31\;\text{J/(mol·K)}$, $\ln 2 \approx 0{,}693$.

Solution

Setup — quantities equal in both cases. The initial state and the final state are identical in the two cases (same gas, same $T$, same $V_A \to V_B$). Since $E_\text{int}$ depends only on $T$ (ideal gas) and is a state function, and $S_\text{gas}$ is also a state function:

$\Delta E_\text{int} = 0 \qquad \Delta S_\text{gas} = nR\ln\frac{V_B}{V_A} = nR\ln 2$

hold in both cases. Numerically $\Delta S_\text{gas} = 1\cdot 8{,}31\cdot 0{,}693 = 5{,}76\;\text{J/K}$.

Case 1 — reversible. The gas pushes the piston with pressure $P = nRT/V$ which decreases gradually. The work is

$L_\text{gas}^\text{rev} = nRT\ln\frac{V_B}{V_A} = nRT\ln 2$

Since $\Delta E_\text{int} = 0$, from the first law $Q^\text{rev} = L_\text{gas}^\text{rev}$. Numerically:

$Q^\text{rev} = L_\text{gas}^\text{rev} = 1\cdot 8{,}31\cdot 300\cdot 0{,}693 = \ev{1728\;\text{J}}$

The thermostat releases this heat at constant $T$, so

$\Delta S_\text{term} = -\frac{Q^\text{rev}}{T} = -nR\ln 2 = -5{,}76\;\text{J/K}$

$\Delta S_\text{tot} = \Delta S_\text{gas} + \Delta S_\text{term} = nR\ln 2 - nR\ln 2 = \ev{0} \quad\checkmark \text{ (reversible)}$

Case 2 — irreversible (free expansion into vacuum). There is no piston, no atmosphere on the other side, no thermostat: the gas pushes against nothing ($P_\text{esterna} = 0$) and exchanges no heat.

$L_\text{gas}^\text{irr} = 0 \qquad Q^\text{irr} = 0$

The temperature stays $300\;\text{K}$ because $\Delta E_\text{int} = Q - L = 0$. With no thermostat, $\Delta S_\text{term} = 0$, hence

$\Delta S_\text{tot} = \Delta S_\text{gas} + 0 = nR\ln 2 = \ev{+5{,}76\;\text{J/K} > 0} \quad \text{(irreversible)}$

Conclusion. Same change of state, but $Q$ and $L$ differ: the reversible route extracts useful work ($1728\;\text{J}$) while keeping $\Delta S_\text{tot} = 0$; the free expansion produces no work and generates entropy. $\Delta E_\text{int}$ and $\Delta S_\text{gas}$ coincide because they are state functions — this is the classic “fictitious reversible path” technique for computing $\Delta S_\text{gas}$ even in an irreversible process.

Links

Topics: Termodinamica · Entropia e secondo principio Concepts: Trasformazioni termodinamiche · Primo principio della termodinamica · Entropia · Energia interna · Secondo principio della termodinamica Skills: Bilancio entropico Objects: Gas ideale