Worked exercise — free expansion is not reversible

Problem

Let us return to the free expansion of $107\;\text{mol}$ of nitrogen from $V_A = 2\;\text{m}^3$ to $V_B = 4\;\text{m}^3$ at constant temperature $T = 300\;\text{K}$, in an isolated container. Calculate $\Delta S_\text{tot}$.

Solution

Setup. The container is isolated, so there is no thermostat to exchange heat with: $\Delta S_\text{atm} = 0$. Only the entropy of the gas remains. In free expansion the temperature does not change (for an ideal gas the internal energy depends only on $T$, and $Q = L = 0$), so $T_B = T_A$ and the temperature term vanishes.

Entropy of the gas (only the volume term survives): $\Delta S_\text{gas} = nC_v\ln\frac{T_B}{T_A} + nR\ln\frac{V_B}{V_A} = 0 + 107\cdot 8{,}31\cdot\ln 2 = 616\;\text{J/K}$

Total change: $\Delta S_\text{tot} = \ev{616\;\text{J/K} > 0}$

The process is highly irreversible. The crucial, and surprising, point is the comparison with the first law: in this transformation $Q = 0$, $L_\text{gas} = 0$ and $\Delta E_\text{int} = 0$, i.e. energy is perfectly conserved and nothing is exchanged with the outside. Yet entropy has grown by $616\;\text{J/K}$. This is proof that the second law adds information that the first does not contain: energy is conserved, but irreversibility is not.

Links

Topics: Entropy and the second law Concepts: Entropy · Second law of thermodynamics · Thermodynamic processes Skills: Entropy balance Objects: Ideal gas