Isothermal expansion: ΔS of the gas

Problem

An ideal gas ($n=1$ mol) expands isothermally at temperature $T=300$ K, doubling its volume. Calculate the entropy change $\Delta S$ of the gas.

Solution

For an ideal gas the entropy change between two states is generally given by $\Delta S = nC_V\ln\frac{T_B}{T_A} + nR\ln\frac{V_B}{V_A}$ In an isothermal transformation $T_B=T_A$, so the first term vanishes and only the volume contribution remains. With $V_B = 2V_A$: $\Delta S = nR\ln\frac{V_B}{V_A} = 1\cdot 8{,}31\cdot\ln 2 = 5{,}76\;\text{J/K}$ $\ev{\Delta S \approx 5{,}76\;\text{J/K}}$ The result is positive: as it expands the gas occupies more accessible microscopic states, so its entropy grows.

Linked atoms

Topics: Entropia e secondo principio Concepts: Entropia · Trasformazioni termodinamiche · Legge dei gas perfetti Skills: Bilancio entropico Objects: Gas ideale