ΔS of an isobaric heating

Problem

One mole of a diatomic gas is heated at constant pressure from $T_1=300\;\text{K}$ to $T_2=600\;\text{K}$. Calculate $\Delta S$ (with $c_p=\tfrac{7}{2}R$).

Solution

For a reversible isobaric process, the entropy change is $\Delta S=n c_p\ln\frac{T_2}{T_1}$. Substituting $n=1$, $c_p=\tfrac{7}{2}R$, $R=8{,}31\;\text{J/(mol·K)}$ and $\frac{T_2}{T_1}=2$: $\Delta S=n c_p\ln\frac{T_2}{T_1}=1\cdot\tfrac{7}{2}\cdot 8{,}31\cdot\ln 2=29{,}1\cdot 0{,}693=20{,}17\;\text{J/K}$ $\ev{\Delta S\approx 20{,}17\;\text{J/K}}$

Links

Topics: Entropy and the second law Concepts: Entropy · Thermodynamic processes · Specific heat and heat capacity Skills: Entropy balance Objects: Ideal gas