A quantity of 1.6 g helium gas is expanded adiabatically 3.0 times and then compressed isobarically — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

Moles of He $n = \frac{1.6}{4} = 0.4$. Step 1 (adiabatic reversible): $\Delta S_1 = 0$. Step 2 (isobaric compression from $3V_1$ to $V_1$): $\Delta S_2 = n C_P \ln\left(\frac{V_1}{3V_1}\right) = -n \frac{5R}{2} \ln 3$. Using $R \approx 2 \text{ cal/mol-K}$, $\Delta S = -0.4 \times 5 \times 1.1 = -2.2 \text{ cal/K}$.