The final volume (in L) of one mole of an ideal gas initially at and 8.21 atm pressure, if it absorb — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

Initial volume $V_1 = \frac{nRT}{P} = \frac{1 \times 0.0821 \times 300}{8.21} = 3\text{ L}$. For a reversible isothermal expansion, $q = -W = nRT \ln(V_2/V_1)$. Substituting the values ($R \approx 2\text{ cal/K-mol}$): $420 = 1 \times 2 \times 300 \ln(V_2/3) \implies 420 = 600 \ln(V_2/3) \implies \ln(V_2/3) = 0.7$. Since $\ln 2 \approx 0.7$, $V_2/3 = 2 \implies V_2 = 6\text{ L}$.