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Chemistry · THERMODYNAMICS
हिन्दीTwo moles of an ideal gas at \( 27^\circ\text{C} \) expand isothermally against a constant external pressure of \( 1.5\text{ atm} \) from \( 5\text{ L} \) to \( 15\text{ L} \). Then, the gas is compressed back to \( 5\text{ L} \) isothermally and reversibly at the same temperature. What is the net work done in the entire cyclic path? (Use \( R = 0.0821\text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\cdot\text{mol}^{-1} \), \( \ln 3 = 1.10 \))
Correct answer · Option 4
\( +39.19\text{ L}\cdot\text{atm} \)
Explanation
Expansion: \( w_1 = -p_{\text{ex}}(V_2 - V_1) = -1.5(15 - 5) = -15\text{ L}\cdot\text{atm} \). Compression: \( w_2 = -nRT \ln(V_1/V_2) = -2 \times 0.0821 \times 300 \times \ln(5/15) = +49.26 \times \ln 3 = 49.26 \times 1.10 = +54.19\text{ L}\cdot\text{atm} \). Net \( w = w_1 + w_2 = -15 + 54.19 = +39.19\text{ L}\cdot\text{atm} \).
