Thermodynamics

Concept:The work done in two-step isothermal compression at constant external pressures is minimized by applying the AM–GM inequality.Explanation:Given $n = 0.5$ mol, initial pressure $P_0 = 2$ bar, constant temperature $T = 600$ K.The gas is compressed in two steps: first against external pressure $P$ ($2 < P < 8$), then against $8$ bar.Using ideal gas law, $PV = nRT$, the initial volume $V_0 = \dfrac{nRT}{2}$, volume after first step $V_1 = \dfrac{nRT}{P}$, and final volume $V_2 = \dfrac{nRT}{8}$.Work done in step 1: $W_1 = P(V_0 - V_1) = P\left(\dfrac{nRT}{2} - \dfrac{nRT}{P}\right) = nRT\left(\dfrac{P}{2} - 1\right)$.Work done in step 2: $W_2 = 8(V_1 - V_2) = 8\left(\dfrac{nRT}{P} - \dfrac{nRT}{8}\right) = nRT\left(\dfrac{8}{P} - 1\right)$.Total work (since both are compression, work is positive): $|W| = W_1 + W_2 = nRT\left(\dfrac{P}{2} + \dfrac{8}{P} - 2\right)$.To minimize $|W|$, minimize $f(P) = \dfrac{P}{2} + \dfrac{8}{P}$ for $2 < P < 8$.By AM–GM: $\dfrac{\frac{P}{2} + \frac{8}{P}}{2} \ge \sqrt{\frac{P}{2} \cdot \frac{8}{P}} = \sqrt{4} = 2$.Thus $\dfrac{P}{2} + \dfrac{8}{P} \ge 4$, with equality when $\dfrac{P}{2} = \dfrac{8}{P}$, i.e., $P^2 = 16$, so $P = 4$ bar.Substitute $P = 4$: $|W|_{\min} = 0.5 \times R \times 600 \times (2 + 2 - 2) = 0.5 \times 600 \times 2R = 600R$ J.Answer:$600R$ J (Option B).