Maxwell Relations And Thermodynamic Identities

For a reversible adiabatic process involving an ideal gas, the temperature and volume are related by: (T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}). This relation arises from combining the First Law (with (q = 0)) and the ideal gas equation for an adiabatic path. Here (T_1 = 400) K, (V_1 = 2) L, (V_2 = 16) L, and (\gamma - 1 = 0.4). Rearranging: (T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma-1} = 400 \times \left(\frac{2}{16}\right)^{0.4} = 400 \times (0.125)^{0.4}). Computing: ((0.125)^{0.4} = (1/8)^{0.4} = 8^{-0.4}). (\ln(8^{0.4}) = 0.4 \times \ln(8) = 0.4 \times 2.0794 = 0.8318), so (8^{0.4} = e^{0.8318} \approx 2.297). Thus (T_2 = 400/2.297 \approx 174) K (\approx 175) K. Option 150 K uses (\gamma - 1 = 0.5) (diatomic monatomic mix). Option 200 K is the isothermal result, ignoring (\gamma). Option 125 K uses an exponent of 0.6 incorrectly. This is a JEE Advanced adiabatic expansion calculation requiring familiarity with the TV relation. Plausibility check: expansion lowers gas temperature (adiabatic cooling), so (T_2 < T_1 = 400) K is mandatory; 175 K is about 44% of the initial temperature for an 8-fold volume increase, which is physically consistent.