Maxwell Relations And Thermodynamic Identities

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For an ideal gas undergoing a reversible adiabatic expansion, starting at $T_1 = 400$ K and $V_1 = 2$ L, expanding to $V_2 = 16$ L with $\gamma = C_p/C_v = 1.4$, what is the final temperature $T_2$?

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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.

Maxwell Relations And Thermodynamic Identities

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