Given below are two statements : Statement I : When \mu amount of an ideal gas undergoes adiabatic change from state (P1, V1, T1) to state (P2, V2, T2), then work done is W = \mu R(T_2 - T_1) \over 1 - \gamma , where \gamma = C_p \over C_v and R = universal gas constant. Statement II : In the above case, when work is done on the gas, the temperature of the gas would rise. Choose the correct answer from the options given below :

Both Statement I and Statement II are true.

Both Statement I and Statement II are false.

Statement I is true but Statement II is false.

Statement I is false but Statement II is true.

Heat and Thermodynamics PYQ — JEE Main 2022 · 28 Jun · Shift 1 · Q49 with Solution

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Correct answer: A

For an adiabatic change, $Q=0$ So from the first law, $\Delta U = Q - W = -W$ where $W$ is the work done by the gas.

For an ideal gas, $\Delta U = \mu C_v (T_2-T_1)$ Hence, $\mu C_v (T_2-T_1) = -W$ $W = -\mu C_v (T_2-T_1) = \mu C_v (T_1-T_2)$

Now use $\gamma = \frac{C_p}{C_v}, \qquad R=C_p-C_v$ So, $C_v = \frac{R}{\gamma-1}$

Substitute into $W$ : $W = -\mu \frac{R}{\gamma-1}(T_2-T_1) = \frac{\mu R (T_2-T_1)}{1-\gamma}$

This matches Statement I.

So, Statement I is true.

In an adiabatic process, if work is done on the gas, then compression occurs.

Since no heat enters or leaves, $Q=0$ If work is done on the gas, the internal energy increases, so temperature rises.

Equivalently, work done on gas means work done by gas is negative: $\Delta U = -W > 0$ Thus, $T_2>T_1$ So the temperature increases.

Therefore, Statement II is true.

Thus, the correct option is: $\boxed{A}$