QJEEM24Ph4a0f: JEE Main 2024 Physics - Class 11 JEEM24Ph4a0f Solution: A sample of 1 mole gas at temperature is adiabatically expanded …

For an adiabatic process, the work done by the gas can be found using the formula:

$W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}$

Given that the volume is doubled ( $V_2 = 2V_1$ ) and the adiabatic constant $\gamma = \frac{3}{2}$ , we can manipulate the ideal gas law and the adiabatic process relationship to find an expression for work done in terms of the initial conditions and $\gamma$ .

Recall, for an adiabatic process, $PV^{\gamma} = \text{constant}$ , so we can write:

$P_1V_1^{\gamma} = P_2V_2^{\gamma}$

Using the fact that $V_2 = 2V_1$ , we can express $P_2$ in terms of $P_1$ and $V_1$ as follows:

$P_1V_1^{\gamma} = P_2(2V_1)^{\gamma}$

$P_2 = P_1 \left( \frac{V_1}{2V_1} \right)^{\gamma}$

$P_2 = P_1 \left( \frac{1}{2} \right)^{\gamma}$

The work done then becomes:

$W = \frac{P_1 V_1 - P_1 \left( \frac{1}{2} \right)^{\gamma} \cdot 2V_1}{\gamma - 1}$

$W = P_1V_1 \frac{1 - \left( \frac{1}{2} \right)^{\gamma} \cdot 2}{\gamma - 1}$

Plugging in $\gamma = \frac{3}{2}$ , we get:

$W = P_1V_1 \frac{1 - \left( \frac{1}{2} \right)^{\frac{3}{2}} \cdot 2}{\frac{3}{2} - 1}$

$W = P_1V_1 \frac{1 - \sqrt{\frac{1}{2}} \cdot 2}{\frac{1}{2}}$

$W = 2P_1V_1 (1 - \sqrt{\frac{1}{2}})$

Since $P_1V_1 = nRT$ for 1 mole ( $n = 1$ ) of gas at temperature $T$ , we can further simplify:

$W = 2RT(1 - \sqrt{\frac{1}{2}})$

$W = 2RT(1 - \frac{1}{\sqrt{2}})$

$W = RT(2 - \sqrt{2})$

Therefore, the correct option is:

Option B $\mathrm{RT}[2-\sqrt{2}]$ .