A sample of 1 mole gas at temperature $T$ is adiabatically expanded to double its volume. If adiab constant for the gas is $\gamma=\frac{3}{2}$, then the work done by the gas in the process is :

<p>For an adiabatic process, the work done by the gas can be found using the formula:</p> <p>$W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}$</p> <p>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 $\gamma$.</p> <p>Recall, for an adiabatic process, $PV^{\gamma} = \text{constant}$, so we can write:</p> <p>$P_1V_1^{\gamma} = P_2V_2^{\gamma}$</p> <p>Using the fact that $V_2 = 2V_1$, we can express $P_2$ in terms of $P_1$ and $V_1$ as follows:</p> <p>$P_1V_1^{\gamma} = P_2(2V_1)^{\gamma}$</p> <p>$P_2 = P_1 \left( \frac{V_1}{2V_1} \right)^{\gamma}$</p> <p>$P_2 = P_1 \left( \frac{1}{2} \right)^{\gamma}$</p> <p>The work done then becomes:</p> <p>$$ W = \frac{P_1 V_1 - P_1 \left( \frac{1}{2} \right)^{\gamma} \cdot 2V_1}{\gamma - 1} $$</p> <p>$W = P_1V_1 \frac{1 - \left( \frac{1}{2} \right)^{\gamma} \cdot 2}{\gamma - 1}$</p> <p>Plugging in $\gamma = \frac{3}{2}$, we get:</p> <p>$$ W = P_1V_1 \frac{1 - \left( \frac{1}{2} \right)^{\frac{3}{2}} \cdot 2}{\frac{3}{2} - 1} $$</p> <p>$W = P_1V_1 \frac{1 - \sqrt{\frac{1}{2}} \cdot 2}{\frac{1}{2}}$</p> <p>$W = 2P_1V_1 (1 - \sqrt{\frac{1}{2}})$</p> <p>Since $P_1V_1 = nRT$ for 1 mole ($n = 1$) of gas at temperature $T$, we can further simplify:</p> <p>$W = 2RT(1 - \sqrt{\frac{1}{2}})$</p> <p>$W = 2RT(1 - \frac{1}{\sqrt{2}})$</p> <p>$W = RT(2 - \sqrt{2})$</p> <p>Therefore, the correct option is:</p> <p>Option B $\mathrm{RT}[2-\sqrt{2}]$.</p>