The pressure and volume of an ideal gas are related as $\mathrm{PV}^{\frac{3}{2}}=\mathrm{K}$ (Constant). The work done when the gas is taken from state $A\left(P_1, V_1, T_1\right)$ to state $B\left(P_2, V_2, T_2\right)$ is :

<p>To find the work done by the gas when it goes from state A to state B, we can look at the definition of work done on or by a gas in a thermodynamic process. For a quasi-static process, the work done $W$ is given by the integral of the pressure $P$ with respect to the volume $V$:</p> <p>$W = \int_{V_1}^{V_2} P dV$</p> <p>Given the relationship $PV^{\frac{3}{2}} = K$, we can solve for $P$:</p> <p>$P = \frac{K}{V^{\frac{3}{2}}}$</p> <p>Now, substitute $P$ into the work integral and evaluate it:</p> <p>$$ W = \int_{V_1}^{V_2} \frac{K}{V^{\frac{3}{2}}} dV = K \int_{V_1}^{V_2} V^{-\frac{3}{2}} dV $$</p> <p>To integrate this, we&#39;ll use the power rule for integration. The integral of $V^{-\frac{3}{2}}$ is $-2V^{-\frac{1}{2}}$, so the work done is:</p> <p>$$ W = K \left[-2V^{-\frac{1}{2}}\right]_{V_1}^{V_2} = K \left(-2V_2^{-\frac{1}{2}} + 2V_1^{-\frac{1}{2}} \right) $$</p> <p>We can rewrite $V^{-\frac{1}{2}}$ as $\frac{1}{\sqrt{V}}$:</p> <p>$W = K \left(-2\frac{1}{\sqrt{V_2}} + 2\frac{1}{\sqrt{V_1}} \right)$</p> <p>Since $P_2V_2^{\frac{3}{2}} = K$ and $P_1V_1^{\frac{3}{2}} = K$, we can express $K$ in terms of $P_1$ and $V_1$ (or $P_2$ and $V_2$, but we’ll use $P_1$ and $V_1$ for now):</p> <p>$$ W = P_1V_1^{\frac{3}{2}} \left(-2\frac{1}{\sqrt{V_2}} + 2\frac{1}{\sqrt{V_1}} \right) $$</p> <p>$$ W = -2P_1V_1\sqrt{V_1}\frac{1}{\sqrt{V_2}} + 2P_1V_1\sqrt{V_1}\frac{1}{\sqrt{V_1}} $$</p> <p>$W = -2P_1V_1\frac{\sqrt{V_1}}{\sqrt{V_2}} + 2P_1V_1$</p> <p>Since $\frac{\sqrt{V_1}}{\sqrt{V_2}} = \sqrt{\frac{V_1}{V_2}}$, and recalling $P_2V_2^{\frac{3}{2}} = K$ once more:</p> <p>$W = -2P_1V_1\sqrt{\frac{V_1}{V_2}} + 2P_1V_1$</p> <p>$W = -2 \frac{P_1V_1^{\frac{3}{2}}}{\sqrt{V_2}} + 2P_1V_1$</p> <p>Putting $P_1V_1^{\frac{3}{2}}$ back as $K$:</p> <p>$W = -2\frac{K}{\sqrt{V_2}} + 2P_1V_1$</p> <p>$W = -2P_2V_2 \sqrt{V_2}\frac{1}{\sqrt{V_2}} + 2P_1V_1$</p> <p>$W = -2P_2V_2 + 2P_1V_1$</p> <p>So the work done is:</p> <p>$W = 2P_1V_1 - 2P_2V_2$</p> <p>Therefore, the correct answer matching the given options is:</p> <p>Option D: $2\left(\mathrm{P}_1 \mathrm{~V}_1-\mathrm{P}_2 \mathrm{~V}_2\right)$</p> <p><b>Shortcut :</b></p> <br/>For $\mathrm{PV}^{\mathrm{x}}=$ constant <br/><br/>If work done by gas is asked then <br/><br/>$$ \begin{aligned} \mathrm{W} & =\frac{\mathrm{nR} \Delta \mathrm{T}}{1-\mathrm{x}} \\\\ & \text { Here } \mathrm{x}=\frac{3}{2} \\\\ \therefore \mathrm{W} & =\frac{\mathrm{P}_2 \mathrm{~V}_2-\mathrm{P}_1 \mathrm{~V}_1}{-\frac{1}{2}} \\\\ = & 2\left(\mathrm{P}_1 \mathrm{~V}_1-\mathrm{P}_2 \mathrm{~V}_2\right) \end{aligned} $$