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 $$ V $$:

$$ W = \int_{V_1}^{V_2} P dV $$

Given the relationship $$ PV^{\frac{3}{2}} = K $$, we can solve for $$ P $$:

$$ P = \frac{K}{V^{\frac{3}{2}}} $$

Now, substitute $$ P $$ into the work integral and evaluate it:

$$ W = \int_{V_1}^{V_2} \frac{K}{V^{\frac{3}{2}}} dV = K \int_{V_1}^{V_2} V^{-\frac{3}{2}} dV $$

To integrate this, we'll use the power rule for integration. The integral of $$ V^{-\frac{3}{2}} $$ is $$ -2V^{-\frac{1}{2}} $$, so the work done is:

$$ W = K \left[-2V^{-\frac{1}{2}}\right]_{V_1}^{V_2} = K \left(-2V_2^{-\frac{1}{2}} + 2V_1^{-\frac{1}{2}} \right) $$

We can rewrite $$ V^{-\frac{1}{2}} $$ as $$ \frac{1}{\sqrt{V}} $$:

$$ W = K \left(-2\frac{1}{\sqrt{V_2}} + 2\frac{1}{\sqrt{V_1}} \right) $$

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):

$$ W = P_1V_1^{\frac{3}{2}} \left(-2\frac{1}{\sqrt{V_2}} + 2\frac{1}{\sqrt{V_1}} \right) $$

$$ W = -2P_1V_1\sqrt{V_1}\frac{1}{\sqrt{V_2}} + 2P_1V_1\sqrt{V_1}\frac{1}{\sqrt{V_1}} $$

$$ W = -2P_1V_1\frac{\sqrt{V_1}}{\sqrt{V_2}} + 2P_1V_1 $$

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:

$$ W = -2P_1V_1\sqrt{\frac{V_1}{V_2}} + 2P_1V_1 $$

$$ W = -2 \frac{P_1V_1^{\frac{3}{2}}}{\sqrt{V_2}} + 2P_1V_1 $$

Putting $$ P_1V_1^{\frac{3}{2}} $$ back as $$ K $$:

$$ W = -2\frac{K}{\sqrt{V_2}} + 2P_1V_1 $$

$$ W = -2P_2V_2 \sqrt{V_2}\frac{1}{\sqrt{V_2}} + 2P_1V_1 $$

$$ W = -2P_2V_2 + 2P_1V_1 $$

So the work done is:

$$ W = 2P_1V_1 - 2P_2V_2 $$

Therefore, the correct answer matching the given options is:

Option D: $$ 2\left(\mathrm{P}_1 \mathrm{~V}_1-\mathrm{P}_2 \mathrm{~V}_2\right) $$

Shortcut :

For $\mathrm{PV}^{\mathrm{x}}=$ constant

If work done by gas is asked then

$$ \begin{aligned} \mathrm{W} & =\frac{\mathrm{nR} \Delta \mathrm{T}}{1-\mathrm{x}} \\\\ & \text { Here } \mathrm{x}=\frac{3}{2} \\\\ \therefore \mathrm{W} & =\frac{\mathrm{P}_2 \mathrm{~V}_2-\mathrm{P}_1 \mathrm{~V}_1}{-\frac{1}{2}} \\\\ = & 2\left(\mathrm{P}_1 \mathrm{~V}_1-\mathrm{P}_2 \mathrm{~V}_2\right) \end{aligned} $$