An ideal gas is taken from the state A (pressure P, volume V) to the state B (pressure P/2, volume 2V along a straight line path in the P-V diagram. Select the correct statement(s) from the following:

i. Work done = area under the curve,

   ${\mathrm{W}}_1=\left(\mathrm{P}+\frac{\mathrm{P}}{2}\right)(2\mathrm{V}-\mathrm{V})=\frac{3}{2}\mathrm{PV}$

Work done under isothermal process,

$\begin{array}{l} {\mathrm{W}}_2=\mathrm{RT}\times 2.3026\log \left(\frac{2\mathrm{V}}{\mathrm{V}}\right)\\ =0.693\mathrm{RT}=0.693\mathrm{PV}\\ \therefore {\mathrm{W}}_1>{\mathrm{W}}_2,\text{ i.e., option (a) is correct. }\end{array}$

ii. Let P₀ and V₀ be the intercepts on the P and V axes. Now the equation of straight line would be

         $\begin{array}{l}\mathrm{P}=-\frac{{\mathrm{P}}_0}{{\mathrm{V}}_0}\times \mathrm{V}+{\mathrm{P}}_0\\ (\because \mathrm{y}=\mathrm{mx}+\mathrm{c})\end{array}$

Further   $\mathrm{PV}=\mathrm{RT}\text{ or }\mathrm{P}=\frac{\mathrm{RT}}{\mathrm{V}}$

(This represents parabola, so option (b) is also correct.)