Onne mole of an ideal gas is taken from state A (pressure P, volume V) to state B (pressure P/2, Volume 2 V) along a straight line in the P-V diagram as shown in the figure. Then.

 

Work done in process A to Bis

$W_1=\text{ area of trapezium }ABCD$

     $\begin{array}{r}=\left(P+\frac{P}{2}\right)V=\frac{3PV}{2}\\ \end{array}$

If the process A to B were isothermal, the work done would be 

$W_2=RT{\log }_e\left(\frac{V_2}{V_1}\right)=RT{\log }_e\left(2\right)=0.69RT$

Thus, $W_1>W_2.$, So choice (a) is correct. 

The isothermal curve passes through the point A and B and from symmetry is is clear that the given process deviates the most from the isothermal process at the mid point of AB i.e. at volume $\frac{3V}{2} and at pressure \frac{3P}{4}.$

Therefore $T_{max }=\frac{{\displaystyle \frac{3P}{4}}.{\displaystyle \frac{3V}{2}}}{R}=\frac{9PV}{8R}$