One mole of an ideal monatomic gas undergoes a linear process from A to B, in which pressure P and its volume V change as shown in figure, where $V_0=8$ litres. As the volume of the gas is increased, in some range of volume the gas expands with absorbing the heat (the endothermic process); in the other range the gas emits the heat (the exothermic process). Then find the volume (in litres) after which if the volume of gas is further increased the given process switches from endothermic to exothermic.

 
$Q=\Delta U+W$
     $p=\left(\frac{-p_0}{V_0}\right)V+p_0$  ---(1)
∴ at point ‘1’
$p_1=\left(\frac{-p_0}{V_0}\right)V_1+p_0$  …..(2)
Therefore, for the process from the initial state with  to the state with P,V,T the heat given to system is
  $$\begin{gathered} \text{Q }=(\text{3}/\text{2})\text{ R }(\text{T }–{\text{ T}}_{\text{1}})+(\text{1}/\text{2})(\text{P }+{\text{ P}}_{\text{1}})(\text{V }–{\text{ V}}_{\text{1}}) \\ = \frac{3}{2} (\text{PV }–{\text{ P}}_{\text{1}}{\text{V}}_{\text{1}})+ \frac{1}{2}(\text{PV }+{\text{ P}}_{\text{1}}\text{V }+{\text{ PV}}_{\text{1}}–{\text{ P}}_{\text{1}}{\text{V}}_{\text{1}}) \mathrm{....}\left(\text{iii}\right) \\ =\text{ 2PV }+\frac{1}{2}{\text{P}}_{\text{1}}\text{V }–\frac{1}{2}{\text{PV}}_{\text{1}}–{\text{ 2P}}_{\text{1}}{\text{V}}_{\text{1}} \end{gathered}$$  
from equation (i), (ii) and (iii), we get
 $Q=2\frac{P_0}{V_0}{V}^2+\frac{5}{2}{P}_{0}V-2P_0{V}_1(\frac{5}{4}-\frac{V_1}{V_0})$
The process switches from endothermic to exothermic as $\frac{dQ}{dV}$  changes from positive to negative,  that is at   $\frac{dQ}{dV}=0$ . Solving we get $V=\frac{5}{8},{\text{ V}}_o=5$