Physics - Heat & Thermo Question with Solution | TestHub

Temperature$T$will be maximum where product of$PV$is max equation of line

$P=-\frac{P_0}{V_0}V+3P_0$

Multiplying both sides by$V$,

$PV=-\frac{P_0}{V_0}{V}^2+3P_{0}V$

For maxima,$\frac{d(PV)}{dV}=0$

$\Rightarrow \frac{-2P_0}{V_0}V+3P_0=0\Rightarrow V=\frac{3V_0}{2}$

$\text{and} P=\frac{3P_0}{2}$

$\Rightarrow T=\frac{PV}{nR}=\frac{9P_0{V}_0}{\mathrm{4\; n}R}$

Alternate method :

Since initial and final temperature are equal.

Hence maximum temperature is at middle of line.

$PV=nR{T}_{\max }$

$\Rightarrow T_{\max }=\frac{\left(\frac{3P_0}{2}\right)\left(\frac{3V_0}{2}\right)}{nR}=\frac{9P_0{V}_0}{4nR}$