$n$  mole of an ideal gas undergo a process$A\to B$ . Maximum temperature of the gas during the process is:

 

 

detailed solution

Correct option is B

Since P−V  graph of the process is a straight line and two points (V0,2P0)  and (2V0,P0)  are known, therefore its equation will be, (P−P0)=(2P0−P0)(V0−2V0)(V−2V0)=P0V0(2V0−V) ∴  P=3P0−P0VV0  (∵y=mx) According to equation for ideal gas,T=PVnR=(3P0−P0VV0)VnR =3P0V0V−P0V2nRV0  …(1) For T  to be maximum,dTdV=0 3P0V0−2P0V=0 or V=3V02  …(2) putting this value in equation (1), we get;Tmax=3P0V0(3V02)−P0(94V02)nRV0 =9P0V04nR