IIT JEE Main 2016 Physics Question No 12

IIT JEE Main 2016 Physics Question No 12

‘n’ moles of an ideal gas undergoes a process A->B as shown in the figure. The maximum temperature of the gas during the process will be:

IIT JEE Main 2016 Physics Question No 12

(1) 9P0V0/(2nR)
(2) 9P0V0/(nR)
(3) 9P0V0/(4nR)
(4) 3P0V0/(2nR)

Solution

The equation of the process is y = mx + c
P = -(P0/V0)V + 3P0
Now PV = nRT
therefore P = nRT/V
=> (nRT/V)=-(P0/V0)V + 3P0
=> nRT=-(P0/V0)V2 + 3P0V
=> T=-(P0/nRV0)V2 + (3P0/nR)V                  … (1)
Now for maximum T,
dT/dV = 0                                                       … (2)
here dT/dV = -(2P0/nRV0)V + (3P0/nR)       … (3)
from (2) & (3) we get,
V=(3/2)V0
putting in (1) we get
T=-(P0/nRV0)(9/4V02) + (3P0/nR)(3/2)V0
=> T=(9/2)(P0V0/nR)-(9/4)(P0V0/nR)
=> Tmax = (9/4)(P0V0/nR)

Answer

Correct answer to given problem No 12 of IIT JEE Main Physics paper of 2016 is Tmax = (9/4)(P0V0/nR) i.e. option number 3.