## 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:

(1) 9P_{0}V_{0}/(2nR)

(2) 9P_{0}V_{0}/(nR)

(3) 9P_{0}V_{0}/(4nR)

(4) 3P_{0}V_{0}/(2nR)

#### Solution

The equation of the process is y = mx + c

P = -(P_{0}/V_{0})V + 3P_{0}

Now PV = nRT

therefore P = nRT/V

=> (nRT/V)=-(P_{0}/V_{0})V + 3P_{0}

=> nRT=-(P_{0}/V_{0})V^{2} + 3P_{0}V

=> T=-(P_{0}/nRV_{0})V^{2} + (3P_{0}/nR)V … (1)

Now for maximum T,

dT/dV = 0 … (2)

here dT/dV = -(2P_{0}/nRV_{0})V + (3P_{0}/nR) … (3)

from (2) & (3) we get,

V=(3/2)V_{0}

putting in (1) we get

T=-(P_{0}/nRV_{0})(9/4V_{0}^{2}) + (3P_{0}/nR)(3/2)V_{0}

=> T=(9/2)(P_{0}V_{0}/nR)-(9/4)(P_{0}V_{0}/nR)

=> T_{max} = (9/4)(P_{0}V_{0}/nR)

#### Answer

Correct answer to given problem No 12 of IIT JEE Main Physics paper of 2016 is T_{max} = (9/4)(P_{0}V_{0}/nR) i.e. option number 3.