'n' moles of an ideal gas undergoes a process $A \rightarrow B$ as shown in the figure. The maximum temperature of the gas during the process will be :

(D) $\frac{9 P_{0} V_{0}}{4 n R}$

(A) $\frac{3 P_{0} V_{0}}{2 n R}$

(B) $\frac{9 P_{0} V_{0}}{2 n R}$

(C) $\frac{9 P_{0} V_{0}}{n R}$

JEE Main · 2016 · 03 Apr 2016 · Offline · Heat & Thermodynamics

'n' moles of an ideal gas undergoes a process

A \to B

as shown in the figure. The maximum temperature of the gas during the process will be :

Answer: (D)

On the

P

V

diagram the process is the straight line from

(V_{0}, 2 P_{0})

(2 V_{0}, P_{0})

, with equation

P = 3 P_{0} - \frac{P _{0}}{V _{0}} V

. The temperature follows from

P V = n R T

, so

T = \frac{P V}{n R} = \frac{P _{0} V ( 3 V _{0} - V )}{n R V _{0}}

, a downward parabola in

V

. It peaks where

\frac{d T}{d V} = 0

, that is

3 V_{0} - 2 V = 0

, giving

V = \frac{3 V _{0}}{2}

and

P = \frac{3 P _{0}}{2}

. The maximum temperature is

T_{m a x} = \frac{1}{n R} \cdot \frac{3 P _{0}}{2} \cdot \frac{3 V _{0}}{2} = \frac{9 P _{0} V _{0}}{4 n R}

JEE Main · 2016 · 03 Apr 2016 · Offline · Single correct option