‘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 :

Options

A. `(P_(0) V_(0))/(nR)`
B. `(3 P_(0) V_(0))/(nR)`
C. `(4 P_(0) V_(0))/(nR)`
D. `(2 P_(0) V_(0))/(nR)`

Answer

C

Explanation

c. `(P - P_(1))/(P_(1) - P_(2)) = (V - V_(2))/(V_(1) - V_(2))`
`(P - P_(1)) (V_(1) - V_(2)) = (V_(1) - V_(2)) = (V - V_(1)) (P_(1) - P_(2))`
`(P - 3 P_(0)) (V_(0) - 3 V_(0)) = (V - V_(0)) (3P_(0) - P_(0))`
`(P - 3 P_(0)) (- 2 V_(0)) = (V - V_(0)) (2 P_(0))`
`- 2 V_(0) P + 6 P_(0) V_(0) = 2 V P_(0) - 2 P_(0) V_(0)`
`2 V P_(0) + 2 V_(0) P - 8 P_(0) V_(0) = 0`
`V P_(0) + (V_(0) nRT)/(V) - 4 P_(0) V_(0) = 0`
`V^(2) P_(0) - 4 P_(0) V_(0) + V_(0) nRT = 0`
`T = (P_(0) (V^(2) + 4 V V_(0)))/(V_(0) nR)`
For maximum of minimum value of `T`,
`(dT)/(dV) = - 2V + 4 V_(0) = 0 implies V = 2 V_(0)`
`(d^(2) T)/(dV^(2)) = -2`
It is negative so `T` is maximum at
`V = 2 V_(0)`
`T_(max) = (P_(0) (- V_(0)^(2) + 8 V_(0)^(2)))/(V_(0) nR) = (4 P_(0) V_(0))/(nR)`