‘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. `(9P_(0)V_(0))/(nR)`
B. `(9P_(0)V_(0))/(4nR)`
C. `(3P_(0)V_(0))/(2nR)`
D. `(9P_(0)V_(0))/(2nR)`

Answer

B

Explanation

T will be max where product of `PV` is max.

equation of line `P = (-P_(0))/(V_(0))V + 3P_(0)`
`PV = (-P_(0))/(V_(0))V^(2) + 3P_(0)V = x` (says)
`{:((dx)/(dV)=0rArrV=(3V_(0))/(2)),(" "rArr P=(3P_(0))/(2)):}}`here `PV "product"` is max.
`implies T = (PV)/(nR) = (9)/(4) (P_(0)V_(0))/(nR)`
Alternative

Since initial and final temperature are equal hence maximum temperature is at middle of line. `PV = nRT`
`((3P_(0))/(2))(3P_(0))/(2)))/(nR) = T_("max") implies (9P_(0)V_(0))/(4nR) = T_("max").`