‘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. `9/4 (p_0 V_0)/(nR)`
B. `3/2 (P_0 V_0)/(nR)`
C. `9/2 (p_0 V_0)/(nR)`
D. `(9 P_0 V_0)/(nR)`

Answer

A

Explanation

As, T will be maximum temperature where product of `p_V` is maximum
Equation of line AB. We have
`y-y_1=(y_2-y_1)/(x_2-x_1)(x-x_1) implies p-p_0= (2p_0-p_0)/(v_0-2v_0)(v-2v_0)`
`implies p-p_0=(-p_0)/(v_0)(v-2v_0) implies p=(-p_0)/(v_0)v+3p_0`
`pv=(-p_0)/(v_0)v^2+3p_0v, nRT= (-p_0)/(v_0)v+3p_0v`
`T=(1)/(nR)((-p_0)/(v_0)v^2+3p_0v)`
for maximum temperature `(del T)/(del V)=0`
`(-p_0)/(v_0)(2v)+3p_0=0, " "(-p_0)/(v_0)(2v)=- 3p_0 implies V=(3)/(2)v_0`
(condition for maximum temperature)
Thus, the maximum temperature of the gasf during the process will be :
`T_(" max ")= (1)/(nR)((-p_0)/(v_0)xx(9)/(4)v_(0)^(2)+3p_0xx (3)/(2)v_0)`
`=(1)/(nR)(-(9)/(4)p_0v_0+(9)/(2)p_0v_0)=(9)/(4)(p_0v_0)/(nR)`.