n' moles of an ideal gas undergoes a process `AtoB` as shown in the figure. The maximum temperature of the gas during the process will be:

Options

A. `(9P_0V_0)/(2nR)`
B. `(9P_0V_0)/(nR)`
C. `(9P_0V_0)/(4nR)`
D. `(3P_0V_0)/(2nR)`

Answer

C

Explanation

(c)
`P=(-P_0)/(V_0)v+3P`
`[slope=(-P_0)/(V_0), c=3P_0]`
`PV_0+P_0V=3P_0V_0…..(i)`
But `pv=nRT`
`:. P=(nRT)/v…(ii)`
From (i) & (ii) `(nRT)/v V_0+P_0V=3P_0V_0`
`:. nRTV_0+P_0V^2=3P_0V_0….(iii)`
For temperature to be maximum `(dT)/(dv)=0`
Differentiating e.q. (iii) by 'v' we get
`nRV_0(dT)/(dv)+P_0(2v)=3P_0V_0`
`nRV_0(dT)/(dv)=3P_0V_0-2P_0V`
`(dT)/(dv)=(3P_0V_0-2P_0V)/(nRV_0)=0`
`V=(3V_0)/2 :. p=(3P_0)/2` [from(i)]
`:. T_(max)=(9P_0V_0)/(4nR)` [From (iii)]