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

Answer

A

Explanation

The `P-V` equation for path AB shown in (figure)
`p= - ((P_(0))/(V_(0)))V+3P_(0)`....(i)
or `pV= - ((P_(0))/(V_(0)))V^(2)+3P_(0)V`
Now, `T=(pV)/(nR)= 1/(nR)(3P_(0)V-(P_(0))/(V_(0))V^(2))`....(ii)
For maximum temoperature, `(dT)/(dV)=0`
`d/(dv)(3P_(0)V-(P_(0)V_(2))/(V_(0)))=0`
`3P_(0)-(2P_(0))/(V_(0))V=0 or V=3/2V_(0)`
From (i) , `p= -((P_(0))/(V_(0)))3/2V_(0)+3P_(0)=3/2P_(0)`
From (ii), `T_(max)=(pV)/(nR)= (3/2P_(0)3/2V_(0))/(nR)`
`T_(max)= (9P_(0)V_(0))/(4nR)`