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

Answer

B

Explanation

b. Since `P - V` graph of the process is a straight line and two point `(V_(0), 2 P_(0))` and `(2 V_(0), P_(0))` are known, its equation will be
`(P - P_(0)) = ((2 P_(0) - P_(0)))/((V_(0) - 2 V_(0))) (V - 2 V_(0)) = (P_(0))/(V_(0)) (2 V_(0) - V)`
`:. P = 3 P_(0) - (P_(0) V)/(V_(0))`
According to equation for ideal gas,
`T = (pV)/(nR)`
`= (3 P_(0) - (P_(0) V)/(V_(0))) (V)/(nR)`
`= (3 P_(0) V_(0) V - P_(0) V^(2))/(nRV_(0))`
For `T` to be maximum, `(dT)/(dV) = 0`
`3 P_(0) V_(0) - 2 P_(0) V = 0`
or `V = (3 V_(0))/(2)`Putting ths value in Eq. (i) we get
`T_(max) = (3 P_(0) V_(0) ((3 V_(0))/(2)) - P_(0) ((9)/(4) V_(0)^(2)))/(nRV_(0)) = (9 P_(0) V_(0))/(4 nR)`