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One mole of an ideal gas undergoes a pro...

One mole of an ideal gas undergoes a process `p=(p_(0))/(1+((V_(0))/(V))^(2))`. Here, `p_(0)` and `V_(0)` are constants. Change in temperature of the gas when volume is changed from `V=V_(0)` to `V=2V_(0)` is

A

`(3)/(4) (P_(0)V_(0))/(R)`

B

`(1)/(2) (P_(0)V_(0))/(R)`

C

`(1)/(4) (P_(0)V_(0))/(R)`

D

`(5)/(4) (P_(0)V_(0))/(R)`

Text Solution

Verified by Experts

The correct Answer is:
D
`V = V_(0) rArr P_(1) = (P_(0))/(2)`
`V = 2V_(0) rArr P_(2) = (7P_(0))/(8)`
`Delta T = T_(2) - T_(1) = (P_(2) V_(2) - P_(1) V_(1))/(nR)`
`= ((7P_(0))/(8) xx 2V_(0) - (P_(0))/(2) V_(0))/(nR) = (5P_(0)V_(0))/(4R)`
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