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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

`-(2P_(0)V_(0))/(5R)`

B

`(11P_(0)V_(0))/(10R)`

C

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

D

`P_(0)V_(0)`

Text Solution

Verified by Experts

The correct Answer is:
B
Ideal gas equation : `PV= nRT` So at `V = V_(0)`, `RT_(1) = ((P_(0))/(2)) (V_(0))` and at `V = 2V`,
`RT_(2) = ((4P_(0))/(5)) (2V_(0)) implies T_(2)- T_(1) = (11P_(0) V_(0))/(10R)`
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