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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 + ((2 V_(0))/(V))^(2)]^(-1)`, where `P_(0) V_(0)` are constants. Change in temperature of the gas when volume is changed from `V = V_(0) to V = 2 V_(0)` is:

A

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

B

`(11R_(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
At `V=-V_(0)'P=(P_(0))/(2)`
`therefore T_(i)=(PV)/(nR)=((P_(0))/(2)(V_(0)))/(R)=(P_(0)V_(0))/(2R)(n=1)`
and at `V=2V_(0),P=(4P_(0))/(5)`
`therefore T_(F)=(PV)/(nR)=((2V_(0))((4P_(0))/(5)))/(R)=(8P_(0)V_(0))/(5R)`
`therefore DeltaT=T_(f)-T_(i)=((8)/(5)-(1)/(2))(P_(0)V_(0))/(R)=(11P_(0)V_(0))/(10R)`
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