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A quantity of 2 mole of helium gas unde...

A quantity of 2 mole of helium gas undergoes a thermodynamic process, in which molar specific heat capacity of the gas depends on absolute temperature` T` , according to relation:
`C=(3RT)/(4T_(0)`
where `T_(0)` is initial temperature of gas. It is observed that when temperature is increased. volume of gas first decrease then increase. The total work done on the gas until it reaches minimum volume is :-

A

`(3)/(2) RT_(0)`

B

`(3)/(4)RT_(0)`

C

`(3)/(8)RT_(0)`

D

`(3)/(10)RT_(0)`

Text Solution

Verified by Experts

The correct Answer is:
B
When it reaches minimum volume, the process can be treated isochoric.
`(3RT)/(4T_(0)) = (3R)/(2)`
` T = 2T_(0)`
` DeltaQ = DeltaU + W`
`int_(T_(0))^(2T_(0))(2)(3RT)/(4T_(0))dT = (2)(3R)/(2)(2T_(0)-T_(0))+W`
` (6R)/(4T_(0))[T^(2)/(2)]_(T_(0))^(2T_(0))= 3RT _(0)+W`
` W=(9RT_(0))/(4)-3RT_(0)`
` W=-(3RT_(0))/(4)`
So work done on the gas `(3RT_(0))/(4)`
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