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5.6 liter of helium gas at STP is adiaba...

5.6 liter of helium gas at STP is adiabatically compressed to 0.7 liter. Taking the initial temperature to be `T_1,` the work done in the process is

A

`9/8RT_1`

B

`3/2RT_1`

C

`15/8RT_1`

D

`9/2RT_1`

Text Solution

Verified by Experts

The correct Answer is:
A
Number of moles of He `=5.6//22.4=1//4`
Now `T(5.6)^(gamma-1)=T_(2)(0.7)^(gamma-1)`
`T_(1)=T_(2)((1)/(8))^(2//3) implies 4T_(1)=T_(2)`
Work done `=-(nR[T_(2)-T_(1)])/(gamma-1)=((1)/(4)R[3T_(1)])/((2)/(3))=-(9)/(8)RT_(1)`
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