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

Options

A. `(9)/(8)RT_(1)`
B. `(3)/(2)RT_(1)`
C. `(15)/(8)RT_(1)`
D. `(9)/(2)RT_(1)`

Answer

A

Explanation

Number of moles of `He = (1)/(4)`
Now `T_(1)(5.6)^(gamma-1) = T(2) (0.7)^(gamma-1), T_(1) = T_(2) ((1)/(8))^(2//3)`
`4T_(1) = T_(2)`, Work done `=- (nR[T_(2)-T_(1)])/(gamma-1)`
`=- ((1)/(4)R[3T_(1)])/((2)/(3))=-(9)/(8)RT_(1)`