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

Answer

A

Explanation

Here, `V_(1)= 5.6 litre, V_(2)= 0.7 litre, W=?`
Helium is a monoatomic gas.
For helium, `gamma= 5//3`
For an adiabatic process,
`T_(1)V_(1)^(gamma-1)= T_(2)V_(2)^(gamma-1) or T_(2)=T_(1)((V_(1))/(V_(2)))^(gamma-1)`
or `T_(2)=T_(1)((5.6)/(0.7))^((5/3-1)) = T_(1)(8)^(2//3)= 4T_(1)`
No. of moles of helium , `n=(5.6 litre)/(22.4 litre)= 1/4`
Work done during adiabatic process is
`W=(nR[T_(1)-T_(2)])/((gamma-1))= (1/4R[T_(1)-4T_(1)])/(5/3-1)= -9/8 RT_(1)`
Here -ve sign shows that work is done on the gas.