Question

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`

Answer 1

Correct Answer - A
(a) Initially
`V_1=5.6l, T_1=273K, P_1=1atm,`
`gamma=5/3 (For monoatomic gas)`
The number of moles of gas is `n=(5.6l)/(22.4l)=1/4`
Finally (after adiabatic compression)
`V_2=0.7l`
For adiabatic compression `T_1V_1^(gamma-1)=T_2V_2^(gamma-1)`
`:. T_2=T_1((V_1)/(V_2))^(gamma-1)=T_1(5.6/0.7)^(5/3-1)=T_1(8)^(2//3)=4T_1`
We know that work done in adiabatic process is
`W=(nRDeltaT)/(gamma-1)=9/8RT_1`