Two moles of a monoatomic ideal gas occupy a volume V at `27^(@)C`. The gas is expanded adiabatically to a volume `2sqrt2V`. The final temperature is 150 K. What is the work done by the gas ? [R = 8.3J/K/mol]

Options

A. 1660 J
B. 3735 J
C. 2490 J
D. 1240 J

Answer

B

Explanation

`DeltaQ=DeltaU+DeltaW`, but for an adiabatic process
`DeltaQ=0" "thereforeDeltaW=-DeltaU" "...(1)`
For a monoatomic gas, `C_(V)=(3R)/2`
and `DeltaU=nC_(V)dT=nC_(V)(T_(2)-T_(1))`
In this case, `n=2,T_(2)=150andT_(1)=273+27=300K`
`thereforeDeltaU=2xx(3R)/2xx(150-300)`
`=3xx8.3xx(-150)=-3735J`
`thereforeDeltaW=-DeltaU=3735J`