Correct Answer - A::B::C
(a) In case of adiabatic change
`TV^(gamma-1)=` constant
So that
`T_1V_1^(gamma-1)=T_2V_2^(gamma-1)` with `gamma=(5/3)`
i.e. `300xxV^(2//3)=T(2V)^(2//3)`
or `T=(300)/((2)^(2//3))=189K`
(b) As `DeltaU=nC_VDeltaT=n(3/2R)DeltaT
So, `DeltaU=2xx(3/2)xx8.31(189-300)`
`=-2767.23J`
Negative sign means internal energy will decrease.
(c) According to first law of thermodynamics
`Q=DeltaU+DeltaW`
And as for adiabatic change `DeltaQ=0`
`DeltaW=-DeltaU=2767.23J`