Correct Answer - A::B::C::D
(A) Process is isobaric
`C_(v)=(5R)/2, C_(P)=(7R)/2`
`|q_(p)|nC_(P)DeltaT=(7RT_(o))/2`
`|DeltaU|=nC_()DeltaT=(5RT_(o))/2`
`|W|=RT_(o)`
(B) Process is isochoric, `C_(v)=(3R)/2`
`|q_(v)|=|DeltaU|=(3RT_(o))/2`
`|W|=0`
(C) `C_(v)=R/(gamma-1)=R/(1.5-1)=2R`
`VT=` constant `PV^(2)=` constant `P/(P^(2))=` constant
`V+(TdV)/(dT)=0, (dV)/(dT)=-V/t, P/n (dV)/(dT)=-(PV)/(nT)=-R`
`|W|=RT_(o), |DeltaU|=2RT_(o), |q|=RT_(o)`
(D) `gamma=1+2/4=3/2`
`C_(v)=r/(gamma-1)=2R`
`PT=` constannt `P^(2)V=` constant
`(P^(2))/(rho)=` constant
`|DeltaU|=2RT_(o)`