One mole of an ideal gas with adiabatic exponent `gamma` undergoes the process
(a) `P=P_0+(alpha)/(V)`
(b) `T=T_0+alphaV`
Find Molar heat capacity of the gas as a function of its volume.

Explanation

From first law `dQ=dU=dW`
`nCdT=nC_VdT+PdV`
`C=C_V+(P)/(n)(dV)/(dT)`
(a) `P=P_0+(alpha)/(V)`
`(nRT)/(V)=P_0+(alpha)/(V)`
`nRT=P_0V+alpha`
`nRdT=P_0dV+0`
`(dV)/(dT)=(nR)/(P_0)`
`C=C_V+(P)/(n)(dV)/(dT)`
`=(R)/(gamma-1)+((P_0+(alpha)/(V)))/(n).(nR)/(P_0)`
`=(R)/(gamma-1)+((P_0+(alpha)/(V))R)/(P_0)`
`=(R)/(gamma-1)+R+(alphaRV)/(P_0)`
`=(gammaR)/(gamma-1)+(alphaRV)/(P_0)`
(b) `T=T_0+alphaV`
`dT=alphadVimplies(dV)/(dT)=(1)/(alpha)`
`C=C_V+(P)/(n)(dV)/(dT)`
`=(R)/(gamma-1)+(RT)/(V).(1)/(alpha)`
`=(R)/(gamma-1)+(R)/(alphaV)(T_0+alphaV)`
`=(R)/(gamma-1)+(RT_0)/(alphaV)+R`
`=(gammaR)/(gamma-1)+(RT_0)/(alphaV)`