One mole an ideal gas whose adiabatic exponent equals `gamma` undergoes a process `p = p_0 + alpha//V`, where `p_0` and `alpha` are positive constants. Find :
(a) heat capacity of the gas as a function of its volume ,
(b) the internal energy of heat transferred to the gas, of its volume increased from `V_1` to `V_2`.

Explanation

Using `2.52`
(a) `C = C_V + (RT)/(V) (dV)/(dT) = C_V + (pdV)/(dT)` (for one mole of gas)
We have `p= p_0 + (alpha)/(V')` or, `(RT)/(V) = p_0 + (alpha)/(V), RT = p_0 V + alpha`
Therefore `R dT = p_0 dV`, So, `(dV)/(dT) = (R)/(p_0)`
Hence `C = C_V + (p_0 + (alpha)/(V)).(R)/(p_0) = (R)/(gamma - 1) + (1 + (alpha)/(p_0 V))R`
=`(R + (R)/(gamma - 1))+ (alpha R)/(p_0 V) = (gamma R)/(gamma - 1) +(alpha R)/(p_0 V)`
(b) Work done is given by
`A = int_(V_1)^(V_2) (p_0 + (alpha)/(V)) dV = p_0 (V_2 - V_1) + alpha 1n (V_2)/(V_1)`
`Delta U = C_V (T_2 - T_1) = C_V((p_2 V_2)/(R) -(p_1 V_1)/(R))` (for one mole)
=`(R)/((gamma - 1)R) (p_2 V_2 - p_1 V_1)`
=`(1)/(gamma - 1)[(p_0 + alpha V_2)V_2 -(p_0 + (alpha)/(V_1)) V_1] = (p_0(V_2 - V_1))/(gamma - 1)`
By the first law of thermodynamics `Q = Delta U + A`
`=p_0(V_2 - V_1)+ alpha 1n (V^2)/(V_1) + (p_0(V_2 - V_1))/((gamma - 1))`
=`(gamma p_0 (V_2 - V_1))/(gamma - 1)+ alpha 1n (V_2)/(V_1)`.