One mole of an ideal gas with heat capacity at constant pressure `C_p` undergoes the process `T = T_0 + alpha V`, where `T_0` and `alpha` are constants. Find :
(a) heat capacity of the gas as a function of its volume ,
(b) the amount of heat transferred to the gas, if its volume increased from `V_1` to `V_2`.

Explanation

(a) Heat capacity is given by
`C = C_V + (RT)/(V)(dV)/(dT) ` (see solution of `2.52`)
We have `T = T_0 + alpha V` or, `V = (T)/(alpha) - (T_0)/(alpha)`
After differentiating, we get, `(dV)/(dT) = (1)/(alpha)`
Hence `C = C_V + (RT)/(V).(1)/(alpha) = (R)/(gamma - 1) +(R(T_0 + alpha V))/(V) .(1)/(alpha)`
=`(R)/(gamma - 1)+ R((T_0)/(alpha V) + 1) = (gamma R)/(gamma - 1)+(RT_0)/(alpha V) = C_V + (RT)/(alpha V) = V_p + (RT_0)/(alpha V)`
(b) Given `T = T_0 + alpha V`
As `T = (pV)/( R)` for one mole of gas
`p = (R)/(V) (T_0 + alpha V) = (RT)/(V) = alpha R`
Now `A = int_(V_1)^(V_2) pdV = int_(V_1)^(V_2) ((RT_0)/(v) + alpha R) dV` (for one mole)
=`RT_0 1n(V_2)/(V_1) + alpha (V_2 - V_1)`
`Delta U = C_v (T_2 - T_1)`
=`C_V[T_0 + alpha V_2 - T_0 alpha V_1] = alpha C_V (V_2 - V_1)`
By the first law of thermodynamics `Q = Delta U + A`
=`(alpha R)/(gamma - 1) (V_2 - V_1) + RT_0 1n (V_2)/(V_1) + alpha R (V_2 - V_1)`
=`alpha R(V_2 - V_1)[1 + (1)/(gamma - 1)] + RT_0 1n (V_2)/(V_1)`
. =`alpha C_p(V_2 - V_1) + RT_0 1n (V_2)/(V_1)`
=`alpha C_p (V_2 - V_1) + RT_0 1n (V_2)/(V_1)`.