Specific heat

For thermally perfect and calorically perfect gases

$$\begin{array}{cc}& C_p=\frac{dh}{dT}\\ & C_v=\frac{de}{dT}\end{array}$$

From the definition of enthalpy and the equation of state $p=\rho RT$

$$h=e+\frac{p}{\rho }=e+RT$$

Differentiate Eqn. \ref{eq:enthalpy} with respect to temperature gives

$$\frac{dh}{dT}=\frac{de}{dT}+\frac{d(RT)}{dT}$$

Inserting the specific heats gives

$$C_p=C_v+R$$

Dividing Eqn. \ref{eq:specificheat:b} by $C_v$ gives

$$\frac{C_p}{C_v}=1+\frac{R}{C_v}$$

Introducing the ratio of specific heats defined as

$$\gamma =\frac{C_p}{C_v}$$

Now, inserting Eqn. \ref{eq:gamma} in Eqn. \ref{eq:specificheat:c} gives

$$C_v=\frac{R}{\gamma -1}$$

In the same way, dividing Eqn. \ref{eq:specificheat:b} with $C_p$ gives

$$1=\frac{C_v}{C_p}+\frac{R}{C_p}=\frac{1}{\gamma }+\frac{R}{C_p}$$

and thus

$$C_p=\frac{\gamma R}{\gamma -1}$$