91Ó°ÊÓ

$\left(C_P-C_V\right)$

The specific heat of a substance can be of two types:
(i) specific heat at constant pressure C_P
(ii) specific heat at constant volume C_V

They are given by

$$\begin{gathered} C_V={\left(\frac{∂U}{∂T}\right)}_V \\ {C}_P={\left(\frac{∂H}{∂T}\right)}_P \end{gathered}$$

Where, U = internal energy, H = enthalpy, V = volume and P = pressure.

Lets consider U=U(V, T), and differentiate above expression with respect to T, we get
$dU={\left(\frac{∂U}{∂V}\right)}_{T}dV+{\left(\frac{∂U}{∂T}\right)}_{V}dT$

Similarly write expression for enthalpy H=U+P V.

Write the expression for d H at constant pressure d H=d U+P d V

Substitute $dU=\left({\left(\frac{∂U}{∂V}\right)}_{T}dV+{\left(\frac{∂U}{∂T}\right)}_{V}dT\right)$

We get,

$$\begin{gathered} dH={\left(\frac{∂U}{∂V}\right)}_{T}dV+{\left(\frac{∂U}{∂T}\right)}_{V}dT+pdV \\ =\left[{\left(\frac{∂U}{∂V}\right)}_T+P\right]dV+{\left(\frac{∂U}{∂T}\right)}_{V}dT \end{gathered}$$

Simplify, divide both sides of the above expression by d T,

${\left(\frac{∂H}{∂T}\right)}_P=\left[{\left(\frac{∂U}{∂V}\right)}_T+P\right]{\left(\frac{∂V}{∂T}\right)}_P+{\left(\frac{∂U}{∂T}\right)}_V...........................\left(1\right)$

Substitute $C_{P}for{\left(\frac{∂H}{∂T}\right)}_{P}and{C}_{V}for{\left(\frac{∂U}{∂T}\right)}_V$

$C_P=\left[{\left(\frac{∂U}{∂V}\right)}_T+P\right]{\left(\frac{∂V}{∂T}\right)}_P+C_V.............................\left(2\right)$

Substitute $-{\left(\frac{∂F}{∂V}\right)}_{T}forP$in equation (2)

$C_P={\left(\frac{∂(U-F)}{∂V}\right)}_T{\left(\frac{∂V}{∂T}\right)}_P+C_V$

Now substitute TS for (U-F)

$C_P=T{\left(\frac{∂S}{∂V}\right)}_T{\left(\frac{∂V}{∂T}\right)}_P+C_{V^{∞}}$

So, the value of $\left(C_P-C_V\right)=\left(T{\left(\frac{∂S}{∂V}\right)}_T{\left(\frac{∂V}{∂T}\right)}_P\right)$