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Found in: Page 159

### An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279

# The formula for ${C}_{P}-{C}_{V}$ derived in the previous problem can also be derived starting with the definitions of these quantities in terms of U and H. Do so. Most of the derivation is very similar, but at one point you need to use the relation $P=-\left(\partial F/\partial V{\right)}_{T}$.

$\left({C}_{P}-{C}_{V}\right)=\left(T{\left(\frac{\partial S}{\partial V}\right)}_{T}{\left(\frac{\partial V}{\partial T}\right)}_{P}\right)$

See the step by step solution

## Given

$\left({C}_{P}-{C}_{V}\right)$

## Explanation

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

They are given by

${C}_{V}={\left(\frac{\partial U}{\partial T}\right)}_{V}\phantom{\rule{0ex}{0ex}}{C}_{P}={\left(\frac{\partial H}{\partial T}\right)}_{P}$

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

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

We get,

$dH={\left(\frac{\partial U}{\partial V}\right)}_{T}dV+{\left(\frac{\partial U}{\partial T}\right)}_{V}dT+pdV\phantom{\rule{0ex}{0ex}}=\left[{\left(\frac{\partial U}{\partial V}\right)}_{T}+P\right]dV+{\left(\frac{\partial U}{\partial T}\right)}_{V}dT$

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

${\left(\frac{\partial H}{\partial T}\right)}_{P}=\left[{\left(\frac{\partial U}{\partial V}\right)}_{T}+P\right]{\left(\frac{\partial V}{\partial T}\right)}_{P}+{\left(\frac{\partial U}{\partial T}\right)}_{V}...........................\left(1\right)$

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

${C}_{P}=\left[{\left(\frac{\partial U}{\partial V}\right)}_{T}+P\right]{\left(\frac{\partial V}{\partial T}\right)}_{P}+{C}_{V}.............................\left(2\right)$

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

${C}_{P}={\left(\frac{\partial \left(U-F\right)}{\partial V}\right)}_{T}{\left(\frac{\partial V}{\partial T}\right)}_{P}+{C}_{V}$

Now substitute TS for (U-F)

${C}_{P}=T{\left(\frac{\partial S}{\partial V}\right)}_{T}{\left(\frac{\partial V}{\partial T}\right)}_{P}+{C}_{{V}^{\infty }}$

So, the value of $\left({C}_{P}-{C}_{V}\right)=\left(T{\left(\frac{\partial S}{\partial V}\right)}_{T}{\left(\frac{\partial V}{\partial T}\right)}_{P}\right)$