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5.15

Expert-verifiedFound in: Page 159

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 $P=-(\partial F/\partial V{)}_{T}$.

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

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

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

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

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)

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

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

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