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Helmholtz Free Energy

Suppose that $ T$ and $ V$ are the are the two independent parameters that specify the system. Because

$\displaystyle T dS = d(T S)-S dT,$ (6.100)

we can rewrite Equation (6.82) in the form

$\displaystyle dF = -S dT - p dV,$ (6.101)

where

$\displaystyle F = E - T S$ (6.102)

is termed the Helmholtz free energy.

Proceeding as before, we write

$\displaystyle F= F(T,V),$ (6.103)

which implies that

$\displaystyle dF= \left(\frac{\partial F}{\partial T}\right)_V dT + \left(\frac{\partial F}{\partial V}\right)_T dV.$ (6.104)

Comparison of Equations (6.101) and (6.104) yields

$\displaystyle \left(\frac{\partial F}{\partial T}\right)_V$ $\displaystyle = -S,$ (6.105)
$\displaystyle \left(\frac{\partial F}{\partial V}\right)_T$ $\displaystyle = -p.$ (6.106)

We also know that

$\displaystyle \frac{\partial^{ 2} F}{\partial V \partial T} = \frac{\partial^{ 2}F}{\partial T \partial V},$ (6.107)

or

$\displaystyle \left(\frac{\partial}{\partial V}\right)_T\left(\frac{\partial F}...
...rac{\partial}{\partial T}\right)_V\left(\frac{\partial F}{\partial V}\right)_T.$ (6.108)

Thus, it follows from Equations (6.105) and (6.106) that

$\displaystyle \left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial p}{\partial T}\right)_V.$ (6.109)


next up previous
Next: Gibbs Free Energy Up: Classical Thermodynamics Previous: Enthalpy
Richard Fitzpatrick 2016-01-25