Is there a way to calculate the coefficient of thermal expansion ($\alpha$) from Helmholtz free energy (A) ?
I use the definition $\alpha = \frac{1}{V}(\frac{\partial V}{ \partial T})_p$.
With Maxwell relations, thermal expansion can be written as a function of Gibbs free energy (G) .
$\alpha = \frac{1}{V} \frac{\partial^{2} G}{\partial T \partial p}$
I want to calculate the coefficient of thermal expansion from Helmholtz free energy.
What i tried:
$G(P) = A(V(P)) + P V(P)$
$\frac{\partial G}{\partial p} = \frac{\partial A}{\partial V}\frac{\partial V}{\partial p} + V + p \frac{\partial V}{\partial p} $
With $\frac{\partial A}{\partial V} = -p$
$\frac{\partial G}{\partial p} = – p \frac{\partial V}{\partial p} + V – p \frac{\partial V}{\partial p} = V$
$\frac{1}{V}\frac{\partial^{2} G}{\partial T \partial p} = \frac{1}{V} \frac{\partial V}{\partial T}$
I could show the relation, but the term with A was cancelled. Is there a way to calculate $\alpha$ directly from A ?
Thanks in advance,
-Edit:
Did i find another way?
Maxwell relations:
$(\frac{\partial V}{\partial T})_p = -(\frac{\partial S}{\partial p})_T$ and $-S = (\frac{\partial A}{\partial T})_V$
$(\frac{\partial V}{\partial T})_p =\frac{\partial}{\partial p} (\frac{\partial A}{\partial T})_V)_T = \frac{\partial}{\partial T} (\frac{\partial A}{\partial V}\frac{\partial V}{\partial p}) $
Or did I miss something ?
Best Answer
Kind of, in differential form we have: $$dA=-SdT-pdV$$ If we Sub in: $$dV=(\frac{\partial V}{\partial T})_p dT+(\frac{\partial V}{\partial p})_T dp$$ We get: $$dA=-SdT-p[(\frac{\partial V}{\partial T})_p dT+(\frac{\partial V}{\partial p})_T dp]$$ So $$(\frac{\partial A}{\partial T})_p=-S-p (\frac{\partial V}{\partial T})_p$$ So: $$\frac{1}{V}(\frac{\partial V}{\partial T})_p=-\frac{1}{pV}[S+(\frac{\partial A}{\partial T})_p]$$