0
$\begingroup$

From wikipedia thermodynamic activity is given as $$ a = \exp(\frac{\mu - \mu_{0}} {RT})$$ which becomes, $$ RT\ln(a) = \mu - \mu_{0}\tag{1}$$

And i have to do a calculation where activity is given as $$ RT\ln(a) = G_{M} + (1-c) \frac{\partial{G_{M}}}{\partial{c}} \tag{2}$$

where $$G_{M} = RT[c\ln(c) + (1-c)\ln(1-c) + c\ln(1-\beta) - \ln(1-\beta c)] + \frac{\omega c(1-c)} {1-\beta c}$$ $\omega$ is the exchange energy

$\beta$ is the related to the volume ratio of mixing of two liquids.

c is the concentration of liquid A and (1-c) is the concentration of liquid B

$\mu$ is the chemical potential and $\mu_{0}$ is the standard chemical potential

How do I get from eqn($1$) to eqn($2$)?

$\endgroup$
1
  • $\begingroup$ Its a guess but perhaps the way is to associate the $RT[\ln(c^c)\cdots]$ term with $RT\ln(a)$ making the derivative $\omega c/(1-\beta c)$ $\endgroup$ – porphyrin Aug 23 '20 at 15:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.