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How to derive the thermodynamic activity?

Physics Asked by Basant Acharya on October 26, 2020

From wikipedia thermodynamic activity is given as
$$ a = exp left(frac{mu – mu_{0}} {RT}right)$$
which becomes,
$$ RTln(a) = mu – mu_{0}tag{1}$$

And i have to do a calculation where activity is given as

$$ RTln(a) = left (frac{partial{G_m}}{partial{N_A}}right)_{T,P,N_{B}} = G_{M} + (1-c) left(frac{partial{G_{M}}}{partial{c}}right)_{T,P,N} tag{2} quad dagger$$
where
$$G_{M} = RT[cln(c) + (1-c)ln(1-c) + cln(1-beta) – ln(1-beta c)] + frac{omega c(1-c)} {1-beta c}$$
$omega$ is the interchange energy

$$ omega = ZN(E_{AB}- frac{E_{AA}+E_{BB}}{2})$$

Z = Coordination number

$E_{AB}$ energy corresponding to bond A-B

$E_{BB}$ energy corresponding to bond B-B

$E_{AA}$ energy corresponding to bond A-A

$beta$ 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
$N_{A}$ and $N_{B}$ are the number of atoms of A and B

$dagger$ https://doi.org/10.1080/13642819008208649 (equation 8)

please somebody can elaborate the steps of the (eqn$2$).

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