The Helmholtz and the Gibbs free energies

Let us now insert the (single microstate) Boltzmann probabilities

in the general definition of the entropy

. We have:

$\displaystyle S = -k_{\rm B}\sum_r p_r\ln \frac{e^{-\beta E_r}}{Z} = -k_{\rm B}\sum_r p_r (-\beta E_r - \ln Z) = \frac{\bar{E}}{T} + k_{\rm B}\ln Z.$

(3.62)

Following the argument given above about the small fluctuations of the energy $E$

around its average value $\bar{E}$ , we write $E$

in place of $\bar{E}$ and rearrange

as:

$\displaystyle F = -k_{\rm B}T \ln Z = E - TS$

(3.63)

where we have defined $F$

as the Helmholtz free energy of the system. Coupling

with

we have:

$\displaystyle dF = d(E - TS) = dE - TdS - SdT = -PdV - SdT + \mu dN,$

(3.64)

from which we see that we can obtain the pressure also as:

$\displaystyle P = -\left ( \frac{\partial F}{\partial V}\right )_{T,N},$

(3.65)

the entropy as:

$\displaystyle S = -\left ( \frac{\partial F}{\partial T}\right )_{V,N}$

(3.66)

and the chemical potential as:

$\displaystyle \mu = \left ( \frac{\partial F}{\partial N}\right )_{V,T}.$

(3.67)

This shows that the same thermodynamic quantity can be obtained in more than one way, depending on the conditions of the ensemble. If we insert the isothermal-isobaric probabilities in

we obtain:

$\displaystyle S = -k_{\rm B}\sum_{V,r} p_{V,r}\ln \frac{e^{-\beta (PV+ E_{V,r})}}{\Delta} =$
$\displaystyle -k_{\rm B}\sum_{V,r} p_{V,r} [-\beta (PV+ E_{V,r}) - \ln \Delta] =$
$\displaystyle \frac{\bar{E}}{T} + \frac{P\bar{V}}{T} + k_{\rm B}\ln \Delta,$	(3.68)

from which we define the Gibbs free energy as:

$\displaystyle G = -k_{\rm B}T \ln \Delta = E + PV - TS,$

(3.69)

where again we have written $E$

for $\bar{E}$ and $V$

for $\bar{V}$ . Coupling

with

we have:

$\displaystyle dG = d(E+PV-TS) = VdP - SdT + \mu dN,$

(3.70)

from which we obtain:

$\displaystyle V = \left ( \frac{\partial G}{\partial P}\right )_{T,N},$

(3.71)

$\displaystyle S = -\left ( \frac{\partial G}{\partial T}\right )_{P,N},$

(3.72)

$\displaystyle \mu = \left ( \frac{\partial G}{\partial N}\right )_{P,T}.$

(3.73)

Since

is extensive, then for fixed $P$

and

it must be proportional to the number of particles in the system and so

$\displaystyle G(P,T,N) = N g(P,T) = N \mu$

(3.74)

with

the Gibbs free energy per particle and the last equality follows from

Let us now refer back to the canonical and the isothermal-isobaric probabilities and , in which we included the degeneracy of the energy as the exponential of the microcanonical entropy. Under the assumption that for large enough systems the fluctuations in the energy and the volume are completely negligible, these systems are, effectively, at constant energy and constant volume, and therefore we can identify the microcanonical entropy with the entropy itself. As a result, we can rewrite the canonical and the isothermal-isobaric probabilities as:

$\displaystyle p(E,S) =\frac{1}{Z} \exp \{ -\beta F\},$