# Deriving Enthalpy from Statistical Mechanics

One can derive all the numerous thermodynamic potentials (Helmholtz, Gibbs, Grand, Enthalpy) by Legendre transformations, but I'm interested in seeing each from Stat Mech instead (ie taking the log of a partition function). I can do this easily for all but the Enthalpy, which has me stumped.

Easy example: Deriving Helmholtz

The Helmholtz energy can be derived by putting our system $S$ in thermal contact with a heat bath $B$ at temperature $T$ and maximizing the total entropy:

$$S_\mathrm{tot}/k_b=\log\sum_{E_S}\Omega_B(E_\mathrm{tot}-E_S)\Omega_S(E_S)$$

If the heat bath is big enough to maintain constant temperature $1/T=\left(\frac{dS_B}{dE_B}\right)_{V_B}$, then we can say $S_B\approx -E_S/T+S_0$ for small $E_S$. Then $\Omega_B\approx \Omega_0 e^{-\beta E_S}$ where $S_0=k_b\log\Omega_0$, $\beta=1/k_bT$, so

$$S_\mathrm{tot}/k_b=\log\sum_{E_S}\Omega_0e^{-\beta E_S}\Omega_S(E_S)=S_0/k_b+\log Z_S$$ where $Z_S=\sum_{E_S}e^{-\beta E_S}\Omega_S(E_S)$. So maximizing the total entropy is just maximizing $Z_S$. If we define the Helmholtz Free Energy $A_S$ as $-\beta A_S=\log Z_S$, and use the Shannon entropy $S/k_b=-\sum p\log p$, we see

$$S_S/k_b=-\sum \frac{\Omega_S(E_S)e^{-\beta E_S}}{Z}\log \frac{\Omega_S(E_S)e^{-\beta E_S}}{Z}$$ $$S_S/k_b=\frac{\beta}{Z}\sum \Omega_S(E_S)E_S e^{-\beta E_S} + \log Z$$ $$S_S/k_b=-\frac{\beta}{Z}\partial_\beta Z + \log Z$$ $$S_S/k_b=\beta \langle E_S\rangle -\beta A_S$$ $$A_S=\langle E_S\rangle -TS_S$$

The other thermodynamic potentials at fixed $T$ are similarly easy to derive.

But now try deriving Enthalpy

The same procedure does not work for enthalpy, because,

$$S_\mathrm{tot}/k_b=\log\sum_{V_S}\Omega_B(V_\mathrm{tot}-V_S,E_{B0}+pV_S)\Omega_S(V_S,E_{S0}-pV_S)$$

...if the bath is big enough to maintain a constant temperature, then its total entropy is constant as a function of $V_S$. That is, if $V_S$ increases, the bath entropy decreases due to less volume, but the bath energy increases by the same amount due to increased energy. So, to first-order, $\Omega_B$ is constant and the total entropy splits into a sum of both individual subsystem entropies.

Is there a way to derive the enthalpy from stat mech considerations, as there is for the other potentials, rather than by Legendre transforming the energy?

By "derive the enthalpy", I mean "derive that the quantity which should be minimized in equilibrium is given by $H=\langle E \rangle+p\langle V \rangle$."

In case you want to compute the ensemble.

The ensemble you are looking for is called the Isoenthalpic-isobaric Ensemble. The paper I read to learn about it is cited in Wikipedia, Andersen, H. C. Journal of Chemical Physics 72, 2384-2393 (1980). Unfortunately, if you aren't at a university, you might not have access or want to pay, so let me summarize.

First off, we want to reproduce thermodynamics, \begin{align} {\displaystyle dH(S,p)=T\,dS+V\,dp.} \end{align}

Lets consider, \begin{align} \Gamma(N, P, H) &= C^{-1}\int dV~d^{3N}q~d^{3N}p~ \delta(\mathcal{H}(q, p; V) + PV - H) \\ &= \frac{1}{i2\pi} C^{-1} \int d\beta~e^{\beta H} \int dV~d^{3N}q~d^{3N}p~ e^{-\beta(\mathcal{H}(q, p; V) + PV)} \\ &= \frac{1}{i2\pi} C^{-1} \int d\beta~e^{\beta H} \mathcal{Z}_G(\beta, P, N) \\ \end{align} where $C^{-1}$ just gets rid of units.

Then, \begin{align} \frac{\partial}{\partial P} k_B \log \Gamma(N, P, H) &= k_B \Gamma^{-1} \left[ \frac{1}{i2\pi} C^{-1} \int d\beta~e^{\beta H} \int dV~d^{3N}q~d^{3N}p~ (-\beta V)~e^{-\beta(\mathcal{H}(q, p; V) + PV)} \right]\\ &= - k_B \langle \beta V \rangle = -\left\langle \frac{V}{T} \right\rangle \end{align}

So we would naturally want to call,

$$S = k_B \log \Gamma(N, P, H)$$

You may want to find the expectation value of V

$\langle V \rangle = \frac{1}{\beta}\frac{\partial \log Z}{\partial p}$

And then just use the normal definition of Enthalpy with respect to the internal energy

$\langle H \rangle = \langle E \rangle + p \langle V \rangle$

If we maximize entropy with respect to another constraint, this time on the expected pressure;

$\langle p \rangle = \sum_{i} P_i p_i$

Then our probability distribution will have the form of

$P_i = \frac{1}{Z}e^{-\beta E_i - \lambda p_i}$

And so the condition on maximum entropy is given by $S_{max} = -k \log(P)$ which give us

$S_{max} = k\log(Z) + k\beta E_i + k\lambda p_i$

Which upon rearrangement gives for the energy

$E_i = \frac{S}{k \beta} - \frac{1}{\beta}\log(Z) - \frac{\lambda p_i}{\beta}$

We also know from the thermodynamic equations that $\langle V \rangle = \langle -\frac{\partial E_i}{\partial p}\rangle$ which gives us

$\langle V \rangle = \frac{\lambda}{\beta}$

And so, $\lambda = \beta \langle V \rangle$. This then tells us that the partition functino is equal to $Z = e^{-\beta H}$ where H is the enthalpy. Taking the log of the partition function from maximizing entropy we have

$E_i + p_i\langle V \rangle = H$

Let me know if I made any mistakes in the logic here.

• Thank you, N. I realize my question was ambiguous. What I'd like to do is derive that the last expression you wrote is the quantity which gets minimized in equilibrium. I updated the question to reflect this. – Sam Bader Mar 15 '16 at 2:14
• Hey Sam, I made some edits, let me know if this jives at all – N. Carrara Mar 16 '16 at 14:43