# Thermodynamic state entirely determined by only two quantities?

1. Jul 11, 2014

### mSSM

I am trying to understand why I can specify the specific volume $v$ of a fluid element as a function of the equilibrium pressure, $p$, and the equilibrium entropy, $s$. This is for example done in this article http://www.sno.phy.queensu.ca/sno/str/SNO-STR-95-051.pdf , on this website: http://wind.mit.edu/~emanuel/geosys/node3.html , and in Landau & Lifshitz fluid mechanics, page 7.

I have spent a while looking some source explaining this, and remembered from statistical physics (quote from Landau & Lifshitz, Statistical Physics, Page 42):
I am trying to understand how the knowledge of $p$ and $s$ would help me to, for example, obtain $T$? How can I get access to other quantities knowing only those two variables?

2. Jul 11, 2014

### WannabeNewton

The statement in Landau is only true if there is only one generalized force associated with the system or equivalently if there is only one generalized coordinate characterizing the system; in the case of Landau's discussion the generalized coordinate is the volume and the conjugate force is the pressure. In more general cases, wherein one has $n$ generalized coordinates or generalized forces, one needs $n+1$ variables in order to completely determine the thermodynamic properties of the system.

With that being said, the enthalpy $H$ of a system is entirely characterized by $p$ and $S$ i.e. $H = H(S,p)$. You can therefore determine the temperature using $T = \frac{\partial H}{\partial S}|_p$ so that $T = T(S,p)$. The enthalpy also allows you to determine the volume as a function of these coordinates i.e. $V = V(S,p)$ using $V = \frac{\partial H}{\partial p}|_S$. All you need to do is determine the enthalpy of the system as a function of the pressure and entropy.

This can always be done in principle by putting the system in thermal and mechanical contact with a heat bath and using the Gibbs canonical ensemble to calculate the Gibbs partition function $\mathcal{Z}$ from which we get the enthalpy $H = -\frac{\partial}{\partial \beta}\ln \mathcal{Z}$. This will tell us $H(T,p)$ since $\mathcal{Z} = \mathcal{Z}(T,p)$ on account of the system being in both thermal and mechanical contact with the heat bath. We can also get $S = S(T,p)$ from $\mathcal{Z}$ and in principle we can invert this to get $T = T(S,p)$ so that $H = H(S,p)$.

Last edited: Jul 11, 2014
3. Jul 11, 2014

### mSSM

Alright, thanks for the explanation. This leaves me with two questions:
• $T$ and $S$ are not viewed as a pair of generalized forces/coordinates in this context?
• What is the argument for needing $n+1$ variables in order to completely determine all thermodynamic properties?

4. Jul 11, 2014

### WannabeNewton

This is really a matter of taste. Some authors will consider $T$ and $S$ as a pair of generalized forces/coordinates whereas others will reserve these qualifications only for coordinates and forces that are related through some form of work $\delta W$ and not through spontaneous heat flow $\delta Q$ as $T$ and $S$ are. I prefer the latter.

Say we have $n$ generalized coordinates for a system. This can be for example volume $V$ and an external magnetic field $B$ if we have a sample of ferromagnetic material and wish to calculate its average magnetization and pressure. The Gibbs partition function will have dependence $\mathcal{Z} = \mathcal{Z}(T,x_1,...,x_n)$ i.e. it will depend on $n+1$ coordinates. As you know, all thermodynamic properties of a system can be determined entirely from $\mathcal{Z}$. For example, going back to the example of the ferrmagnetic, we have $p = \frac{1}{\beta}\partial_V \ln \mathcal{Z}$ and $\langle M \rangle = \frac{1}{\beta}\partial_B \ln\mathcal{Z}$ as well as $H = -\partial_{\beta}\ln\mathcal{Z}$.