# Thermodynamic variables

1. Mar 3, 2014

### aaaa202

So I have had related questions over the past month, but I would like to ask this question to clarify my understanding.
In thermodynamics you work with certain potentials, which are a function of the thermodynamic variables, i.e.:

U(S,T,V,N,P)

Now for U one has the identity:

dU= TdS+SdT-pdV+VdP etc etc.

From these one figure out relations like:

T = dU/dS at fixed V,T,P...

It is this thing about the thermodynamic variables being fixed that has always confused me. In general are the thermodynamic variables S,P,T,V,N not correlated? How am I to understand then the derivative if I am keeping the other thermodynamic variables fixed. Consider for instance including particles of different kinds:

U = ... + μ1N1 + μ2N2

Now we have that:

μ1 = dU/dN1 at fixed T,S,V, N2

But how can I keep N2 fixed if I am in a resevoir, where adding a particle to one phase with N1 particles, actually takes away a particle from the other i.e. dN1=-dN2

Similarly, if I change for instance V, don't I change S or P etc etc.

2. Mar 3, 2014

### vanhees71

Usually you use functions with arguments that are independent from each other. In the grand-canonical ensemble there are three independent degrees of freedom.

Different potentials have different "natural" independent variables. For the internal energy you have
$$\mathrm{d} U=T \mathrm{d} S - p \mathrm{d}V + \mu \mathrm{d} N.$$
The natural independent variables for $U$ are thus the entropy, the volume and the particle number, and you have the relations
$$\left (\frac{\partial U}{\partial S} \right )_{V,N}=T, \quad \left (\frac{\partial U}{\partial V} \right )_{S,N}=-p, \quad \left (\frac{\partial U}{\partial N} \right )_{S,V}=\mu.$$
For other combinations of independent variables other potentials are more convenient. E.g., the enthalpy. It's given by a socalled Legendre transformation of the internal energy
$$H=U+p V.$$
$$\mathrm{d} H= \mathrm{d}U + p \mathrm{d} V+V \mathrm{d} p=T \mathrm{d} S + V \mathrm{d} p + \mu \mathrm{d} N.$$
The natural independent variables for the enthalpy are thus the entropy, pressure, and particle number. From this you get
$$\left (\frac{\partial H}{\partial S} \right )_{p,N}=T, \quad \left (\frac{\partial H}{\partial p} \right )_{S,N}=V, \quad \left (\frac{\partial H}{\partial N} \right )_{S,p}=\mu.$$
As you see, it is important to note, which independent variables are to be held fixed when taking a partial derivative.

Other important relations, socalled Maxwell relations, can be found from the 2nd mixed derivatives. E.g., for the internal energy you have
$$\frac{\partial^2 U}{\partial V \partial S}=\left (\frac{\partial T}{\partial V} \right )_{S,N}=-\left (\frac{\partial p}{\partial S} \right )_{V,N}.$$
For more details, see Wikipedia:

http://en.wikipedia.org/wiki/Maxwell_relations

3. Mar 3, 2014

### hilbert2

In thermodynamics, you only need a certain number of "independent" variables to fully define the state of the system. The other variables are then fixed by the equation of state. The number of independent variables, $F$ is given by the Gibbs phase rule:

$F=C-P+2$,

where $C$ is the number of chemical components in the system and $P$ is the number of phases in the system.

For example, the equation of state of an ideal gas is $PV=nRT$, and let's suppose that it consists of a single chemical component. You only need to specify three of the four variables $P,V,n,T$ and the fourth is fixed by the equation of state. We can then denote partial derivatives of the variables with a notation like

$\left(\frac{\partial P}{\partial V}\right)_{n,T}=-\frac{nRT}{V^{2}}$,

where we specify that $n$ and $T$ are treated as constants in the differentiation. Here we fix enough variables to make sure that $P$ and $V$ are the only variables we don't know. You can easily imagine a process where we increase the volume of a closed ideal gas system while keeping temperature constant (a container with heat conducting walls), and the measure the corresponding change of pressure.