# Given constant T, why does P affect internal energy?

It has always bugged me that tables for water (and other) properties have the capability to look up internal energy as a function of both temperature and pressure. If we limit the discussion to liquid below the saturation temperature, then what is the qualitative argument to say that $u(T)$ is inaccurate and that the multivariate function $u(P,T)$ is needed?

From Wikipedia Internal Energy:

In thermodynamics, the internal energy is the total energy contained by a thermodynamic system. It is the energy needed to create the system, but excludes the energy to displace the system's surroundings, any energy associated with a move as a whole, or due to external force fields.

I understand that internal energy is not fully a proxy for temperature, so what thermodynamic property could we define (in $J/kg$) that would be a fully 1-to-1 relationship with temperature with no influence from pressure? If a liquid was fully incompressible would internal energy then not be a function of pressure?

If my physics understanding is correct, temperature has a definition that stems from the concept of thermal equilibrium. Quantitatively, I thought that temperature was proportional to the average kinetic energy the molecules, but I doubt that as well (in fact, I think this is wrong). The zeroth law of thermodynamics is necessary for formally defining temperature but it, alone, is not sufficient to define temperature. My own definitions for temperature and internal energy do not have the rigor to stand up to scrutiny. What qualitative arguments can fix this?

Symbols

• $u$ - internal energy
• $P$ - pressure
• $T$ - temperature
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I doubt you could define anything except $\int X dT$, where X is a specific heat sort of thing. But all the specific heats are pressure dependant/process dependant anyway for real systems :/ – Manishearth♦ Apr 5 '12 at 3:51

## 2 Answers

While there are many variables that characterize a thermodynamic system, such as volume $V$, pressure $P$, particle number $N$, chemical potential $\mu$, temperature $T$ and entropy $S$, these are not all independent of each others! In fact, any thermodynamic potential (such as internal energy, free energy, enthalpy) can be written as functions of either three of these variables.

Thus, in the most general case, you will get something like $$U(T,P,N)$$ where you specify temperature, pressure, and number of particles.

I think it's easier to understand if you realize that pressure and volume are intimately linked, and then think about the effect of interactions: These should get stronger if you reduce the volume of the system so particles are closer together and thus (typically) have a higher interaction energy.

In an ideal gas, there are no interactions, so volume doesn't really have an effect on the internal energy: $$U = \frac{3}{2} N k T$$

But if you have interactions, they will give you a contribution that depends on volume and, thus, on pressure.

EDIT: As an example, the van-der-Waals equation describes a gas of weakly interacting particles. There, Wikipedia gives the internal energy as $$U = \frac{3}{2} N k T - \frac{a' N^2}{V}$$ where $a'$ is a parameter describing the interaction.

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To start on a positive, your $U$ equation is a fantastic insight.  Moving on, shouldn't $(T,P)$ imply $\rho$ which would then imply $N$?  One factoid I constantly constantly fall back on is the claim that a thermodynamic state is a function of 2 independent variables, not 3.  Implicit in this argument is that you already know the composition, i.e. H2O.  As far as I can tell, if you specify temperature and pressure, you would not then specify # of particles, so let's get that argument out of the way first. – AlanSE Apr 5 '12 at 5:22
I'm pretty sure you definitely need three independent variables. If you have an equation of $\rho$ in terms of $T$ and $P$, then to get $N$ you also have to know $V$, since $\rho = N/V$. In many situations, one variable is trivially fixed, so then of course there are only two variables left to specify. – Lagerbaer Apr 5 '12 at 5:27
Ok, this is reconciled by the fact that you include a macroscopic $V$, which is fine.  My statements about 2 variables should be qualified as applying to a differential volume element. – AlanSE Apr 5 '12 at 5:33

The temperature is the average kinetic energy for classical nonrelativistic particle admixtures. The reason is that the temperature is what you multiply the energy by to get the probability of a given microstate. Kinetic energy is nonrelativistically always quadratic, and the probability distribution is Gaussian.

So the average KE of the atoms/molecules is your quantity in J/atom (or J/kg if you convert) and it is just ${3\over 2}kT$. To prove this, you just have to note that the expected value of $x^2$ for a gaussian of variance $\sigma$ is just $\sigma$, and the probability distribution for atoms having positions x and momentum p is

$$\rho(x,p) = e^{-\beta(\sum_i {p_i^2\over 2m} + V(x_1,x_2,...,x_N))}$$

which is Gaussian of the same width in p for all values of x, so the momentum variables are always distributed according to a Gaussian (Maxwell-Boltzmann) distribution.

If you imagine particles which have a potential energy function which is just proportional to the density (this is a grossly nonlocal potential energy) then if you increase the pressure at constant temperature, you will increase the density, and increase the internal energy.

So in classical statistical mechanics, assuming that the interaction is potential type, the pressure dependence tells you how the potential energy contributes.

In quantum statistical mechanics, at cold enough temperatures that the atomic motion energy levels are further apart then kT, these motions do not contribute to the specific heat, and the kinetic energies for these motions do not average to {1\over 2} kT in each direction. But the classical description is accurate for those motions which are classical, and that's most of the gross motions at room temperature.

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The OP is about water, which has a qualitatively different set of features than the ideal gas. – Jerry Schirmer Apr 5 '12 at 12:45
You presented the definition that average KE = $3/2 kT = 1/N \sum{ m v_i^2 /2 } = 1/N \sum{ p_i^2 / 2 m}$.  Does this satisfy the requirement that gases of two different molecular masses at the same $T$ can't transfer heat?  Even in the presence of a potential function? @JerrySchirmer I think the presence of the term $V(x_1,..)$ breaks the assumption of an ideal gas. – AlanSE Apr 5 '12 at 14:59
@JerrySchirmer: Where do you see an ideal gas? The answer that the average kinetic energy of each center of mass motion is 1.5kT is correct for any classical system. – Ron Maimon Apr 5 '12 at 16:18
@AlanSE: Yes it does, in the thermodynamic limit (which is the only case in which it is valid that two systems at the same temeperature interacting don't transfer heat--- otherwise the interaction changes the ensemble), because the average kinetic energy is just the ordinary thermodynamic temperature. – Ron Maimon Apr 5 '12 at 16:21
A free classical system of non-interacting particles with no internal degrees of freedom, yes.  That is hardly a description of anything but a gas.  Even an ideal gas of molecules with a sufficently high temperatures to excite their rotational modes will have $U=nkT$ with $n>\frac{3}{2}$. – Jerry Schirmer Apr 5 '12 at 16:54
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