Is there a dynamic prelude to the state equation?

If I'm not mistaken, the state equation $$PV=nRT$$ also known as the 'ideal gas law' serves as a mathematical model relating pressure-volume-mass-temperature states of an assumed ideal gas. In other words it generally is applied as a 'static' transformation - providing a way to predict what values one or more of the other variables will change to at steady state.

But in practical situations, and perhaps even when one holds to the ideal gas assumptions, there are transients occurring when gases transition from one set of states to another.

So my question - is (or can) the state equation (be) derived from a more general dynamic physical law for gases?

To further illustrate the motivation for this question, consider a system where the number of gas molecules and the volume remain constant, but where pressure and temperature can vary. So from the state equation we predict what resulting temperature we would have in state $2$ starting with the temperature in state $1$ by $$T_2=\frac{P_2}{P_1}T_1$$ Again this equation has no dynamic components, and therefore a step change in pressure leads to an instant step change in temperature. But we know nothing happens instantly. Since the volume and number of particles assumed constant in this example, the only way to get to $P_2$ from $P_1$ is to increase the internal energy, so heat must be input (or if $T_2 < T_1$ , taken) from the system.

So then are the limitations of how fast you are able to move energy into the system the factor that determines the dynamic nature - and thus the differential heat transfer equations that I should be considering to model this physical behavior?

• – Kyle Kanos Oct 1 '15 at 16:17

The dynamic version of equilibrium thermodynamic is fluid dynamics, and the dynamic version of equilibrium statistical mechanics is kinetic theory.

In fluid dynamics the stress (pressure) tensor of the gas is expanded in time and space derivatives of the thermodynamic variables. This expansion converges if the state variables vary sufficiently slowly. In the limit of infinitely slow variation you get equilibrium thermodynamics. If the variation is too fast for any kind of expansion to converge, you have to switch to a more microscopic description like kinetic theory.

At leading order in the derivative expansion you get the Euler equation of (ideal) fluid dynamics. In this case the pressure in the local rest frame of the fluid is given by the thermodynamic relation $P=P(n,T)$. For a weakly interacting gas $P=nT$. Ideal fluid dynamics already takes into account that thermodynamic variables throughout the fluid cannot change instantaneously without violating causality. If I push a piston to increase pressure, a pressure wave moves into the fluid with the speed of sound. If I heat the bottom of the container, convection has to occur for energy to be transfered and the temperature to change.

At next-to-leading order I get the Navier-Stokes equation of (non-ideal) fluid dynamics. At the NS level the pressure tensor in the local restframe is no longer equal to the equation of state, but has a correction that depends on the rate at which the fluid is moving. For example, the pressure in an expanding gas cloud is reduced relative to its equilibrium value by an amount proportional to the rate of expansion and the bulk viscosity. Energy can move into the fluid by thermal diffusion.

This continues. At next-to-next-to-leading (Burnett) order the stress tensor is not instantaneously equal to the Navier-Stokes form, and the heat current is not instantaneously equal to $-\kappa\nabla T$, but they lag by viscous and thermal relaxation times.

All of these effects are controlled by transport coefficients. In kinetic theory, these coefficients are related to the rate at which atoms and molecules scatter, which determines the rate at which local equilibrium can be re-established if any of the thermodynamic variables suddenly changes.

Also note that there is a subtlety here: $P=nT$ is the equation of state of a non-interacting gas. But a non-interacting gas can never equilibrate $\ldots$.

• So then my hunch was right (see last paragraph)? And the ideal gas equation is only one equation is a system of equations, and to treat dynamics you need something that looks at the time rate of flow of energy into/out of/ and within the system? – docscience Oct 1 '15 at 16:45
• Yes, except that in ideal fluid dynamics the form of the energy current is fixed by the equilbrium equation of state (so there is no extra parameter). Extra parameters appear if you want to achieve higher accuracy. Also added a small comment above: No dynamical system can have an equation of state $P=nT$. – Thomas Oct 1 '15 at 16:56

The ideal gas law can be derived from the fundamental differential equation of thermodynamics:

First, consider a lattice model with M sites and N particles. Then the multiplicity of the system is:

$W = \frac{M!}{N!(M-N)!}$

Thus the entropy of the system $S = klnW = kln(\frac{M!}{N!(M-N)!}).$

Using Stirling's approximation: $n! \approx (\frac{n}{e})^n$

$S \approx kln(\frac{M^M}{N^N(M-N)^{(M-N)}}) = k(MlnM-NlnN-(M-N)ln(M-N))$

$\frac{\partial S}{\partial M} = -kln(1-\frac{N}{M})$

Next we want to bring in pressure and temperature.

Starting from first principles, $dE = \frac{\partial E}{\partial V}dV + \frac{\partial E}{\partial S}dS$ when we fix $N$. By definition, $T = \frac{\partial E}{\partial S}$ and $P = - \frac{\partial E}{\partial V}$. Therefore we arrive at the fundamental thermodynamical equation:

$dE = TdS - PdV$

Assuming conservation of energy, $dE = 0$ Therefore, we can move terms around and say $\frac{P}{T} = \frac{\partial S}{\partial V}$

In this lattice model, it is obvious that $M$ (number of sites) is proportional to $V$ (volume of the system). Therefore we can say $V = aM$ where $a$ is basically volume per site. By this definition it follows that $\frac{dV}{dM} = a$

Therefore, $\frac{P}{T} = \frac{\partial S}{\partial V} = \frac{\partial S}{\partial M} \frac{dM}{dV} = \frac{-kln(1-\frac{N}{M})}{a}$

Ideal gases have low densities. Thus we assume that $N$ is much smaller than $M$, in which case we can approximate the term $ln(1-\frac{N}{M})$ as a taylor series around 1 and keep only the first order term $-\frac{N}{M}$

After this simplification, $\frac{P}{T} = k\frac{N}{Ma}$

Since $Ma = V$, this expression further simplifies to: $PV = NkT$ which is the ideal gas law!

As you can see, the statistical description of entropy and its relationship to volume, pressure and temperature through differentiation allowed a derivation of ideal gas law from first principles (the kind of dynamical laws that the OP was referring to). In fact, the more complicated Van Ders Waals Law can be derived just by keeping more terms in the taylor expansion of $ln(1-\frac{N}{M})$. So statistical thermo is pretty neat!

Edit: I am adding a bit of my own derivation of what OP referred to as time-dependent differential heat equation when volume and number of particles is fixed. Correct me if I am wrong...

So you know that $P = \frac{nRT}{V}$ and thus $dP = \frac{nR}{V} dT$ Since $$dE = dq (heat) + dw(work)$$ and $$dE = \frac{3nR}{2} dT$$

No work has been done since volume remains constant. Therefore we can claim that: $$dq = \frac{3nR}{2} dT$$

Comparing this equation with $dP = \frac{nR}{V} dT$ We get $$dP = \frac{2}{3V}dq$$

• That's very nice, but I don't think this answers the question. The question (I think) is what is the dynamic (time dependent) version of $P=P(n,T)\simeq nT$. – Thomas Oct 1 '15 at 15:41
• @Thomas Yes that's ultimately what I'm looking for, but part of my question does ask if the ideal gas law can be derived from dynamic equations. I believe that's what Zhengyan and airguru have produced in their answers. All answers so far, great and informative. – docscience Oct 1 '15 at 16:47
• Does my edit help somewhat? – Zhengyan Shi Oct 1 '15 at 17:58
• @ZhengyanShi nearly so. You just need to divide each side of your last equation by dt. Whether it's correct; not sure. From my original example looking at going from one temperature to another based on pressure changes, and combining your equations I get $$\frac{dT}{dt}=\frac{V}{nR}\frac{dP}{dt}$$. Which is no different than writing the state equation and dividing each side by dt. And heat wasn't even involved. – docscience Oct 1 '15 at 18:21
• yeah but the point is that you can relate dq (heat) with dT(change in temperature) in this special scenario right? I guess I don't know exactly what you mean by a "differential heat equation"... – Zhengyan Shi Oct 1 '15 at 18:40

First of all, both pressure and temperature are quantities statistically connected to averages (average force on the wall exerted by the particles and their average energy). So using them in cases, when the system is not homogeneous is not too appropriate. If we want to look at, let's say, gas in the box receiving heat from one side, and see how the temperature and pressure propagates, it makes more sense to divide the box into subsystems (smaller boxes) that are sufficiently homogeneous for that.

Regarding the derivation of state equation, the rough overview can go like this: Pressure comes from particles colliding with the wall. Take a look here: http://homepage.smc.edu/gallogly_ethan/files/ideal%20gas%20law%20derivation.pdf and return when you reach the middle of page 5.

At that point in the pdf, they say that energy of a particle in one dimension can be related to the system temperature with the relation: $E = \frac{1}{2}kT$. What they really mean is the average energy, and basically this can be taken as a definition of the temperature. So when they arrive at the state equation a few lines afterwards, it comes out the way we know it, because we chose to define the temperature that way (as proportional to the average energy of particles).

The equivalent, and probably more elegant definition of temperature goes like this:

1) We put bunch of particles in a box, give them some velocities and let them randomly collide. Make sure the sum of velocities is zero, so the box doesn't get moved from the inside.

2) Over time, the distribution of their momentums (in all directions) stabilizes on normal distribution with zero mean $\overline{p} = 0$ and some, non-zero variance $\sigma^2$. This can be rigorously proven, but sounds reasonable. You know, everything somehow converges to normal distribution when subject to random additions and substractions. This is precisely the case of particle collisions.

3) We define temperature as proportional to this variance: $\sigma^2 = kT$. The $k$ constant is just for convenience to keep the $T$ value in nice ranges.

4) From this it immediately follows, that average kinetic energy in each dimension is $\overline{E} = \frac{1}{2}kT$. Because $\overline{p^2} = \sigma^2 + \overline{p}^2$ (see https://en.wikipedia.org/wiki/Variance)

When the gas particles have more degrees of freedom than three, like vibrations and rotations between atoms in the molecule, or anything else, as long as the energy is transferable between these degrees of freedom, the normal distribution convergence usually stil holds. This is a statement of Equipartition theorem (see https://en.wikipedia.org/wiki/Equipartition_theorem for detailed conditions when it applies).

EDIT: I changed velocities to momentums, because for velocities it does not hold. When colliding, the particles randomly exchange parts of their momentums and the total momentum is conserved.

• Thanks - the class notes easy to follow. And I couldn't find fault with any of the assumptions/steps. – docscience Oct 3 '15 at 17:24