The Ideal Monatomic Gas

Today at the Topos Institute, Sophie Libkind, Owen Lynch and I spent some time talking about thermodynamics, Carnot engines and the like. As a result, I want to work out for myself some basic facts about the ideal gas. This stuff is all well-known, but I’m having trouble finding exactly what I want—and no more, thank you—collected in one place.

Just for background, the Carnot cycle looks roughly like this:

This is actually a very inaccurate picture, but it gets the point across. We have a container of gas, and we make it execute a cyclic motion, so its pressure $P$ and volume $V$ trace out a loop in the plane. As you can see, this loop consists of four curves:

• In the first, from a to b, we put a container of gas in contact with a hot medium. Then we make it undergo isothermal expansion: that is, expansion at a constant temperature.

• In the second, from b to c, we insulate the container and let the gas undergo adiabatic reversible expansion: that is, expansion while no heat enters or leaves. The temperature drops, but merely because the container expands, not because heat leaves. It reaches a lower temperature. Then we remove the insulation.

• In the third, from c to d, we put the container in contact with a cold medium that matches its temperature. Then we make it undergo isothermal contraction: that is, contraction at a constant temperature.

• In the fourth, from d to a, we insulate the container and let the gas undergo adiabatic reversible contraction: that is, contraction while no heat enters or leaves. The temperature increases until it matches that of the hot medium. Then we remove the insulation.

The Carnot cycle is important because it provides the most efficient possible heat engine. But I don’t want to get into that. I just want to figure out formulas for everything that’s going on here—including formulas for the four curves in this picture!

To get specific formulas, I’ll consider an ideal monatomic gas, meaning a gas made of individual atoms, like helium. Some features of an ideal gas, like the formula for energy as a function of temperature, depend on whether it’s monatomic.

As a quirky added bonus, I’d like to highlight how certain properties of the ideal monatomic gas depend on the dimension of space. There’s a certain chunk of the theory that doesn’t depend on the dimension of space, as long as you interpret ‘volume’ to mean the n-dimensional analogue of volume. But the number 3 shows up in the formula for the energy of the ideal monatomic gas. And this is because space is 3-dimensional! So just for fun, I’ll do the whole analysis in n dimensions.

There are four basic formulas we need to know.

First, we have the ideal gas law:

$PV = NkT$

where

$P$ is the pressure.
$V$ is the n-dimensional volume.
$N$ is the number of molecules in a container of gas.
$k$ is a constant called Boltzmann’s constant.
$T$ is the temperature.

Second, we have a formula for the energy, or more precisely the internal energy, of a monatomic ideal gas:

$U = \frac{n}{2} NkT$

where

$U$ is the internal energy.
$n$ is the dimension of space.

The factor of n/2 shows up thanks to the equipartition theorem: classically, a harmonic oscillator at temperature $T$ has expected energy equal to $kT$ times its number of degrees of freedom. Very roughly, the point is that in n dimensions there are n different directions in which an atom can move around.

Third, we have a relation between internal energy, work and heat:

$dU = \delta W + \delta Q$

Here

$dU$ is the differential of internal energy.
$\delta W$ is the infinitesimal work done to the gas.
$\delta Q$ is the infinitesimal heat transferred to the gas.

The intuition is simple: to increase the energy of some gas you can do work to it or transfer heat to it. But the math may seem a bit murky, so let me explain.

I emphasize ‘to’ because it affects the sign: for example, the work done by the gas is minus the work done to the gas. Work done to the gas increases its internal energy, while work done by it reduces its internal energy. Similarly for heat.

But what is this ‘infinitesimal’ stuff, and these weird $\delta$ symbols?

In a minute I’m going to express everything in terms of $P$ and $V.$ So, $T, N$ and $U$ will be functions on the plane with coordinates $P$ and $V.$ $dU$ will be a 1-form on this plane: it’s the differential of the function $U.$

But $\delta W$ and $\delta Q$ are not differentials of functions $W$ and $Q.$ There are no functions on the plane called $W$ and $Q.$ You can not take a box of gas and measure its work, or heat! There are just 1-forms called $\delta W$ and $\delta Q$ describing the change in work or heat. These are not exact 1-forms: that is, they’re not differentials of functions.

Fourth and finally:

$\delta W = - P dV$

This should be intuitive. The work done by the gas on the outside world by changing its volume a little equals the pressure times the change in volume. So, the work done to the gas is minus the pressure times the change in volume.

One nice feature of the 1-form $\delta W = -P d V$ is this: as we integrate it around a simple closed curve going counterclockwise, we get the area enclosed by that curve. So, the area of this region:

is the work done by our container of gas during the Carnot cycle. (There are a lot of minus signs to worry about here, but don’t worry, I’ve got them under control. Our curve is going clockwise, so the work done to our container of gas is negative, and it’s minus the area in the region.)

Okay, now that we have our four basic equations, we can play with them and derive consequences. Let’s suppose the number $N$ of atoms in our container of gas is fixed—a constant. Then we think of everything as a function of two variables: $P$ and $V.$

First, since $PV = NkT$ we have

$\displaystyle{ T = \frac{PV}{Nk} }$

So temperature is proportional to pressure times volume.

Second, since $PV = NkT$ and $U = \frac{n}{2}NkT$ we have

$U = \frac{n}{2} P V$

So, like the temperature, the internal energy of the gas is proportional to pressure times volume—but it depends on the dimension of space!

From this we get

$dU = \frac{n}{2} d(PV) = \frac{n}{2}( V dP + P dV)$

From this and our formulas $dU = \delta W + \delta Q, \delta W = -PdV$ we get

$\begin{array}{ccl} \delta Q &=& dU - \delta W \\ \\ &=& \frac{n}{2}( V dP + P dV) + P dV \\ \\ &=& \frac{n}{2} V dP + \frac{n+2}{2} P dV \end{array}$

That’s basically it!

But now we know how to figure out everything about the Carnot cycle. I won’t do it all here, but I’ll work out formulas for the curves in this cycle:

The isothermal curves are easy, since we’ve seen temperature is proportional to pressure times volume:

$\displaystyle{ T = \frac{PV}{Nk} }$

So, an isothermal curve is any curve with

$P \propto V^{-1}$

The adiabatic reversible curves, or ‘adiabats’ for short, are a lot more interesting. A curve $C$ in the $P V$ plane is an adiabat if when the container of gas changes pressure and volume while moving along this curve, no heat gets transferred to or from the gas. That is:

$\delta Q \Big|_C = 0$

where the funny symbol means I’m restricting a 1-form to the curve and getting a 1-form on that curve (which happens to be zero).

Let’s figure out what an adiabat looks like! By our formula for $Q$ we have

$(\frac{n}{2} V dP + \frac{n+2}{2} P dV) \Big|_C = 0$

or

$\frac{n}{2} V dP \Big|_C = -\frac{n+2}{2} P dV \Big|_C$

or

$\frac{dP}{P} \Big|_C = - \frac{n+2}{n} \frac{dV}{V}\Big|_C$

Now, we can integrate both sides along a portion of the curve $C$ and get

$\ln P = - \frac{n+2}{n} \ln V + \mathrm{constant}$

or

$P \propto V^{-(n+2)/n}$

So in 3-dimensional space, as you let a gas expand adiabatically—say by putting it in an insulated cylinder so heat can’t get in or out—its pressure drops as its volume increases. But for a monatomic gas it drops in this peculiar specific way: the pressure goes like the volume to the -5/3 power.

In any dimension, the pressure of the monatomic gas drops more steeply when the container expands adiabatically than when it expands at constant temperature. Why? Because $V^{-(n+2)/n}$ drops more rapidly than $V^{-1}$ since

$\frac{n+2}{n} > 1$

But as $n \to \infty$,

$\frac{n+2}{n} \to 1$

so the adiabats become closer and and closer to the isothermal curves in high dimensions. This is not important for understanding the conceptually significant features of the Carnot cycle! But it’s curious, and I’d like to improve my understanding by thinking about it until it seems obvious. It doesn’t yet.

20 Responses to The Ideal Monatomic Gas

1. domenico says:

From the formula
$\delta Q=dU-\delta W=\frac{n}{2}(VdP+PdV)+PdV$
the work is negligible, then the infinitesimal heat is a differential in a n-dimensional space, when n is an infinity.
It is not possible each cycle in a infinitesimal space, because of the heat is a differential and the work must be a differential form (a difference of differential form): there is not more a heat engine, because isothermal heat (same temperature, same energy, no heat flow) is equal to the adiabatic heat (no heat flow).
Very, very interesting.
It is also possible to study motors in two dimensions, that might be technologically feasible.

2. scentoni says:

Typo:

From this and our formulas $dU = \delta W + \delta Q, \delta W = -pdV$ we get

Should be

From this and our formulas $dU = \delta W + \delta Q, \delta W = -PdV$ we get

• John Baez says:

Thanks! One book I have uses $p,$ another uses $P$, but I think $P$ is standard.

3. I’ll admit to not reading all of this post, as I worked my way through Fermi’s Thermodynamics years ago.

Anyway, here’s my understanding of your question on the adiabats and isothermals. Adiabates are steeper than isothermals because as you compress the gas, the energy injected increases the temperature (as you’d expect) and so you’re moving to higher isothermals.

Now what happens as n→∞? The increased energy goes entirely into kinetic energy (ideal gas), but more degrees of freedom means that it’s “diluted” more and more. So it doesn’t increase the temperature as much, and the effect becomes less and less significant.

4. Toby Bartels says:

Besides δ (which is rather overloaded), another symbol that I’ve seen used in place of a traditional d when it's not actually a differential of anything, is đ.

• John Baez says:

Yeah, I didn’t know how to get that in LaTeX and I’m not so sure what I like best. I kinda like $\delta$, but if I were really in charge of this I’d pick names for non-exact 1-forms that look like 1-forms, not like differentials of functions, e.g. $\omega$ for work (a pun) and $\eta$ for heat (also a pun, since capital $\eta$ is H).

5. Toby Bartels says:

Their own symbols would be best; of course the traditional symbols come from when people thought that W and Q were state functions. The European Computer Modern (ec) fonts have ⟨đ⟩, because it's a letter in the Croatian alphabet (and also Vietnamese). I have an old plain TeX macro that probably works in LaTeX (with a tiny modification) but would probably break MathJax. On the other hand, iTeX will simply pass through the Unicode character; maybe MathJax will do that too.

Tests:

Unicode character directly: $đ$

Decimal numerical character reference: $&273;$

Hexadecimal character reference: $&x0111;$

My TeX macro to see how far MathJax can handle it: $\mathrm {\mkern 3 mu \mathchar "0016 \mkern - 12 mu d\mkern 1 mu }$

My macro clumsily modified not to use an upright font (which I prefer), in case MathJax likes the simpler code, even though the spacing probably won't be correct anymore: $\mkern 3 mu \mathchar "0016 \mkern - 12 mu d\mkern 1 mu$

• Toby Bartels says:

Wow, the methods that I thought had the least chance is working seem to have worked beautifully; even the slanted spacing that I never tested looks right!

6. Toby Bartels says:

You say that the isotherms follow $P \propto V^{-1}$, but the curves in the picture don't look like that.

The one on the bottom is the most egregious; it actually has a positive slope on the right side.

• John Baez says:

As I said, “this is actually a very inaccurate picture, but it gets the point across”.

Hyperphysics has a more accurate looking picture, but it felt a bit busy to me:

Here’s one from Lumen Learning:

I was too lazy to draw my own.

7. I guess that is because of the equipartition theorem:
$\delta U=\frac{n}{2}N k\delta T$.
The adiabatic process moves from one isotherm to another. If $n\rightarrow\infty$, the variation in temperature becomes negligible. So the adiabats become closer and and closer to the isothermal curves in high dimensions.

• John Baez says:

That sounds right. To understand an adiabat we need to imagine an insulated container of gas expanding. As it expands, its internal energy goes down like this:

$dU = - P dV$

Since

$dU = \frac{n}{2} N k dT$

its temperature drops. But when the number $n$ of degrees of freedom of each molecule is huge, it only takes a small change in temperature to create a given change in internal energy. So as the insulated container expands, the temperature only drops a little. So in this case, the adiabats are almost isotherms!

• Toby Bartels says:

Just checking here … T is a state variable, so we can write dT instead of δT (or đT or whatever), right?

• Yeah, to be precise I should write
$dU=\frac{n}{2}NkdT$, both $U$ and $T$ are state variables. I use $\delta$ to denote the variation in the quantity, kinda sloppy notation.

• John Baez says:

I wrote $\delta U$ just because Wei-Xiang did. I’ll fix that.

• Perhaps it is not so relevant to this topic. The adiabatic index is crucial to explain why black hole can form automatically from a self-gravitating monatomic fluid (or gas) sphere if it is sufficiently relativistic in (3+1).

To determine the stability of a self-gravitating fluid sphere, one has to perturb the equilibrium solution and obtain the so-called radial pulsation equation. In order to check the instability, we need to know the compressibility of the fluid, which is characterized by the adiabatic index of the fluid, $\gamma$ from $P\sim V^{-\gamma}$.

In Newtonian gravity, the pulsation equation goes as:
$\delta\ddot{R}+\omega^2 \delta R=0$, where $\omega$ is the oscillation frequency after the perturbation of the star radius R. And $\omega^2 \sim\gamma_{\rm cr}-\langle\gamma\rangle$, where $\gamma$ is the average adiabatic index over the whole fluid, the $\gamma_{\rm cr}$ is the critical value when the fluid becomes unstable. If $\gamma_{\rm cr}>\langle\gamma\rangle$ the perturbation is unstable and we expect it would collapse into a black hole.

Interestingly, $\gamma_{cr}=2(1-1/n)$, where $n$ is the space dimension. (There is a simple derivation from Misner, Thorne and Wheeler (1973) in (3+1) but it is easy to generalize to (n+1).)

For ideal gas, $\gamma$ goes from $1+2/n$ to $1+1/n$ as the velocity dispersion $(v/c)$ of the fluid goes from 0 (non-relativistic) to 1 (ultra-relativistic). Solving the inequality, we get $n<3$ or $n<4$ is the dimension that makes the fluid stable. For example, $\gamma_{\rm cr}=4/3$ and $4/3<\gamma<5/3$, it is still too stable in Newtonian context.

However, in GR (from Chandrasekhar 1964), $\gamma_{cr}=4/3+$ correction terms of $mathcal{O}(v/c)$, that is the reason why in (3+1) the fluid can collapse into a BH if the fluid is sufficiently relativistic.

8. 1Writer says:

Interesting! How could we visualize ‘adiabats’ in some kind of geometrical format? Or how adiabats are geometrically related to the isotherms? I am trying to think through Robert Kiehn’s suggestion of the resemblance of Cartan’s magic formula to the first law. Ali

9. westy31 says:

It is fun to consider adiabatic expansion in a rocket, where the work does convert to displacing a piston, but to kinetic energy of the gas.
On an atomic scale, all energy remains in the form of kinetic energy of the atoms, but the kinetic energy goes from random motion to all in the same direction.
To make a rocket, we can imagine connecting the gas to a long tube, with the open end of the tube far away at vacuum. To simplify things, we make the container very heavy, so that the kinetic energy that is “lost” to the space ship is negligible. (The designers of the rocket would define “loss” in the opposite sense!) Note that we could always reconvert the kinetic energy into other forms by colliding the rocket jet with some turbine blades.
Although the total kinetic energy of the atoms remains constant, we can expand until the temperature (a measure of the random component of kinetic energy), becomes almost zero. Where does all the entropy go? To the larger volume that the gas occupies! The volume of phase space remains constant during the expansion.
The higher the dimension of space, the faster the velocity of the rocket gas, because the kinetic energy of the other directions get focused into the direction of the rocket.

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