# Why would a ~1 cm thick layer of argon be a significantly poorer conductor of heat than air?

In yesterday's new Periodic Video, Argon (new) - Periodic Table of Videos, after about 07 min 00 sec, Sir Martyn Poliakoff says:

The final, and I have to admit perhaps slightly boring application of argon, is in double glazing (of windows) to keep our houses warm. Because argon has a heavy atom, the atoms don’t move very fast in the gas phase, and therefore they’re bad at conducting heat (emphasis added).

So if you have double glazing (that’s two sheets of glass in your window) if you put argon between the glass, it is much more effective at insulating the inside of the house, keeping cold out, than if you use air, because air has a much higher conductivity of heat.

So, if you look out of the window, you may find that you are looking through argon gas.

At a given temperature or kinetic energy per molecule, velocity will roughly scale with the inverse square-root of the mass, so I estimate that the argon atoms would be moving

$$\sqrt{\frac{28}{40}} \approx 0.84$$

only 16% slower than nitrogen molecules would be moving.

Why then would a trapped ~1 cm thick layer of argon gas be a significantly poorer conductor of heat than nitrogen gas or air?

I'm wondering if it has more to do with the fewer internal degrees of freedom of a monatomic gas, than it does with the mass of the molecule. There would be a similar collision rate, but each collision of an atom of Ar could transfer less energy to the glass surface than a molecule of N2 could transfer.

• Ar has no molecular vibrations in which to store heat. Oxygen and nitrogen do. At impact diatomic molecules can transfer kinetic energy of linear motion as well as vibrational energy to the substrate. – MaxW Jul 31 '18 at 6:33
• @MaxW sounds like you agree with the last paragraph? – uhoh Jul 31 '18 at 6:34
• A problem with air insulation, as opposed to argon or even N2 - O2 mix, is that air contains water vapor, which efficiently passes heat, particularly over the temperature range where condensation and evaporation are possible (e.g. winter in temperate or colder climates). – DrMoishe Pippik Jul 31 '18 at 7:53
• $\mathrm{C_p[\ce{Ar}] ≈ \frac{3}{2}R}$. $\mathrm{C_p[\ce{O2}] ≈ C_p[\ce{N2}] ≈ \frac{5}{2}R}$ – A.K. Jul 31 '18 at 15:13

The thermal conductivity coefficients of some gasses in units of W/(m•K) are helium 0.142, argon 0.016, air 0.026, methane 0.03, propane 0.015, bromine 0.004, and steam 0.018. These are all rather similar except for helium and bromine.

Using the kinetic theory of gasses the thermal conductivity coefficient is given by $\kappa \sim n\bar c\lambda C_V$ where $n$ is concentration, $\bar c$ the mean speed, $\lambda$ the mean free path and $C_V$ the heat capacity. Using definitions $n\lambda \sim 1/\sigma^2$ where $\sigma$ is the diameter of a gas atom or molecule; $\bar c \sim 1/\sqrt{m}$ and the heat capacity is $C_V \sim sk_\mathrm B$ where $k_\mathrm B$ is Boltzmann constant and $s$ is a multiplier depending on the type of species, atom, diatomic molecule, etc.

The conductivity coefficient at constant temperature is therefore $\displaystyle \kappa \sim \frac{sk_\mathrm B}{\sigma^2\sqrt{m}}$.

Thus we can see that helium will have high $\kappa$ as it has a small diameter and mass, and bromine will have a small value of $\kappa$; large diameter and mass. The other gasses are similar to one another because although the heat capacity increases between an atom and molecule and increases further as the molecule has more vibrational modes, the diameter and mass also increase when the molecule has more atoms and these terms cancel out to some extent.

So it is not just about speed, but also size, mass and number of vibrational modes. Using argon over dry air would seem to give a small advantage, but it may not be cost effective, i.e. heat saved vs. initial cost and energy needed to use argon.

(The molecular heat capacity increases above that for translational motion with the number of squared terms in the energy, potential and kinetic, each term contributes $k_\mathrm BT/2$ to the energy and so $k_\mathrm B/2$ to the heat capacity. Each vibration populated thus contributes $2k_\mathrm B/2$ and rotations the same for a linear molecule, but $3k_\mathrm B/2$ for a not linear one.)

• Thank you for your answer! I'm curious what the multiplier $s$ would be (approximately) for argon versus nitrogen. – uhoh Jul 31 '18 at 10:17
• When you say the thermal conductivity coefficients are all "rather similar" (argon .016, air .026) you may find people will disagree. Argon is 40% lower than air which I would call "significantly different". – matt_black Jul 31 '18 at 10:43
• @matt-black Of course it is a matter of opinion what is similar or not, but compared to He and Br they are similar and each differ by a lot more to these gasses than to one another. – porphyrin Jul 31 '18 at 11:59
• A monatomic gas has $3k_B/2$ ( 1/2 in each of x, y and z). A diatomic has 2 rotation directions because rotation along the bond does not change anything so rotations add another $2k_B/2=k_B$. Then vibrations add another $k_B/2$ because of their kinetic energy and the same for their potential energy making an additional $k_B$ per vibrational. The vibrations only count if vibrational quanta are populated at the prevailing temperature. Same applies for rotations but these are usually populated at room temperature, so for a diatomic $5k_B/2$ or $7k_B/2$ if the vibration is included. – porphyrin Jul 31 '18 at 12:06