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Heat capacity of gases again

Dear reader,

When I was teaching about the heat capacity of gases I went through the ways in which a gas can store energy. Now we will consider the gas to be an ideal gas, which is a series of hard spheres which have next to no volume, who do not interact with each other (unless they hit each other) which wizz around in a box. The gas molecules can bounce off the walls of the container thus exerting a force.

These moving gas molecules can move in three different dimensions thus we need to consider the kinetic energy stored in the movements in three different directions at 90 degrees to each other

If we consider nitrogen gas, then the mass of each molecule will be 28 atomic mass units, so one mole will have a mass of 28 grams.

Now at constant volume the heat capacity of a diatomic ideal gas is 5/2R, the reason it is higher than the 3/2R for a monoatomic gas (like argon) is that the molecule can spin and thus store energy spinning on the x axis, and the y axis. As it is linear it can spin all it likes on the z axis but it will never store any energy by doing so !

Energy of a rotating object = E = ½ Iw2

As the value for I for rotation on the z axis is zero, you should be able to see that on one axis the molecule is never going to be able to store any energy.

For rotational energy of a diatomic molecule

E = B J(J+1)

J is the number of the spinning state (it can only be an integer), while B is a constant for the molecule. If we take B as being given by.

B = h / (8 pi2 c I)

Lets do some dimensional analysis on this

As for a diatomic molecule, I = [(m1m2)/(m1+m2)]r2

Where r is the bond length, then the units of I will be kg m2

Thus the units for B will be Js / (m s-1 kg m2) = N m s / (m s-1 kg m2) = N s2 kg-1 m-2

This looks like a bit of a mess, then we read it a second time and see that it says under that equation (on the wavenumber scale). So it is clear we have the wrong equation.

At a university physics site it gives a different expression for B

B = h2 / (2 x 4 pi2 x I)

If we do the dimensional analysis for this then we get

J2 s2 / kg m2 = N2 m2 s2 / kg m2 = N2 s2 / kg = N m s-2 s2 = N m = J

Much nicer !

Then the moment of inertia for the nitrogen molecule will be given by

{(2.32481 x 10-26 kg x 2.32481 x 10-26 kg) / (2.32481 x 10-26 kg + 2.32481 x 10-26 kg)} x (1.1 x 10-10 m)2 = 1.41 x 10-46 kg m2

So as planks constant (h) is equal to 6.62606957 x 10-34 J s, we can calculate B

B = (6.62606957 x 10-34 J s)2 / (8 x pi2 x 1.41 x 10-46 kg m2)

B = 3.95 x 10-23 J

Now if J changes from 0 to 1 we will need 2B of energy

This will be  7.9 x 10-23 J

The frequency for this photon energy is 1.19 x 1011 Hz (119 GHz)

This puts these photons in the upper end of the microwave spectrum, the highest frequency I am currently licensed for is 250 GHz but I do not know if anyone has managed to build equipment for this frequency. I suspect that this frequency allocation is something which is there for the future.

But back to our problem of nitrogen, based on the energy of the photons and the number of molecules in a mole, we can say that one mole of 119 GHz photons will have an energy of 47.6 joules.

This corresponds to about 4 cm-1 radiation.

As Cv for argon is equal to 3/2 R = 8.3144621 J K-1 mol-1 x 2/3 = 12.47169315 J K-1 mol-1 then it is very clear that at room temperature gas molecules will have the energy required for them to start to spin.

At room temperature a mole of an ideal monoatomic gas has 3717 joules of internal energy.

On the other hand if we consider a typical vibrational frequency for a triple bonded diatomic then it will be about 2200 cm-1.

A mole of an ideal diatomic gas will have at room temperature will have 6194 joules of energy, the inverse wavelength (wavenumber) for the nitrogen corresponds to 66000 GHz (66 THz) which means that one mole of these photons will have an energy of 26335 joules.

Thus as a typical molecule only has about 25 % of the energy required to start to store energy by vibrating, as a result very few nitrogen molecules will be able to get more than the zero point energy for their vibrations.

Thus we only consider the translational energy and the rotational energy of a typical diatomic gas when we try to work out Cv and Cp.


2 Responses

  1. I like the way you simplify things and make them easy to understand. Thanks for Thermodynamic topics you are actually developping on your blog!!

    • Dear Jeremie,

      Thanks for the comment, I hope that I am doing some good my explaining things on my blog.

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