Do rotational degrees of freedom contribute to temperature?

[Philip Koeck]

I'd like to pose a related question: Given a hypothetical (or maybe not?) system consisting of diatomic molecules that cannot be excited vibrationally (because the vibrational energy levels are too far apart) and cannot move away from the fixed locations they are in, but can rotate freely. Is it possible to increase the temperature of this system by adding heat to it?

 

[rude man]

The rotational degrees of freedom contribute to the internal energy and not to the temperature.
I am not even sure it make sense to say that "it contributes to the temperature".

For an ideal gas there is a 1-1 correlation between internal energy and temperature: U = U(T) only.

 

[nasu]

Are you sure it's a 1 to 1? You can have two gases with same temperature but different internal energies. Is this what you call 1-1 correlation?

Not that I really see how is relevant for the passage you quote.

 

[DrDu]

I'd like to pose a related question: Given a hypothetical (or maybe not?) system consisting of diatomic molecules that cannot be excited vibrationally (because the vibrational energy levels are too far apart) and cannot move away from the fixed locations they are in, but can rotate freely. Is it possible to increase the temperature of this system by adding heat to it?

Something like this is observed in solids containing nearly spherical molecules, like e.g. methane. Then, near the melting point, the heat capacity increases because the methane molecules can rotate at their lattice positions.

 

[Mark Harder]

I'd like to pose a related question: Given a hypothetical (or maybe not?) system consisting of diatomic molecules that cannot be excited vibrationally (because the vibrational energy levels are too far apart) and cannot move away from the fixed locations they are in, but can rotate freely. Is it possible to increase the temperature of this system by adding heat to it?

Rotational energy levels can be promoted by absorption of microwave energy, if that helps. I'm not so sure about whether temperature will increase as a result of rotational excitations. The way I think of it, translational motion increases temperature by transferring momentum from molecules in the sample to those in a thermometer. There would have to be some other kind of coupling between rotational states in sample and thermometer in order to case the latter to respond. Same would apply to translational states, I suppose.

 

[lychette]

When you measure temperature you measure temperature and not any energy, translational or other kind.
The fact that the temperature is related to the motion of the molecules does not mean that temperature is the energy of this motion.

You are starting with some unsound assumptions in the OP.

One is:
"The temperature is the average kinetic energy of the particles"
which is obviously wrong. Temperature and energy are different quantities, with different units.

And then you say that "only includes translational degrees of freedom: velocity" which I suppose you mean to refer to the energy and not the temperature. (the sentence as written is confusing). Definitely the energy includes all degrees of freedom that are thermally excited, so it is not true if you mean it to refer to energy.
And to say that "the temperature includes" does not make sense. So what do you actually mean?
Is there some reference or definition that you are confused about?

Sure, the average energy per degree of freedom is proportional with the temperature but this does not mean that each degree of freedom (translational or not) contributes to temperature. The temperature is not a sum of terms from various degrees of freedom, as the word contribution would (I think) imply.

I think the statement is that 'temperature' is a measure of the translational KE of molecules, doubling the temperature means that the average KE of molecules has also doubled.
Does anybody know what 'temperature' is other than the average translational KE of molecules or something that thermometers measure?

 

[nasu]

Does anybody know what 'temperature' is other than the average translational KE of molecules or something that thermometers measure?

Yes, I suppose some people know that temperature can be defined in terms of the derivative of entropy versus energy. See a textbook or the wiki page on temperature.

 

[Nugatory]

Yes, I suppose some people know that temperature can be defined in terms of the derivative of entropy versus energy. See a textbook or the wiki page on temperature.

And that definition is essential to understanding how additional degrees of freedom affect the relationship between temperature and heat/energy...

 

[lychette]

And that definition is essential to understanding how additional degrees of freedom affect the relationship between temperature and heat/energy...

How do these ideas contribute to the understanding of temperature (T) that arise from the behaviour of ideal gases (PV = nRT)

 

[Mark Harder]

Yes, I suppose some people know that temperature can be defined in terms of the derivative of entropy versus energy. See a textbook or the wiki page on temperature.

Yes, the thermodynamic, phenomenological definition is quite useful. It predicts the possibility of negative temperature, which has been observed. I think what the posters are groping for is a statistical mechanical definition of temperature.

 

[Philip Koeck]

Something like this is observed in solids containing nearly spherical molecules, like e.g. methane. Then, near the melting point, the heat capacity increases because the methane molecules can rotate at their lattice positions.

Interesting! I wonder, however, whether that answers my question. There are two ways to interpret this. One could say that at the melting point rotations become available and suddenly energy is distributed over two extra DOFs so we can add energy without increasing the temperature (like a phase change). Therefore the rotations have nothing to do with the temperature. Or we could say that energy is simply distributed over 8 DOFs now instead of the 6 vibrational ones and the rotations do correspond to temperature just like the translational energy in the vibrations. In the latter case many textbooks are a bit inaccurate in their formulation.

A decisive experiment would require a system where molecules can only rotate, i.e. molecules held in place by very strong forces (no vibration) but free to rotate.
Would T be proportional to the average rotational energy?
Is it even meaningful to speak of temperature in such a system?

 

[DrDu]

You can define temperature also for spin systems which have no translational degree of freedom.

 

[sophiecentaur]

If you have two objects that are measured to have the same temperature and you bring them into contact. Neither will change its measured temperature because they are in equilibrium. This has nothing to do with the internal energy with the objects.
You complained, earlier on, that you were told what temperature is not. Fair enough. Well, the above paragraph gives a way into a definition in as far as the temperature difference is what tells you which way Energy will flow and the rate of that flow.

 

[Philip Koeck]

You can define temperature also for spin systems which have no translational degree of freedom.

I was looking for something closer to the everyday experience. I don't suppose you can measure the temperature of a spin system with a thermometer. The same is true for negative Kelvin temperatures that have been mentioned. For example a pumped laser is described with negative T simply because an exited state has higher occupancy than the ground state, which can be described by a Boltzman factor with negative T, but nobody will try to measure this with a gas thermometer for example. The question remains as I see it: Can pure rotations of molecules "carry" temperature? How would you even measure it (refering to Mark Harder's earlier post)?

 

[DrDu]

I was looking for something closer to the everyday experience. I don't suppose you can measure the temperature of a spin system with a thermometer. The same is true for negative Kelvin temperatures that have been mentioned. For example a pumped laser is described with negative T simply because an exited state has higher occupancy than the ground state, which can be described by a Boltzman factor with negative T, but nobody will try to measure this with a gas thermometer for example. The question remains as I see it: Can pure rotations of molecules "carry" temperature? How would you even measure it (refering to Mark Harder's earlier post)?

I suppose you can excite molecules like HCl in the gas phase with a microwave field with a Boltzmann spectral distribution. If the field is intense and the gas dilute, only the rotational degrees of freedom will be excited immediately after switching off the microwave field. So after excitation, the gas can be described by two temperatures, one rotational and one translational. Collisions of the molecules will equilibrate this difference.
This multiple temperature distributions are also common in the outer atmosphere of Earth where the highly energetic UV light from the sun yields electronically excited atoms who nevertheless have a relatively low translational temperature.

 

[Death eater]

I couldn't understand this?

 

[haruspex]

I couldn't understand this?

The internal energy resides partly in the motions of the molecules and partly in the vibrations of atoms within the molecules.
When you measure the temperature you are measuring the motions of the molecules.
If a high proportion of the energy is in motions within molecules then you will need to raise the internal energy more for the same measured rise in temperature.

 

[Death eater]

The internal energy resides partly in the motions of the molecules and partly in the vibrations of atoms within the molecules.
When you measure the temperature you are measuring the motions of the molecules.
If a high proportion of the energy is in motions within molecules then you will need to raise the internal energy more for the same measured rise in temperature.

Please check the photo I posted?

 

[haruspex]

Please check the photo I posted?

I did, and my previous reply paraphrases what the book says.