# What temperature actually a measure of?

1. Jan 15, 2012

### ovais

Hello guys
In my previous tags come up with some molecular description of substances. And it has been suggested that temperature, a degree of hotness or coldness is basically, a direct indication of the average kinetic energy of the molecules the body made of. After a deeper look toward this saying it creates some confusion to me. The theory behind the confusion lies in three things.
1. Temperature is a direct measure of average kinetic energy of the molecules a body has.
2. different things may have different degrees of freedom and in each degree of freedom molecule has posses same energy equal to 1/2κT.
3. Gases having different degree of freedom have different specific heats.
By assuming that temperature is a measure of the average kinetic energy of the molecules, we must say the different gasses at the same temperature must have same average kinetic energy. TRUE?
Now consider the following-
I take two gasses one monoatomic ( having 3 degree of freedom) and other diatomic (having 5 degree of freedom) in two separate and closed containers initially at the same temperature, they therefore will have same number of moles(avagadro's law). As such the monoatomic gas has specific heat 3/2R and the diatomic one has 5/2R. Because ΔT=Q/(nCv), it means when same amount of heat Q (energy) is given to these gases the rise in temperature in the gasses will be different. TRUE?
Also as such the volume of the gas remains constant, no work is done by the gas and as per the first law of thermodynamic the entire heat must be used to raise the internal energy.As such this energy now get stored as kinetic energy of the gas.TRUE?
Drawn conclusion- by giving same heat ,same amount of energy per molecule has been increased TRUE? these gasses initially had same temperature and therefore initially there molecules must same average kinetic energy now equal amount of kinetic energy has been raised by giving same heat.But from ΔT=Q/(nCv), ΔT is found to be different for the two gasses and thus their final temperature must be different(as initially they have same temperature. So the molecules of the two gasses at the second state have same average kinetic energy but different temperature. From this it appears as the excess (2 vibrational) degrees have just consumed heat and their increases in kinetic energy contribute nothing to temperature.This very observation forces me not to accept that temperature as a measure of the average kinetic energy of molecules of the body. i will be very thankfull if somebody clear this to me. REGARDS!

2. Jan 15, 2012

### Staff: Mentor

Realize that temperature is a measure of the average translational KE. It does not reflect other forms of internal energy, such as rotational and vibrational energy. (Otherwise everything would have the same specific heat capacity!)

(Edit: In light of maverick_starstrider's comments, I realize that this is a somewhat misleading way of expressing my point. )

Last edited: Jan 15, 2012
3. Jan 15, 2012

### maverick_starstrider

Uh... what? The equipartition theorem states that each quadratic degree of freedom adds 1/2kT to the energy, the energy is averaged over ALL quadratic degrees of freedom INCLUDING vibrational and rotational ones when relating it to temperature (in the classical limit). Also, if you run the math, each linear degree of freedom contributes kT.

Classically the specific heat of a substance is just nk/2 where n IS the numbers of (quadratic) degrees of freedom (again INCLUDING rotational and vibrational).

4. Jan 15, 2012

### Staff: Mentor

I agree with everything you say here, but I don't see how it contradicts what I said. (Perhaps I didn't express it very well.)

If two substances (with the same mass) have the same temperature, it means the molecules have the same average translational KE. Their internal energy is not necessarily the same.

5. Jan 15, 2012

### maverick_starstrider

Your description below is correct (classically). Adding the same amount of energy leads to different increases of temperatures and the two system are no longer in thermal equilibrium. That's what specific heat is, water for example, has a very high specific heat so one can take a lot of energy from it before it drops a little bit in temperature (which is why we have a climate on earth).

However, your 1. is not correct, temperature is classically the average energy PER (QUADRATIC) DEGREE OF FREEDOM, not just kinetic. Also you need to remember that this is only a classical result, you may notice that if the equipartition theorem were really true that the specific heat is a constant which means that the entropy should be infinite as you go to absolute zero, thus violating the third law. This is because you're working in the classical limit.

The absolutely most correct definition of temperature (that is equally true in quantum) is that it is the THING which is the same when two systems that are free to exchange energy are in thermal equilibrium. It is the derivative of the energy with respect to entropy keeping volume and particle number constant

6. Jan 15, 2012

### maverick_starstrider

What about system that don't have kinetic energy and do have, say, quantized vibrational energy? There's nothing special about KE, classically temperature is equivalent to the average energy per QUADRATIC degree of freedom, that could be 1/2m v^2, sure. It could just as easily also be 1/2 k x^2, 1/2 I omega^2, etc.

7. Jan 15, 2012

### Staff: Mentor

True.
False. (At least not solely translational KE--it gets distributed among all degrees of freedom.)

Last edited: Jan 15, 2012
8. Jan 15, 2012

### Staff: Mentor

I was restricting my comments to the example used by the OP--gases which may or may not have additional degrees of freedom beyond translational. (But you make a good point.)

9. Jan 15, 2012

### klimatos

For gases, molecular mass is irrelevant. If two gases have the same temperature, their molecules will automatically have the same mean KE of translation. What one lacks in mass, it will make up for in the square of the translational velocity.

On a separate issue, I seem to detect an assumption that just because a gas possesses a theoretical degree of freedom, that that degree of freedom will be exercised--or if exercised, all molecules of the gas will exercise it.

For real gases, this is simply not true. That is why specific heats vary with temperature.

At temperatures normally encountered in the free atmosphere, nitrogen (N2) has a CV of 4.6 out of a theoretical value of 5.0. This suggests that not all nitrogen molecules have both rotational modes excited. At the other end, you will never see H2 or O2 molecules with their vibrational modes excited because those gases disassociate long before the necessary gas temperatures can be reached.

Please correct me if I am wrong, but I thought that the failure of equipartition theory to apply to specific heats was one of the problems that led to the development of quantum thermodynamics.

10. Jan 15, 2012

### klimatos

All my references define vibrational energy as kinetic energy.

11. Jan 15, 2012

### klimatos

When applied to monatomic gases at temperatures around 300°K, the equipartition theorem gives theoretical specific heat values (CV) that are in close agreement with observed values. When applied to diatomic gases, the agreement is good, but not great. When applied to polyatomic gases, some gases have good agreement and some have poor agreement.

As you depart from 300°K, agreement deteriorates for all gases, and generally fails at very low and very high temperatures.

12. Jan 15, 2012

### Ken G

One cannot necessarily associate T with the mean kinetic energy of a gas. In addition to issues like what the constant of proportionality should be (involving issues like dimensionality and translational vs. internal degrees of freedom), there are more basic problems like quantum mechanical effects (such as in white dwarf stars, which can have essentially zero T, yet vast mean kinetic energy per particle).

I think it is better to understand what temperature really is, rather than how it is manifested in specific situations. First of all, temperature is not really a property of the gas itself, it is actually a property of a thermal reservoir that can be associated with the gas by virtue of the assumption of "thermal equilibrium" with that reservoir. Anything that is in thermal equilibrium with a reservoir at T will be at T, whether it is a gas or a solid or anything, and being at T won't necessarily tell you about the mean kinetic energy of the gas, that depends on other details.

What T means for the reservoir is, if you multiply by k, the energy that, once removed from the reservoir, causes the number of configurations the reservoir has access to to be reduced by a factor of e. If the system can compensate for that by increasing its own access to new configurations, then the energy will come across until that is no longer the case. At that point, the system is also at T.

13. Jan 15, 2012

### maverick_starstrider

I'm not quite sure what the point of this post is. Are you trying to tell me what a classical approximation is? Thanks I'm well aware, which is why I put parenthetical (classical) statements everywhere.

14. Jan 15, 2012

### maverick_starstrider

Um... I'd get new references then. Quantized vibrations, a la a harmonic oscillator, are a result of potential wells, this is why they count as two degrees of freedom, a kinetic component and a potential component. Both contribute in the classical equipartition theorem. Or one could simply be looking at a model like the Einstein solid, in which case their is no kinetic component at all.

15. Jan 16, 2012

### Ken G

Yes, any mode that allows for the system to reach new configurations will draw energy from the reservoir, not just translational modes. Temperature doesn't care what type of mode it is, all it cares is that the reservoir is going to lose a factor of e in number of accessible configurations if it parts with energy kT, so it will only continue to part with that energy if the system can reach a factor of e more configurations in the process. This is the fundamental meaning of T, and it does put kT of energy into every mode that corresponds to a quadratic term in the system Hamiltonian, as mentioned above. In some cases, that connects with translational kinetic energy of a particle, sometimes vibrational, but that's not what T actually is in a more general sense.

16. Jan 16, 2012

### klimatos

Alright, everybody, slow down!

Let's go back to the OP by Ovais. In it, he is trying to make sense of the notions that two gases may have different specific heats and the same temperature.

Firstly, we can assume that we are talking about real gases and not ideal gases, since his examples are not identified as ideal gases. Secondly, we can assume room temperatures, since no unusual temperatures have been specified.

This means that discussions should be focused on real gases at normal temperatures. Liquids, solids, harmonic oscillators, and white dwarfs should be considered to be outside the limits of discussion.

Kinetic gas theory defines gas temperatures as T = 2/3 u/kB. Here, u is the mean kinetic energy of translation along the gas molecules' true paths and kB is Boltzmann's Constant. This is an ideal gas equation, but both observation and experiment have shown that it can be applied to real gases under the conditions specified.

Note that only the kinetic energy of translation is measured. Kinetic energies of rotation and kinetic energies of vibration/libration are ignored. They play no role in gas temperatures under the specified conditions.

For the real gases of the atmosphere at normal temperatures, not all possible degrees of freedom are always exercised. All of the atmospheric gases exercise all three degrees of translational freedom at the range of temperatures normally encountered in the free atmosphere.

For diatomic gases, some molecules will not exercise both degrees of rotational freedom, and it is likely that some may not exercise any. Some will exercise both rotational degrees and one or two (Cl2) vibrational modes as well. For polyatomic gases, the picture is extremely murky.

All of this is useful in explaining the different specific heats of real gases, but play no role in explaining gas temperatures. For real gases, degrees of freedom and specific heats are completely irrelevant to the measured temperatures of those gases.

Conductive thermometers measure the mean impulse energy transmitted to their sensing surfaces (measured normal to the surface) of that subset of molecules that impact upon their sensing surfaces in some measure of time. That's all that they are designed to measure and that is all that they are capable of measuring.

Last edited: Jan 16, 2012
17. Jan 16, 2012

### maverick_starstrider

Again, I'm not really sure the point of your post. You've come onto a thread where the main question seems to have been completely answered then write these post that have an air of correcting but actually take more general and fundamentally correct answers and repeat them in some oddly specialized, less accurate and I'd almost say engineery way. Your agenda seems to be to furiously pick at nits but you haven't actually found any. Temperature and thermal equilibrium is so conceptually beyond Van Der Waals gases and pressure gauges and it is important to understand what is generally true and what is approximately true and one should really avoid jumbling the two up without clarification. Such as your description which takes the classical result of the equipartition theorem jumbled up with the quantum notion of frozen degrees of freedom without sufficient clarification.

18. Jan 16, 2012

### Ken G

I agree the issues with the different specific heats was resolved early on, all the modes where energy can go will draw energy at T (if T is high enough to get around quantum mechanical issues-- not just in white dwarfs, but all types of gases), so there is no fixed proportionality between internal kinetic energy and T (unless one restricts to 3D translational modes). What I was saying is that all of that is very specific to the situation mentioned in the OP-- but if people want to know what T really is, it is something that transcends those kinds of details, and has more to do with the reservoir that it characterizes than the specific system that is in contact with that reservoir.

19. Jan 17, 2012

### ovais

Sorry guys for being so late in response. Actually i didn't expect such of long and idverse discussion on the seemingly simple concept of temperature. But temperature appears to be a very difficult property to unerdstsnd at sub-particle level and in that term is really cery difficult to define. However as is being said that temperature is that property which is same when two bodies are in thermal equilibrium. this defintion does not create problems. However what i wanted to know actually was answered by some ones.When kineric theory says that average kinetic energy of molecules as a meaure of temperature it gives wrong impression when eeling with two systems with ifferent egrees of freedom. This situation was solved by maverick_starstrider when he says that temperature is a measure of average kinetic energy of moleculrs PER dEGREE OF FREEdOM. I going over this very thing my question is NOT towards why do theoritical an practical value of specific heat varies in case of multiedgrees of freedom, neither quantum issues regadring the deeper insight to temperature. I again want to verify that classically is it true that temperature is related the average kinetic energy of molecules PER dEGREE OF FREEdOM? as suggested by maverick, are all satisfy? for me i think this solves the problem an classically valid. THANKS

20. Jan 17, 2012

### Ken G

Yes, you got the answer early on-- in the cases you are interested in, you can connect temperature to the energy per accessible degree of freedom. This won't work in general because of how quantum mechanical issues affect accessibility, or because in general T is a property of a thermal reservoir (and that's where it has a clear definition, where kT is the energy that causes the thermal reservoir to access a factor of e more configurations). But the main point is that a system acquires the meaning of T from the reservoir it is in thermal contact with, and from that it follows, under the circumstances you are interested in, that kT/2 of energy goes into each accessible quadratic degree of freedom (whether quadratic in velocity, like kinetic energy for free particles, or quadratic in position, which must be included if the particles are bound).