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I have a doubt about radiation absorption and temperature increase. I was taught that the larger heat capacity of diatomic molecules over monoatomic molecules is the result of the possibility that the diatomic molecules may absorb some photon and begin vibrating or rotating instead of getting an increase in translational speed (which is proportional to the measured temperature). So, some of the energy that impinges on a molecule goes into vibration instead of raising the temperature, but in a monoatomic molecule the only interaction can be the increase of translational speed because there are no vibrational or rotational modes.

But now I learned that water molecules absorb the microwaves in an oven into the vibrational mode and this is what increases the temperature. See Charles H Martin's comment here.

Can anyone explain the apparent contradiction? If absorbing in the vibrational mode does increase temperature, then what is the proper explanation for the higher heat capacity of diatomic molecules over monoatomic molecules?

Thanks

user3113647
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  • Because there is coupling between the various modes, particularly through collisions. VIbrational energy can become translational energy and vice-versa. – Jon Custer Apr 16 '20 at 20:25
  • I'm not sure what to make of that, Jon. You are saying that the diatomic molecules have a higher heat capacity for some reason other than the idea that some of the absorbed energy goes into vibration?

    What is the correct explanation for the higher heat capacity of diatomic molecules, then?

     – user3113647 Apr 16 '20 at 20:31
  • (Also, why do people just comment on questions anymore instead of answering them??) – user3113647 Apr 16 '20 at 20:32
  • Changes to work practices while telecommuting play a role for me. As to why the specific heat changes, since some of the energy can go into vibrations, not translations, you have to put more energy in to get the average translational velocity to increase the same as it would for a monoatomic species. But, that doesn't prevent vibrational and translation energy exchange - that only means that both modes will thermalize relative to each other. – Jon Custer Apr 16 '20 at 21:21

1 Answers1

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Mathematically it’s easy to see. Specific heat is defined as the temperature derivative of internal energy. And since there’s translational, vibrational and rotational energy contributing in polyatomic molecules, the specific heat is scaled similarly. See here for details.

Intuitively speaking, consider a box filled with polyatomic gas. Let us try to measure the temperature of the gas by sticking in a thermometer. The temperature of the gas can only be measured when there are gas particles colliding with the thermometer to transfer the energy. But if part of the energy is stored as vibrational and rotational, these don’t contribute much to the energy transfer. It is in this sense we say that polyatomic gases can store more heat without contributing to temperature which implies higher specific heat.

Now coming to your question, when we microwave water, it absorbs the photons and this is converted to the rotational (not vibrational, that’s IR) energy if the water molecules. This by definition is increasing the energy of the system, it is heating the system. And that’s why the temperature rises. Now because polyatomic systems have higher specific heat this just means that we need to heat more to increase the temperature by a unit as compared to a monoatomic gas.

In other words, if you took two systems (one mono and one poly) that absorb a given microwave wavelength at a temperature $T$ and microwave them for a fixed amount of time, the monoatomic gas would have a higher final temperature than the polyatomic gas. Because the polyatomic has more ways to store the energy. This is the effect of specific heat.

Superfast Jellyfish
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  • Nice answer, I do see an issue here, also with the link provided however: It seems counter-intuitive that the low-energy modes (rotation and vibration) are frozen-out at low temperatures. Shouldn't those be carrying all the energy, if mode populations follow a Maxwell-Boltzmann distribution? – AtmosphericPrisonEscape Apr 16 '20 at 21:36
  • @AtmosphericPrisonEscape, I’m not quite sure I understand your question. What is implying low-energy modes being frozen at low temperature? – Superfast Jellyfish Apr 16 '20 at 21:51
  • You would expect low-energy modes to NOT be frozen out at low energies – AtmosphericPrisonEscape Apr 16 '20 at 22:07
  • Thank you, but I find part of your answer confusing. You say in paragraph 2 that if part of the E is stored in rotation, it isn't contributing much to the heat transfer and thus to the temperature. But then in paragraph 3 you seem to say the exact opposite - that storing in rotation causes the temperature to rise.

    Just real simply: Take two identical samples and same energy input to each. S1 absorbs all E in translation. S2 absorbs all E in rotation. Temperature of S1 increases but not S2. Yes or no?

     – user3113647 Apr 16 '20 at 22:07
  • @user3113647 temperature in both increases. But the temperature rise in S1 is higher than that in S2. This is because even when stored in vibration or rotation there are still energy transfer happening from collision. But the probability of the collision is much lower. What I said was “ these don’t contribute much“ which is not the same as no transfer at all. – Superfast Jellyfish Apr 16 '20 at 22:12
  • @AtmosphericPrisonEscape, I’m sorry for being unclear. What I meant to ask is where am I suggesting that the low-energy modes are frozen at low temperature? – Superfast Jellyfish Apr 16 '20 at 22:14
  • I am sorry for also having been unclear. You do not suggest that in your answer, but it is in the link you have provided and as well as its applications in your question. The issue is, that with the thermal distribution given as $P(E)\sim exp(-\Delta E/k_B T)$, one would expect low energy levels to be excited easily. Those levels with small $\Delta E$ are the rotational modes first, then vibrational, then electronic. Hence it seems strange, that at low temperatures the high-energy modes should be populated first. If I am misremembering something fundamental here, please point it out. – AtmosphericPrisonEscape Apr 16 '20 at 22:39
  • I see. Thanks. So there is more to the story than what my p-chem professor let on.

    Let me ask a final (hopefully) question. The higher temperature of S1 is easy to see: the molecules are bumping into the thermometer harder (transferring more momentum) since they have a higher translation energy than before. For S2, the higher temperature comes from the rotating molecules banging into the thermometer and losing some of their rotational energy because the thermometer applies a torque impulse that slows the rotation? Something like that?

     – user3113647 Apr 16 '20 at 22:40
  • @AtmosphericPrisonEscape, yes the low energy modes get filled first. However there’s also density of states we need to factor in. Higher energy levels have many more states with the same energy. Roughly speaking if energy goes as $E=n_x^2+n_y^2+n_z^2$ where the n’s are integers, then the number of ways to get $E=100$ is much greater than $E=5$ say. This means that higher energy states contribute more to the integral than the lower ones. Also truly speaking lowest energy mode is the translational one as the spectrum is continuous there. – Superfast Jellyfish Apr 16 '20 at 22:46
  • @user3113647, yup exactly like that. But remember that they don’t just lose rotational energy but also translational. Each molecule is in general translating, rotating and vibrating simultaneously. – Superfast Jellyfish Apr 16 '20 at 22:48
  • Thank you so much. That makes sense, I just was confused because my p-chem professor said that in diatomic molecules some of the energy added goes into vibrational and rotational modes instead of raising the temperature. But it makes sense that there could still be energy transfer to a thermometer from these modes, even if it isn't as large as the transfer from translation. Can you recommend a work that quantifies these energy transfers (i.e., to a thermometer) from a molecular mechanics point of view? It would be interesting to see why they are less quantitatively. – user3113647 Apr 16 '20 at 22:57
  • @user3113647, you can look up any stat mech books for this. Particularly the section on canonical partition function. An easy introduction to this can be seen in Daniel Schroeder’s Thermal Physics book (the chapter on stat mech, I forget the number). It is highly readable. And I think what your prof meant when he said instead of raising temperature is that the raise in translational energy is equal to the energy you’re giving (because some go to other modes). And since mainly it is translation causing the transfer of energy, the temperature is lower. – Superfast Jellyfish Apr 16 '20 at 23:12