# In a real gas is heat carried by molecular vibrations as well as translational velocity?

It's clear that the mean velocity of gas particles determine what we measure as temperature, but is there also a significant amount of energy carried in molecular vibrations?

For example, at the same temperature and pressure would you expect water molecules to carry more heat than nitrogen molecules since the water molecules have more modes of vibration?

The simple answer is "look at the heat capacity".

For a simple mono-atomic gas, the heat capacity is $\frac32 R$ per mole. For a diatomic gas, it's $\frac52 R$ - because two new modes (rotations) appear that can be excited and contain energy ... after which the equipartition principle does the rest.

Note that the diatomic gas does not have a heat capacity of $\frac72 R$ as you would expect if rotation about its axis or vibration along its axis were involved in storing energy. The reason is that these modes are quantized - and that the thermal energies are insufficient to excite a reasonable fraction of the molecules into the higher energy state. Thus the assumptions that go into the equipartition principle don't apply - and the heat capacity reflects that.

If you have more complex molecules, you can get certain bending modes that will be excited more easily - these are capable of acting as reservoirs of thermal energy, and will affect the heat capacity. For such gases the answer would be "yes".

Note that for water, a significant additional complication is the formation of hydrogen bonds between molecules at lower temperatures; this affects the heat capacity of water in the liquid form in particular (heat capacity changes with temperature for liquid water; and to a lesser degree, for steam). According to this table, the heat capacity of steam actually increases with temperature. This suggests that the vibration modes become more important as temperature goes up. This is even more evident in the heat capacity for liquid water: while it is initially almost flat at 4.2 kJ/Kg/K, around 260°C it rises to 5 kJ/kg/K and then quickly to 10 kJ/kg/K around 250 °C.

There is a good write-up of all this on wikipedia - where it is stated that the energy spacing of the vibrational modes is inversely proportional to the reduced mass of the molecule: this implies that "heavy" diatomic molecules (like $\rm{Br_2}$) have their vibrational mode excited, while the lighter ones do not. An idealized plot of the heat capacity of hydrogen as a function of temperature (from http://theory.physics.manchester.ac.uk/~judith/stat_therm/node81.html which credits P. Eyland from the University of New South Wales ... but I could not locate the original) looks like this:

showing that the heat capacity increases first when the rotational modes are accessible for equipartition of energy, and that vibrational modes follow at some (much) higher temperature.

EDIT

The heat capacity for steam at 100°C is 27.5 J/mol/K, which is about 3.3 R. It stays near that value until you get to T=700°C, where it rises to 3.5 R, and then to 1000°C where it is almost 4.0 R. In the case of a molecule like water, there are degrees of freedom associated with bending. There is a nice overview of the bending modes of water here; IR spectroscopy is the usual way to probe these modes, and the page gives diagrams of the vibration modes as well as the wavenumbers of these modes:

The wavenumbers of the three modes are given as 3657/cm, 1595/cm, 3756/cm (values change with pressure, state - so these are very approximate). Now $\rm{1 ~cm^{-1} = 1.24\cdot 10^{-4} ~eV}$ so the energy associated with these modes is 0.45, 0.20, 0.47 eV respectively. At room temperature, the average energy per degree of freedom is $\rm{\frac12 kT = 0.0125 ~eV}$ (rule of thumb: kT = 1/40 eV at room temperature). That means the probability of exciting these modes is pretty small - and that the $\nu_2$ (symmetrical bending) mode will be the first to be excited as temperature rises.

• This. Many thermodynamics books will have a figure showing data on specific heat for some real gas as a function of temperature. The steps and plateaus that are a signature of the model and are present in the data. – dmckee Feb 9 '17 at 1:10
• @dmckee - after your comment I went and found just such a diagram. Thanks for the suggestion. – Floris Feb 9 '17 at 16:17
• @Floris thanks for an excellent answer. My more focused interest is comparing the heat transfer differences between diatomic molecules like N2 and O2 and water vapor. Since the mass of hydrogen is about 1/16 that of O, my first thought would be that the energy carrying capability of H2O would be closer to monotomic molecules rather than diatomic. Or does the presence of the H and the H-bonds provide more significant capacity? In this case you have one heavy atom and two much lighter. Does the polar nature of H2O perhaps provide more 'springiness' to carry more resonant energy, heat? – docscience Feb 9 '17 at 18:48
• @docscience Thanks. I explored the specific question in your comment more in an edit to my answer. – Floris Feb 9 '17 at 19:49

Assuming the system is in thermal equilibrium, what you are looking for is the equipartition of energy which states that each possible degree of freedom in a molecule contains the same amount of energy once equilibrium has been reached. This is only true for degrees of freedom which depend on quadratic terms.

If you visit a nice question over on Chemistry.SE, they probe whether this applies to vibrational modes. The vibrational mode contains two quadratic terms, and therefore it does, in fact, contain a portion of the energy when things are in equilibrium. In fact, it contains more of the energy than the translational and rotational modes because it has two quadratic terms instead of one.

• This is not entirely accurate for a "true gas". If the quantum energy spacing of a particular mode is large compared to the thermal energy, you will not excite this mode, and will not be included in the equipartition theorem. This is the case for vibrational modes of diatomic gases at room temperature: en.wikipedia.org/wiki/Heat_capacity#Diatomic_gas – anon01 Feb 8 '17 at 21:42
• @ConfusinglyCuriousTheThird That's a good point. I think the equipartition of energy is as close as we can get analytically to what really happens, but it is at the end of the day just an approximation. – tpg2114 Feb 8 '17 at 21:48
• No, it really isn't... this can be treated with statistical models. The accepted answer gives a good summary of this. – anon01 Feb 17 '17 at 4:55