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Consider some liquid in a beaker. I am measuring its temperature using a thermometer. When i put my thermometer's bulb in the liquid, the particles of the bulb gain the same kinetic energy as of the liquid particles. Now, my question is that Do the particles of thermometer's bulb attain the same translational, rotational and vibrational energy (vibration of atoms within a particle) as of the particles of the liquid? I have read somewhere that there is no effect of increase of rotational energy or vibration of molecules (as a whole) on temperature readings. Why so? Do the vibrations of the molecules don't change kinetic energy of the thermometer's bulb's particles?

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    $\begingroup$ Near the room temperature, many of the vibrational degrees of freedom in common molecules are frozen. They don't get any energy. The rest get $kT/2$ per each. $\endgroup$ – Ivan Neretin Sep 11 at 9:31
  • $\begingroup$ i am not specifying the temperature of the liquid.It is at any temperature. So, is rotational energy of the molecules or intramolecule vibrations get transferred to the bulb?? $\endgroup$ – user36956 Sep 11 at 11:38
  • $\begingroup$ It surely does. $\endgroup$ – Ivan Neretin Sep 11 at 11:42
  • $\begingroup$ Let me suppose that the molecules of substance under measurement has 0 translational energy. So, the molecules of the substance which are very very near to the bulb having only intramolecular vibrations hit the particles of the bulb, wouldn't there be increase in the translational kinetic energy of those particles? In this way, if the substance under measurement has no translational kinetic energy of its molecules, but intramolecular vibrations, then too there will be some temp reading on the thermometer. Theoretically, its temp was to be 0( as translational kinetic energy is 0) $\endgroup$ – user36956 Sep 11 at 11:55
  • $\begingroup$ That's right (except the last sentence). Also, when the atoms of those molecules hit the atoms of other such molecules, they give them some translational kinetic energy, so in a very short order all energy in the sample will be distributed equally between all possible degrees of freedom. $\endgroup$ – Ivan Neretin Sep 11 at 12:06
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1) You are correct that the thermodynamic temperature is a measure only of the translational kinetic energy. Intramolecular vibrations do not contribute to the temperature.

2) You are also correct that rotational and vibrational KE can be converted to translational KE in a collision (as long as total momentum and kinetic energy of the system are conserved).

To reconcile these two statements, we need to remember that temperature is a bulk property of the population, not a property of individual molecules. If we say that a mole of gas is as 273 K, it does not mean that every molecule of gas has the same amount of translational KE. It only tells us the population average.

We also need to remember that the average distribution of energy between translation and other movements (eg vibration and rotation) in the ground state is a fixed property of a substance in a given phase, represented empirically by the heat capacity.

If we look at a single collision between two molecules, we might see a change in the net translational KE if some energy is transferred to a rotational mode (resulting in a loss of translational KE). However, that means that the molecule in question is now in an excited rotational state. Across the entire population, that excitation will be offset by a different molecule dropping from an excited rotational state to the ground state by transferring some rotational KE to translational KE in a collision. The number of molecules in the excited state will on average remain at the value we calculate using stat mech.

So just as there are fluctuations in the distribution of translational KE among molecules within a bulk population due to changes in speed, there are also fluctuations due to transfers of energy between different types of motions, but the population average remains essentially constant if the population is large enough that stat mech approaches are relevant.

In your example, the local translational KE at one point in the thermometer bulb might increase, but it will be offset by a decrease somewhere else, such that the measured temp is an accurate reflection of the translational KE of the combined thermometer+material system.

In the specific case of solids, the idea of translational KE is confusing, since the molecules seem stationary. However, there are small movements of the centers of mass of the molecules relative to each other even if the molecules are confined. This is distinct from vibrations, in which the center of mass remains fixed as the atoms move relative to each other.

Consider, for example, $\ce{CO2}$. If the C atom remains in a fixed position and the C=O bonds stretch asymmetrically, there is a net movement of the center of mass. Likewise if the O's remain stationary during the same asymmetric stretch movement. Likewise for a bending mode. In reality, the motion of the atoms at any given moment is a complicated mix of the possible vibrational, rotational and translational movements.

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Do the particles of thermometer's bulb attain the same translational, rotational and vibrational energy (vibration of atoms within a particle) as of the particles of the liquid?

No, but they will have the same temperature. Solids don't have any translational energy, the atoms time-averaged positions are constant. If you want to link the temperature to the energy of the particle, it is easiest to start out with a monoatomic ideal gas because here, there is a linear relationship.

Relationship between change in temperature and energy

Although under some ideal circumstances, particles have an energy of $\frac{1}{2} k T$ per degree of freedom, how much energy is needed to raise the temperature of a sample depends on the sample (and the temperature). The heat capacity is a measure of change in energy per change in temperature. So the heat capacity tells you the total energy going into all degrees of freedom when the temperature increases. The splitting up into translational, rotational and vibrational energy depends on the nature of the sample (for example, the physical state) and the temperature.

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    $\begingroup$ I think you mean $\frac{3}{2}kT$. $\endgroup$ – Buck Thorn Sep 12 at 6:47
  • $\begingroup$ if the particles of solids have 0 translational energy, then why do they have non-zero temp? isn't the temp of a substance the average translational energy of particles? $\endgroup$ – user36956 Sep 12 at 8:09
  • $\begingroup$ @user36956 No. The vibrational energy of the particles in a solid also matters. Heat isn't just translational energy. $\endgroup$ – matt_black Sep 12 at 11:10
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    $\begingroup$ Vibration is movement of atoms within a molecule relative to each other such that the center of mass does not move. If the center of mass moves, it’s translation. When you’re dealing with a metal instead of a covalent solid, the independent unit is an atom instead of a molecule. When a metallic solid is melted, the atoms move freely from each other, so it makes sense. $\endgroup$ – Andrew Sep 14 at 10:32
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    $\begingroup$ @KarstenTheis - It doesn't have to "know" anything. The termperature is a measure of the ability to transfer heat through collisions (ignoring for this discussion radiative transfers). At the bulk population level, that's a measure of the translational kinetic energy only. I'll try to explain that in a longer answer. $\endgroup$ – Andrew Sep 14 at 11:11

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