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My book (book link) has this question:

The kinetic energy of molecules at constant temperature in gaseous state is:

  1. more than those in the liquid state
  2. less than those in the liquid state
  3. equal to those in the liquid state
  4. None of these

The answer given is (3). It gives the reason as KE per molecule is still $=3/2\cdot k\cdot T$. So, the KE are both same!

But, I wonder, how can the kinetic energy of atoms in gaseous state be equal to that in the liquid state? Being in the gaseous state implies rapid and random motion of molecules, with little to no intermolecular attraction, spreading in all directions of any container. A liquid, however, is bound by intermolecular forces and is not so free to move.

In fact, I even wonder whether the formula $\text{KE per molecule }=3/2\cdot k\cdot T$ is valid in the liquid state, as it's derived from the kinetic theory of gases.

What exactly is correct in this situation here?

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – orthocresol
    Mar 22 '18 at 12:32
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The average translational kinetic energy of a molecule is $3kT/2$ irrespective of whether the molecule is in the gas, liquid, or solid phase. In the liquid the motion giving rise to kinetic energy is restricted to a narrower range about the potential energy minimum than it is in the gas phase. The equipartition theorem is quite general. A derivation is given here; Derivation of mean kinetic energy

For ideal gases, the result may also be derived via kinetic theory. However, the most general derivation comes from the equipartition theorem.

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    $\begingroup$ Woah! That math passed over my head too quick! Though I'll accept the truth that (I still don't know too much physical chemistry :P) average KE only depends on temperature and is independent of phase of the substance. And that this is based on the equipartition theorem. $\endgroup$ Jan 17 '18 at 14:12
  • $\begingroup$ I don't see how particles in a solid could have translational kinetic energy. All their kinetic energy would be vibrational, no? There are exceptions, such as the free electrons in metals—but those are exceptions. $\endgroup$
    – theorist
    Jan 3 at 4:25
  • $\begingroup$ Lattice vibrations $\endgroup$
    – porphyrin
    Jan 3 at 9:28
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    $\begingroup$ A vibration is not the same degree of freedom as translation $\endgroup$
    – Maurice
    Jun 14 at 13:09
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A deep dive for the sceptics

Since this is a question about which there seems to be a lot of misunderstanding, it seems worthwhile to take a deeper look, even though the given answers are perfectly satisfactory.

First, some terminology clarification. When people say something like “the kinetic energy of molecules in a liquid is the same as in a gas at constant temperature”, the “kinetic energy” in question is very specifically the average molecular translational kinetic energy. Mathematically, this is

$$K.E. = \frac12m\langle v_x^2\rangle + \frac12m\langle v_y^2\rangle + \frac12m\langle v_z^2\rangle$$

Where $\langle v_x^2\rangle$ is the average of the square of the x-component of the velocity of the molecule’s center of mass for a large population of molecules.

Note that rotations and vibrations in which the molecular center of mass does not move do not contribute to this total, even though there is atomic kinetic energy in those movements. Those kinetic energies are part of the total energy of the molecule, but are not part of the kinetic energy that we are referring to in the context of thermal equilibrium between phases.

So the claim that is being made is that this specific sum of translational kinetic energies is the same for molecules at the same temperature regardless of state. As I’ll explain below, this is true for systems that behave classically, which is a good approximation for most cases, but at low temperatures the quantum behavior of actual molecular systems causes deviations from the classical result.

Equipartition

You can find much more thorough information on equipartition elsewhere. What we need to know here is that the equipartition theorem tells us that in a classical system at thermal equilibrium, energy is evenly distributed between all “quadratic” energy terms. Our kinetic energy terms represent some of these quadratic energies, since the velocity terms are squared. The other type of quadratic term that is of interest is the potential energy of a harmonic oscillator, which has the form $\frac12 k \langle\Delta x^2\rangle$, where $ \Delta x$ is the deviation from the equilibrium position.

In this classical framework, if a gas and a solid are in thermal equilibrium, then the three molecular translational KE terms of the molecules in the gas must have the same amount of energy as the three molecular translational KE terms of the molecules in the solid. Thus the sum of the three terms is the same for a gas as for a solid. This sum is $\frac32 k_B T$ per molecule or $\frac32 RT$ for a mole of molecules.

Equipartition is directly observable in the heat capacity of substances. Because all of the quadratic energies must remain equal to each other, one cannot increase the molecular translational energies without simultaneously increasing the energies of all of the other quadratics. Since the temperature is proportional only to the molecular translational kinetic energy, the total number of these quadratic terms determines the amount of total heat that has to be added for a given increase in the molecular translational KE, ie the heat capacity of a substance.

Quantum effects

Of course, molecules are not purely classical systems, so we need to consider how quantum effects change this result. A main difference between classical and quantum systems is the quantization of energy levels. These energy levels can be spaced far enough apart that there is not enough energy at a given temperature to excite the system from the ground state to the next higher energy level. This results in some quadratic energy modes not containing as much energy as would be predicted by equipartition. A commonly given example is the intramolecular vibration of dihydrogen, which is not observed to contribute to the heat capacity except at high temperatures. We say that this vibrational mode is “frozen out” at lower temperatures.

In the gas phase, translational modes can be treated as classical continuous energies as long as the container in which the gas is held is sufficiently large. We need now to determine if the same is true for solids.

So the question we must now answer is whether molecular translations can be frozen out in solids, and if so, at what temperatures?

The nature of translations in solids

Within a molecular solid crystal, each molecule typically cannot move very far from its equilibrium position. When it does move from that position, it is pulled and pushed back towards the equilibrium position by interactions with adjacent molecules. Because the center of mass of the molecule moves a short distance and then moves back, these movements are typically described as vibrations rather than translations. It is for this reason that some textbooks will (confusingly) say that solids do not have translations, only vibrations. Because the center of mass of the molecule is in motion, however, these motions contribute to the molecular translational kinetic energy terms as we have defined them above. So with respect to equipartiation, they provide the three KE terms that sum to $\frac32 k_B T$. Mathematically, though, they can be modeled quite well as quantum harmonic oscillators just like intramolecular vibrations (with stationary COM) can. To distinguish between the two, the translational vibrations are often called “external vibrations” or “lattice vibrations”, while the intramolecular vibrations are “internal vibrations.” [That means that these translations also provide potential energy terms to the equipartition, but those aren’t of concern here.]

Because the energy levels of quantum harmonic oscillators can be widely spaced (as in the $\ce{H2}$ vibration example above), it is entirely possible that these translational vibrations in solids can be frozen out at sufficiently low temperature, so we need to know more about the frequencies involved.

Conceptually, we can anticipate that since intermolecular forces in molecular solids are weaker than intramolecular forces and the molecular masses are greater than atomic masses, the frequencies (and therefore vibrational temperatures) of the external (translational) vibrations are expected to be lower than those of intramolecular vibrations, so they would require lower temperatures in order to be frozen out. This is indeed the case.

Example

For crystals with Z molecules in each unit cell, there will necessarily be a total of 3Z orthogonal external vibrational modes spread across a range of frequencies. The highest of these tend to be in the far infrared region with wavenumbers around 100 cm$^{-1}$ or less. For example, solid carbon disulfide has two IR-active modes with wavenumbers around 65 cm$^{-1}$[1]. Solid iodine has three modes between 40 and 60 cm$^{-1}$ and three at even lower wavenumbers.

Using carbon disulfide as a test case, let’s look at the contribution of these modes to the heat capacity at the melting temp, when the solid and liquid phases are in thermal equilibrium. Carbon disulfide has two molecules per unit cell, so there are six external vibration modes, and the two at 65 cm$^{-1}$ are the highest frequency. The melting temperature is 161 K. Using the standard formula, we find that a vibration with wavenumber of 65 cm$^{-1}$ contributes 0.97 R to the molar heat capacity at constant volume at 161 K, only 3% less than the value of 1R predicted by equipartition in a classical system. [Only $\frac12 R$ is translational for each mode. The other half is in the mentioned potential energy of the "vibration". So the contribution to $C_v$ is 3% less than a gas phase translational mode.]

In order to reduce this contribution to less than 90 % of the classical value, we need to go down to nearly 80 K, and this, remember, is for the highest frequency translational vibrations. The contribution of the lower frequencies will remain higher, and so the overall translational KE will be higher than 90% of the predicted value. In this example, therefore, we can conclude that the statement that the translational KE is equal between two phases at thermal equilibrium is a good approximation except at very low temperature.

The same analysis can be conducted with other molecular solids with similar results.

[1] Ishi, K. and Takahashi, S. (1985) Temperature dependence of the infrared active translational modes of solid carbon disulphide. Journal of Chemical Physics 82, 1476. doi: 10.1063/1.448422

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  • $\begingroup$ When you perform a vibrational analysis you remove modes that are associated with translation of the CM or rotation of the inertial axes from consideration of the vibrational normal modes. $\endgroup$
    – Buck Thorn
    Jun 21 at 20:49
  • $\begingroup$ Then in the harmonic treatment half of the energy is potential, the other half of the vibrational energy is kinetic (expectation value), with motion about the CM. At least that's how I remember it. $\endgroup$
    – Buck Thorn
    Jun 21 at 20:51
  • $\begingroup$ @BuckThorn You're right about removing translation from vibrational analysis in the case of the gas phase normal mode analysis, but not for modeling of solid lattices. And you are correct that for any harmonic oscillator, half of the energy is potential and half kinetic (hence R per mode rather than 1/2R), but that KE is not counted towards the KE in the kinetic temperature. Hence the specific definition in the beginning of molecular translational KE. $\endgroup$
    – Andrew
    Jun 21 at 21:19
  • $\begingroup$ @BuckThorn - here's a good summary reference on modeling motions in solid lattices: aip.scitation.org/doi/10.1063/1.1678617 $\endgroup$
    – Andrew
    Jun 21 at 21:21
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    $\begingroup$ @BuckThorn - Right. And I'd add that although the Debye temp for atomic solids is often quite high, so are the melting and boiling points, so if we are considering two phases of the same substance in thermal equilibrium, the temperatures required to have two phases are high enough that the assumption of approximate equivalence of kinetic energy between phases is often still valid. $\endgroup$
    – Andrew
    Jun 22 at 12:16
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But, I wonder, how can the kinetic energy of atoms in gaseous state be equal to that in the liquid state? Being in the gaseous state implies rapid and random motion of molecules, with little to no intermolecular attraction, spreading in all directions of any container. A liquid, however, is bound by intermolecular forces and is not so free to move.

You would probably have no objection to saying that the temperature of a gas can be equal to the temperature of a liquid. In fact, over time (when there is nothing else going on), two phases in contact will approach temperature equilibrium, i.e. will have the same temperature.

In general, this happens through heat exchange, which could be through radiation, convection or conductance. The latter happens through collisions between particles of the two phases at their interface (let's say there is no particle exchange to make it simple). What happens in a collision depends on the speed and mass of the particles, and on what kind of intermolecular interactions they have. No matter what the details, though, after many collisions between particles of two phases, the kinetic energy will be distributed evenly (equi-partition theory mentioned in answer by porphyrin).

The reason that the same kinetic energy "looks" different in a gas, liquid and solid are the different intermolecular interactions. In a solid, there are strong and persistent intermolecular interactions. A given amount of kinetic energy allows the particle to "wiggle" a bit, against a strong intermolecular force toward the equilibrium position. In a liquid, the particle will diffuse a bit fast at higher temperature, bumping against other particles and making and breaking non-covalent bonds. In a gas, the particle moves pretty fast. Some of the kinetic energy is also present in bond vibrations (all phases) and in molecule rotation (gas phase).

In fact, I even wonder whether the formula KE per molecule =3/2⋅k⋅T is valid in the liquid state, as it's derived from the kinetic theory of gases.

Because of the equipartition theorem, answer (3) of the exercise posted in the question is correct in a rigorous way. In fact, the new 2019 SI definition of the Kelvin temperature scale is based on it, relating temperature to kinetic energy and the Boltzmann constant, which is now fixed to a constant value.

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  • $\begingroup$ I've been wondering about this recently: Does the same T for different phases of the same substance mean the same KE? We know this isn't generally the case for different substances—e.g., at the same T, a diatomic ideal gas will have more KE than a monatomic ideal gas, because of the former's rotational dof's. But what about two phases of the same substance? Consider my answer here: chemistry.stackexchange.com/questions/129290/… For a liquid and gas, say, to have the same KE at the same T, they would need to have the same total dof's.. $\endgroup$
    – theorist
    Jan 12 at 6:37
  • $\begingroup$ ...weighted by their fractional accessibility. Would this always be the case? $\endgroup$
    – theorist
    Jan 12 at 6:41

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