# Kinetic energy of molecules in liquid state?

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?

• Comments are not for extended discussion; this conversation has been moved to chat. – orthocresol Mar 22 '18 at 12:32

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.

• 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. – Gaurang Tandon Jan 17 '18 at 14:12
• 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. – theorist Jan 3 at 4:25
• Lattice vibrations – porphyrin Jan 3 at 9:28

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.

• 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.. – theorist Jan 12 at 6:37
• ...weighted by their fractional accessibility. Would this always be the case? – theorist Jan 12 at 6:41