4
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – orthocresol Mar 22 '18 at 12:32
7
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Gaurang Tandon Jan 17 '18 at 14:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.