The first law of thermodynamics states that $\Delta U=q+w$ where $ΔU$ is the increase in internal energy of the system, $q$ is the thermal energy supplied to the system and $w$ is the work done on the system.

I have a few questions regarding this law due to my elementary understanding of internal energy.

  1. If an ideal gas is being compressed without supplying heat, $q=0$ and $w$ is positive. Now $ΔU$ must also be positive. This means that the microscopic kinetic energy of the particles $K_{E}$ is increased and hence temperature increases. But since the gas is being compressed i.e. the intermolecular distance is being decreased, does this play a role in decreasing $ΔU$?

  2. If the gas above were being compressed at constant temperature then $q$ would have been negative so as to maintain a constant temperature. In this case does not the decreasing of intermolecular distances make $ΔU$ negative since $K_{E}$ remains constant this time?

  3. Consider ice being melt into water at a constant temperature of 273K. Now the intermolecular distance of the molecules is being decreased here since water is denser than ice. So why isn't $ΔU$ negative here? How exactly do the microscopic potential energies of the water molecule come into play here?


2 Answers 2


If you compress an ideal gas adiabatically, the average kinetic energy of the particles will increase because collisions with the moving piston will increase the kinetic energy of the colliding particle. For an ideal gas, the intermolecular distances have no bearing on the internal energy.

More specifically, suppose the piston is moving inwards at velocity $u$ and a gas particle hits the piston wall at velocity $v$. Because the piston is much more massive than the particle, you can assume that it gets completely reflected, so the particle will come away with velocity $v+2u$. In practice, $v$ is much bigger than $u$, so this amounts to only a small increase in $K_E$ per collision, but the piston is moving slowly and there will be many such collisions over the course of the compression.

If the compression is being done isothermically, then you are assuming that any excess kinetic energy is being dumped into the heat reservoir. The flow of energy is then from the moving piston to the kinetic energy of the gas particles through collisions with the piston, and from there to the reservoir, and this is assumed to be at such small increments that the mean kinetic energy does not change. Again, for an ideal gas, the intermolecular distances have no bearing on the internal energy.

Regarding water ice, the process of melting is obviously complicated and you cannot simply wrap it up into a simple function of the mean interparticle distance. There are complex interactions between the molecules and these depend quite sensitively on their relative orientation, with the end result being the crystal structure of ice. Your question is, basically, "what is it about the molecular interactions in water that makes the lowest-energy crystal form have the structure it does?", which is obviously a much more complicated question in and of itself (though, of course, it's very well studied).


For an ideal gas, we neglect all intermolecular interactions (except for the trivial case of elastic collisions). $\Delta U$ therefore depends purely on the kinetic energies of the particles. The 'decreasing of intermolecular distances' has no role to play here. However, these things are taken care of to some extent by the van der Waals equation

Melting is a slightly different thing - intermolecular forces play a big role here. In a very non-detailed treatment, we say that melting requires the 'latent heat of melting'. This is in fact a form of internal energy that, to use the example of melting of ice to water, represents the stronger bonding effects (less energy) in ice as opposed to water. It is not so much about the particle density as about the intermolecular interactions themselves. In this case, $\Delta U$ would be positive as the latent heat is supplied to the ice (and this is why we need to heat ice to get water, as opposed to the other way around, as your suggestion implies).

  • $\begingroup$ Your answer makes things much clearer. By the way, during melting of ice, is it correct to say that work done by/on ice is zero? If yes, how can I conclude this? $\endgroup$
    – PdX
    Apr 20, 2015 at 19:11
  • $\begingroup$ There is no (or negligible) expansion involved, and therefore no work is done. $\endgroup$
    – AV23
    Apr 20, 2015 at 19:20

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