My textbook says translational energy level is so closed, and it basically covers the whole EMR spectrum, so any temperature beyond absolute zero is high enough for molecules to have translational energy. Why does ice only has vibrational energy? And if water molecules are all vibrating, do they have kinetic energy?


All such things as 'translational energy' or 'vibrational' are more of extrapolating concepts. Yes, there is rough approximation of $C_v$, but that works only for gases with low density.

When density in your material increases, the pressure which you 'expect' is defined with not only 'impulse' part, but also 'force' part of pressure.

$P = nkT + \frac{1}{V_{elementary}} \sum r_i F_i$

Imagine lots of springs connected together. When you press it, springs are stressed and atoms are getting closer. But nothing moves in this case.

Why does ice only has vibrational energy?

Models of $H_2 O$ which are known to me all include full accounting of 3d motion of molecule. So throwing it away could work for some cases, only if desired result is expected to be rough estimation $\pm$ one order of magnitude.

PS. "tip3p" for more.


Yes translational energy levels are very close together. These can be calculated with particle in a box type quantum mechanics for molecules in the gas phase, which are free to move about.

In ice the molecules are held together next to each other and cannot move about - they can vibrate about the positions that they are held in. So their average position is fixed, but they can move about around that position a little.

Vibrating molecules have kinetic energy. They also have stored potential energy - it is a bit like a mass on a spring described by simple harmonic motion where the energy moves between kinetic energy ${1\over 2}mv^2$ to potential energy ${1\over 2}kx^2$ where $k$ is the spring constant, and $x$ the displacement from the equilibrium position.

  • $\begingroup$ Ok, I got it. It's not like completely fixed. There is room for a little of translational motions. $\endgroup$
    – user40003
    Nov 11 '14 at 18:33
  • $\begingroup$ @user40003 - yes that's it $\endgroup$
    – tom
    Nov 11 '14 at 20:17

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