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If temperature is defined as the average kinetic energy per particle, and heat energy is defined as the total kinetic energy of all the particles (or more strictly, heat transferred is the total kinetic energy transferred to those particles), surely we can get heat by just multiplying temperature by the number of particles? If so, shouldn't molar heat capacity be constant across all substances?

I'm fairly sure I'm missing something that should be obvious here...

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before we even start with anything else, you should consider that not all kinetic energy is rotational. So, molecules with a nonzero moment of inertia will necessarily have more places to put their kinetic energy... –  Jerry Schirmer May 11 at 16:09
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In addition, state of a body cannot be ascribed "heat energy". Heat transfer is just a way how energy can be transferred. When heat is added to the system, its internal energy increases. This has contribution also due to potential energy of the particles, which is not connected to temperature the way kinetic energy is. –  Ján Lalinský May 11 at 16:14
    
Maybe for an ideal gas, but the interactions between molecules would influence the heat capacity of the gas. –  LDC3 May 11 at 17:46
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@JerrySchirmer - Perhaps you meant to write "not all kinetic energy is translational"? (Despite both statements being true, the latter seems to fit better with your last sentence. (looks like a lapsus digitae) –  Peltio May 12 at 9:45

2 Answers 2

up vote 10 down vote accepted

Temperature is not the average kinetic energy of a particle it is the average energy per mode.1

In very simple models (i.e. the monoatomic ideal gas) the number of modes per particle is fixed and can not vary, so that the heat capacity of these simple models is indeed fixed. And in fact, good approximations to mono-atomic, ideal gases (noble gases, other simple gases at low temperature but still low pressure) do all have nearly the same heat capacity.

More complicated real systems however have more modes. Depending on the system and the temperature molecular rotational and vibration modes may be present. At still higher temperature molecular and atomic excitation modes come into play. In crystal system phonon excitations are available.

So of these modes are only available when the temperature gets high enough, and an interesting thing to do is observer the heat capacity of a gas go through step-like increases as new modes become occupied.

This looks roughly like enter image description here (figure converted from the Wikipedia image at http://en.wikipedia.org/wiki/File:DiatomicSpecHeat1.png). The steps represent temperature where the mean energy per mode increases to the ground state energy of the newly accessible mode. They are not sharp because the actually energy in any particular microscopic mode is not guaranteed to be exactly the mean energy per mode, but could be a little more or a little less.


1 What is a mode? I'm glad you asked...

Without being too precises a mode is any distinct way for energy to be stored in microscopic physics inside a bulk material. That is the all the ways there can be "internal energy". For instance the translational motion of particles in a mono-atomic gas is three modes ($mv_x^2/2$, $mv_y^2/2$ and $mv_z^2/2$; and we distinguish them because we do study one and two dimensional systems and it matters). More complicated molecules can also rotate (a $I\omega^2/2$ contribution for every direction it can rotate) or vibrate ($kx_{max}^2/2$ for each vibration) and so on.

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Thanks for that, excellent answer! Does this make a difference to the derivation of pressure? It's usually explained only with reference to translational kinetic energy, but if kinetic energy is split between several modes, shouldn't that be reflected in pV=nRT? Or doesn't it make a difference because vibration and rotation can be transmitted to boundary surfaces just the same as if they'd been translation? –  Oolong May 11 at 20:32
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On the microscopic scale, pressure is the cumulative impulse of a lot of atoms bouncing off the container, and that depends only on the temperature (for kinetic energy) and volume (for frequency of collisions), so I believe that the ideal gas law continues to hold for gasses that remain diffuse and weakly interacting. But it will take more energy (heat or work) to change the temperature as much. –  dmckee May 11 at 22:35
    
I'm slightly confused about something else entirely. Average kinetic energy per mode would be in units of Joules/mode or whatever. Temperature is Kelvin, so how can you even claim they're the same? –  Mehrdad May 12 at 8:47
    
@Mehrdad It is not obvious that they are until you have studied the problem for a while. The kinetic theory of gasses is one of the nicest ways to go about convincing yourself of this because the internal energy is so simple. Consider what properties go with high and low temperature in a simple system like the ideal gas, how temperature fits into the laws of thermodynamics and how work is done on/by a gas trapped in a cylinder by a piston. Once you are convinced that they are two aspects of the same thing, notice the units of the Boltzmann constant. –  dmckee May 12 at 13:15
    
@Mehrdad, the fact that a unit can be expressed one way doesn't mean that's the only way it can be expressed! For example, gravitational field strength can be written as m/s^2 or as N/kg; both are perfectly correct. For a fun project, you could try thinking of as many alternative ways as you can to write the units for energy! –  Oolong May 12 at 13:16

If equipartition holds, temperature corresponds to average energy per degree of freedom, which - besides kinetic ones - include internal ones like vibrational and rotational degrees of freedom.

Even in cases of structurally similar molecules with the same degrees of freedom, because of energy quantization (in particular vibrational energy), heat capacity may still differ - see eg the heat capacity of diatomic gases.

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