# (Average) Kinetic Energy of Molecules

1. Nov 26, 2012

### Leoragon

I'm confused with this topic. However, I think I know a bit. There's something to do with the temperature and it affects the energy of the molecules.

Can someone help?

2. Nov 27, 2012

### AbsoluteZer0

Temperature is proportional to the kinetic energy of the molecules of a substance. Kinetic energy is the energy of an object due to its motion.

Heat is generated during the process of transferring energy from one body to another. Take note that the direction of transfer is always from hot to cold. (Second law of thermodynamics.) When two objects that are in thermal contact reach the same temperature they reach thermal equilibrium.

There are various types of energies such as rotational energy, translational energy, etc. All of these energies of a specific body combined form the internal energy of that body.

The three laws of thermodynamics are:

1) When heat is added to a system, it is transformed to an equal amount of some other form of energy. In the simplest sense, energy can't be created or destroyed. Rather, it can only be transformed.

2) Heat always flows from hot to cold. (There are other less friendly definitions of the second law.)

3) Absolute Zero temperature cannot be reached (In classical mechanics. In quantum mechanics it can, but you don't need to worry about this yet.) Absolute zero is -273.15°C or 0°K (Kelvin.) The reason it can't be reached is because when a substance is at absolute zero, the atoms of that substance come to a stand still. This is not possible as kinetic molecular theory states that all atoms are perpetually vibrating.

There is another law known as the Zeroth Law which states that if System A is in thermal equilibrium with System B and System B is in thermal equilibrium with System C, then System A is in thermal equilibrium with System C.

Last edited: Nov 27, 2012
3. Nov 27, 2012

### the_emi_guy

Nice summary, but here is a nit-pick:

Temperature is *proportional* to average kinetic energy. Specifically for monatomic gases:

$$\bar {E} = \frac{3}{2}k_bT$$

4. Nov 27, 2012

### Rap

Another nit-pick - the zeroth law holds for any system, not just a gas.

5. Nov 27, 2012

### the_emi_guy

I don't see where AbsoluteZer0 implied that the zeroth law only applied to a gas.

6. Nov 27, 2012

### AbsoluteZer0

I edited that section and corrected the mistake.

7. Nov 27, 2012

### Leoragon

What does that formula mean?

8. Nov 27, 2012

### the_emi_guy

Temperature is directly proportional to average kinetic energy of molecules.
In other words if I double the temperature (kelvin scale), then I have also doubled the average kinetic energy of the molecules.

The constant of proportionality depends on the particular substance, but there is a whole class of substances called monatomic gasses that have the same constant.

$$\bar {E}$$
Average kinetic energy of molecules.

$$k_b$$
Boltzmann constant

$$T$$
Temperature

9. Nov 27, 2012

### Leoragon

Sorry, but can you put in an example for me? One with a monoatomic gas and another that's diatomic or whatever.

10. Nov 27, 2012

### haruspex

It's the energy per state rather than the energy per molecule. Diatomic gases have more states than monatomic ones in which to store energy, so pack more energy per molecule for the same temperature.

11. Nov 27, 2012

### the_emi_guy

An example of a monatomic gas is helium. Helium atoms live solitary lives, bouncing around all by themselves.

An example of a diatomic gas is oxygen. Oxygen atoms find another oxygen atom to pair up with to form an oxygen molecule O2.

The thermal energy in monatomic gasses is made up of the translational kinetic energy of the atoms, in other words the energy that they have because they are moving around.

The thermal energy in diatomic gasses consists of the same translational kinetic energy, but has some of its energy in the form of internal kinetic energy within the molecule itself. Imagine the two oxygen molecules vibrating back and forth with a spring connecting them.

Turns out that 20% of the thermal energy in a diatomic molecule like O2 is this internal energy.

Therefore for diatomic molecules:

$$\bar {E} = \frac {5}{2} k_b T$$

12. Nov 27, 2012

### Leoragon

What does that mean? Sorry for all these questions, I'm unfamiliar with this.

13. Nov 27, 2012

### K^2

Kinetic energy includes center of mass motion only. So it doesn't matter how many other degrees of freedom the molecule has. The average kinetic energy will always be 3/2 kbT.

The total internal energy will scale differently depending on molecular structure, and that has to be taken into account if you are considering heat capacities. But if you are interested in kinetic energy of molecules only, then you don't have to worry about any of it.

14. Nov 27, 2012

### Rap

Right - except very close to absolute zero temperature.

15. Nov 27, 2012

### the_emi_guy

Perhaps this is just semantics, but I was taught that all of the thermal energy was kinetic. The motion of the center of mass being the translational kinetic energy, plus the internal kinetic energy in the form of lattice vibrations/rotations etc. Quantum mechanics was required in order to get the right count of degrees of freedom, but that it was all kinetic energy.

16. Nov 27, 2012

### Leoragon

So, in a nut shell, temperature is proportional to the kinetic energy? And there's that formula that shows the average kinetic energy of a monoatomic gas. For diatomic, its 5/2?

17. Nov 28, 2012

### K^2

Vibrational DoF includes kinetic and potential contributions. That's where the double-count of these DoF comes from. Rotational is purely kinetic, yes, but usually when people talk about "average kinetic energy," they are referring to translational part only. Otherwise, it becomes unclear what you mean. Including kinetic term, but not potential term from vibrations would be just silly, for example. And looking at just translational and rotational doesn't make much sense either.

There is sense in considering just translational kinetic energy, however. For example, when you consider pressure of an ideal gas, only translational kinetic energy is relevant. So the formula is exactly the same for all gasses. This is the typical context in which the term "average kinetic energy" is most frequently used. And if you look at Wikipedia article on kinetic theory, you'll notice that it's most frequently referred to as just "kinetic energy" and twice as "(translational) kinetic energy".

So in context of gas kinematics, "average kinetic energy" will almost always refer to just the translational part.

It might be just semantics, but I also wouldn't say QM was necessary to count DoF. Just to figure out why not all of them contribute, and why some of them contribute only partially. That, of course, has to do with quantization of rotational and vibrational energies.

If you talk only about translational kinetic energy, which is what is usually meant by "average kinetic energy", then it's always 3/2.

If you look at total mechanical energy of a diatomic gas, you get 3/2 from translational DoF, 2/2 from rotational, and 2/2 from vibrational, of which 6/2 total is kinetic and 1/2 is potential energy. However, some of these will be "frozen out". Specifically, rotational degrees of freedom are usually inaccessible because the quantum of energy is much higher than available amount of energy at room temperatures. So you end up with roughly 5/2 total mechanical energy for diatomic gases.

Last edited: Nov 28, 2012
18. Nov 28, 2012

### Leoragon

What? So average kinetic energy is 3/2, and the total mechanical energy is 5/2? What happens when the molecule is triatomic? Is it the same?

19. Nov 28, 2012

### the_emi_guy

K^2
Ok,
Thanks for the correction on this.

Leoragon,
Its (3/2)kT and (5/2)kT.

I believe that when the molecules get bigger it depends on their structure, for example whether they are linear or non-linear.

Last edited: Nov 28, 2012
20. Nov 28, 2012

### K^2

With triatomic it depends on whether all three are different, and if not, how they are arranged. For CO2, H2O, and similarly structured molecules, you get something very close to 7/2, because rotational modes are still frozen out, but you pick up an extra vibrational mode.

You can look up these values easily. The quantity most directly associated with average mechanical energy of molecules in a gas is specific heat capacity at constant volume. So when average mechanical energy per molecule is 3/2 kbT, the CV = 3/2 R per mol of gas. R is the ideal gas constant, of course. You can also measure heat capacity of gas at constant pressure, allowing volume to expand as gas heats up. CP = CV + nR. Finally, the quantity that's most commonly used and measured is the heat capacity ratio, γ = CP/CV. For light monatomic gases, γ=5/3=1.67 almost perfectly. For diatomic, it's closer to γ=7/5=1.4, but here you'll start seeing significant temperature dependence. It's slightly higher at lower temperatures. For triatomic gasses with structure of H2O and CO2, it's close to γ=9/7=1.29. You can compare these to values on this page. The simple theoretical prediction works out very close at about 100°C, where it's hot enough to excite vibrational modes, but cold enough to leave rotational more or less alone.

As the molecule gets more complicated, this approach gets worse. At some point, you need to honestly consider QM and see what the probabilities are of exciting states at given temperature.