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(Average) Kinetic Energy of Molecules

by Leoragon
Tags: average, energy, kinetic, molecules
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Leoragon
#1
Nov26-12, 11:07 PM
P: 41
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?
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AbsoluteZer0
#2
Nov27-12, 12:01 AM
P: 126
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 -273.15C or 0K (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.
the_emi_guy
#3
Nov27-12, 01:04 AM
P: 589
Quote Quote by AbsoluteZer0 View Post
Temperature is the average kinetic energy of the molecules of a substance.
Nice summary, but here is a nit-pick:

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

[tex] \bar {E} = \frac{3}{2}k_bT [/tex]

Rap
#4
Nov27-12, 10:28 AM
P: 789
(Average) Kinetic Energy of Molecules

Another nit-pick - the zeroth law holds for any system, not just a gas.
the_emi_guy
#5
Nov27-12, 11:32 AM
P: 589
Quote Quote by Rap View Post
Another nit-pick - the zeroth law holds for any system, not just a gas.
I don't see where AbsoluteZer0 implied that the zeroth law only applied to a gas.
AbsoluteZer0
#6
Nov27-12, 12:08 PM
P: 126
Quote Quote by the_emi_guy View Post
I don't see where AbsoluteZer0 implied that the zeroth law only applied to a gas.
I edited that section and corrected the mistake.
Leoragon
#7
Nov27-12, 06:32 PM
P: 41
Quote Quote by the_emi_guy View Post
Nice summary, but here is a nit-pick:

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

[tex] \bar {E} = \frac{3}{2}k_bT [/tex]
What does that formula mean?
the_emi_guy
#8
Nov27-12, 07:00 PM
P: 589
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.

[tex] \bar {E} [/tex]
Average kinetic energy of molecules.

[tex] k_b[/tex]
Boltzmann constant

[tex] T [/tex]
Temperature
Leoragon
#9
Nov27-12, 07:22 PM
P: 41
Quote Quote by the_emi_guy View Post
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.

[tex] \bar {E} [/tex]
Average kinetic energy of molecules.

[tex] k_b[/tex]
Boltzmann constant

[tex] T [/tex]
Temperature

Sorry, but can you put in an example for me? One with a monoatomic gas and another that's diatomic or whatever.
haruspex
#10
Nov27-12, 09:04 PM
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Quote Quote by the_emi_guy View Post
Temperature is directly proportional to average kinetic energy of molecules.
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.
the_emi_guy
#11
Nov27-12, 09:10 PM
P: 589
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:

[tex] \bar {E} = \frac {5}{2} k_b T [/tex]
Leoragon
#12
Nov27-12, 10:37 PM
P: 41
Quote Quote by haruspex View Post
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.
What does that mean? Sorry for all these questions, I'm unfamiliar with this.
K^2
#13
Nov27-12, 10:46 PM
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Quote Quote by the_emi_guy View Post
Nice summary, but here is a nit-pick:

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

[tex] \bar {E} = \frac{3}{2}k_bT [/tex]
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.
Rap
#14
Nov27-12, 10:52 PM
P: 789
Quote Quote by K^2 View Post
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.
Right - except very close to absolute zero temperature.
the_emi_guy
#15
Nov27-12, 10:59 PM
P: 589
Quote Quote by K^2 View Post
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.
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.
Leoragon
#16
Nov27-12, 11:39 PM
P: 41
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?
K^2
#17
Nov28-12, 12:26 AM
Sci Advisor
P: 2,470
Quote Quote by the_emi_guy View Post
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.
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.

Quote Quote by 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?
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.
Leoragon
#18
Nov28-12, 12:57 AM
P: 41

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.
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?


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