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(Average) Kinetic Energy of Molecules 
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#1
Nov2612, 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? 


#2
Nov2712, 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 273.15°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. 


#3
Nov2712, 01:04 AM

P: 589

Temperature is *proportional* to average kinetic energy. Specifically for monatomic gases: [tex] \bar {E} = \frac{3}{2}k_bT [/tex] 


#4
Nov2712, 10:28 AM

P: 789

(Average) Kinetic Energy of Molecules
Another nitpick  the zeroth law holds for any system, not just a gas.



#5
Nov2712, 11:32 AM

P: 589




#6
Nov2712, 12:08 PM

P: 126




#7
Nov2712, 06:32 PM

P: 41




#8
Nov2712, 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 


#9
Nov2712, 07:22 PM

P: 41

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


#10
Nov2712, 09:04 PM

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#11
Nov2712, 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] 


#12
Nov2712, 10:37 PM

P: 41




#13
Nov2712, 10:46 PM

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P: 2,470

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
Nov2712, 10:52 PM

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#15
Nov2712, 10:59 PM

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#16
Nov2712, 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?



#17
Nov2812, 12:26 AM

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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 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. 


#18
Nov2812, 12:57 AM

P: 41




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