# I (2/3) as constant factor In Kinetic theory of gases

#### TheColector

Hi
I'm in high school but what I'm going to ask you is probably being teached in college.
General formula: p=(2/3)*(N/V)*Ek
p- pressure
N- amount of molecules
V- volume of the container
Ek - AVERAGE kinetic energy

I've been told by my physics teacher, that 2/3 constant factor in kinetic theory of gases is the result of using mathematical statistics. As there're lots of molecules moving all the time with high velocity and different directions, kinetic energy of each is different. Therefore we use AVERAGE kinetic energy of molecules. In order to calculate this average Ek we use mathematical statistics(which with I'm not acquired at all). All of this seems logical to me. Can you possibly tell me if all of this is correct ?

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#### lychette

your expression is a version of the equation for the pressure of a gas derived from kinetic theory
P = 1/3(Nmc2)/V where N = number of molecules, m is the mass of 1 molecule and c is the average (rms) velocity of a molecule and V is the volume
this can be written 2/3 N/V(mc2/2)
or P = {2/3N/V} x(average KE of molecule)

#### haruspex

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The 3 comes from the fact that we are in 3 dimensions.
You can get the flavour of it by considering a cubic box, and a molecule bouncing back and forth along one axis.
Each time it hits a wall it imparts an impulse 2mv (do you see why?)
If the length of the box is L then it hits that wall each period of time 2L/v. So the average force on that wall from that molecule is mv2/L.
Of course, the molecules are bouncing around in all directions, but it turns out that you get the right result if you just consider the three axes of the box, i.e. take it as though a third are bouncing along each axis.

#### TheColector

I see now. That's pretty clever way of thinking about it. What does the "2mv" stand for ?

#### haruspex

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I see now. That's pretty clever way of thinking about it. What does the "2mv" stand for ?
Mass x velocity = momentum. If it hits with velocity v, head on, and bounces back with velocity v, the net change in momentum is m(v-(-v))=2mv.

#### lychette

I see now. That's pretty clever way of thinking about it. What does the "2mv" stand for ?
2mc is the change in momentum due to an elastic collision of a molecule with the walls....rate of change of momentum = force on wall. ie number of collisions per second x 2mc = force
You have mc2 in the expression so it is 'nice' to have 2 x 1/2mc2 so that 1/2mc2 appears as average KE....this is where the 2 in the top line comes from.....good stuff

#### TheColector

I totallly forgot about the fact that all the collisions are elastic. Thanks for enlightening me.

#### mpresic

You will probably see all this again if you take a physics or chemistry course in college. As you can see by the posts above, analyzing physical systems using simple models can be fascinating and successful and fun. This is the best advertising for these college courses.

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