# How to derive the kinetic gas equation?

1. Oct 14, 2006

### loom91

Hi,

The kinetic molecular theory of ideal gasses is a topic that we have to study in both our Physics and Chemistry syllabus. A part is deriving the kinetic gas equation $PV = \frac {1}{3}mN<v^2>$ froom the postulates of KMT.

However, of the 3 Physics and 2 Chemistry textbooks I've consulted, all give a slightly different version of the postulates and the derivation. Every derivation seems to have a different inaccuracy or hidden assumption.

Can you please tell me or give me a link to the actual postulates of KMT for ideal gasses and a rigorous derivation of the kinetic gas equation from them? Thanks.

Molu

2. Oct 14, 2006

### Andrew Mason

You might have a look at the http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/kinthe.html#c1"

The derivation for pressure from Newton's laws of motion should answer your question.

AM

Last edited by a moderator: Apr 22, 2017
3. Oct 14, 2006

### loom91

Last edited by a moderator: Apr 22, 2017
4. Oct 14, 2006

### Andrew Mason

Perhaps you would care to explain why...

AM

5. Oct 14, 2006

### tim_lou

ok, first of all, imagine you have a wall, and a gas hitting the wall continuously.

let n be the number of molecules per unit volume, let the wall be the yz plane.
for time dt, the molecules that may hit the wall must be within v_x*dt meters from the wall. the number of molecules that may possibly hit the wall is:
$$nAv_xdt$$
since half the molecules are moving away from the wall, and half are moving toward, the number of molecules that hit the wall is:
$$nAv_xdt/2$$

v_x is the root mean velocity in the x direction.

the average momentum of each particle in the x direction is:
$$mv_x$$

the momentum each particle transfered to the wall is
2mv_x since they do not stop after they hit the wall, they bounce back.

hence the total momentum transfered to the wall in dt is:
$$2nAdt*mv_x^2/2$$

force is dp/dt, so
$$F_x=nA*mv_x^2$$

the pressure is therefore (P=F/A):
$$P=nmv_x^2$$

so,
$$PV=n*Vmv_x^2=NVmv_x^2$$

since there is no perferred direction, let <> denote the mean value:
$$<v^2>=<v_x^2>+<v_y^2>+<v_z^2>$$
$$<v_x^2>=\frac{<v^2>}{3}$$

hence:
$$PV=\frac{mN<v^2>}{3}$$

edit: fixed some typoes

Last edited: Oct 14, 2006
6. Oct 16, 2006

### loom91

It makes a common mistake that was exposed in the classic HC Verma's Concepts of Physics: it derives the expression for the time between two sucessive collisions of a particle with a wall by assuming it travels between the two walls with constant velocity, implicitly disregarding collisions between gas molecules themselves, though this is definitely not a postulate of KMT. This is one of the inaccuracies in the textbooks I was talking about.

The correct way to handle this is to assume a time-independent velocity distribution, which then shows that the system would behave AS IF the molecule under consideration went to the opposite wall and then came back at constant velocity.

7. Oct 16, 2006

### loom91

I can't see how you get this. What you are effectively assuming here is that the molecules obey a discrete velocity distribution with 0.5 on -v_x and 0.5 on v_x, where v_x is root mean square velocity in the actual velocity distribution. What is the justification of this assumption?

v_x was already a measure of central tendency, the rms velocity, how can you take its mean again?

8. Oct 16, 2006

### Andrew Mason

The premise is that pressure is the same throughout the whole volume. If you take an element of length dL (volume AdL) - in which the probability of a collision with another molecule approaches 0 - and do the same analysis, you get the same result for pressure.

AM

Added comment: The Boltzmann distribution results from thermodynamic equilibrium which necessarily implies random energy transfer among all molecules. So when you use $<v_x>$ or the rms speed, you are implicitly taking into account the collisions that occur.

Last edited: Oct 16, 2006
9. Oct 16, 2006

### Andrew Mason

The http://en.wikipedia.org/wiki/Equipartition_theorem" [Broken]. The validity of this principle is borne out by its agreement with the observed result.

AM

Last edited by a moderator: May 2, 2017
10. Oct 17, 2006

### loom91

An element of volume AdL is not a box with two walls from which the molecule repeatedly rebounds, so this analysis can't be applied directly to such an element.

You are mixing two derivations. The assumption of no collision is in the hyperphysics derivation (also found in many textbooks), which uses individual molecular velocity.

Tim_lou's derivation is a slightly different one, it does not make the assumption of no collision but instead assumes that half the molecules have +v_x and the other have -v_x velocity where v_x is rms velocity (not individual). How is this related to equipartition of energy or the Boltzman distribution? Further, tim_lou's derivation takes the mean of rms velocity, which seems meaningless.

11. Oct 17, 2006

### Andrew Mason

It uses the individual molecular velocity to show that the force from one molecule is proportional to mv^2 (ie to energy) for that molecule. It then shows that the force for N molecules with various speeds is proportional to the sum of the squares of their individual speeds, (which is N times the average of the sum of the squares of their individual speeds - or N times the square of the square root of the mean of the sum of the squares of their individual speeds: $F \propto Nv_{rms}^2$

Force, hence pressure, is proportional to $N \times v_{rms}^2$ AND, of course, total energy is proportional to the sum of the squares of the speeds of all the molecules (ie. $N \times v_{rms}^2$).

So if pressure on all walls is equal, energy is equipartitioned among the x, y and z directions: $v_{rmsx}^2 = v_{rmsy}^2 = v_{rmsy}^2$.

AM

12. Oct 17, 2006

### tim_lou

ok, let's assume that it IS NOT rms velocity, instead, it is the mean of the absolute value of the velocity in the x direction. in terms of mathematics, the rms of a set of positive numbers is always GREATER or EQUAL to the arithmetic mean. however, since we assume that each particle has the same speed, the rms of the velocity is the SAME as the average speed. This is the equality case of the RMS and AM inequality.

you can check out this website... but it has limited information regarding this inequality:
http://www.artofproblemsolving.com/Wiki/index.php/RMS-AM-GM-HM [Broken]

edit: actually, it is easily seen (assume that all particle travels at an average speed, v):
$$v_{rms}=\sqrt{\frac{n*v^2}{n}}=\sqrt{v^2}=<|v|>$$

Last edited by a moderator: May 2, 2017
13. Oct 18, 2006

### loom91

And it is in this step that the false assumption is made.

An actual molecule will rarely make the trip in exactly this time.

14. Oct 18, 2006

### loom91

But the postulates of KMT does not assume that all molecules move with the same speed! It would be a very lame theory if it did.

Last edited by a moderator: May 2, 2017
15. Oct 18, 2006

### Andrew Mason

If I understand your objection correctly, you take issue with the analysis that leads to:

$$F = Nmv_{x}^2/L$$

where L is the length of the chamber, m is the mass of molecule, N is the number of molecules and $v_x$ is the x component of the rms speed of the molecules ($v_x^2$ being the mean of the square of the speeds of all the molecules).

I agree that this analysis is somewhat oversimplified. Maxwell-Boltzmann statistics are not simple. You might find http://user.mc.net/~buckeroo/MXDF.html" [Broken] to be more to your liking.

AM

Last edited by a moderator: May 2, 2017
16. Oct 20, 2006

### loom91

I can't understand equation 2. How does he turn a product of summations into summation of products?

Last edited by a moderator: May 2, 2017
17. Oct 20, 2006

### Andrew Mason

He is just replacing N with equation 1.

AM

18. Oct 21, 2006

### loom91

When he writes equation (1), he (correctly) assumes that the particles are moving with different velocities, but in the rest of the derivation he implicitly assumes that all particles are moving with same velocity, otherwise he can't write $$\delta p = N2mv_x$$. Further, he uses the same label to identify this velocity and the variable velocity in the sum, effectively rendering the expressions logically meaningless. I never imagined that a derivation of such a common equation would be so difficult to find.