Deriving Ideal Gas using Molecular Flux Equation

1. Mar 4, 2015

Seydlitz

So basically I was wondering whether it's possible to get the expression of ideal gas using molecular flux equation which is $\phi = \frac{1}{4}\bar{v}n$. The derivation should be straightforward. I need to get the expression of pressure. Because the flux by definition already gives the rate of number of particles per area. I multiplied $\phi$ by the change of momentum when a particle hits a wall and rebounds which is $\frac{2}{3}m\bar{v}$.

Eventually I get.
$$P = \phi \, dp = \frac{1}{2}\frac{1}{2}m\bar{v}^2\frac{2}{3}n$$
Notice however that I get extra $\frac{1}{2}$ factor which makes the final expression smaller by a half than the correct. $PV = NkT$

What should be fixed in this case?

Thank You

2. Mar 5, 2015

Staff: Mentor

Can you show the details of your derivation?
Are the units consistent?

3. Mar 5, 2015

Seydlitz

Sure. So basically instead of using the usual ideal gas derivation, I just take the molecular flux, which is the rate of the number of molecules per area and multiplied it by the change of momentum of one particle to get the pressure. The reasoning is pretty similar to the classic superman bullet problem if I can find the analogue, where you are given the rate of the bullet hitting superman chest and you need to find the average force hitting the man.

The unit is also consistent. Molecular flux has $1/m^2s$ units, which will become pascal or $N/m^2$ if multiplied by momentum change.

$$\phi = \frac{1}{4}\bar{v}n \\ P = \phi \, dp = \frac{1}{4}\bar{v}n \cdot \frac{2}{3}m\bar{v}$$

After re-arranging,

$$P = \frac{1}{2}\frac{1}{2}m\bar{v}^2\frac{2}{3}n \\ P = \frac{1}{2}\frac{3}{2}kT\frac{2}{3}n \\ P = \frac{nKT}{2} \\$$

Now that's where the problem lies. I somehow need to get that half factor fixed.

4. Mar 5, 2015

Staff: Mentor

Doesn't $m\bar{v}^2/2$ look like the average kinetic energy? You're not recovering a pressure. In the last equation, you're missing a volume, so you need to introduce it somehow.

5. Mar 5, 2015

Seydlitz

This might be the reason why there's that strange factor. I thought I can replace that term that looks like average kinetic energy with the usual $3kT/2$. Maybe I can't? But I don't see any problem. The mass of the particles are all the same and I'm using average velocity. So it's the average kinetic energy right?

The one with volume is fine, I can happily substitute $n = N/V$ and get $PV$ on one side. But the half factor remains regardless. I also need to add the fact that $n$ is not the number of molecules but molecules per volume. (This is what makes the unit consistent)

6. Mar 5, 2015

Staff: Mentor

Sorry, that should have been obvious. Forget my post #4.

One thing that I now see that doesn't work is that $\bar{v}^2 \neq \overline{v^2}$, and it is the latter that appears in the equation for kinetic energy.

7. Mar 5, 2015

Seydlitz

Ah yes. It makes sense, thanks for noticing that. But can I find any workaround though if I still want to use the derivation, or it's just not possible to get the ideal gas equation back?

8. Mar 6, 2015

nasu

You don't have a flux of molecules in a gas at equilibrium. On the average, half of them will move in the opposite direction, for any axis you choose.
Is this included in that formula? Where does it come from? I mean the one with the 1/4.

9. Mar 7, 2015

Seydlitz

Here is one way of deriving it. (Page 3 Molecular flux)
http://www.lehman.edu/faculty/dgaranin/Statistical_Thermodynamics/Molecular_theory.pdf

The derivation sums molecules moving in one direction per surface I think but you can re-check it. If we need to half the molecules, the final formula will be a quarter less then.

10. Mar 7, 2015

nasu

Then why don't you follow the derivation of pressure, which is in the next section?
It seems that the flux described by that formula is not a flux of particles with speeds normal to the wall but with an angular distribution.
The momentum transferred by one atom is not just 2mv but 2mv cos(theta), as you can see in their formula. You need to integrate over this angle to get the total momentum transfer.

11. Mar 7, 2015

Seydlitz

There exist indeed the derivation of pressure already, but I'm trying to do the same using only the molecular flux formula created beforehand without integrating all of the quantities from the beginning.

I think I understand now, what you said makes sense. All of the particles in the flux considered can collide with the wall in different manner and hence their momentum change is not trivial. This is not the same with the ordinary pressure derivation, as at the beginning only the molecules with one velocity component is considered, hence we can use 2mv change directly.

Do you think I get the reasoning right?

12. Mar 8, 2015

Staff: Mentor

That sounds right.