# B What is pressure when there are no container walls?

1. Jun 4, 2017

### Edison Bias

Hi!

I wonder what pressure is when there are no walls.

The reason for me wondering this is that gas pressure actually is defined by change of particle momentum, like

$$p=\frac{F}{S}=\frac{1}{S}\frac{dp}{dt}=\frac{1}{S}\frac{d2mv}{dt}$$

So how can there be pressure without change of momentum?

It's like Heisenberg almost because the pressure is only there when you measure it, when the walls are gone there is no pressure.

In my amateur world I would say that this wall-less pressure is made of intermolecular collissions, it has to be because there can be no pressure without collisions.

This equation can be shown to be derived from particles hitting a surface:

$$p=\frac{2}{3}nEk_p$$

where Ekp is the particle kinetic energy and n the particle density.

It is a very interesting equation, not to mention the fact that it is derived from particles actually hitting a surface.

It says that pressure actually is proportional to kinetic enegy density, it is a fascinating description of pressure which's I have never understood but right now I think I understand because pressure is the sum of all the particle's kinetic energy in a certain volume where the quadratic velocities of the different particles is directly reversely proportional to the mass, due to

$$Ek_p=\frac{3}{2}kT$$

While the common gas law may be written

$$pV=n'RT=\frac{N}{N_A}RT=N\frac{R}{N_A}T=NkT$$

This is the equation many physisists rely on but remember these things:
1) Normal gases do interact/collide, Dalton's Law is a joke
2) Boyle's law relies on isothermal changes that are very smal with respect to To
3) The thermal expansion koefficient of volume and pressure may not be the same (at a larger scale).

Best regards, Edison

2. Jun 4, 2017

### Staff: Mentor

You seem to have an analytical mind and a curious nature. Good for you.

Why be satisfied with a quickie answer on an Internet forum? I think you would enjoy the following and benefit from it. Leonard Susskind's video course on statistical mechanics. It not only answers your question, but probably the question after that and the next few. It shows you how the perfect gas law was derived. In addition, Susskind make it all entertaining.

Only 10 lectures, all available free on youtube. It sounds like you already have the math background needed.

3. Jun 5, 2017

### Jamison Lahman

You answer this question yourself. There can't be pressure without change of momentum.
Why does the surface have to be a wall? Why can't it be another molecule? How would you measure pressure without some reference surface?

Last edited: Jun 5, 2017
4. Jun 5, 2017

### Edison Bias

It seems like we can agree on something, there can be no pressure without change of momentum.

This means that particles have to collide with either themselves or with some kind of wall.

If there is no wall, particles then have to collide with themselves.

I have above mentioned one way to calculate pressure using the fact that pressure is kinetic energy density (derived from change of momentum, though)

Personly I would prefere to calculate pressure without walls as being a function of particle collisions, with the assumption that there actually are lots of collisions within a gas of "average" particle density.

Let's try to calculate the the impact number, ns, which gives the number of collisions per time and area unit:

The infinitesimal volume is

$$dV=Svdt\cdot cos\theta...1$$

where theta is the angle with respect to the normal of the surface S.

The number of particles within this volume is

$$n(v, \theta)dvd\theta dV...2$$

but it can be shown (comparing space angles with full space angle) that this is equal to

$$n(v)dv\frac{1}{2}sin\theta dV...3$$

so that the number of molecules within the volume is

$$dn_s=n(v)dv\frac{1}{2}sin\theta Svdt\cdot cos\theta...4$$

Deviding this with dt and S gives the number of particles per time and surface unit, thus

$$n_s=\int_{-\infty}^{\infty}n(v)vdv\int_0^{\pi/2}\frac{1}{2}sin\theta \cdot cos\theta...5$$

where

$$\int_{-\infty}^{\infty}n(v)vdv=n<v>...6$$

where <v> is the mean velocity, it can further be shown that the angle-integral equals 1/4 so that the number of collisions per area and time unit is:

$$n_s=\frac{1}{4}n<v>\propto n<v>...7$$

Now, let's try to calculate mean free length för a particle to pass within a gas

The volume it consumes each second it "flies" is

$$dV=\frac{\pi}{4}d^2vdt...8$$

While distancies is measured centre-to-centre the area of the sweeping volume is actually

$$dV=\pi d^2 vdt...9$$

then the number of molecules wthin dV (and therefore subject to collisions) per unit time is

$$N=\pi n d^2 v...10$$

I've been told that this should be corrected with sqrt(2) due to Maxwell so let's put that in also

$$N=\sqrt(2) \pi n d^2 v...11$$

this is a probability function where N should not exceed unity for no collisions which means that the mean free length is

$$l=\frac{v}{\sqrt(2) \pi n d^2 v}=\frac{1}{\sqrt(2) \pi n d^2}...12$$

where v simply is shorted out due to not being relevant for mean free length.

Now we have two equations: 7 & 12.

7 tells us that the number of collisions on a surface per area and time unit is

$$n_s=\frac{1}{4}n<v>\propto n<v>...7$$

so if we know n and v we can calculate the number of collisions per area and time unit.

n of ordinary air may be calculatet by

$$n=\frac{p}{kT}...13$$

and using p=1,013E5 Pa and T=293K I have recently calculated n as around E25.

v may be calculated as

$$Ek=\frac{3}{2}kT=\frac{mv^2}{2}...14$$

thus

$$v=\sqrt{\frac{3kT}{m}}...15$$

where m can be estimated as 2/10*2*16mp+8/10*2*14mp, reflecting the fact that air consists of 80 nitrogen and neutrons weigh almost equal to protons.

mp=1,67E-27, putting this into the equation gives v=502m/s, thus 7 gives E27... per unit area...

so how wide is a proton?

I have never heard of that kind of data in a physics handbook, lets try to estimate it:

Lets' say that it's mass density corresponds to a known metal in the middle of the periodic table, say copper, the density of copper is 9g/cm^3=900kg/m^3~1000/m^3

We know that it weighs around E-27 kg and its volume is 4pi/3R^3~piR^3.

$$\rho=\frac{m}{V}...16$$

which gives

$$V=\frac{m}{\rho}...17$$

which gives R as E-10, giving a cross sectional area of some E-20, putting this in the equation 7 above gives ns=E7, thus the numer of particle-particle collisions per second is E7 in ordinary air.

The mean free length (equation 12) for air is

$$l=\frac{1}{\sqrt(2) \pi n d^2}...12$$

Using d as approximatelly R (i.e E-10) and n as E25 this gives 1um

Lot's of collisions, indeed :D

Best regards, Edison

5. Jun 5, 2017

### Jamison Lahman

The pressure without walls is the same as with walls. It is easier to calculate with walls because it is an arbitrarily large surface, so we can easily calculate the average force per area of the gas. If you are familiar with Gauss's Law, the wall can be thought of as a Gaussian surface. There does not have to be an actual wall to use a wall to calculate the pressure.

The classical radius of a proton is of the order E-18m (for context, the Bohr radius is is of order E-11m) but that is not the number you want. The protons are tightly bound in the nuclei of the gas particles. It is the gas particles that would be colliding therefore it is the size of the gas particles you would want.

Last edited: Jun 5, 2017
6. Jun 7, 2017

### Edison Bias

I just want to say that a radious for a proton equalling something around E-18 is astonishing!

Just for convenience:

$$R^3\approx \frac{m}{\rho}$$

Putting m=E-27 and rho=E3 gives R=E-10.

Observe that the mass density already is quite high, almost as high as Copper.

To reduce this proton radious to E-18 we will have to increase rho with a factor of close to one thosand.

Thus the proton mass density is almost 1000 times higher than Copper, that is quite dense!

Best regards, Edison
PS
I did miscalculate the molecular width using one proton only, I should have considered at least a volume of 14 protons, too tired :)

7. Jun 7, 2017

### Staff: Mentor

Isn't that what you expect? Atoms (like copper) are made of protons, neutrons, electrons, plus lots of empty volume.

8. Jun 7, 2017

### Khashishi

You are over-analyzing this.
Your title asks "What is the gas pressure when there are no power inputs (Watts)?"
The pressure depends on the temperature and density. The gas still has a temperature if you don't give it a power input.

The next question is "I wonder what pressure is when there are no walls"
Pressure is defined on an imaginary surface element. You don't need a real surface to calculate the pressure. It is a momentum flux passing through some hypothetical surface.

No. Collisions are not necessary to have pressure. A gas of photons, for example, has a pressure and no collisions. If you put a wall in a box of photons, the photons will exert a force on the wall. But, you don't actually have to put the wall in there to define the pressure. It's imaginary.

9. Jun 7, 2017

### Staff: Mentor

Minor quibble with that last paragraph; I think you meant collisions with walls are not necessary; collisions between gas molecules are.

But broadly, I agree; to @Edison Bias ; what is it you want to know? You seem to have kind of asked multiple questions that are only sort of related and then gone on some sort of stream of consciousness thought/math exercise. Could you please more concisely state what you are after?

10. Jun 7, 2017

### Jamison Lahman

As Rutherford described the gold foil experiment, it was "as if you had fired a 15-inch [artillery] shell at a piece of tissue and it came back and hit you." The atom is very spacious and the nucleus is very dense.

11. Jun 7, 2017

### Jamison Lahman

Hmm. I did not know that. The photons are still colliding with the wall, no, or does collision imply a mass interaction?
Edit: Per Wikipedia - Collision, "Collisions in ideal gases approach perfectly elastic collisions, as do scattering interactions of sub-atomic particles which are deflected by the electromagnetic force." This is curious to me.
Edit: Per Wikipedia - Photoelectric Effect, "electron emission when collisions happen with higher energy photons."

12. Jun 7, 2017

### Khashishi

That's not what I meant, which is why I mentioned a photon "gas". Photons hardly collide with other photons, but still have a pressure.

13. Jun 7, 2017

### Staff: Mentor

I'm not sure that they are equivalent. In either case, the difference between a gas an a really really fine pile of dust on the floor is the fact that the particles bounce around off each other with a certain momentum, causing the collection of particles to have a pressure. Gases need to be constrained by *something*, to do this, which is either a container or gravity, and with gravity it is molecules constraining other molecules.

I suppose you could describe a single molecule in a box as having momentum and therefore a pressure (very unusual), but again in order to have pressure it has to be constrained and bounce off the walls of the box. If not constrained by the walls or other molecules, it will just fly off into space and not have an identifiable control volume.

14. Jun 8, 2017

### Edison Bias

Someone has changed my title, I want it back.

Best regards, Edison
PS
To russ_watters I just want to say that I am trying very hard to understand gases, pressure and temperature because it interests me.

15. Jun 9, 2017

### Edison Bias

I think this answer is interesting and addresses my (original) thread title. I have a really hard time grasping the concept of pressure, it is unclear. One way of vewing it, that is kind of easy to understand, is the fact that pressure is kinetic energy density, it is thus the sum of all particles with thermal speed per volume unit, inside this volume particles flies around with the thermal speed and this speed is dependant on temparature and mass of particle only (due to Ek=3/2kT).

So we have a situation where pressure is made out of particles flying around in a volume at a certain speed and particle density.

This is pressure.

The problem is that the formula used (p=2/3nEk) is derived from the assumption that there are hits on a surface, i.e there has been a change of momentum for the particles, I am questionizing if this formula is correct when there are no walls and therefore no change of momentum but perhaps inter-molecular collisions takes over to such a degree that it "replaces" the hits on a actual wall?

Best regards, Edison

16. Jun 9, 2017

### Staff: Mentor

I thought I watched the thread from the beginning and didn't see a title change; what title do you want?

17. Jun 9, 2017

### Staff: Mentor

Yes; the math reflects the fact that there is no functional difference between the particles bouncing off each other and bouncing off a wall.

18. Jun 9, 2017

### Staff: Mentor

I changed the title soon after the thread was created, I think before there were any replies. The original thread title made no sense to me given the text of the first post, but if the thread has developed far enough now that the original title makes sense, we could change it back.

It was: 'What is pressure when there are no walls?'

19. Jun 9, 2017

### Edison Bias

Interesting, so my statement is correct that the physics of it all changes from wall-impulses to particle-impulses just due to the fact that there are no walls anymore which we would need to measure pressure?

I have learned and above derived an equation that gives the probability for collisions per molecule which I repeat

$$a=\sqrt{2}\pi nd^2<v>$$

the number of molecules that collides per second are then

$$na=\sqrt{2}\pi n^2d^2<v>$$

and devided by two this gives the number of collisions and for ordinary air I have (using p=nkT) calculated n as E25 and v (using Ek=3/2kT) as around E3 and if d=E-10 for an air molecule this equals somewhere around E33 collsions per second.

Quite many collisions, thus there are obviously lots of momentum changes in gases, even without walls :)

On the other hand, if we add walls should not the pressure get even higher?

Best regards, Edison

20. Jun 9, 2017

### Jamison Lahman

Calculating the number of collisions allows us to assume that the changes in momentum are frequent enough that the pressure is uniform over the whole volume. Adding walls gives a surface which we can use to find the exact pressure on that surface. The only difference with the walls is we no longer have to rely on the assumption that there a sufficient number of collisions. Furthermore, the pressure on the surface would be known, and then the pressure within the surrounding gas would be assumed given a high number of collisions.

21. Jun 10, 2017

### Edison Bias

Interesting, thanks for that explanation.

Now I wish to try to calculate the pressure without walls knowing particle density and collision density per time unit only.

I will recalculate the number to try to get everything right (please tell me if I'm more than a magnitude wrong)

First we have (for ordinary air)

$$n=\frac{p}{kT}\approx \frac{E5}{E-23\cdot E3}...1$$

which makes n=E25 and this is the particle density in ordinary air at normal pressure and temperature.

The collision density per time unit is

$$n_{ct}=\frac{1}{2}\sqrt{2}\pi n^2d^2...2$$

Here we need to estimate d for an air molecule.

The mass for an air molecule might be estimated by

$$m=(2/10*16*2+8/10*14*2)m_p...3$$

where mp is the mass of a proton and we have 20 percent Oxygen and 80 percent Nitrogen, thus the mass per molecule is 29mp~E2mp.

Knowing from above that the radious of a proton is around E-18 we have a cross sectional area of about E-36.

Taking E2 times E-36 we have E-34 as area for the air molecule , thus E-17 might be considered its diameter.

Getting back to equation 2, we now have

$$n_{ct}=\frac{1}{2}\sqrt{2}\pi n^2d^2\approx E50*E-34=E16$$

So we now suddenly have only E16 collisions per volume and time unit.

The thermal velocity comes from

$$E_k=\frac{3}{2}kT$$

thus

$$v=\sqrt{\frac{3kT}{m}}=\sqrt{\frac{3kT}{29m_p}}\approx \sqrt{\frac{E-23*E3}{E-26}}=E6$$

Now to the exciting part, what pressure does this mean?

$$p=\frac{F}{S}=\frac{2mv/dt}{S}=\frac{2n\cdot 29m_p\cdot v}{n_{ct}d^2}\approx \frac{E25 \cdot E-26\cdot E6}{E16\cdot E-17}=E6$$

I can't believe it because this is only one magnitude from the truth (E5 Pa)!

Best regards, Edison

22. Jun 10, 2017

### rcgldr

As an example of pressure without container walls, consider a sphere of gas in space contained only by gravity.

23. Jun 10, 2017

### Edison Bias

Interesting.

Let's take it closer to earth.

I was about to start a new thread about a related thought titled "What makes the atmosphere end?"

First, we may roughly estimate the hight of the atmosphere as

$$h=\frac{p}{\rho g}$$

using rho as 1, g as E1 and p as E5 we get E4 and thus 10km.

I don't know how far from the truth this is but it sounds reasonable, it sounds reasonable in spite of the fact that rho can hardly be the same all the way up to the end of the atmosphere.

So what happens up there?

The mass density (rho) will probably get less and less.

Another way to view the familiar pressure equation p=nkT is actually

$$p=\frac{\rho}{m}kT$$

But here we can see that temperature is mixed into this also so we can't blame a lesser pressure for lesser rho only, temperature is probably changing also.

And way up there pressure should equal zero, I think, because further out we have space and vacuum.

And the only way for pressure to equal zero is if rho (or n) is zero.

In other words, there may still be temperature there but... wait a minute, isn't temperature defined by speed of particles?

So how can there be temperature (I believe that the tempereture at the end of the atmosphere is not 0K) when there are no particles?

Best regards, Edison
PS
What should end the atmosphere, I believe, is gravity but the way gravity works in this respect is unknown to me. One way of thinking is that, while 10km is so much less than 6370 km as I think the radious of the Earth is, the g-force is almost constant and working hard on the air molecules and while their temperaure and thus speed is low they do not have the momentum to get any higher, please correct me from this pathetic attempt in understanding :)

24. Jun 10, 2017

### rcgldr

As the altitude increases, the temperature decreases part of the way, but then starts increasing as altitude increases beyond some point.

Due to the randomness of collisions, some of the hydrogen molecules at the edge of the atmosphere achieve escape velocity.

25. Jun 11, 2017

### Jamison Lahman

Molecules are most certainly not 10-17m meters in diameter. Molecules (unless completely ionized) have a cloud of electrons. The molecular collisions we have been discussing are not with the molecules' nuclei. The inter-molecular repulsion is due to the electrons which is also how we define the size of the molecule.

You are certainly mathematically proficient and you have a genuine curiosity which is probably most important. As @anorlunda suggested, it might serve you more to get a textbook and go through that. Understand the underlying principles and then you'll easily be able to apply them to a plethora of scenarios.