# What is pressure when there are no container walls?

• Edison Bias
In summary: 12 tells us that the mean free length isl=\frac{v}{\sqrt(2) \pi n d^2 v}=\frac{1}{\sqrt(2) \pi n d^2}...12
You don't need to have a wall to have pressure. Now you have two Mentors telling you this.

Imagine that you have a thin rigid barrier between the two sides of a compartment, and a gas at the same pressure on both sides. Molecules on each side of the barrier bounce off the barrier and impose a momentum flux on each side of the barrier (which is interpreted as a force). Now, remove the barrier, and replace it with an imaginary plane. In this case, there are no molecules bouncing off of each side, but now there is a momentum flux on each side from molecules passing through the plane from the other side. So the effect is exactly the same as if there were a real solid barrier present. The molecules passing through the plane are not bouncing off the plane, but there are molecules passing through the plane from the other side just as if they had; it's just that they are not the exact same molecules. Otherwise, everything is exactly the same.

In the case of a liquid, the picture is even more compelling because the molecules on the two sides of the plane are actually touching each other.

Khashishi said:
A gas of photons, for example, has a pressure and no collisions.
Does photon gas pressure correspond to the pressure perfect mirror walls would experience?

For a gas at atmospheric pressure, a typical value of the mean free path of a molecule (Wikipedia) is 68 nm. So there are lots of collisions of molecules within any practical volume of gas, even as ideal gas behavior is approached. This is how pressure variations are transmitted through a gas is cases where gas is accelerating or where the gas is in a gravitational field (which, according to equivalence principle, is equivalent to acceleration).

Edison Bias said:
So it seems like the atmosphere ends due to gravity and thus air molecules having too little momentum to move any further, right?
The first part is trivially true (a pile of books is only so high because it only has so many books), but the momentum part isn't. The momentum is related to the temperature and it certainly isn't absolute zero at the outer reaches of the atmosphere.
Please don't give me any links to read or tell me to go and read a textbook (which I am already doing, by the way), let's just dicuss this like grown and passionate human beings.
Please understand that everyone here is volunteering their time, so it isn't always reasonable or efficient to meander toward answers to questions that have well known answers you could just google.

Chestermiller said:
You don't need to have a wall to have pressure. Now you have two Mentors telling you this.

Imagine that you have a thin rigid barrier between the two sides of a compartment, and a gas at the same pressure on both sides. Molecules on each side of the barrier bounce off the barrier and impose a momentum flux on each side of the barrier (which is interpreted as a force). Now, remove the barrier, and replace it with an imaginary plane. In this case, there are no molecules bouncing off of each side, but now there is a momentum flux on each side from molecules passing through the plane from the other side. So the effect is exactly the same as if there were a real solid barrier present. The molecules passing through the plane are not bouncing off the plane, but there are molecules passing through the plane from the other side just as if they had; it's just that they are not the exact same molecules. Otherwise, everything is exactly the same.

In the case of a liquid, the picture is even more compelling because the molecules on the two sides of the plane are actually touching each other.
This is an very interesting answer, than you!

I will give you an analogy to what you so nicely just have said, I like using emitter followers with my tube amps and the current direction has always puzzled me becuase you can always source current (the limit is only how high in base voltage you can go) but the sinking of current has been an enigma, it turns out that while current goes from the load (to swing negatively) it can not go to ground, because it started there, so it goes to plus, so it actually goes against emf but it does so only in the way of decreasing the collector current and this can be interpreted as current going the "wrong" way, actually the maximum negative swing is the DC over (and through) the emitter resistor.

But how about my attempt above in trying to calculate pressure without walls?

I will not be satisfied before I have reached normal air pressure without walls and I hope you have the lust into helping me, one thing that has struck my mind is that I have used normal air pressure and temperature using The Ideal Gas Law to calculate the particle density so I have already used the pressure I want to calculate, which can't be right but is there another way i calculating the particle density in ordinary air without using the normal air pressure?

Anyways, these are the formulas I have used:

The Ideal Gas Law
$$n=\frac{p}{kT}\approx \frac{E5}{E-23\cdot E3}=E25...1$$
The collision density per time unit
$$n_{ct}=\frac{1}{2}\sqrt{2}\pi n^2d^2\approx n^2d^2=E50*d^2...2$$
An estimation of the air molecular mass
$$m=(2/10*16*2+8/10*14*2)m_p...3$$
The relationship between temperature and kinetic energy
$$E_k=\frac{3}{2}kT...4$$
The speed of the air molecules due to temperature
$$v=\sqrt{\frac{3kT}{m}}=\sqrt{\frac{3kT}{29m_p}}\approx \sqrt{\frac{E-23*E3}{E-26}}=E6...5$$
The pressure without walls
$$p=\frac{F}{S}=\frac{2mv/dt}{S}=\frac{2n\cdot 29m_p\cdot v}{n_{ct}d^2}\approx \frac{E25 \cdot E-26\cdot E6}{E28\cdot E-22}=E-1...6$$

In equation 2 I need the diameter of the air molecule which I estimated as E-17 and someone here in this nice forum told me that that is totally incorrect but I don't know how to estimate it, someone here said E-11 as the Bohr-radious for a proton and I imagine this is what I should calculate with, if this is true the diameter of an air molecule still is E-11 i magnitude.

If d=E-11 and n=E25 then equation 2=E28, and equation 6=E-1...

This did not go so well :)

May I ask what I did wrong?

But let's do some reverse engineering:

$$p=\frac{F}{S}=\frac{2mv/dt}{S}=\frac{2n\cdot 29m_p\cdot v}{n_{ct}d^2}\approx \frac{E25 \cdot E-26\cdot E6}{n_{ct}d^2}=E5...7$$

and when

$$n_{ct}d^2 \approx n^2d^4...8$$

then if n^2=E50, d should be 3E-13 (where 3 is crucial due to ^4).

But as I've already said, I am cheating with the calculation of the pressure by already using the normal air pressure :)

Edison
PS
I apologize for my earlier behaviour, I am a social disaster.

Last edited:
Chestermiller said:
For a gas at atmospheric pressure, a typical value of the mean free path of a molecule (Wikipedia) is 68 nm. So there are lots of collisions of molecules within any practical volume of gas, even as ideal gas behavior is approached. This is how pressure variations are transmitted through a gas is cases where gas is accelerating or where the gas is in a gravitational field (which, according to equivalence principle, is equivalent to acceleration).
This is also interesting.

Let's use what I've recently read, the mean free path is

$$l=\frac{1}{\sqrt{2}\pi nd^2}$$

and as I've recently calculated, n for ordinary air in room temperature is around E25 and d seem to be 3E-13 for air, we have l=2,5dm.

If my old teacher isn't totally wrong the "correct" answer is 2,5dm but there is probably a systematic error here from my side.

Is the diameter of an air molecule close to 3E-13 or not?

Edison
PS
I love these kind of answers (without wikipedia links), this is discussing physics in my book.

Edison Bias said:
This is also interesting.

Let's use what I've recently read, the mean free path is

$$l=\frac{1}{\sqrt{2}\pi nd^2}$$

and as I've recently calculated, n for ordinary air in room temperature is around E25 and d seem to be 3E-13 for air, we have l=2,5dm.

If my old teacher isn't totally wrong the "correct" answer is 2,5dm but there is probably a systematic error here from my side.

Is the diameter of an air molecule close to 3E-13 or not?

Edison
PS
I love these kind of answers (without wikipedia links), this is discussing physics in my book.
When you get an answer that does not match the wikipedia number like this, you should at least consider the possibility that you made a mistake.

Chestermiller said:
When you get an answer that does not match the wikipedia number like this, you should at least consider the possibility that you made a mistake.
Of course I do understand that I've made a mistake but what is that mistake?

According to my former teacher and his self-written litterature the formula mentioned should get the correct answer but it obviously doesn't so my simple question is why and how I may correct it, I typed it several times on my mini-counter so it should be correct unless I have totally misunderstood it all.

Edison

Jamison Lahman said:
Per https://en.wikipedia.org/wiki/Molecule, "The smallest molecule is the diatomic hydrogen (H2), with a bond length of 0.74 Å." So it can't possibly be 3E-13.

It is easily google-able questions like this that @russ_watters was referring to.
So I was wrong again, interesting to hear about the H2 with diameters close to 1Å (a swedish unit btw), I find this very interesting because two protons seem to consume about 1Å of diameter, right?

Let's say that one proton consumes 0,5Å of diameter.

29 protons, as in air, may be lumped together in a ball or a cube, then the increase of diameter and thus the new diameter is approximately

$$D=29^{1/3}*0,5$$

Which gives us 3Å, so how about 3E-10? :)

Thanks for your help so far!

Edison

Edison Bias said:
So I was wrong again, interesting to hear about the H2 with diameters close to 1Å (a swedish unit btw), I find this very interesting because two protons seem to consume about 1Å of diameter, right?
H2 is close to 1 Angstrom because the size of H is about 1 Angstrom. Again, the size of a proton is much smaller than an atom/molecule.

Bird, Stewart, and Lightfoot, Transport Phenomena, give the characteristic collision diameter of oxygen and nitrogen in the Chapman-Enscog theory as 3.5 Angstroms. What do you get if you use this in your equation for the mean free path?

Chestermiller said:
Bird, Stewart, and Lightfoot, Transport Phenomena, give the characteristic collision diameter of oxygen and nitrogen in the Chapman-Enscog theory as 3.5 Angstroms. What do you get if you use this in your equation for the mean free path?
So I have hit the spot, this was very much fun!

But I couldn't have done it without you guys, thank you!

Let's recalculate:

$$l=\frac{1}{\sqrt{2}\pi nd^2}$$

Again, using n=E25 and now d=3E-10 I get 0,25um!

How stupid was I not before to get 2,5dm? :)

Checking with what you said before...you said 68nm, well at least 250nm is within the same magnitude.

I rest this part of my case :)

But let's calculate pressure without walls again:

The Ideal Gas Law
$$n=\frac{p}{kT}\approx \frac{E5}{E-23\cdot E3}=E25...1$$

The collision density per time unit
$$n_{ct}=\frac{1}{2}\sqrt{2}\pi n^2d^2v\approx n^2d^2v=E50*E-19*E3=E34...2$$

An estimation of the air molecular mass
$$m=(2/10*16*2+8/10*14*2)m_p=29m_p\approx E-26...3$$

The relationship between temperature an kinetic energy
$$E_k=\frac{3}{2}kT...4$$

The speed of the air molecules due to temperature
$$v=\sqrt{\frac{3kT}{m}}=\sqrt{\frac{3kT}{29m_p}}\approx \sqrt{\frac{E-23*E3}{E-26}}=E3...5$$

The pressure without walls
$$p=\frac{F}{S}=\frac{2mv/dt}{S}=\frac{2n\cdot 29m_p\cdot v}{n_{ct}d^2}\approx \frac{E25 \cdot E-26\cdot E3}{E34\cdot E-19}=E-13...6$$

E-14 as normal air pressure, I can obviously not use my equations :)

But which ones are wrong?

Equation 1 is the cheating part but should be right, giving E25 as particle density at normal air pressure and temperature, this one ought to be right.

Equation 2 comes from my teacher's litterature and I have also derived it above so it should be right.

Equation 3 is my amature estimation of the mass for the air molecule, should be right within a magnitute though, I think.

Equation 4 comes from the Ideal Gas Law via p=2/3nEk which I think I have derived above by using momentum changes on an "imaginary" surface, commonly accepted.

Equation 5 is a direct consequence of equation 4 using Ek=mv^2/2, see no reason why this would be terribly wrong.

Equation 6 is the tricky one and obviously terribly wrong, but what can you expect when it is my own composition (using known relationships)?

Edison

Edison Bias said:
So I have hit the spot, this was very much fun!

But I couldn't have done it without you guys, thank you!

Let's recalculate:

$$l=\frac{1}{\sqrt{2}\pi nd^2}$$

Again, using n=E25 and now d=3E-10 I get 0,25um!

What do you get if you use n = 2.68E25 and d = 3.5E-10?

Chestermiller said:
What do you get if you use n = 2.68E25 and d = 3.5E-10?
So you are a curious guy for details :)

Well, let's try it once more with your data:

Writing the equation again just to burn it into my memory:

$$l=\frac{1}{\sqrt{2}\pi n d^2}=68nm$$

68nm, just as you said!

Edison
PS
Now I know that both n and d are almost correct in my formulas, thank you for that! This means that Eq. 1 is correct and if Eq. 2 is faulty only v (which comes from equation 4, an established equation), is wrong or the whole equation is wrong, so as I see it, only Eq. 2 and/or Eq. 6 may be wrong. But Eq. 2 comes directly from my litterature and it is easily derived, so it's got to be Eq. 6 that is wrong.

Another way to define the number of collisions per volume and time unit is

$$n_{ct}\approx \frac{nv}{l}...7$$

this while

$$n_{ct}=\frac{1}{2}\sqrt{2}\pi n^2d^2v...8$$

and

$$l=\frac{1}{\sqrt{2}\pi nd^2}...9$$

And while we now have correct values for n, l and I am quite confident that Ek=3/2kT holds at least for three dimensions of freedom such that v actually is approximatelly E3.

This then means that the result of Eq. 7 is approximatelly E25*E3/E-7=E35.

E35 collisions per cubic meter and second is amazing!

But now when we have n, n_ct, m, v and d we should be able to calculate pressure without walls.

I repeat my amateur formula:

$$p=\frac{F}{S}=\frac{2mv/\Delta t}{S}=\frac{2n\cdot 29m_p\cdot v}{n_{ct}d^2}...10$$

And once again evaluate it: p=E25*E-26*E3/(E35*E-19)=E-14 Pa

Still uncorrect obviously but now we know that all the parameters are right which makes my number 10 formula incorrect.

But, I am summing up all particles and their change of momentum when colliding with each other, then I am dividing with the area sum of all colliding particles.

The numer of collisions per volume unit is also per time unit, so dt vanishes from the equation.

I really wish to calculate pressure without walls so please someone help me.

If I can't get this right I will go on believing that pressure without walls is not the same as pressure with walls.

Edison
PS
I love the english language, it is so much fun to be able to express oneself in a foreign language, in spite of all my wrong spelling :)

Set up the analysis for a control volume with walls .

Determine whether the analysis gives the same or varying answers for pressure when walls are separated by small , large or infinite distances .

Nidum said:
Set up the analysis for a control volume with walls .

Determine whether the analysis gives the same or varying answers for pressure when walls are separated by small , large or infinite distances .
A very good suggestion, I will try to do that.

Thank you!

Edison

Deriving pressure without walls

Suppose we have a surface with gas where the surface is denoted with S and the infinitesmal hight of the surface is vdt.

Gas molecules hit this surface with an angle relative the normal of the surface called theta.

While we're only interested in impulses normal to the surface the volume can be written

$$dV=S\cdot vdt\cdot cos\theta...5.1$$

where a molecule with theta as impact angle is moving from one side of S to the other side of S within the time dt.

n denotes the number of molecules per volume unit but if we want to consider the number of molecules per volume, speed and angle unit the differential of n may be written as

$$dn(v, \theta)=n(v, \theta)\cdot dv \cdot d\theta...5.2$$

then we have the molecule density as a function of speed and impact angle change

The number of molecules within theta+d_theta and v+dv is then

$$dN(v, \theta)=dV\cdot dn(v,\theta)=S\cdot vdt\cdot cos\theta\cdot n(v, \theta)\cdot dv \cdot d\theta...5.3$$

then we have the number of molecules within the volume V+dV.

The impulse change per molecule with impact angle theta can then be written

$$P(v, \theta)=2mv\cdot cos\theta...5.4$$

which enables all molecules impulses within N+dN or P+dP to be written

$$P(v, \theta)\cdot dN(v, \theta)=dP(v, \theta)=2mv\cdot cos\theta\cdot S\cdot vdt\cdot cos\theta \cdot n(v, \theta)\cdot dv \cdot d\theta...5.5$$

or

$$dP(v, \theta)=2mv^2\cdot S\cdot dt\cdot n(v, \theta)\cdot cos^2\theta \cdot dv \cdot d\theta...5.6$$

Now, pressure is defined as

$$p=\frac{F}{S}=\frac{1}{S}\frac{dP}{dt}...5.7$$

Then

$$p=2mv^2\cdot n(v, \theta)\cdot cos^2\theta \cdot dv \cdot d\theta...5.6$$

or

$$p=4\frac{mv^2}{2}\cdot n(v, \theta)\cdot cos^2\theta \cdot dv \cdot d\theta=4E_k\cdot n(v, \theta)\cdot cos^2\theta \cdot dv \cdot d\theta...5.7$$

One thing with n is actually

$$\frac{1}{A}\frac{dn(v)}{dv}=e^{\frac{mv^2/2}{kT}}...5.8$$

where the integration of dn over all velocities gives the constant A and thus n (the integration is not trivial).

The interesting thing here is that we get n by integrating this Maxwellian distribution function, but what to do with the theta dependency?

I have learned this solution from my old teacher's nice litterature:

$$\frac{n(v,\theta)dvd\theta}{n(v)dv}=\frac{2\pi sin\theta d\theta}{4\pi}...5.9$$

Here I am sceptical, the space angle for the denominator might be R in sprerical radial length, 2pi in circumfrerence and R sin(theta) in polar radial level such that

$$\frac{n(v,\theta)dvd\theta}{n(v)dv}=\frac{2\pi R^2 sin\theta d\theta}{4\pi R^2}...5.10$$

I can't use a small volume and increase that volume because the radius (R) cacelles out, so if I am right so far I need to find an expression for n(v, theta) that does use actual radial size or something like that.

To make things complete 5.10 makes

$$p=2\int_{-\infty}^{\infty} n(v)\frac{mv^2}{2}dv \int_0^\frac{\pi}{2}cos^2(\theta) sin(\theta) d\theta=2\int_{-\infty}^{\infty} n(v)\frac{mv^2}{2}dv \int_{0}^\frac{\pi}{2} cos^2(\theta)sin(\theta)d\theta=\frac{2}{3}nEk...5.11$$

Edison
PS
On the other hand, look at 5.10 and what do we see? We see that R cancelles out but wasn't that the kind of proof I was after all the time :D

Why do we need walls when we have gravity?
If we "needed" walls then we would not have atmospheric Pressure. Isn't gravity holding Nitrogen and Oxygen at those varying densities the reason we are still living under an ocean of air Ans it hasn't floated away

Just a thought from an HVAC Contractor

russ_watters
Then it is fortunate that we have gavity as "walls" for our gases, but if we did not have gravity what happens then?

What happens if there is an amount of gas but the amount is not large enough for a gravity of its own (or gravity from a planet)?

Maybe it just disappears into space.

But somewhere along the line there should have been pressure beyond ambient (and thus without walls).

Just a playful thought from Edison.

I still wonder how to calculate pressure without walls.

Instead of using known pressure formulas which are derived from molecular collisions with surfaces I wish to calculate normal air pressure by only considering molecular collisions, thus p=2/3nEkp can not be used while it is derived from molecular collisions with surfaces.

p=nkT may be used but then n must be determined in another way, on the other hand we know p and T we can calculate n but in that case I would prefere another expression for p otherwise it is just a circumferance argument.

I have my first expression for p but as Ekp=3/2kT it is yet another circumferance argument.

I need a new expression to determine pressure.due to molecular collisions only.

Edison

Alternatively Pressure at a point in a fluid can also be related with energy density around that point. Then the fundamental difficulty of wall real orimaginary will evaporate. It is more correct also as pressure is a scalar quantity. Some one needs to follow this idea and derive.

Jamison Lahman said:
Hmm. I did not know that. The photons are still colliding with the wall, no, or does collision imply a mass interaction?
Even without a wall and without collisions, there is a rate at which photons (or better, just "stuff") passes through any surface that you choose to define. The rate at which momentum is transferred through that surface per unit area per unit time amounts to a pressure.

You do not have to have a force or collisions to transfer momentum.

jbriggs444 said:
The rate at which momentum is transferred through that surface per unit area per unit time amounts to a pressure.
Isn't there a factor of 2, because elastic collisions would reverse the momentum?

A.T. said:
Isn't there a factor of 2, because elastic collisions would reverse the momentum?
You have stuff flowing from one side to the other. That's a factor of one. But you also have stuff flowing from the other side to the one. There's your factor of two.

From the point of view of side A, both flows amount to a momentum transfer in the same direction. You have a loss of material with a toward-the-boundary momentum and a gain of material with a from-the-boundary momentum. Same for side B, of course.

Chestermiller
jbriggs444 said:
Even without a wall and without collisions, there is a rate at which photons (or better, just "stuff") passes through any surface that you choose to define. The rate at which momentum is transferred through that surface per unit area per unit time amounts to a pressure.

You do not have to have a force or collisions to transfer momentum.
If they just pass through there is no change of momentum so they transfer nothing. You need to have a hard surface or you need to imagine it there is no escape if you wish to apply Newton's Laws. It is another matter if photons get reflected from an imaginary mirror how much momentum they transfer. I think they transfer nothing as they lose no energy. In certain sense photons are unlike classical particle we have long chain of replies on this aspect of photon on Physics Forum

How did photons get into this discussion?

Let'sthink said:
If they just pass through there is no change of momentum so they transfer nothing.
If moving material leaves a system, it carries momentum away with it. If moving material enters a system, it carries momentum with it. A transfer of material transfers momentum.

Chestermiller said:
How did photons get into this discussion?
A photon gas was mentioned at least once as a situation with pressure but without collisions.

jbriggs444 said:
A photon gas was mentioned at least once as a situation with pressure but without collisions.
This was the OP's original intent?

Chestermiller said:
This was the OP's original intent?
It seems very much on-point. The original post ponders pressure without a change of particle momentum.

67 posts, all trying to nail down what is only an idealised behaviour of an idealise model.
Pressure is 'the force per unit area that would be there IF a wall was in position. Just 'this side' of the virtual wall, there would be no way of telling if it were there (by looking at the passing molecules) or not. It's just an abstraction that's useful for deriving gas laws from some simple basic kinetics - that's all. Physics is full of such very basic models and people keep believing they're real life. Some things you just have to suck up and go along with them. If you do that, you have a chance of understanding what it's all about.

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