What is pressure when there are no container walls?

[Edison Bias]

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 :)

 

[Nidum]

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 .

 

[Edison Bias]

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

 

[Edison Bias]

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

But how about the relevance?

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

 

[Mike Bergen]

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

 

[Edison Bias]

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.

 

[Edison Bias]

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

 

[Let'sthink]

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.

 

[jbriggs444]

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.

 

[A.T.]

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?

 

[jbriggs444]

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.

 

[Let'sthink]

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

 

[Chestermiller]

How did photons get into this discussion?

 

[jbriggs444]

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.

 

[jbriggs444]

How did photons get into this discussion?

A photon gas was mentioned at least once as a situation with pressure but without collisions.

 

[Chestermiller]

A photon gas was mentioned at least once as a situation with pressure but without collisions.

This was the OP's original intent?

 

[jbriggs444]

This was the OP's original intent?

It seems very much on-point. The original post ponders pressure without a change of particle momentum.

 

[sophiecentaur]

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.