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.