Pressure exerted by an ideal gas

  • Thread starter PainterGuy
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  • #1
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Hi

Please have a look on the attachment. Suppose that a 0.1 kg rubber ball having velocity of 60 m/s is moving between two walls A and B, and the distance between the walls is 1 m. It is having elastic collisions with the walls.

Let's focus on what is happening at wall B. The ball is moving toward wall B. The momentum of ball is 0.1 x 60 = 6 kgm/s. It strikes wall B, has elastic collision, and reflects back and starts traveling toward wall A. Let's assume that it spends an infinitesimal time, dt=10-3s, during the collision. The force exerted by the wall on ball is F=(mv-mu)/dt=(-6-6)/10^-3=-12000 N. The force exerted by the ball on wall B is 12000 N.

After having a collision with wall B, the ball starts traveling toward wall A and it would cover a distance of 2 m before it has another collision with wall B. It will take almost 0.034 seconds to strike wall B again and the frequency of collisions with wall B is 30 collisions per second. The average force might measure almost 12000 N or little less than that on some sensor placed at wall B while this process of the ball movement between two walls continues.

Do you agree with what I'm saying above? Thank you for the help.
 

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  • #2
russ_watters
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The average force might measure almost 12000 N or little less than that on some sensor placed at wall B while this process of the ball movement between two walls continues.

Do you agree with what I'm saying above? Thank you for the help.
There is no need to guess about the average force; you gave us the time of and between collissions.
 
  • #3
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There is no need to guess about the average force; you gave us the time of and between collissions.

Thank you for the reply.

I believe that I understand what you are saying. Let's assume that the force sensor is an analog one. When the first collision happens, the sensor will read 12000 N but as soon as the ball starts traveling toward wall A, the sensor needle will start falling toward 0 N and the falling movement will continue until another collision occurs but if the frequency of collisions is high enough the needle won't have enough to move much from the mark of 12000 N.
 
  • #4
russ_watters
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Thank you for the reply.

I believe that I understand what you are saying. Let's assume that the force sensor is an analog one. When the first collision happens, the sensor will read 12000 N but as soon as the ball starts traveling toward wall A, the sensor needle will start falling toward 0 N and the falling movement will continue until another collision occurs but if the frequency of collisions is high enough the needle won't have enough to move much from the mark of 12000 N.
Again: you are guessing based on the functionality of a hypothetical force sensor. Again: you gave us the times, so there is no need to guess. Calculate!
 
  • #5
Lord Jestocost
There is no need to assume that the ball spends an infinitesimal time during the collision (Edit: regarding the pressure exerted on walls by an ideal gas)!

The ball moves horizontally with speed |vx| back and forth between two walls, which are a distance L apart. When it hits the right wall and bounces back, its momentum changes by Δpball = -2m|vx| (here, m is the ball's mass). As the time interval between successive hits on the right wall is Δt = 2L/|vx|, the average force the wall exerts on then ball is Fball = Δpball/Δt = -mvx2/L. The average force that the ball exerts on the wall is thus Fwall = mvx2/L.
 
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  • #6
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Again: you are guessing based on the functionality of a hypothetical force sensor. Again: you gave us the times, so there is no need to guess. Calculate!

Let me try.

Total time spent during 30 collisions is 30x10^-3=0.03s, and the ball exerts 0 N force during 0.97 s. I believe that the average force should be almost 360 N. Please let me know if I have it correct. Thanks.
 
  • #7
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Thank you.

(Edit: regarding the pressure exerted on walls by an ideal gas)!

Your comment was on point. Now I understand that the force used in kinetic theory of ideal gases is an average force.

While calculating the pressure exerted by gas molecules, the most important unstated assumption, in my opinion, made is that after striking the wall a molecule travels toward the other wall and then back toward the same wall without colliding any other molecule on its way. There would be billions of molecules and I'm sure it won't be possible to travel back and forth without colliding with any other molecule. In my opinion, it would be a huge achievement of a molecule even if it could reach the other wall!

What really troubles me is that if a molecule makes collisions with other molecules on its journey back and forth, it would affect the force exerted on wall drastically. Could you please guide me? How is this a reasonable assumption to assume that a molecule doesn't collide with other molecules when the physical reality is quite different? Thank you.
 
  • #8
Lord Jestocost
Thank you.

Your comment was on point. Now I understand that the force used in kinetic theory of ideal gases is an average force.

While calculating the pressure exerted by gas molecules, the most important unstated assumption, in my opinion, made is that after striking the wall a molecule travels toward the other wall and then back toward the same wall without colliding any other molecule on its way. There would be billions of molecules and I'm sure it won't be possible to travel back and forth without colliding with any other molecule. In my opinion, it would be a huge achievement of a molecule even if it could reach the other wall!

What really troubles me is that if a molecule makes collisions with other molecules on its journey back and forth, it would affect the force exerted on wall drastically. Could you please guide me? How is this a reasonable assumption to assume that a molecule doesn't collide with other molecules when the physical reality is quite different? Thank you.

To calculate the pressure for an ideal gas with a huge number of particles, you consider averages in velocity. Have a look at http://hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/kinthe.html.

Edit: A good description can be found in The Feynman Lectures on Physics (for online reading): The Feynman Lectures on Physics, Volume I, Chapter 39. The Kinetic Theory of Gases (http://www.feynmanlectures.caltech.edu/I_toc.html)
 
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  • #9
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Thank you.

To calculate the pressure for an ideal gas with a huge number of particles, you consider averages in velocity. Have a look at http://hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/kinthe.html.

I'm sorry but I don't see how it relates to my question.

Edit: A good description can be found in The Feynman Lectures on Physics (for online reading): The Feynman Lectures on Physics, Volume I, Chapter 39. The Kinetic Theory of Gases (http://www.feynmanlectures.caltech.edu/I_toc.html)

I skimmed through Chapter 39 but I couldn't find any answer to my question.

The text I'm using is attached. I would request you to check my question again and guide me. Thank you.

PS: Please check the following link if the attached image isn't clear.
http://imageshack.com/a/img922/6356/nghW13.jpg
 

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  • #10
Lord Jestocost
I'm sorry but I don't see how it relates to my question.


The point is (simplified): due to the huge number of collisions all molecules acquire finally the same average velocity. The collisions balance themselves in such a way that at the end half of the molecules move with a certain velocity +vx towards the right wall, and half of the molecules with -vx to the left wall. A collision can, for example, change the direction of movement of a certain molecule from +x to –x, but there will be always another molecule whose direction of movement is at the same time changed from –x to +x by a collision. Based on this, you can calculate the pressure; see, for example, https://smartsite.ucdavis.edu/access/content/user/00002774/Sears-Coleman Text/Text/C16-20/20-3.html
 
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  • #11
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Thank you.

I think that I understand your point. The text I'm using says at the beginning that the number of molecules is large enough to allow a meaningful statistical treatment. Let me put it into my own wording.

Let's start from the moment when a molecule_1 has completed its collision with with right wall and is on its way toward left wall. The molecule_1 collides with another molecule within an infinitesimal amount of time, dt, and starts traveling in some other direction but not in its original direction. But as there are going to be trillions and trillions of molecules so there is going to be some other molecule_2 which at very same instant starts traveling with same average velocity in the same direction as molecule_1. Again, after an infinitesimal amount of time, dt, the molecule_2 collides with another molecule and starts traveling in some other direction. Again, at the very same instant, some other molecule_3 starts traveling in the very same direction as molecule_2 with same average velocity, and this indirect 'tag' process goes on indefinitely. I hope that I have it correct.
 
  • #12
Lord Jestocost
Thank you.

......... the number of molecules is large enough to allow a meaningful statistical treatment. Let me put it into my own wording.


I would describe it in a slightly different way: Consider a small volume element ΔV inside your gas-filled box (the box of volume V contains N molecules). In case that you are able to watch the molecules inside ΔV in course of time, you would find that molecules enter or leave ΔV, but on the average you have at every point in time approximately ΔN = NV/V molecules in ΔV. You would also find that molecules in ΔV suddenly change there direction of movement owing to collisions, but on the average, at every point in time, roughly half of the molecules in ΔV are moving in +x direction and the other half in –x direction. Some of the molecules have a higher speed, some a lower speed, but when averaging you can say that roughly half of the molecules in ΔV move with an average velocity <vx> in +x direction and roughly half of the molecules with an average velocity <vx> in -x direction..
 
  • #13
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Thank you for the help.

The r.m.s speed of air molecules is more of a result of few mathematical equations and it's value at standard atmospheric pressure is almost 503 m/s but 'more real' speed for the air molecules is the average speed which is taken to be almost 450 m/s. I hope that I have it right.

The sound in air travel due to the motion of air molecules where the disturbance is passed one molecule to the next and sound speed is about 340 m/s. I'm not much worried about the difference between speed of sound and r.m.s. speed of air molecules because r.m.s. is not really the actual speed for the molecules, at least to me. But why is the speed of sound only about 340 m/s when the average speed for the air molecules is 450 m/s?

Please also read the circled text in red at the bottom of attached image. If you are not able to see the attached image properly, please check the link below.
http://imageshack.com/a/img923/2963/PF4t04.jpg

Thank you.
 

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  • #14
Lord Jestocost
Qualitatively, you can say: "The speed of sound in a gas will always be less than vrms since the sound propagates through the gas by disturbing the motion of the molecules. The disturbance is passed on from molecule to molecule by means of collisions; a sound wave can therefore never travel faster than the average speed of the molecules (http://teacher.pas.rochester.edu/phy121/lecturenotes/Chapter18/Chapter18.html)."

Quantitatively, you need the Boltzmann equation to calculate the speed of sound in gases. Have a look at https://physics.stackexchange.com/q...lation-between-speed-of-sound-and-r-m-s-speed.
 
  • #15
russ_watters
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Sorry I missed this before:
Let me try.

Total time spent during 30 collisions is 30x10^-3=0.03s, and the ball exerts 0 N force during 0.97 s. I believe that the average force should be almost 360 N. Please let me know if I have it correct. Thanks.
That is correct, though the way you did it is a bit odd, and I'm not sure how you didn't get the wrong answer: you used a weighted average force over time, which is fine, but it is easier to just plug the collision rate itself into the momentum/force equation (f=delta-mv/t). But there is an error in your initial use in that the change in velocity is twice the speed. So you should have been low by a factor of 2....but whatever, I think you basically get it now.
 
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  • #16
russ_watters
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Thank you.

Your comment was on point. Now I understand that the force used in kinetic theory of ideal gases is an average force.

While calculating the pressure exerted by gas molecules, the most important unstated assumption, in my opinion, made is that after striking the wall a molecule travels toward the other wall and then back toward the same wall without colliding any other molecule on its way. There would be billions of molecules and I'm sure it won't be possible to travel back and forth without colliding with any other molecule. In my opinion, it would be a huge achievement of a molecule even if it could reach the other wall!

What really troubles me is that if a molecule makes collisions with other molecules on its journey back and forth, it would affect the force exerted on wall drastically. Could you please guide me?
It's the Prego Principle: it's in there. Or the way you put it, it's flipped over:
How is this a reasonable assumption to assume that a molecule doesn't collide with other molecules when the physical reality is quite different?
It's not in there. That isn't being assumed.
 
  • #17
russ_watters
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But why is the speed of sound only about 340 m/s when the average speed for the air molecules is 450 m/s?
I'm not certain, but I think it is because the molecules have both an average speed and an average direction. In essence, means the sound is zig-zagging toward you.
 
  • #18
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Thank you for your help.

Yes, the statement "a sound wave can therefore never travel faster than the average speed of the molecules" is correct but it's not complete. It does make sense that a sound wave cannot travel faster than the average speed of molecules but it can still travel at the average speed of molecules.


For some time I did agree with the statement "I think it is because the molecules have both an average speed and an average direction. In essence, means the sound is zig-zagging toward you" but later observed that it needs more explanation.

Suppose that we could draw all the arrows possible from the center of unit sphere toward its surface. There would be an infinite number of such arrows and each arrow represents a direction. Each of the air molecule could have any direction and like the arrows of unit sphere the number of such directions is infinite, and the direction of each molecule gets changed continuously as a result of collisions with other molecules.

To begin with, in my humble opinion, I don't think that there is an average direction. I think if we could add up the velocities of all molecules, the resultant velocity would be zero. I believe that I might be misinterpreting the use of word "average direction" above.


Please have a look on the attached image. Please also have a look on longitudinal wave motion on the following page: http://www.edu.pe.ca/gray/class_pag...l and Transverse Wave Motion.htm#longitudinal. The air molecules in the pipe are moving with average speed of 450 m/s (1620 km/h or 1007 miles/h) in all possible directions. As the speaker moves, it imparts momentum to the right and right of screen (momentum imparted toward the right in case of compression and toward the left in case of rarefaction). I hope that I have it right. Please have a look on the attached image. The horizontal and vertical components of projectile motion are independent of each other. Likewise, the momentum imparted by speaker to the air molecules is independent of their random motion and therefore zig-zagging shouldn't affect the speed of sound. I mean that if the speaker is imparting concerted velocity of 2 m/s toward the right of screen then the molecules should be able to transfer this velocity thru their collisions with each other with their average speed, i.e. 450 m/s.


In post #16, Prego Principle was mentioned. What is it? Thank you.
 

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  • #19
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Could you please comment on the post above? Thank you.
 
  • #20
Lord Jestocost
Yes, the statement "a sound wave can therefore never travel faster than the average speed of the molecules" is correct but it's not complete. It does make sense that a sound wave cannot travel faster than the average speed of molecules but it can still travel at the average speed of molecules.

The speed of sound must be somewhat less than the “average” molecular speed because not all molecules are moving in exactly the same direction as the sound wave.
 
  • #21
Charles Link
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The OP @PainterGuy previously asked (post #7) if collisions would affect the pressure on the walls. The number of collisions on the wall per unit area per unit time is given by the effusion rate formula ## R=\frac{n \bar{v}}{4} ## where ## n ## is the number of molecules per unit volume and ## \bar{v} ## is the average speed of the molecules. This formula is most readily derived by considering collisionless molecules, but it holds reasonably accurately in most cases, and so does the ideal gas equation ## PV=nRT ##. Effectively, collisions make it like the walls of the container are closer together for some of the molecules. In cases where there are huge deviations from the ideal gas law, (so that Van der Waals gas equation ## (P+\frac{a n^2}{V^2})(V-nb)=nRT ## is a better approximation), the collisions could affect the resulting pressure, but generally these deviations are fairly insignificant. e.g. attractive forces between the molecules would have the effect of lowering the pressure slightly.
 
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  • #22
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The speed of sound must be somewhat less than the “average” molecular speed because not all molecules are moving in exactly the same direction as the sound wave.

Thank you.

I think that I could understand it to some extent. Let me try. Suppose that we have a circular cylinder 2100 m in length and its diameter being 1 m. There is a piston at one end which can be moved and the other end is open. The speed of sound is 350 m/s in air and average speed of air molecules is 450 m/s.

[Just to give you an idea: Say a speaker is making a 60 Hz sound and travels 2 cm from crest to trough. Then its maximum speed is 2pi 60 Hz times 1 cm = 377 cm/s = 3.77 m/sec. The mach number of about 0.01. Such a speaker, assuming it was large enough, would make a very loud sound.]

So, let's say that the piston moves at speed of 3 m/s toward the right of screen, along x-axis, just for 1 second and then it's stopped suddenly. Please note that 3 m/s speed is almost 0.5% of average speed of air molecules which is 450 m/s! Suppose that we are able to see the movement of piston during 1-second duration at every instant.

Suppose that horizontally right of your screen represents positive x-axis and the left of screen negative x-axis and with origin being at the center of screen.

Very few molecules would be moving directly into the piston with their other velocity components, y and z, being zero. Likewise, few molecules would be moving entirely in y and z directions. Most of them would have a resultant velocity with components x, y, and z.

Only the molecules whose only velocity component is along negative x-axis could transfer piston's movement at 450+3 m/s after colliding into the piston. Furthermore, any molecules which are already moving in positive x-axis direction and get in touch with the piston momentarily while the piston is already moving in positive x-axis will also transfer the movement of piston at 450+3 m/s. Please remember that piston's speed is just almost 0.5% of the molecules speed. All other molecules have x-component as a part of their resultant velocity. For example, one of the molecules might have these component c_x=80, c_y=299, and c_z=315, and resultant velocity c=sqrt(c_x^2 + c_y^2 + c_z^2)=442 m/s. If we were able to divide the cylinder into thin slices each having infinitesimal dx thickness, most probably we would see that all the collinear molecules (along the x-axis) would have different result velocity. It means that a molecule from dx_1 slice whose only velocity component is along positive x-axis might collide with another molecule from dx_2 whose resultant velocity consists x, y, and z components. Likewise, another molecule from slide dx_1 whose resultant velocity consists of only y and z components might collide another molecules from slide dx_2 whose resultant velocity consists of only y component. So, we can see that the compression due to piston's motion cannot be transferred at the speed of air molecules. The speed of compression transfer along x-axis should be less. Furthermore, the compression transfer along all points of individual slices, like dx_1, dx_2, etc., takes place at the same average speed. What it means that it's not that the compression transfer moves at different speeds along different individual infinitesimal sections of each individual slices.
 
  • #23
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I can't see whether the point below has been mentioned.

With ideal gases the collisions are assumed to be perfectly elastic the result being that the contact time during a collision is zero. Using F = dmv/t we may write F= dmv/o which some people may interpret as being an infinite force. But if it was an infinite force it would be an infinite force acting for zero time so it all gets messy and meaningless. To get some sense out of it we don't consider the contact time of collisions but the time between collisions.
 
  • #24
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Hi again,

I understand that why the speed of sound, 340 m/s, is lower than the speed of air molecules, i.e. 450 m/s. In contrast, the average of water molecules is almost 600 m/s but the speed of sound in water is almost 1500 m/s! Why isn't the speed of sound lower than the average speed of molecules in this case? I understand that the almost incompressibility of water would also play role.

The same goes for the case of speed of sound in iron. In iron the speed of sound is almost 5000 m/s. I couldn't find the average speed of atoms in iron or metals. Could you please help me with it?

Thank you.
 

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