kinetic particle theory's gas pressure


by sgstudent
Tags: kinetic, particle, pressure, theory
sgstudent
sgstudent is offline
#1
Mar15-12, 12:27 AM
P: 636
When I calculate the pressure in an enclosed cylinder, I have to find the average force of the collisions of the air molecules. So that force will be the pressure per unit area.

When they make this statement is there a time variable to it? Since increasing the temperature in a fixed volume means that there is a greater frequency of the collisions then it equates it to having a greater pressure exerted. But if there's no time variable then the frequency of the collision doesn't matter.

Lastly, when a helium balloon is released into the atmosphere what happens to it? My guess is that it increases in size as the air pressure decreases as it goes up. So the gas pressure in the balloon is greater than the atmospheric pressure. Thus, the collisions of the helium molecules will result in the increase of volume such that the pressure in the balloon will be equal to the air pressure at the outside of the balloon. Hence, as it goes higher it will keep increasing in size until the point that it bursts.

Thanks for the help! :-)
Phys.Org News Partner Physics news on Phys.org
Physicists consider implications of recent revelations about the universe's first light
Vacuum ultraviolet lamp of the future created in Japan
Grasp of SQUIDs dynamics facilitates eavesdropping
Philip Wood
Philip Wood is offline
#2
Mar15-12, 03:47 AM
P: 863
First para. Pressure means force (at right angles to a surface) per unit area, so "pressure per unit area" doesn't make sense!

Second para. The kinetic theory equation for gas pressure of an ideal gas involves crms2 = [itex]\overline{c^2}[/itex], the mean square speed of the molecules. This mean square speed may itself be a function of time. Time doesn't need to be included in the pressure equation as an explicit variable.

Third para. I agree.
sgstudent
sgstudent is offline
#3
Mar15-12, 09:41 AM
P: 636
)I'm sorry. I didn't really understand how they attained the pressure in a enclosed cylinder. It's stated that you find the average of all the forces of the collision. So wouldn't that be the force of only one particle? So if I want the pressure of the whole cylinder then do I add up all those forces? Or are they talking about the forfeit of one particle per unit area of just one particle? So that's the pressure of one particle?

Oh so the more frequency means for force in a way? Thanks for the help Mr Wood, I really appreciate all the help you've given me :)

sgstudent
sgstudent is offline
#4
Mar15-12, 09:46 AM
P: 636

kinetic particle theory's gas pressure


Quote Quote by Philip Wood View Post
First para. Pressure means force (at right angles to a surface) per unit area, so "pressure per unit area" doesn't make sense!

Second para. The kinetic theory equation for gas pressure of an ideal gas involves crms2 = [itex]\overline{c^2}[/itex], the mean square speed of the molecules. This mean square speed may itself be a function of time. Time doesn't need to be included in the pressure equation as an explicit variable.

Third para. I agree.
But if I have more frequency of collisions does it mean more force? Thus more pressure?
Ken G
Ken G is offline
#5
Mar15-12, 09:52 AM
PF Gold
P: 3,075
Quote Quote by sgstudent View Post
But if I have more frequency of collisions does it mean more force? Thus more pressure?
Yes. The pressure depends on both the average momentum of the gas particles, and the average rate that they strike the wall. Increasing temperature increases both. Both the momentum, and the rate, increase like the square root of temperature, so combining them gives proportional to T, as in the ideal gas law.
sgstudent
sgstudent is offline
#6
Mar15-12, 10:02 AM
P: 636
Quote Quote by Ken G View Post
Yes. The pressure depends on both the average momentum of the gas particles, and the average rate that they strike the wall. Increasing temperature increases both. Both the momentum, and the rate, increase like the square root of temperature, so combining them gives proportional to T, as in the ideal gas law.
Thanks for the help! :) could you explain my first question in pressure too? I'm unsure if how it works (I dont really understand Wikipedia's explanation). Greatly appreciate the help!
Philip Wood
Philip Wood is offline
#7
Mar15-12, 10:02 AM
P: 863
1. It's total normal force due to all collisions divided by area of surface, so the more molecules in the container the more the pressure, for a given rms speed.

2. If the rms speed ([itex]c_{rms}[/itex]) of molecules is increased (that is if the temperature rises) this has two effects: (a) on average the hits with the wall are harder (larger momentum change) and (b) the hits are more frequent. Both these effects increase the pressure, which is why the pressure is proportional to [itex]c_{rms}^{2}[/itex].

The kinetic theory formula can be established in a page and a half or so of mathematical physics, but you need to have a good understanding of momentum, and good algebra to appreciate it.
sgstudent
sgstudent is offline
#8
Mar15-12, 10:42 AM
P: 636
Quote Quote by Philip Wood View Post
1. It's total normal force due to all collisions divided by area of surface, so the more molecules in the container the more the pressure, for a given rms speed.

2. If the rms speed ([itex]c_{rms}[/itex]) of molecules is increased (that is if the temperature rises) this has two effects: (a) on average the hits with the wall are harder (larger momentum change) and (b) the hits are more frequent. Both these effects increase the pressure, which is why the pressure is proportional to [itex]c_{rms}^{2}[/itex].

The kinetic theory formula can be established in a page and a half or so of mathematical physics, but you need to have a good understanding of momentum, and good algebra to appreciate it.
Oh then what does it mean by average force acting on the walls of the container? That's just the force of the individual air particle right? So I can't get the pressure this way?

The text in my textbook: when an air molecule hits the inner wall of a container, a force is exerted the wall. At any one time, there are numerous such collisions taking place between the air molecules and the wall. If we add up and find the average of all the forfes from such collisions we can find the averse force that is exerted on the wall. Since pressure is force per unit area, the gas is exerting a pressure on the container's wall.

I don't quite comprehend this text though. Thanks so much for the help! :)
Philip Wood
Philip Wood is offline
#9
Mar15-12, 10:53 AM
P: 863
Over any minute patch of wall, the pressure will fluctuate with time, as, for example, there will be intervals when no molecules happen to hit it. However with an 'ordinary' area of wall, these local effects aren't noticed – a sort of averaging.

You'll get a much better feel for this from the mathematical treatment I mentioned earlier, but, as I said, you do need some preparation for this, and the preparation can't be rushed.
sgstudent
sgstudent is offline
#10
Mar15-12, 11:02 AM
P: 636
Quote Quote by Philip Wood View Post
Over any minute patch of wall, the pressure will fluctuate with time, as, for example, there will be intervals when no molecules happen to hit it. However with an 'ordinary' area of wall, these local effects aren't noticed a sort of averaging.

You'll get a much better feel for this from the mathematical treatment I mentioned earlier, but, as I said, you do need some preparation for this, and the preparation can't be rushed.
Oh, I think I understand more now. Are they just trying to imply that each particle exerts a force. And if I divide that force with the contact area of particle, I'll get the pressure of that single gas particle, so if I multiply that by the number of collisions ill get the pressure of the gas? Really sorry but I don't understand what they are trying to imply for that part. Thanks for the help!
Philip Wood
Philip Wood is offline
#11
Mar15-12, 11:08 AM
P: 863
You don't do the pressure calculation using the contact area of the molecule. If you did, you'd no doubt find a huge local pressure over a very short time interval. But that's probably not what you're interested in. What you probably want is the steady pressure over the wall as a whole, so you divide the total force exerted by the molecules (there's a special way of finding this in terms of momentum) by the total wall area.
sgstudent
sgstudent is offline
#12
Mar15-12, 11:16 AM
P: 636
Quote Quote by Philip Wood View Post
You don't do the pressure calculation using the contact area of the molecule. If you did, you'd no doubt find a huge local pressure over a very short time interval. But that's probably not what you're interested in. What you probably want is the steady pressure over the wall as a whole, so you divide the total force exerted by the molecules (there's a special way of finding this in terms of momentum) by the total wall area.
Oh ok thanks! Then why does the pressure in a tyre is equal at all points of the walls, cos the pressure should be different at all points due to the different forces. Thanks so much for all the help!
Philip Wood
Philip Wood is offline
#13
Mar15-12, 11:21 AM
P: 863
It is different at different points, but you don't notice this as it averages out over time and wall area!
sgstudent
sgstudent is offline
#14
Mar15-12, 11:29 AM
P: 636
Quote Quote by Philip Wood View Post
It is different at different points, but you don't notice this as it averages out over time and wall area!
Could u explain this concept to me? I don't really understand the concept of it, really sorry! And thanks for the help!
Rap
Rap is offline
#15
Mar15-12, 11:44 AM
P: 789
Quote Quote by sgstudent View Post
Lastly, when a helium balloon is released into the atmosphere what happens to it? My guess is that it increases in size as the air pressure decreases as it goes up. So the gas pressure in the balloon is greater than the atmospheric pressure. Thus, the collisions of the helium molecules will result in the increase of volume such that the pressure in the balloon will be equal to the air pressure at the outside of the balloon. Hence, as it goes higher it will keep increasing in size until the point that it bursts.

The pressure inside the balloon is always higher than atmospheric because the balloon skin is compressing it. The size of the balloon tells you the difference in pressure. As the balloon goes up, the atmospheric pressure goes down, the difference in pressure increases, and the balloon expands. The expansion drops the pressure inside, but not to atmospheric. I don't remember the equations exactly, but its something like

[tex]P_b=\frac{NkT}{4 \pi r^3/3} = P_a +P_s(r)[/tex]

where [itex]P_b[/itex] is the pressure inside the balloon, the middle term is the ideal gas law, [itex]4 \pi r^3/3[/itex] is the volume of the balloon as a function of radius [itex]r[/itex], [itex]P_a[/itex] is atmospheric pressure, and [itex]P_s(r)[/itex] is the inward force of the skin divided by the surface area of the skin - in other words, the extra pressure provided by the skin. I forget what [itex]P_s(r)[/itex] looks like, I think its something like [itex]/alpha/r^s[/itex] where [itex]/alpha[/itex] relates to the springiness of the balloon.

The inward force of the balloon skin depends on the radius of curvature. The higher the radius of curvature, the higher the force, but the higher the radius of curvature the smaller the area, so its a combination of the two. For a tire, where it sits on the ground, the radius of curvature is big, because the tire is about flat. Its the ground that supplies the force there, not the balloon (tire), and since the balloon is not supplying a force, its radius of curvature is large (almost a flat surface).

Quote Quote by sgstudent View Post
Oh, I think I understand more now. Are they just trying to imply that each particle exerts a force. And if I divide that force with the contact area of particle, I'll get the pressure of that single gas particle, so if I multiply that by the number of collisions ill get the pressure of the gas? Really sorry but I don't understand what they are trying to imply for that part. Thanks for the help!
A single particle does not exert a force, it transfers momentum. Particles hit an area at a certain rate, like 1000 hits in 2 seconds. The force is the amount of momentum transferred by 1000 hits divided by 2 seconds, and the pressure is that force divided by the area.
Philip Wood
Philip Wood is offline
#16
Mar15-12, 12:11 PM
P: 863
A molecule hitting the wall does exert a (varying) force over a very small time interval. How could it not do so? When it comes to calculating the mean pressure on the wall (that is the observed pressure) it is, as Rap nicely explains, the momentum change of the molecules which we use, not the forces of single particles.
Rap
Rap is offline
#17
Mar15-12, 03:51 PM
P: 789
Quote Quote by Philip Wood View Post
A molecule hitting the wall does exert a (varying) force over a very small time interval. How could it not do so? When it comes to calculating the mean pressure on the wall (that is the observed pressure) it is, as Rap nicely explains, the momentum change of the molecules which we use, not the forces of single particles.
That's right - the molecules do exert a force over a very small time interval. What I should have said was that the size of that force and the duration of that time interval isn't really what's important, its their product that's important, which is equal to the momentum that they transfer. That's because the time interval we are averaging over is many, many times bigger than the collision time interval.


Register to reply

Related Discussions
Is there a Conceptual problem here? (Pressure in the Kinetic theory of gases) Classical Physics 15
Kinetic Theory- Calculating Kinetic Energy Biology, Chemistry & Other Homework 3
Kinetic theory: gas/pressure Introductory Physics Homework 1
Kinetic Theory - Pressure? Advanced Physics Homework 0
Help Needed in Pressure and Kinetic theory of GAses! Introductory Physics Homework 12