Pressure on an open piston (derivation)

[rohanlol7]

Homework Statement

http://imgur.com/a/QY1Fs

Homework Equations

F=dp/dt
P=F/A

The Attempt at a Solution

I do not have the answer to this problem. Its not for homework so i won't be getting it either. I don't think my solution is correct though
Let us consider a particle with speed V(x) perpendicular to the piston.
As it collides with the piston we conserve both momentum and energy
mV(x)+MU(x)=mV(1)+MU(1)
V(x)-U(x)=U(1)-V(1)
solving for V(1) we get V(1)= 2U(x)-V(x) in the approximation M>>m.
then I'm not sure whether i should assume V(x)>>U(x) and simplify this further to V(1)=-V(x) but i did use this for further calculations.
So the change in momentum of that particle is -2mV(x)
Now we calculate the average time a particle will take to collide with the piston as the average distance of particles from the piston divided by V(x) = x/2*(V(x))
the average force on the piston is thus 4m(V(x))^2/x
the average pressure is thus 16mn(V(x))^2/(x*pi*d^2)
since the particle can be moving in any direction we get V(x)^2=V^2/3
and then we replace that back in the above.
Now the next part i just use F=Ma and obtain an expression. However this seems very simplistic for an olympiad level problem so I'm not sure about this

 

[lychette]

you are on the right track with the change in momentum of 1 particle = 2mV.
You also recognise that force = rate of change of momentum. Do you see that the force due to this one particle = change in momentum x number of collisions per second? (I think your calculation of average time for a collision is misleading)

 

[rohanlol7]

you are on the right track with the change in momentum of 1 particle = 2mV.
You also recognise that force = rate of change of momentum. Do you see that the force due to this one particle = change in momentum x number of collisions per second? (I think your calculation of average time for a collision is misleading)

Yes I thought about that but I could not see how to calculate that in terms of the values

 

[rohanlol7]

you are on the right track with the change in momentum of 1 particle = 2mV.
You also recognise that force = rate of change of momentum. Do you see that the force due to this one particle = change in momentum x number of collisions per second? (I think your calculation of average time for a collision is misleading)

Thinking about it more maybe I could consider that n particles will collide in x/2v seconds and 2vn/x collisions occur per unit time ?

 

[lychette]

the time taken for the particle to travel the length, l, of the tube and return to the wall is 2l/v...so how many collisions per second

 

[rohanlol7]

the time taken for the particle to travel the length, l, of the tube and return to the wall is 2l/v...so how many collisions per second

I guess that would be nv/2l but why are we considering the full length of the tube despite there being a vacuum?

 

[lychette]

It will be v/2L for one particle.
I must admit that I do not fully understand the context of the question...you are correct the space is marked vacuum and my logic is that the pressure on the left hand of the piston will be the same as the pressure on the right hand side if it were filled with n molecules each of mass m moving with a velocity v .
so if you get an expression for the pressure in terms of n, m, L and v that will be the relevant pressure...! It is easy to get the expression with these considerations.

 

[haruspex]

perpendicular to the piston

The trouble is that vanishingly few will be.
Do you understand how to derive the $\frac 1{3V}nmv^2$ equation? It involves considering particles at angle theta to the normal and integrating. Thetricky part is that you have to account for the fact that a particle at a shallow angle to the surface will have a relatively long return time, i.e. it will tend to be a while before it strikes again.
Now, I see that you divided by three to account for the fact that there are three dimensions, but that is really just hopeful. E.g. you could equally argue that there are six directions, and only one of those hits the piston. And it is not clear to me how you would bring u_x into it in a cogent manner.

Anyway, the way I would solve this question is to go through that derivation from scratch, but including the velocity u_x in the algebra. You should get something like $\frac{nm}{V}(\frac 13v^2-\alpha u_xv)$. Not sure what they are expecting for $\alpha$ though. A crude approach gives 1, while invoking the Maxwell-Boltzmann distribution gives $\sqrt{\frac{8}{3\pi}}$.

 

[rohanlol7]

It will be v/2L for one particle.
I must admit that I do not fully understand the context of the question...you are correct the space is marked vacuum and my logic is that the pressure on the left hand of the piston will be the same as the pressure on the right hand side if it were filled with n molecules each of mass m moving with a velocity v .
so if you get an expression for the pressure in terms of n, m, L and v that will be the relevant pressure...! It is easy to get the expression with these considerations.

I should have posted the full question. This is basically a model for the propulsion for a pingpong ball in a vacumm tube.
Okay, maybe u are right since the question does not define what n is exactly. Yep i can get an expression for the pressure from there.
Okay but now if that's the pressure i' not sure how to get the acceleration as a function of x

 

[rohanlol7]

The trouble is that vanishingly few will be.
Do you understand how to derive the $\frac 1{3V}nmv^2$ equation? It involves considering particles at angle theta to the normal and integrating. Thetricky part is that you have to account for the fact that a particle at a shallow angle to the surface will have a relatively long return time, i.e. it will tend to be a while before it strikes again.
Now, I see that you divided by three to account for the fact that there are three dimensions, but that is really just hopeful. E.g. you could equally argue that there are six directions, and only one of those hits the piston. And it is not clear to me how you would bring u_x into it in a cogent manner.

Anyway, the way I would solve this question is to go through that derivation from scratch, but including the velocity u_x in the algebra. You should get something like $\frac{nm}{V}(\frac 13v^2-\alpha u_xv)$. Not sure what they are expecting for $\alpha$ though. A crude approach gives 1, while invoking the Maxwell-Boltzmann distribution gives $\sqrt{\frac{8}{3\pi}}$.

Yeah actually the previous part requires me to derive this equation ( without calculus though ) I basically used v^2=v(x)^2 + v(y)^2 + v(z)^2 and from there used the fact that the random motion and concluded that on average V^2=3v(x)^2.
Do you have any idea about how to get an expression for the pressure in terms of the qunatities given (l,n,v,d only )?
I don't think it requires calc though they usually mention calc when its needed.

 

[haruspex]

Yeah actually the previous part requires me to derive this equation ( without calculus though ) I basically used v^2=v(x)^2 + v(y)^2 + v(z)^2 and from there used the fact that the random motion and concluded that on average V^2=3v(x)^2.
Do you have any idea about how to get an expression for the pressure in terms of the quantities given (l,n,v,d only )?
I don't think it requires calc though they usually mention calc when its needed.

In the equation, n is the number of molecules in some volume V and m is the mass of each. So in the problem nm/V is the density of the air. But they do not give the volume to the left. To apply the equation, without being told V or the density, the length you need is from the open end of the tube to the piston. I.e., it should say to relate the pressure to n, m, v, x and d. I suggest you assume that.

Also, it says "when the piston is moving at speed u_x". That implied to me you should get a pressure that depends on the current speed of the piston. It certainly will reduce slightly as the piston gains speed, since we are told to assume that the air molecules keep traveling at the same speed (relative to the tube, I assume). The equation you quote does not take that into account.
But if you are not expected to use calculus then probably you are supposed to ignore the effect of u_x on the pressure. That speed will be small compared to v. Maybe there's a non-calculus way of figuring it out but I can't think of one.

 

[lychette]

The trouble is that vanishingly few will be.
Do you understand how to derive the $\frac 1{3V}nmv^2$ equation? It involves considering particles at angle theta to the normal and integrating. Thetricky part is that you have to account for the fact that a particle at a shallow angle to the surface will have a relatively long return time, i.e. it will tend to be a while before it strikes again.
Now, I see that you divided by three to account for the fact that there are three dimensions, but that is really just hopeful. E.g. you could equally argue that there are six directions, and only one of those hits the piston. And it is not clear to me how you would bring u_x into it in a cogent manner.

Anyway, the way I would solve this question is to go through that derivation from scratch, but including the velocity u_x in the algebra. You should get something like $\frac{nm}{V}(\frac 13v^2-\alpha u_xv)$. Not sure what they are expecting for $\alpha$ though. A crude approach gives 1, while invoking the Maxwell-Boltzmann distribution gives $\sqrt{\frac{8}{3\pi}}$.

This is easy to deal with by considering the x, y and z components of the particles velocity. If the number of particles is large and the behaviour is random then the mean square velocities in the x,y and z directions will be equal and = 1/3 of the mean square velocity.

 

[rohanlol7]

thats exactly what i assumed. Let's ignore U(x) and try to think about this. since the molecules speeds relative to the piston isn't changing can't we argue that the number of collisions per unit time is a constant? ( collisions of molecules with the piston ) if so, shouldn't our answer be independent of x?

Yep i thought Ux should be included too, but it gets complicated from there so i decided to work with v>>Ux.
Does lychette's approach take that into account ?

 

[lychette]

I did not take x into account at all ! in the derivation of the original expression. This is a standard derivation and is found in many textbooks.
I think the 'x' part of the question comes into the second part of the question.

 

[rohanlol7]

I did not take x into account at all ! in the derivation of the original expression. This is a standard derivation and is found in many textbooks.
I think the 'x' part of the question comes into the second part of the question.[/QUOTE
this is weird as the previous part of the question is literally about the standard derivation of P=nmv^2/3V

 

[lychette]

It would be interesting to see the full question. This is part (c), what was the reference to a ping pong ball?

 

[rohanlol7]

It would be interesting to see the full question. This is part (c), what was the reference to a ping pong ball?

here is the question http://www.physics.ox.ac.uk/olympiad/Downloads/PastPapers/BPhO_Paper3_2005_QP.pdf is question 2.

 

[haruspex]

This is easy to deal with by considering the x, y and z components of the particles velocity. If the number of particles is large and the behaviour is random then the mean square velocities in the x,y and z directions will be equal and = 1/3 of the mean square velocity.

There's a little more to it than you have spelled out there, which perhaps you realize. E.g. it would not work if you were trying to find the average momentum imparted per collision. But I do see now how to make that line of argument work for finding the pressure, thanks.

 

[haruspex]

shouldn't our answer be independent of x?

It would be if you could include the density of the air in the equation, but you are told to use n, the number of molecules in some volume, and the mass m of each molecule, and the diameter of the tube. The only way for you to express the density is by specifying the volume for n, and to do that you need to mention x.

 

[lychette]

here is the question http://www.physics.ox.ac.uk/olympiad/Downloads/PastPapers/BPhO_Paper3_2005_QP.pdf is question 2.

Thanks for this, it makes it a really nice question.
I would say the speed of the ping pong ball would be the speed of sound (given in the data sheet) which makes the KE easy to calculate.

 

[rohanlol7]

Thanks for this, it makes it a really nice question.
I would say the speed of the ping pong ball would be the speed of sound (given in the data sheet) which makes the KE easy to calculate.

I would have put the speed of sound as the maximum speed too, however I'm not 100% convinced.

 

[lychette]

I would have put the speed of sound as the maximum speed too, however I'm not 100% convinced.

the speed of sound in a gas is essentially the speed of the gas molecules. If you use the expression for pressure of a gas to calculate the speed of molecules (it is the root mean square speed) for air at atmospheric pressure you get essentially the speed of sound in air.
I think this is why the tubes in the ping pong ball experiment were evacuated, the expansion of the gas into a vacuum tube is 'free'

 

[haruspex]

Thanks for this, it makes it a really nice question.
I would say the speed of the ping pong ball would be the speed of sound (given in the data sheet) which makes the KE easy to calculate.

It is stated that one should assume it is very much less than the speed of sound.

 

[haruspex]

the speed of sound in a gas is essentially the speed of the gas molecules

It's a bit less. For a diatomic ideal gas it would be $\sqrt{\frac{\gamma}3}=\sqrt{\frac 7{15}}$ of the rms speed.