# Homework Help: Pressure on an open piston (derivation)

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1. Aug 27, 2016

### rohanlol7

1. The problem statement, all variables and given/known data
http://imgur.com/a/QY1Fs

2. Relevant equations
F=dp/dt
P=F/A

3. 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

2. Aug 27, 2016

### 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)

3. Aug 27, 2016

### rohanlol7

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

4. Aug 27, 2016

### rohanlol7

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 ?

5. Aug 27, 2016

### 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

6. Aug 27, 2016

### rohanlol7

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

7. Aug 27, 2016

### 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.

8. Aug 27, 2016

### haruspex

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 ux 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 ux 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 $\sqrt{\frac{8}{3\pi}}$.

Last edited: Aug 27, 2016
9. Aug 28, 2016

### rohanlol7

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

10. Aug 28, 2016

### rohanlol7

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.

11. Aug 28, 2016

### haruspex

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 ux". 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 travelling 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 ux 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.

12. Aug 28, 2016

### lychette

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.

13. Aug 28, 2016

### rohanlol7

thats exactly what i assumed. Lets ignore U(x) and try to think about this. since the molecules speeds relative to the piston isnt 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 ?

14. Aug 28, 2016

### 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 text books.
I think the 'x' part of the question comes into the second part of the question.

15. Aug 28, 2016

### rohanlol7

16. Aug 28, 2016

### lychette

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

17. Aug 28, 2016

### rohanlol7

18. Aug 28, 2016

### haruspex

There's a little more to it than you have spelled out there, which perhaps you realise. 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.

19. Aug 28, 2016

### haruspex

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.

20. Aug 28, 2016

### lychette

21. Aug 28, 2016

### rohanlol7

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

22. Aug 28, 2016

### lychette

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'

23. Aug 28, 2016

### haruspex

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

24. Aug 28, 2016

### haruspex

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.