Year 12: Cambridge Physics Problems (Joule's classification on molecule factor)

[johnconnor]

Derive an expression for the number of impact of gas molecules on unit area in unit time in terms of the number of molecules per unit volume, n, and their mean speed, <c>. Explain the assumptions you make.

A disc of radius "a" rotates with constant angular velocity "omega" in a gas at low pressure so that the molecules of the gas strike the disc from random directions. If the molecules are momentarily attached to the disc and leave in random directions relative to the disc, derive an expression for the torque exerted on the disc by the gas in terms of <c>, the mean speed of the gas molecules, rho, the density of gas, "a" and omega.

The guide says such:

The Joule classification of velocities is normally acceptable at this level of application of the kinetic theory of gases. Quite by chance, the Joule classification does lead to the correct numerical factor (1/3) in the formula for the pressure of a gas but it does not give the correct factor for the number of molecules striking unit area in unit time. The Joule classification gives n<c>/6 whereas the correct expression is n<c>/4. Try working out the correct expression by first considering those molecules in a hollow cone of angle between θ and (θ+ ∂θ) and then integrating over all values of θ from 0 to ∏/2 rad.

Whoa. Whadahell just happened? Is this something beyond a CIE A Level Physics student's reach?

 

[RyanCG]

This is pretty similar to one of the questions I got two years ago. Since the guidelines say I can't just tell you the answer I'll give you the advice that helped me most.

Brush up on Brownian motion(although not exactly what you're doing here, it's similar) and make sure you aren't just picking bits from the question to put in an equation. Some of the most important stuff here is given by the way the question is written.

 

[Infinitum]

I get n<c>/6.

I don't know about Joule classification of velocities, but how I solved the problem is assumed that the unit volume where there are 'n' molecules is in form of a cube(this is the standard kinetic theory assumption while calculating velocities), and so the number of molecules dashing out of every face of the cube is n<c>. But only one face of the cube with unit area is facing the disc...

I can't make sense of the other hollow cone method.

 

[johnconnor]

In the second part of the question, the molecules will strike both sides of the disc with random speeds c. When they leave the disc, they will have a systematic velocity v superimposed on c due to the motion of the disc. The momentum given to each molecule will cause a dragging effect on the disc. The drag on a molecule that strikes the disc between radii r and r + δr is mωr (hence drag increases with increasing radius). Remembering that there are two sides to the disc, the total number of molecules striking this annulus in unit time is 2 \times 2\pi r \delta r \times n&lt;c&gt;/4. The torque due to this annulus is the moment of this force about the centre of the disc, so the total torque involves an integral of the form \displaystyle\int kx^3 dx.

This is AWESOME! I would have never thought of something like this. But what I don't get is the part

The drag on a molecule that strikes the disc between radii r and r + δr is mωr

and

the total number of molecules striking this annulus in unit time is 2 \times 2\pi r \delta r \times n&lt;c&gt;/4

. And why do we have to integrate \displaystyle\int kx^3 dx?

 

[johnconnor]

Anyone?

 

[johnconnor]

Come on guys let's get pumping! XD