Speed distribution for a sample of N gas particles

[esoteric deviance]

Homework Statement

http://www.fileden.com/files/2006/11/15/381656/19.39.png

The figure shows a hypothetical speed distribution for a sample of N gas particles (note that P(v) = 0 for speed v > 2v₀).

(b) What is the value of v_avg/v₀?

Homework Equations

v_avg = int[vP(v)]dv

The Attempt at a Solution

The solutions manual says that, for the triangular portion of the distribution, P(v) = av/v₀.
This may be a stupid question, but how does one get that?

(only need help with that portion of the problem for now)

 

[Dick]

Well, for the given function P(v), P(0)=0, P(v0)=a, and the function is linear in v. So it looks like your graph.

 

[m2003]

The value of av/v0 can be obtained by using the equation for the area under a triangle. Since the speed distribution in this case is a triangular shape, we can use the equation for the area of a triangle, which is 1/2 * base * height. In this case, the base is v0 and the height is av. Therefore, the area under the triangle is 1/2 * v0 * av = av/2.

Since we know that the area under a probability distribution curve must be equal to 1, we can set av/2 = 1 and solve for av. This gives us av = 2, which means that P(v) = 2v/v0.

To find the value of vavg/v0, we can use the equation given in the problem, vavg = int[vP(v)]dv. Plugging in the value of P(v) that we just found, we get vavg = int[v(2v/v0)]dv.

Integrating this expression gives us vavg = 2v^2/v0. We can then divide this by v0 to get vavg/v0 = 2v^2/v0^2. Simplifying this gives us vavg/v0 = 2v^2/4v0 = v^2/2v0.

Therefore, the value of vavg/v0 is equal to v^2/2v0, where v is the speed at which the triangular portion of the distribution reaches its maximum value.