Finding average velocity in a velocity distribution

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The discussion focuses on finding the average velocity of gas molecules given a speed distribution described by N(v) = Cv^2 for 0 < v < V0. The first part involves integrating N(v) to derive C in terms of N and V0, resulting in C = 3N/(V0^3). The second part calculates the average speed of the particles, yielding Vav = (3/4)V0. Participants express confusion over the notation used, particularly the dual use of N to denote both the number of molecules and the speed distribution. Overall, the calculations presented are deemed correct, with a note on the importance of clear variable definitions.
Woolyabyss
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Homework Statement


A gas has a hypothetical speed distribution for N gas molecules with N(v) = Cv^2 for 0 < v < V0. Find (i) an expression for C in terms of N and v0 (ii) the average speed of the particles

Homework Equations



N/A

The Attempt at a Solution


(i)
integrating N(v) with respect to v from 0 to V0 gives
N = (C*V0^3)/3 (where is N is total number of molecules)
rearranging gives C =3*N/(v^3)
(ii)
integrating N(v)*v with respect to v gives
sum(v) = (C*V0^4)/4
subbing in for C gives
sum(v) = (3N/4)*V0

sum(v)/N = Vav = (3/4)*V0

I've been having difficulty with this question for a while and can't seem to find any similar question online so I have no idea if what I did is correct. Any help would be appreciated.
 
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Woolyabyss said:
N(v) = Cv^2 for 0 < v < V0
What does that mean?
In the preceding sentence we were told that ##N## is the number of molecules in the system. Now all of a sudden ##N## is being used to denote something completely different.
From the fact that later on it is integrated to obtain an expected value, I presume that ##N(v)## is supposed to be a probability density function (pdf) for the random variable ##V## that is the speed of a randomly selected molecule. But if that's the case it needs to use a different letter. For instance it would be common practice to write ##F## for the cumulative distribution function so that

$$Prob(V\leq v)\equiv F(v)$$
and then
$$f(v)=\frac{dF(v)}{dv}$$
and then specify the distribution by the equation ##f(v)=Cv^2##.

But I don't know whether that is what was intended.
 
andrewkirk said:
What does that mean?
In the preceding sentence we were told that ##N## is the number of molecules in the system. Now all of a sudden ##N## is being used to denote something completely different.
From the fact that later on it is integrated to obtain an expected value, I presume that ##N(v)## is supposed to be a probability density function (pdf) for the random variable ##V## that is the speed of a randomly selected molecule. But if that's the case it needs to use a different letter. For instance it would be common practice to write ##F## for the cumulative distribution function so that

$$Prob(V\leq v)\equiv F(v)$$
and then
$$f(v)=\frac{dF(v)}{dv}$$
and then specify the distribution by the equation ##f(v)=Cv^2##.

But I don't know whether that is what was intended.
according to my lecture notes the dimensionless product N(v) dv gives the number of molecules having speeds in the range v to v +dv.
 
In that case, they mean by ##N(v)## the same thing as I have written above as ##f(v)##. It's bad practice to use the same symbol ##N## for two different things, but there's not much you can do about the notation your lecturer uses.
Using that interpretation, your calcs look correct. The only flaw is that in your final conclusion you have written 'sum(v)' where one would expect you have to written the average v0, since they have asked for an expression for C in terms of N and v0.
 
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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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