Why Is Average Relative Speed 4/3v in a Uniform Gas Speed Scenario?

AI Thread Summary
In a uniform gas where all molecules travel at the same speed v, the average relative speed is calculated to be 4/3v, contrasting with the sqrt(2)v derived from actual molecular speed distributions. The discussion critiques a method from a linked source, arguing that it leads to incorrect conclusions about average relative speed due to misapplication of mathematical principles. It emphasizes that the average speed perceived by any molecule regarding others remains consistent across the system. Participants suggest that calculating the average speed V from a chosen molecule's perspective is essential for determining the overall average relative speed. The conversation highlights the importance of understanding the geometric and statistical implications of molecular motion in gases.
Oojee
Messages
6
Reaction score
0

Homework Statement



For a gas in which all molecules travel with the same speed v, show that average relative speed = 4/3v (rather than sqrt(2) v which is the result obtained when we consider the actual distribution of molecular speeds.)


Homework Equations





The Attempt at a Solution



http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/menfre.html#c5
 
Physics news on Phys.org
How do you approach this problem? Show us your work so far.
By the way, I doubt the correctness of the method provided in the link.
 
hikaru1221 said:
How do you approach this problem? Show us your work so far.
By the way, I doubt the correctness of the method provided in the link.

I used the method given in the link & for the case of same speed for each particle this method gives zero average relative speed. I cannot think of any other approach to the problem :confused: that is why I posted the question here.
 
IMHO, the method in the link is a fallacy, or at least, it's a source of misunderstanding. The wrong step is that from this equation:
v_{rel}=\sqrt{\vec{v_1}^2-2\vec{v_1}\vec{v_2}+\vec{v_2}^2
we CANNOT go straight to this equation:
<v_{rel}>=\sqrt{<\vec{v_1}^2>-<2\vec{v_1}\vec{v_2}>+<\vec{v_2}^2>
simply because these two are different:
<\sqrt{\vec{v_1}^2-2\vec{v_1}\vec{v_2}+\vec{v_2}^2}>\neq \sqrt{<\vec{v_1}^2>-<2\vec{v_1}\vec{v_2}>+<\vec{v_2}^2>
That this method yields the correct result, I think, is just a mathematical coincidence.

We have the relative speed between 2 particles: v_{rel}=|\vec{v_1}-\vec{v_2}| = 2v|cos\phi | where \phi is one-half of the angle occupied by 2 vectors \vec{v_1} and -\vec{v_2}. Now some questions:
1. Because of the randomness of the gas, in the viewpoint of each molecule, the average speed of all other molecules relative to it (let V denote this average speed) is the same for every molecule; that is, the average speed in the viewpoint of each molecule doesn't depend on which molecule considered. Do you agree?
2. From that, what can you deduce about the relation between the average relative speed of the whole system and V? Do we only need to compute V in order to calculate the needed average relative speed?
3. Calculating V: Pick an arbitrary molecule with velocity \vec{u}. You can use \vec{u} as a fixed axis and calculate <|cos\phi |>. Notice the uniform distribution of velocities in every direction, which leads to spherical symmetry.

P.S.: The < > sign means average.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top