Maxwell Distribution of speeds

  • Thread starter Qwurty2.0
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  • #1
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Homework Statement


A gas consisting of 15,200 molecules, each of mass 2.00 x 10-26 kg, has the following distribution of speeds, which crudely mimics the Maxwell distribution:

Number of Molecules - Speed (m/s)
1600 - 220
4100 - 440
4700 - 660
3100 - 880
1300 - 1100
400 - 1320

(a) Determine vrms for this distribution of speeds.

Homework Equations


vrms = √(2 * Ek/m)
Ek = (1/2) * n * M * v2

The Attempt at a Solution


Weighted Average
((1600 * 220) + (4100 * 440) + (4700 * 660) + (3100 * 880) + (1300 * 1100) + (400 * 1320)) / 15200
= (9944000 m/s) / 15200
= 654.32 m/s

vrms = √(2 * Ek/m)
Ek = (1/2) * n * M * v2
n = number of moles
M = molar mass
m = mass of single molecule

n = 15200 molecules / 6.02x1023 molecules/mole
= 2.525x10-20 mole

M = 2.00x10-26 kg / molecule * 6.02x1023 molecules / mole
= 0.01204 kg / mol

Ek = (1/2)(2.525x10-20moles)(0.01204 kg / mole)(654.21 m/s)2
= 6.506x10-17kg⋅m/s

vrms = √(2 * (6.506x10-17 kg⋅m/s) / 2.00x10-26 kg)
= 80659.78 m/s ...

The correct answer is 710 m/s so obviously I am either grossly overcomplicating this, or I am using the wrong equation.
 

Answers and Replies

  • #2
vela
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RMS means "root mean square". It's the square root of the mean of the square, so take the velocities and square them, find the weighted average of the squares, and then take the square root of the average.
 
  • #3
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I did what you said: I squared the velocities, then I calculated the weighted average (same way as I did above), and then took the square root of the weighted average. I ended up getting 82914.03 m/s, which is close to the answer I originally got but not the correct answer.

What am I missing?
 
  • #4
vela
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I got 707 m/s. I gather you're still messing around with calculating the kinetic energy, etc. That's unnecessary. The rms speed depends only on the velocity distribution of the molecules.

Your mistake seems to be that you're calculating the total internal energy of the gas, but ##E_k## is supposed to be the average kinetic energy of a single molecule.
 
  • #5
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My bad, I calculated it wrong. I tried again and got the right answer.

Thank you!
 

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