Why Does My Calculation of Maxwellian Gas Velocity Yield an Incorrect Result?

[Saptarshi Sarkar]

Homework Statement

Calculate the value of ##<(v-\bar v)²>## for a Maxwellian gas

Relevant Equations

##<(v-\bar v)²> = <v²> + <\bar v²> - 2<v><\bar v>##

I expanded it as shown above and got

##<v²> + <\bar v²> - 2<v><\bar v>## = ##v_{rms}^2 + \bar v^2 = \frac {3kT} m+\frac {8kT} {πm}##

I used ##<v> = 0## as the velocity is equally likely to be positive as it's likely to be negetive.

From the above I get the answer ##\frac {kT} m(3+\frac 8 π)## but, the answer should be ##\frac {kT} m(3-\frac 8 π)##

Please help me understand what I did wrong.

 

[vela]

I don't understand either answer. What does ##\bar v## represent here? Isn't it the same as ##\langle v \rangle## so should be 0?

 

[Saptarshi Sarkar]

I don't understand either answer. What does ##\bar v## represent here? Isn't it the same as ##\langle v \rangle## so should be 0?

##\bar v## is the average velocity of the molecules in the gas and it is the same as ##<v>##, it would be 0 if the difference was not squared. The term ##<(v - \bar v)^2>## is the Variance of the velocity distribution.

 

[vela]

So you're saying ##v_{\rm rms}^2 = \frac {3kT}{m} + \frac{8 k T}{\pi m}##? Can you explain how you got that?

 

[Saptarshi Sarkar]

So you're saying ##v_{\rm rms}^2 = \frac {3kT}{m} + \frac{8 k T}{\pi m}##? Can you explain how you got that?

##\frac {3kT}{m}## & ##\frac{8kT}{\pi m}## are respectively the formulas for the square of the RMS and Average velocities of a molecule in a gas according to Maxwell-Boltzmann distribution.

 

[vela]

##\frac {3kT}{m}## & ##\frac{8kT}{\pi m}## are respectively the formulas for the square of the RMS and Average velocities of a molecule in a gas according to Maxwell-Boltzmann distribution.

How do you reconcile this statement with your earlier statement that the average velocity is 0?

 

[Saptarshi Sarkar]

How do you reconcile this statement with your earlier statement that the average velocity is 0?

I think I understood! ##v## cannot be negetive as it is velocity in 3D (##\vec v = v_x\hat i + v_y\hat j + v_z\hat k##) and it's magnitude must be positive! Only the average of ##v_x,v_y,v_z## would be 0.

Thanks a lot!

 

[vela]

I think I understood! ##v## cannot be negative as it is velocity in 3D (##\vec v = v_x\hat i + v_y\hat j + v_z\hat k##) and its magnitude must be positive! Only the average of ##v_x,v_y,v_z## would be 0.

I think you're working with velocities, not speeds, in this problem. That is, ##v## and ##\bar v## both represent vectors, not the magnitudes. If this is the case, you generally have
$$\langle (v-\bar v)^2 \rangle = \langle (v-\bar v)\cdot (v-\bar v) \rangle = \langle v^2 \rangle - 2\langle v \cdot \bar v \rangle + \langle \bar v^2 \rangle.$$ You can show that
$$\langle v \cdot \bar v \rangle = \langle v \rangle \cdot \bar v = \bar v \cdot \bar v$$