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
$$\langle v \cdot \bar v \rangle = \langle v \rangle \cdot \bar v = \bar v \cdot \bar v$$