We are given moles of gas let say $n$. Temperature is increased from $T_1$ to $T_2$. Volume is held constant at $V_0$. We need to find the average speed of gas molecule when the temperature $T_2$ is achieved.

I am confused to use $v= \sqrt{\dfrac{3k_BT}{m}}$ or to use $v= \sqrt{\dfrac{8k_BT}{\pi m}}$ ; we are given monatomic gas having atomic $m = 6.4 \times 10^{-25} \:\rm kg$.

Please explain me which formula is applicable here and why.

  • 1
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    – garyp
    Apr 15 '15 at 16:29
  • $\begingroup$ @user36790 Nice work, I didnt understand anything before the edit $\endgroup$
    – jaromrax
    Apr 15 '15 at 18:47

The distribution of speeds in an ideal gas is given by the Maxwell-Boltzmann distribution. There are a variety of average speeds e.g. the most probable speed, the mean speed and the root mean square speed. Which one you use will depend on the application. The two equations you give are for the RMS speed:

$$ \sqrt{\langle v^2 \rangle} = \sqrt{\frac{3kT}{m}} $$

and the mean speed:

$$ \langle v \rangle = \sqrt{\frac{8kT}{\pi m}} $$

The question doesn't make it clear what speed is required, but I would guess that it is the mean.

  • $\begingroup$ I am asked to find the average speed of a gas molecule after the final temperature is achieved. Even I am confused to take which one of these? $\endgroup$ Apr 15 '15 at 17:15
  • $\begingroup$ RMS seems to me more like variation, not the average... $\endgroup$
    – jaromrax
    Apr 15 '15 at 18:46
  • $\begingroup$ @jaromrax: no, root mean square is a commonly used average. I wonder if you're mixing it up with the standard deviation. $\endgroup$ Apr 15 '15 at 19:24
  • $\begingroup$ @JohnRennie - well, I dont want to argue with you, ($\sqrt{v^2}$ really resembles $v$), I just feel uncomfortable when I look at $root$ histogram and I see mean and RMS fields with a bit different meaning...that was my point... $\endgroup$
    – jaromrax
    Apr 15 '15 at 20:48
  • $\begingroup$ @jaromrax - actually the RMS will tell you something about the mean energy of the atoms, since each atom has energy proportional to $v^2$. If I have one atom with v=0 m/s and another with v=2 m/s, the RMS value is $\sqrt{2}$ m/s but the mean velocity is 1 m/s. You get the same mean when both velocities are 1 m/s, but then you would get a smaller RMS velocity (also 1 m/s) - reflecting that the energy of the 0+2 is greater than the 1+1. $\endgroup$
    – Floris
    May 8 '15 at 18:30

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