9
$\begingroup$

In chemistry we are learning about kinetic molecular theory (KMT) of gasses, and I just couldn't help being surprised when I saw $\pi$ in the equation of mean speed. I know that whenever $\pi$ is involved in an equation, it somehow involves circles, but KMT assumes that the molecules are points. The mean speed is $\sqrt{(8RT)/(\pi\,M)}$, where $R$ is the universal gas constant (8.3144 J/K mol), $T$ stands for the temperature in kelvins, and $M$ is the molar mass of the molecule.

It just fascinates me that $\pi$ is everywhere, especially here and I would like to know why.

$\endgroup$
3
  • $\begingroup$ Also the Kinetic molecular theory of gases make the assumption that the molecular sizes are much smaller than the intermolecular distances so the assumption of considering them as points isn't a strong assumption (I haven't come across, I could be wrong). This is evident from the fact that when accounting for vibrational or rotational modes from which energy terms are obtained they are not considered as points. And in that case perhaps the spherical nature of atoms is still acceptable. $\endgroup$ Oct 22, 2019 at 15:18
  • $\begingroup$ Small note: I changed "velocity" to "speed" in your question, since we are technically looking at the magnitude of the velocities here (i.e. speed). I think technically the average velocity would be $0$ since we wouldn't expect any certain direction to be favored. $\endgroup$ Oct 22, 2019 at 15:29
  • $\begingroup$ You get factors of $\sqrt{\pi}$ whenever you integrate any function containing $e^{-kx^2}$ over all space. For an intuitive reason why that's true, the Math StackExchange might be a better place to look. $\endgroup$ Oct 22, 2019 at 15:36

3 Answers 3

8
$\begingroup$

This can be obtained from the Maxwell Distribution for the speeds of a molecule in an ideal gas $$P(v)=\left(\frac{m}{2\pi kT}\right)^{3/2}4\pi v^2e^{-mv^2/2kT}$$ where $m$ is the mass of the molecules, $k$ is Boltzmann's constant, $T$ is the temperature, and $v$ is the speed. $P(v)\,\text dv$ tells us the probability of observing a molecule with a speed between $v$ and $v+\text dv$

  • The $e^{-mv^2/2kT}$ term is just the Boltzmann factor that tells us the probability of observing a molecule with speed $v$ (or technically energy of $\frac12mv^2$).
  • The $4\pi v^2$ factor can be thought of as the the number of velocity vectors $\mathbf v$ with a speed $v$. The idea is that you can think of the set of these vectors as lying on a "sphere", so the number will be proportional to the surface area of this "sphere". (Technically the important part here is the $v^2$, and the $4\pi$ could be left out and then brought back upon normalization, but I think an interpretation like this is the main thing you are looking for here).
  • The $\left(\frac{m}{2\pi kT}\right)^{3/2}$ is just a constant term so that the entire distribution integrates to $1$, i.e. there is a probability of $1$ of observing any speed. A different view of this normalization constant can be found in Ezze's answer.

To obtain the expression for the average speed, we find the average through usual means $$\bar v=\int_0^\infty vP(v)\,\text dv=\sqrt{\frac{8kT}{\pi m}}$$


So where do the $\pi$'s come into play here? Well, we saw it in our velocity "sphere", and therefore it also comes into play in the constant factor. Then we end up with an overall $\pi^{-1/2}$, which carries over into the average speed. Since you were looking for circles, I would say the mostly likely culprit is just from that idea of the velocity "sphere".

$\endgroup$
0
2
$\begingroup$

The reason for the appearance of Pi in the Maxwell-Boltzmann distribution can be rationalized by the following steps:

1) The probability of a given state in a thermal ensemble is proportional to the Boltzmann factor $\exp{(-E/kT)}$.

2) The energy of the point gas particle is expressed by its momentum $p$, the Boltzmann factor is then $\exp{(-p^2/2mkT)}$.

So we know that the probability of a molecule having a total momentum of $p$ is proportonal to $\exp{(-p^2/2mkT)}$. To make a proper distribution $f_p$ that tells you the probablity of a molecule having a momentum between $p$ and $p+\delta p$ we must make sure that the integral of $f_p$ gives 1. That means we got to normalize it by the integral of the Boltzmann factor:

$ \int \int \int_{-\infty}^{\infty} \mathrm{d}p_x \mathrm{d}p_y \mathrm{d}p_z \exp{(-(p_x^2+p_y^2+p_z^2)/2mkT)} $

Now, know that this integral is the famous Gaussian integral:

$ \int_{-\infty}^{\infty} \exp{(-x^2)} \mathrm{d}x = \sqrt{\pi} $

Which, when applied for the three spatial dimensions, gets you the $\pi ^{-3/2} $ factor that you must have encountered in your lecture somewhere.

Side note: $\pi$ always appears when you are dealing with 3D problems simply because you almost always 'partition' the space into small spheres and you get the volume of the sphere in your results.

$\endgroup$
2
  • 1
    $\begingroup$ So perhaps the answer to why $/pi$ shows up in the ideal gas average speed is because it shows up in the value of the Gaussian integral. And why it shows up in the Gaussian integral is the deeper question $\endgroup$ Oct 22, 2019 at 16:14
  • $\begingroup$ I do not think the question should be "why it shows up in the Gaussian integral" because frankly there is no real physical intuition behind that. Mathematically it's nice and can be understood well, but fr a physics point of view, that is the value of the integral, stop. For me, the question could be: why does the Gaussian integral show up in the speed distribution, and for that, the answer is that the Boltzmann distribution is exponential. $\endgroup$
    – user112876
    Oct 23, 2019 at 9:15
0
$\begingroup$

This has to do with the probability distribution of the velocities of the molecules. Without knowing the distribution, one can only calculate the rms of the velocity (see e.g. here). Which leads to a different equation without $\pi$:

$v_{rms} = \sqrt{3RT/M}$

If you know the distribution, you can integrate the probability distribution of velocities over the volume (this is where the factor of $\pi$ comes in). You then arrive at the equation stated in your question. See e.g. here.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.