Maxwell-Boltzmann distribution

[arccosinus]

Hello, I have a question regarding the Maxwell-Boltzmann distribution.

As you know, the distribution basically looks like
$$n(v) = constant \cdot v^2e^{-\frac{mv^2}{2kT}}$$
, where v is the speed, m is the particle mass, is k the Boltzmann constant and T is the absolute temperature.

Now, one can calculate the mean value of the squared velocity $$<v^2>$$ by evaluating the integral
$$\frac{\int_0^\infty v^2 n(v) dv}{\int_0^\infty n(v) dv} = \frac{3kT}{m}$$

From here, we can calculate the average (translational-)kinetic energy of particles,
$$<\frac{mv^2}{2}> = \frac{m}{2} <v^2>=\frac{3kT}{2}$$

Here comes the question, say we have air of temperature T. How can one easily show that the fraction of (for example) oxygen molecules with kinetic energy greater than $$\frac{3kT}{2}$$ is less than 50%?

My idea is basically to create a new distribution, say $$n_1(v^2)$$ by substituting $$v$$ with $$v^2$$ in the original distribution, integrating from 0 to $$\frac{3kT}{m}$$ and then from $$\frac{3kT}{m}$$ to infinity. One can then compare these integrals and conclude whether the fraction of molecules with kinetic energy greater than $$\frac{3kT}{2}$$ is less than 50%. However, this approach seems extremely superflous and I am not even sure the integrals converge (even though I think so). There has to be an easier argument! Can anyone help please?

 

[Gokul43201]

What does the area under the normalized Boltzmann distribution curve (say between v1 and v2) tell you ?

 

[astrokreng]

Hello, thank you for your question about the Maxwell-Boltzmann distribution. The Maxwell-Boltzmann distribution is a probability distribution that describes the distribution of speeds for a group of particles at a given temperature. It is derived from the kinetic theory of gases and is based on the assumptions of classical mechanics and the Boltzmann statistics.

To answer your question, we can use the concept of the cumulative distribution function (CDF). The CDF for the Maxwell-Boltzmann distribution is given by F(v) = 1 - e^{-\frac{mv^2}{2kT}}. This function represents the probability that a particle has a speed less than or equal to v.

To find the fraction of molecules with kinetic energy greater than \frac{3kT}{2}, we can integrate the CDF from \frac{3kT}{m} to infinity. This represents the probability that a particle has a speed greater than \frac{3kT}{2}. The integral can be evaluated as follows:

\int_{\frac{3kT}{m}}^\infty F(v) dv = \int_{\frac{3kT}{m}}^\infty (1 - e^{-\frac{mv^2}{2kT}}) dv = 1 - \int_{\frac{3kT}{m}}^\infty e^{-\frac{mv^2}{2kT}} dv

Using the substitution u = \frac{mv^2}{2kT}, the integral can be rewritten as:

1 - \int_{\frac{3kT}{2kT}}^\infty e^{-u} \frac{du}{2kT} = 1 - e^{-\frac{3}{2}} = 1 - 0.2231 = 0.7769

Thus, the fraction of molecules with kinetic energy greater than \frac{3kT}{2} is approximately 77.69%, which is greater than 50%. This means that the majority of molecules in a gas at temperature T have kinetic energy greater than \frac{3kT}{2}.

I hope this helps answer your question and provides a simpler approach to understanding the fraction of molecules with a certain kinetic energy in a Maxwell-Boltzmann distribution.