Maxwell-Boltzmann distribution

1. arccosinus

1
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?

2. Gokul43201

11,141
Staff Emeritus
What does the area under the normalized Boltzmann distribution curve (say between v1 and v2) tell you ?