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How do I derive the specific heat of an ideal monatomic gas from the Maxwell-Boltzmann distribution?

I think I understand the derivation based on degrees of freedom, but I'm supposed to be able to derive it from the Maxwell-Boltzmann distribution.

My understanding on the M-B distribution equation is that it simply finds the probability that a single molecule has a particular speed, but I'm not sure how to connect that to specific heat.

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The Maxwell-Boltzmann distribution is also a tool to compute average values out of physical quantities written in terms of the velocity of the particles. For instance, the average kinetic energy is given by $$\langle K\rangle=\int\frac{1}{2}mv^2F(v)dv=\frac{3}{2}kT,$$ where $F(v)$ is the Maxwell-Boltzmann distribution.

Now since the gas is monoatomic this is the only contribution to its internal energy.

The last step is to compare the thermodynamic internal inergy $dU=nc_vdT$ with the internal energy obtained by the kinetic theory. This would give you the specific heat $$c_v=\frac{3}{2}R.$$

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