Me and my friend came across this derivation is some lecture notes of our thermal physics module. We have been trying to calculate the partial differential of the internal energy but cannot get the given answer. I tried doing it using the quotient rule but cannot see where the $\left(\frac{hv_E}{K_BT}\right)^2$ term comes from. $$U=\frac{3Nhv_E}{\exp\left(\frac{hv_e}{K_BT}\right)-1}$$ $$C_V=\left(\frac{\partial U}{\partial T}\right)_V = \frac{3NK_B\left(\frac{hv_E}{K_BT}\right)^2\exp\left(\frac{hv_E}{K_BT}\right)}{\left[\exp\left(\frac{hv_E}{K_BT}\right)-1\right]^2}$$ Any help would be much appreciated, thanks!

  • $\begingroup$ As far as I can see it, only a factor of $h$ is missing. But I guess, it is missing in the expression for $U$ already. Just use the chain rule, note $\partial_T 1 /T = -1/T^2$. $\endgroup$ – Sebastian Riese May 24 '15 at 12:31

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