I am calculating the kmolar entropy of a Debye solid at low temperature ($T << \theta_D$) and high temperature ($T >> \theta_D$). For the low temperature, I use $U = U_0 + \frac{9 k_B T^4}{\theta_D^3} \int_0^{\theta_d / T}{\frac{x^3}{e^x-1}}dx$ to evaluate $$C_V=\dfrac{dU}{dT}=\frac{12 \pi^4 R_0}{5}(\frac{T}{\theta_D})^3$$ since for $T<<\theta_D$ we can approximate the upper bound of the integral as infinity.

Then, $S=\int_0^T{C_V \frac{dT}{T}}=\frac{4 \pi^4}{5}R_0 (\frac{T}{\theta_D})^3$. This is the answer I expect. When I try this for the high temperature case, however, I use the fact that $T >> \theta_D$ to expand the exponential in the integrand $e^x \approx 1 + x$ and evaluate, but get $C_v = 3R_0$, yielding $S=\int_0^T{3 R_0 \frac{dT}{T}}$, which is not defined. The textbook suggests I should get $S=3 R_0 \left[\ln{\frac{T}{\theta_D}}+ \frac{4}{3}\right]$.

I think my misunderstanding is in how the entropy S is related to $C_V$. How is the entropy of the Debye solid at high temperature ($T >> \theta_D$) related to its entropy $S$?

## Best Answer

No, you are doing it correctly. The problem is only with the limits of your integration.

Start with an indefinite integral like: $$ \int \ dx \ \frac{1}{x} = \ln \lvert x \rvert + C $$ Then use your limits to define $x \rightarrow \frac{T}{\theta_{D}}$ and $C$ to give the proper constant when $\lim_{T \rightarrow 0} S$.

The problem arises because the temperature cannot actually get to zero in this approximation (or rather, this approximation is probably not valid when $T \sim 0$).

Side NoteI would recommend defining $\theta_{D}$, $T$, etc. explicitly in your question to help others more clearly understand your question.