# Third law of thermodynamics; residual entropy

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 3.9.

The entropy is related to temperature by

$\displaystyle \frac{1}{T}=\left(\frac{\partial S}{\partial U}\right)_{N,V} \ \ \ \ \ (1)$

Using the chain rule, and keeping everything at constant ${N}$ and ${V}$, we can measure the change in entropy due to a change in temperature as

$\displaystyle dS=\frac{dU}{T}=\left(\frac{\partial U}{\partial T}\right)_{N,V}\frac{dT}{T}=C_{V}\frac{dT}{T} \ \ \ \ \ (2)$

where ${C_{V}}$ is the heat capacity at constant volume:

$\displaystyle C_{V}=\left(\frac{\partial U}{\partial T}\right)_{N,V} \ \ \ \ \ (3)$

If we know ${C_{V}\left(T\right)}$ as a function of temperature, we can therefore find the change in entropy for a finite change in temperature by integration:

$\displaystyle \Delta S=S_{f}-S_{i}=\int_{T_{i}}^{T_{f}}\frac{C_{V}\left(T\right)}{T}dT \ \ \ \ \ (4)$

The total entropy in a system at temperature ${T_{f}}$ could theoretically be found by setting ${T_{i}=0}$ in the integral

$\displaystyle S_{f}-S\left(0\right)=\int_{0}^{T_{f}}\frac{C_{V}\left(T\right)}{T}dT \ \ \ \ \ (5)$

In theory, at absolute zero, any system should be in its (presumably) unique lowest energy state so the multiplicity of the zero state is 1, meaning that ${S\left(0\right)=0}$, and this integral does in fact give the actual entropy in a system at temperature ${T_{f}}$. It’s also obvious that for this integral to be finite (and positive) ${C_{V}\rightarrow0}$ as ${T\rightarrow0}$ at a rate such that the integral doesn’t diverge at its lower limit. Thus we must have ${C_{V}\left(T\right)\propto T^{a}}$ where ${a>0}$ as ${T\rightarrow0}$. Either of these conditions is a statement of the third law of thermodynamics, which basically says that at absolute zero, the entropy of any system is zero.

In practice, as a substance is cooled, its molecular configuration can get frozen into one of several possible ground states, so that there is a residual entropy even when ${T=0\mbox{ K}}$.

Example Carbon monoxide molecules are linear and in the solid form, they can line up in two orientations: OC and CO. Thus at absolute zero, the collection of molecules can be considered as a frozen-in matrix of molecules oriented randomly, so for a sample of ${N}$ molecules, there are ${2^{N}}$ possible structures. For a mole, the residual entropy is therefore

$\displaystyle S_{res}=k\ln2^{6.02\times10^{23}}=\left(1.38\times10^{-23}\right)\left(6.02\times10^{23}\right)\ln2=5.76\mbox{ J K}^{-1} \ \ \ \ \ (6)$

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