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I've been reading that unlike enthalpy, entropy has an absolute value that we obtain by setting it equal to zero at absolute zero temperature, but I'm having a hard time understanding how to avoid a divide-by-zero when actually calculating it.

I'm thinking that if you wanted to calculate the absolute entropy of an ideal gas, you'd start at absolute zero with zero entropy, then heat it up at constant volume using

$\Delta S = c_v\ln\frac{T_2}{T_1}$

After that, you could isothermally expand it to whatever state you want, but if we're starting at absolute zero, then $T_1$ is 0K. Am I making some kind of mistake somewhere? Thanks.

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    $\begingroup$ Yes, the mistake is that you assumed the heat capacity $c_v$ was constant. In reality, the very observation you just made, along with the fact that entropy should be finite at absolute zero, tells us that $c_v$ must vanish as $T$ goes to zero. This is called the Third Law of Thermodynamics. $\endgroup$ – knzhou Jul 14 '16 at 1:51
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    $\begingroup$ From Quantum Statistics one can show that the specific heat goes to zero as a power law of the temperature. $\endgroup$ – Diracology Jul 14 '16 at 2:18
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I think your mistake comes from assuming that there is an absolute zero temperature. Your formula should be corrected as below: $$\Delta S =\lim _{{T_1}\to 0}\int_{T_1}^{T_2}\frac{c_v(T)\mathrm dT}{T}$$ And as it is said in comments this is not a divided by zero. This includes an indeterminate form. Because when $T_1\to 0$, $c_v(T)\to 0$.

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  • $\begingroup$ Thanks! That makes sense, but how do you calculate absolute entropy if you don't do it that way? I thought about trying to calculate it from the enthalpy and Gibbs free energy of formation but that gave a negative number for water vapor, which doesn't make sense. $\endgroup$ – John Stanford Jul 14 '16 at 3:31
  • $\begingroup$ $G=H-TS$, (-228.6 kJ/mol)=(-241.82 kJ/mol) - (298K)S gives S = -0.044 kJ/mol $\endgroup$ – John Stanford Jul 14 '16 at 3:35
  • $\begingroup$ What I'm ultimately trying to do is write some code that calculates the Gibbs free energy for any ideal gas given the necessary thermodynamic properties and state information, but even if there's a way to bypass the need to calculate absolute entropy, I'm still curious to know how to do it for it's own sake. $\endgroup$ – John Stanford Jul 14 '16 at 3:41
  • $\begingroup$ @JohnStanford Sorry, sufficient discussion about absolute entropy (entropy with respect to entropy at zero temperature) needs to use quantum theories in statistical mechanics and I have no expertness in that topic. I hope there will be more posted answers by competent users. $\endgroup$ – lucas Jul 14 '16 at 4:26
  • $\begingroup$ @JohnStanford In addition, as far as I know all existent tables include the amounts of properties with respect to a conventional state. $\endgroup$ – lucas Jul 14 '16 at 4:29

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