# Avoiding a Divide by Zero with Absolute Entropy

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.

• 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. – knzhou Jul 14 '16 at 1:51
• From Quantum Statistics one can show that the specific heat goes to zero as a power law of the temperature. – Diracology Jul 14 '16 at 2:18

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 $T_1\to 0$, $c_v(T)\to 0$.
• $G=H-TS$, (-228.6 kJ/mol)=(-241.82 kJ/mol) - (298K)S gives S = -0.044 kJ/mol – John Stanford Jul 14 '16 at 3:35