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
ΔS=cvlnT2T1ΔS=cvlnT2T1
After that, you could isothermally expand it to whatever state you want, but if we're starting at absolute zero, then T1T1 is 0K. Am I making some kind of mistake somewhere? Thanks.
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1 Answer
I think your mistake comes from assuming that there is an absolute zero temperature. Your formula should be corrected as below:
ΔS=limT1→0∫T2T1cv(T)dTTΔS=limT1→0∫T1T2cv(T)dTT
And as it is said in comments this is not a divided by zero. This includes an indeterminate form. Because when T1→0T1→0, cv(T)→0cv(T)→0.
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