Third principle of thermodynamics and the unattainability of absolute zero

Question

Consider a $S-T$ diagram (entropy-temperature) and consider cooling a substance by doing a series of succesive isothermal and reversible adiabatic processes between two volumes $V_{1}$ and $V_{2}$. Now when cooling the substance from $T_{1}$ to $T_{2}$ in the reversible adiabatic process we can write: $$S(0, V_{1})+\int_{0}^{T_{1}}\frac{C_{V}}{T}dT = S(0, V_{2})+\int_{0}^{T_{2}}\frac{C_{V}}{T}dT$$ letting $T_{2}=0$ will lead to: $$\underbrace{\int_{0}^{T_{1}}\frac{C_{V}}{T}dT}_{>0} = \underbrace{S(0, V_{2})-S(0, V_{1})}_{=0}$$ a contradiction showing that the third principle of thermodynamics implies that absolute zero cannot be achived. Is this reasoning correct?

Answer 1

Your reasoning proves that we cannot reach absolute zero by reversible adiabatic process from a non zero temperature. I prefer to reach this conclusion as follows:

Your first equation applies to any reversible adiabatic process between states $(T_1,V_1)$ and $(T_2,V_2)$. If we set one temperature to absolute zero the other must also be zero because both states must have the same entropy –zero. In other words, we cannot connect $T=0$ to any finite temperature via an isentropic path.

Answer 2

Another perspective: The equation is simply not defined for $T_2=0$, since this would result in dividing by 0. You could at most let $T_2$ go against 0 for both sides, but then you have to prove that you can commute the limit with the integral. (In general e.g. by using Lebesgue's convergence theorem).

tl;dr: Correct math is important.

PS: Another example where such intuitive "proofs" can go wrong are infinitesimal rotations, but I can't re-find the paper which I have in my mind.