Here I follow calculations made by Andrew Dessler in this twitter thread. Given that it is an interesting calculation and that reading math in twitter is not ideal, I decided to replicate it here.
Start with 1 mole of gas, divided into a section with 80% $N_2$ and 20% $O_2$ and another section with $CO_2$. This is all at 293K and 1 atm of pressure. $CO_2$ makes up .0004 of the initial volume, such that when mixed, it will have 400ppm.
Now we let the gases mix, and we have an entropic increase. This increase is equal to the free expansion of each compartment into a vacuum. $$\Delta S = nR\log{\frac{V_f}{V_i}}$$ For $CO_2$, we have $n=.0004$ and $\frac{V_f}{V_i} = \frac{1}{.0004}$, so we get $$\Delta S = .0004 * 8.3 * \ln{2500} = .026 J / K$$ And similarly for the $N_2$ and $O_2$ section, we get $$\Delta S = .003 J /K$$ These entropy changes are added in the system, so we have $$\Delta S = .029 J / K$$ Now we get the Gibbs free energy by: $$\Delta G = -T\Delta S$$ $$\Delta G = -293*.029 = -8.4 J$$ Which means we our limit is using 8.4 J to remove .0004 of a mol of $CO_2$ from the atmosphere. This means a mole takes $8.4*2500 J$ to remove. A mole of $CO_2$ has a mass of 44 grams, so a kilogram takes $$E = 8.4*2500*(1000/44) = 477 kJ$$ Note: This applies to ground level, what happens if we work higher in the atmosphere, probably not engineering feasible
Q: Earth isn't an insulated volume is it Q: What about the reversal logic in thermodynamics Q: What about Joule expansion not being a reversible process, but Gibbs free energy being based on that A: You can't do better than the process being reversible, so this is an optimistic limit calculation