You're asking an important question, since removing CO
2 from the atmosphere and storing it as a liquid deep underground ("sequestering") is one way of reducing global warming. Obviously, it would not be an effective way if the energy required were greater than the energy obtained from burning the fuel which generated the CO
2 in the first place.
The specific form of the question you ask, however, is not actually answerable. At sea level pressure, CO
2 does not have a liquid form, but converts directly from gas to dry ice upon cooling. The liquid form is stable under higher pressure.
With a little care, we could calculate the net amount of free energy thermodynamically required to cool and pressurize the CO
2, but we would need to look up the detailed properties of CO
2 which is far from an ideal gas at high pressure. We won't bother, because you are probably more interested in the (larger) free-energy required with practical devices, not the ideal thermodynamic values.
I do not immediately know how to find those values. However, there is a relevant fact. I've heard that coal-burning power plants would require something like roughly 25% more fuel per energy output in order to sequester most of their CO
2 output.
Here's how I would go about doing a very rough calculation of what you want to know.
1. Sites on coal sequestration say that to sequester the 3.7 kg of CO
2 produced by burning 1kg of coal, you need about 20% of the electrical energy obtained from burning that coal.
2. I guess that burning coal, like most carbohydrates in foods, releases about 5kCal of energy per gram, or about 2*10
7 J per kg.
3. The efficiency of a typical power plant is about 30%, so that kg of coal makes about 6*10
6 J of electrical energy.
4. So 3.7 kg of CO
2requires something like 20% of that, or roughly10
6 J to sequester.
5. Liquid CO
2 is probably only a bit denser than water (you could look this up), so the cubic meter you want has mass of roughly 2000 kg.
6. Sequestering that cubic meter then seems to require around 5*10
8 J of electrical energy.
That answer could easily be off by a factor of two, but it should get you started.
Mike W.
(published on 06/03/2010)