Here's my attempt to answer that question:
The first stop is to get a sense of what the realistic energy density of these approaches might be. A quick search lands you here:
http://en.wikipedia.org/wiki/Compressed_air_energy_storage#E...
My take-away: 1 m3 of air = about 300,000 J
How much energy does a typical house in the US use per day?
http://wiki.answers.com/Q/How_much_electricity_does_an_avera...
I'll use 50KWh per day
1kWh = 1,000W x 3,600s = 3,600,000J
This typical house, then, consumes 180,000,000J per day
How much air do we need to compress to provide all of the energy needs of this one house (per day)?
180,000,000J / 300,000J = 600 m3
How many homes in California?
http://quickfacts.census.gov/qfd/states/06/06037.html
Let's say it's about 14,000,000 homes
How much air do we have to compress every day to service these homes:
600 m3 x 14,000,000 homes = 8,400,000,000 m3
OK, there's a number, whatever it means.
Hmmm. How much of the available air are we using?
What's the volume of air of the atmosphere?
Tough question to answer. I think the number we'd want would be that of the Troposphere.
http://en.wikipedia.org/wiki/Atmosphere_of_Earth
I am going to use an over-simplification (you know, "assume a cow is a uniform sphere of milk" type stuff) to try to get a number. Sphere within a sphere to get the volume of the troposphere.
Average Earth diameter: 12,742km
http://www.universetoday.com/15055/diameter-of-earth/
Troposphere thickness: 17km
http://en.wikipedia.org/wiki/Troposphere
Troposphere volume: 4,341,334,943,758,290,000 m3
That means that California would use 0.00000019% of the troposphere per day if every single home was powered using compressed air energy storage.
Put a different way: It would take nearly 1.5 million years to process all of the air in the troposphere.
I'm not sure if the above is complete nonsense or not. The problem is far more complex than these quickie calculations might suggest. On first inspection it sounds like we have plenty of air to go around.
Would there be any environmental and/or air quality issues stemming from this approach? Do we end-up with cleaner air locally because of the process?
Interesting stuff.
.
EDIT: A few more data points.
How big of a container is required to store all of this air?
The original assumption was that 1 m3 of air would compress into a 5L bottle, or 0.005 m3.
Storage cube side length: 348m
Storage sphere diameter: 431m
How much would this much air weigh?
1 m3 of air at 20C = 1.204 kg
8,400,000,000 m3 = 10,113,600,000 kg
The question, for me, begins to be about how realistic it might be to construct enough smaller storage vessels to capture this volume safely.
The article mentions something about 40ft standard shipping containers. Assuming that the storage vessel has the internal dimensions of a standard 40ft container:
http://en.wikipedia.org/wiki/Intermodal_container
Container volume: ~ 67 m3
Containers required to store enough compressed air to supply homes in California: ~627,000 units.
That's a lot of containers, even if the calculations are off by 100%.