# volume comparison of compressed air and gas duster

I'm currently using 400ml Dust Off 67 gas dusters and want to replace them with compressed air and a blow-out gun.

In order to select the right equipment and get a better understanding I would like to compare the amount of air the gas duster contains to the compressed air solution. I know the gas duster does not actually contain compressed air, but for me it's all about the result. I know how much I can work with the gas duster and want to have a comparison.

So considerung the 400ml Dust Off 67 (see linked datasheet for additional details, I don't know what might be relevant there). If I had a gas container which I can fill up with a pressure of up to 8 bar, how much volume (in liters) would the gas container need to get the same amount of air out of it as I would get the with the gas duster?

• My first thought is that this is a hard question to answer without knowing the chemical composition of the product. You could use the Ideal Gas Law but would need to know the number of the number of moles of the gas. Commented May 15, 2017 at 15:56
• @Drew_J thanks for the comment, any way to estimate this? Commented May 16, 2017 at 8:12
• the cans contain liquid flourocarbons. Example: en.wikipedia.org/wiki/Hexafluoroethane you can see the gas volume is approx 300x the liquid volume. For what you are doing you probably dont need more accuracy than that. A 200ml can is like 2ft^3 .. You can experimentally determine how long it takes to empty a can. Commented May 16, 2017 at 14:34

According to to the SDS found here the product contains tetrafluoroethane as an ingredient. I looked around and found other similar products with the same ingredient.

For sake of simplicity lets say that all the $400 ml$ are tetrafluoroethane and the compressed density is $1.01 g/cm^3$ according to your table. With this information we know that one can contains $404 g$ of tetrafluoroethane by multiplying the density by the volume. ( $1 ml = 1 cm^3$ ) According to Wikipedia the molar mass of tetrafluoroethane is $102.03 g/mol$. With this information we know that bottle contains 3.96 moles of tetrafluoroethane. ( $404 g / 102.03 g/mol = 3.96 mol$ )

Now we can use the Ideal Gas Law to solve for volume at STP. In our case a handy rule of thumb is 1 mole = $22.4 L$ at STP. So you canister contains 88.7 liters of tetrafluoroethane ( $3.96 * 22.4$ ) if it were not compressed.

Now we reverse this process for air. This time around the handy rule is PV( initial ) = PV ( final ). 8 bar = 7.9 atmmospheres. $1 atm * 88.7 L = 7.9 atm * 11.2L$. So the answer to your question is that you will need 11.2 liters of compressed air at 8 bar to equal to your .4 liters of Dust Off! The reason that Dust Off is so efficient is that it liquefies when compressed where air does not.

If the process seems confusing read up on the Ideal Gas Law. It is very powerful when dealing with gasses at different conditions.

• Nicely done :) you got my vote. Commented May 16, 2017 at 17:03
• note my quick and dirty estimate in comment was only off 20%.. Commented May 16, 2017 at 20:40
• your answer did take me some time to understand, but - just great, thank you! Commented May 19, 2017 at 16:54