I would like to know how many liters of gas there would be in a gram of liquid petroleum. I know there is a method to use involving the weight of a gram and the boiling point, and the energy the molecules need to stay at a vaporous state, but if it is only a liquid because it is dense, and boils at 28 degrees c, they would stay at that phase, at any temperature over 28, right?
July 9, 2009
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Trevor has given you the formula for density and volume assuming your boiling point (b.p.) value for octane of 28 C, but the actual b.p. is ~126 C. If you want to know the equilibrium vapor pressure (evp) at any temperature you can go to the ref., which uses the Antoine parameters of various compounds to calculate evp. Once you have the evp at the temperature you are interested in, you can find the density from:
n = no. of moles, mw = molecular weight ~= 114
PV = nRT (gas law) ==> n/V = P/(RT)
density = 0.001*mw*n/V = 0.001*mw*P/(RT) = 0.001*mw*evp/(RT).
From the ref., the evp of octane at 20 C is only 10.44 mmHg or 1392 Pa. From this I get density at 20 C = 0.065 kg/m^3 or g/L, so 1 g takes up 15.4 L.
Of course gasoline is compounded of many substances probably including some with lower b.p., so the evp of gasoline at 20 C will probably be higher and possibly the vapor density will be higher as well, but the vapor will not consist of the same mixture ratios as the liquid.
Comment by kirchwey — July 9, 2009 @ 5:04 pm
I assume that you are talking about what I call petrol and the Americans call gasoline. Let us call this octane, C8H18
Molar mass C8H10 = 8*12.011+18*1.008 = 114.232g/mol
1gram = 8.75*10^-3 mol
Now use the Universal gas equation:
PV = nRT
Take
P = 1 atmosphere gas pressure = 101.3kPa
V = volume in litres = what you want
n = no of moles = 8.75*10^-3
R = 8.314
T = 28°C = 301K
101.3*V = 8.75*10^-3 * 8.314*301
V = 21.91/101.3
V = 0.216 litres
Comment by Trevor H — July 9, 2009 @ 5:04 pm