-22C is well below the boiling point. Only a small amount will be present as vapor. You can estimate that amount using the Clausius-Clapeyron equation and the 1 atm vapor pressure at -0.4 C.
You can also use the C-C equation for the vapor pressure at the higher temperature, 25 C. It might all be vapor but start by finding out how much of the mass can be in the vapor state in that volume at that temperature.
You can also look up the vapor pressures at both temperatures, but they probably want you to calculate them from the information provided.
A graph of log of vapor pressure vs 1/T is usually a very straight line.
Butane (C4H10) has a heat of vaporization of 22.44 kJ/mol and a normal boiling point of -0.4 C. A 250 mL sealed flask contains 0.5 g of butane at -22 C.
How much butane is present as a liquid?
If the butane is warmed to 25 C, how much is present as a liquid?
4 answers
Use the Clausius-Clapeyron equation to calculate pressure at -22. Then PV = nRT should get n for the gas and you can go from there.
0.204 gm
a) is around 2.0
b) is zero, all the substance has vaporized at 25 degress there for none of it will be left as a liquid
b) is zero, all the substance has vaporized at 25 degress there for none of it will be left as a liquid
1 answer