An aerosol spray can with a volume of 250 mL contains 2.30 g of propane gas (C3H8) as a propellant.?
A. If the can is at 25 ∘C, what is the pressure in the can?
B. What volume would the propane occupy at STP?
C. The can says that exposure to temperatures above 130 ∘F may cause the can to burst. What is the pressure in the can at this temperature?
moles propane = 2.30 g x 1 mole/44 g = 0.052 moles
From PV = nRT, P = nRT/V = (0.052)(0.0821)(298)/0.25 = 5.1 atm
At STP 1 mole of a gas = 22.4 L, thus 0.052 moles x 22.4 L/mole = 1.16 liters
AT 130ºF = 403K
P = nRT/V = (0.052)(0.0821)(403)/0.25 = 6.9 atm
An aerosol spray can with a volume of 250 ml contains 2.30 g of propane gas (c3h8) as a propellant.