An aerosol spray can with a volume of 456mL contains 3.18g of propane gas as a propellant. If the can is at 23C, and 50atm, what volume would the propane occupy at STP?

21.03L

Explanation:

V1 = 456mL = 0.456L

T1 = 23°C = (23 + 273.15)K = 296.15K

P1 = 50atm

V2 = ?

T2 = 273.15K

P2 = 1.0atm

Note : P2 and T2 are at STP which are 1.0atm and 273.15K

To find V2, we have to use the combined gas equation,

(P1 × V1) / T1 = (P2 × V2) / T2

P1 × V1 × T2 = P2 × V2 × T1

V2 = (P1 × V1 × T2) / (P2 × T1)

V2 = (50 × 0.456 × 273.15) / (1.0 × 296.15)

V2 = 6227.82 / 296.15

V2 = 21.029L

Final volume of the gas is 21.03L

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