Ask Question
28 June, 16:33

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?

+2
Answers (1)
  1. 28 June, 16:52
    0
    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
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “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 ...” in 📙 Chemistry if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers