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

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be

the answer is b. dilution