Question

An aerosol spray can with a volume of 456mL contains propane gas as a propellant. If the can is at 23.00C, and 0.500 atm, what volume would the propane occupy at STP?

Answer 1

Step-by-step explanation:

To solve this problem, we can use the ideal gas law, which states:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

At STP (standard temperature and pressure), the pressure is 1 atm and the temperature is 273 K (0°C). We can use these values to find the number of moles of propane gas in the can:

P1V1 = nRT1

n = P1V1 / (RT1)

= (0.500 atm) * (0.456 L) / (0.0821 Latm/(mol*K) * 296 K)

= 0.0103 mol

Next, we can use the molar volume of a gas at STP (22.4 L/mol) to find the volume that the propane would occupy at STP:

V2 = n * Vm

= 0.0103 mol * 22.4 L/mol

= 0.231 L

Therefore, the volume that the propane would occupy at STP is approximately 0.231 L (or 231 mL).