An aerosol can filled with propane gas occupies a volume of 250 ml at 23 c and 1.54 atm. if the conditions are changed to stp, what is the new volume in the can?

The new volume is 388 mL

Explanation:

Step 1: Data given

Volume = 250 mL = 0.250 L

Temperature = 23 °C = 296 Kelvin

Pressure = 1.54 atm

Step 2: Calculate number of moles

p*V =n*R*T

n = (p*V)/(R*T)

⇒ n = the number of moles = ?

⇒ V = the volume: the gas occupies = 0.250 L

⇒ The pressure of propane gas = 1.54 atm

⇒ Temperature = 23.0 °C = 296 Kelvin

 ⇒ R = The gas constant = 0.08206 L*atm/K*mol

n = (1.54 * 0.205 ) / ( 0.08206 * 296 )

n = 0.013 moles

Step 3: Calculate new volume

(P1*V1)/T1 = (P2*V2)/T2

⇒ with P1 = the initial pressure =  1.54 atm

⇒ with V1 = the initial volume = 0.250 L

⇒ with T1 = the initial temperature = 296 K

⇒ with P2 = 1.00 atm

⇒ with V2 = The new volume at STP

⇒ with T2 = 298 Kelvin

(1.54*0.250)/296 = (1.00 * V2)/298

V = 0.388 L = 388 mL

The new volume is 388 mL

4d^55s^1 ( third choice)