Question

What is the volume of 28.0g of propane at 15C degrees, and 1.2 atm?

Answer 1

Answer: Volume of 28.0 g of propane is 15.0 Liters

Step-by-step explanation:

According to ideal gas equation:

`PV=nRT`

P = pressure of gas = 1.2 atm

V = Volume of gas = ?

n = number of moles=
`\frac{\text {given mass}}{\text {Molar mass}}=(28.0g)/(44.1g/mol)=0.635mol`

R = gas constant =
`0.0821Latm/Kmol`

T =temperature =
`15^0C=(15+273)K=288K`

`V=(nRT)/(P)`

`V=(0.635* 0.0821L atm/K mol* 288K)/(1atm)=15.0L`

Thus volume of 28.0 g of propane is 15.0 Liters