Answer: Volume of 28.0 g of propane is 15.0 Liters
Explanation:
According to ideal gas equation:
[tex]PV=nRT[/tex]
P = pressure of gas = 1.2 atm
V = Volume of gas = ?
n = number of moles=[tex]\frac{\text {given mass}}{\text {Molar mass}}=\frac{28.0g}{44.1g/mol}=0.635mol[/tex]
R = gas constant =[tex]0.0821Latm/Kmol[/tex]
T =temperature = [tex]15^0C=(15+273)K=288K[/tex]
[tex]V=\frac{nRT}{P}[/tex]
[tex]V=\frac{0.635\times 0.0821L atm/K mol\times 288K}{1atm}=15.0L[/tex]
Thus volume of 28.0 g of propane is 15.0 Liters
