An `LPG` cylinder weighs `14.8 kg` when empty. When full it weighs `29.0 kg` and the weight of the full cylinder reduces to `23.2 kg`. Find out the volume of the gas in cubic metres used up at the normal usage conditions and the final pressure inside the cylinder. Assume `LPG` to be `n`-butane with normal boiling point of `0^(@)C`.

Explanation

Decrease in the amount of PG`=29.0-23.2=5.8" kg"=(5800)/(58)" kg moles"=100" moles"`
Volume of 100 moles at 1 atm `300" K "(nRT)/(P)=(100" moles"xx0.0821" L atmK^(-1)xx300 K)/(1 atm)`
`=2463" L "=2463xx10^(-3)m^(3)=2.463m^(3)`
Final pressure inside the cylinder. As the cylinder contains liquefied petroleum gas in equibrium with its vapours, therefore, so long as temperature remains constant and some LPG is present, pressure with remain constant. As the cylinder still contains LPG=23.2-14.8=8.4 kg, pressure inside the cylinder will be same, i.e., 2.5 atm.