An LPG cylinder weighs 18.4 kg when empty when full, it weighs 29.0 kg and shows a presure of 2.5 atm. In the course of use `27^(@)C`, the mass of the full cylinder is reduced to 23.2 kg. find out the volume of the gas is cubic meters used up at the normal usage conditions, and the final pressure inside the cylinder. assume LPG to be n-butane with normal boiling pont of `0^(@)C`.

Explanation

Mass of the gas in full cylinder`=(29.0-18.4)kg=10.6kg`
Mass of the gas after some of it is used up`=(23.2-18.4)=4.8kg`
Mass of the gas used up`=(10.6-4.8)=5.8kg`.
So, 5.8 kg of gas=5.8 kg of n-butane`=(5.8xx10^(3))/(58)=100`
mol of n-butane [molar mass of n-butane=`58g*mol^(-1)`]
`V=(nRT)/(P)=(100xx0.0821xx300)/(1)=2463L`
[at normal usage condition, P=1 atm]
Therefore, the volume of used gas at normal usage conditoins=`2463L=2.463m^(3)" "because[1L=10^(-3)m^(3)]`
In the cylinderr, LPG exists in liquid state which remains in equilibrium with its vapour. so long LPG exist in liquid state in the cylinder, the pressure in the cylinder remains fixed. so, the pressure of the remaining gas in the cylinder will be 2.5 atm.