An LPG (liquefied petroleum gas) cylinder weighs 14.8 kg when empty. When full it weighs 29.0 kg and shows a pressure of 2.5 atm. In the course of use at `27^(@)C`, 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

Step I. Mass of LPG left in the cylinder
Mass of LPG in the cylinder before use `= 29.0 - 14.8 = 14.2 kg`
Mass of LPG used up `= 29.0 - 23.2 = 5.8 kg`
Mass of LPG left in the cylinder `= 14.2 - 5.8 = 8.4 kg`
Step II. Final pressure inside the cylinder
According to ideal gas equation `PV = nRT`
`P_(1)V = n_(1)RT`
`P_(2)V = n_(2)RT` , (' `:'` Volume is constant )
`:. (P_(1))/(P_(2)) = (n_(1))/(n_(2))`
`n_(1) = ("Mass of butane initially present")/("Molar mass of butane") = ((14.2 xx 1000 g))/((58 g mol^(-1)))`
`n_(2) = ("Mass of butane left")/("Molar mass of butane") = ((8.4 xx 1000g))/((58 g mol^(-1)))`
Now, `(P_(1))/(P_(2)) = (n_(1))/(n_(2))` or `P_(2) = P_(1) xx (n_(2))/(n_(1))`
`P_(2) = ((2.5 atm) xx (8.4 xx 1000g) xx (58 g mol^(-1)))/((58 g mol^(-1)) xx (14.2 xx 1000g)) = 1.48 atm`
Step III. Volume of the gas used
Mass of the (butane) used up `= 5.8 kg = 5800 g`
No. of moles of butane `= ((5800 g))/((58 g mol^(-1)) = 100 mol`.
Normal pressure `= 1 atm`.
Normal temperature `= 27 + 273 = 300 K`
According to ideal gas equation, `PV = nRT`
`:. V = (nRT)/(P) = ((100 "mol") xx(0.082 L "atm" K^(-1) mol^(-1)) xx (300 K))/((1 atm))`
`= 2463 L = 2.463 m^(3)`