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

We know,
Weight of butane in cylinder `= 29.0 -14.8 = 14.2 kg`
` =14.2 xx 10^(3) g`
`P =2.5 atm, T =300K, M` wt of butane `=58`
`:' PV = (w)/(m) RT`
`2.5 xx V = (14.2 xx 10^(3))/(58) xx 0.0821 xx 300`
`:. V = 2.4 xx 10^(3) litre = 2.4 m^(3)`
This is volume of cylinder or volume of gas Now the wt.of gas left after use ` =23.2 -14.8 = 8.4 kg`
` =8.4 xx 10^(3) g`
The volume remains constnat
Again using `PV = (w)/(m) RT`
`P xx 2.412 xx 10^(3) = (8.4 xx 10^(3))/(58) xx 0.0821 xx 300`
`:.` Pressure `(p)` of gas left in cylinder ` =1.50 atm` Now pressure of gas given out `=1`
Wt of gas given out ` =29.0 - 23.2`
` = 5.8 kg = 5.8 xx 10^(3) g`
Thus volume of gas given out under these condition is
` 1 xx V = (5.8 xx 10^(3))/(58) xx 0.0821 xx 300`
`V =2.4 xx 10^(3) "litres" = 2.4 m^(3)` .