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An LPG cylinder weighs 14.8 kg when empt...

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`.

Text Solution

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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)`
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