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Ideal Gas Equation

An LPG cylind...

Question

An LPG cylinder weighs $14.8 kg$ when empty. When it is full, it weighs $29 kg$ and after consumption, the weight of the cylinder reduces to $23.2 kg$ . Find out the volume of n-butane in cubic metres used up at $27^{\circ} C$ and $1 a t m$ .
[Molecular mass of n-butane = 58 g/mol]

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Solution

Mass of n-butane gas used $29 - 23.2 = 5.8 kg = 5800$ g
Number of moles of butane used $= \frac{5800}{58} = 100$ mol
Using the ideal gas equation, PV = nRT
T = $27^{o} C$ = 300 K, R = 0.0821 atm.L/ (mol.K)
So, $V = \frac{n R T}{P} = \frac{100 \times 0.0821 \times 300}{1}$
On solving, V $= 2463$ L
Since, 1 L = $10^{- 3} m^{3}$
Volume of the n-butane gas used is V $= 2.463 m^{3}$

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