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Standard XII

Chemistry

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]

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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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Ideal Gas Equation

Standard XII Chemistry

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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∘C and 1 atm. [Molecular mass of n-butane = 58]

Mass of n-butane gas used 29−23.2=5.8 kg=5800 g Number of moles of butane used =580058=100 mol Using the ideal gas equation, PV = nRT T = 27oC = 300 K, R = 0.0821 atm.L/ (mol.K) So, V=nRTP=100×0.0821×3001 On solving, V=2463 L Since, 1 L = 10−3m3 Volume of the n-butane gas used is V =2.463 m3

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]