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Standard XII

Chemistry

Ideal Gas Equation

An LPG cylind...

Question

An LPG cylinder weighs 14.8 kg when empty. When full it weighs 29.0 kg and after consumption, the weight of the full cylinder reduces to 23.2 kg. The volume of the gas consumed in cubic metres at the normal usage conditions is:
[Assume LPG to be $n$ -butane with a normal boiling point of $0^{\circ} C$ .]

V = 2.642 m^{3}

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V = 2.956 m^{3}

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V = 2.115 m^{3}

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V = 2.447 m^{3}

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Solution

The correct option is D $V = 2.447 m^{3}$
$Weight of full cylinder = 29 k g$

$Weight after usage = 23.2 k g$

$Gas consumed = 29 - 23.2 = 5.8 k g o r 5800 g$

$For butane (C_{4} H_{10}), M = 58$

$Normal pressure = 1 a t m$

$Room temperature = 298 K$

$P V = n R T$

or $1 \times V = \frac{5800}{58} \times 0.0821 \times 298$

or $V = 2446.58 L \approx 2.447 m^{3}$

Hence, the correct option is $D$

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An LPG cylinder weighs 14.8 kg when empty. When full it weighs 29.0 kg and after consumption, the weight of the full cylinder reduces to 23.2 kg. The volume of the gas consumed in cubic metres at the normal usage conditions is:[Assume LPG to be n-butane with a normal boiling point of 0∘C.]

The correct option is D V=2.447m3Weight of full cylinder=29kgWeight after usage=23.2kgGas consumed=29−23.2=5.8kgor5800gFor butane (C4H10),M=58Normal pressure=1atmRoom temperature=298KPV=nRTor 1×V=580058×0.0821×298or V=2446.58L≈2.447m3Hence, the correct option is D