States of Matter - Result Question 78
78. An LPG (liquefied petroleum gas) cylinder weighs $14.8 kg$ when empty. When full it weighs $29.0 kg$ and shows a pressure of $2.5 \hspace {1mm} atm$. In the course of use at $27^{\circ} C$, 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 the $n$-butane with normal boiling point of $0^{\circ} C$.
(1994, 3M)
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Answer:
Correct Answer: 78. $2.46 \hspace {1mm} m^3$
Solution:
Weight of butane gas in filled cylinder $=29-14.8 kg$
$ =14.2 kg $
$\Rightarrow$ During the course of use, weight of cylinder reduces to $23.2 kg$
$\Rightarrow$ Weight of butane gas remaining now
$ =23.2-14.8=8.4 kg $
Also, during use, $V$ (cylinder) and $T$ remains same.
Therefore, $\quad \dfrac{p _1}{p _2}=\dfrac{n _1}{n _2}$
$ \Rightarrow \quad p _2=\left(\dfrac{n _2}{n _1}\right) p _1=\left(\dfrac{8.4}{14.2}\right) \times 2.5 \quad\left[\text { Here, } \dfrac{n _2}{n _1}=\dfrac{w _2}{w _1}\right] $
$ =1.48 \hspace {1mm} atm $
Also, pressure of gas outside the cylinder is $1.0 \hspace {1mm} atm$.
$\Rightarrow \quad p V =n R T $
$\Rightarrow \quad V =\dfrac{n R T}{p}=\dfrac{(14.2-8.4) \times 10^{3}}{58} \times \dfrac{0.082 \times 30}{1} L $
$ =2460 L=2.46 m^{3}$
BETA
