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 $2.5 atm$. 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. $\left(1020 g mol^{-1}\right)$ 83. $(5.23 L) \quad$ 85. (10)
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, $\quad \frac{p _1}{p _2}=\frac{n _1}{n _2}$
$$ \Rightarrow \quad p _2=\left(\frac{n _2}{n _1}\right) p _1=\left(\frac{8.4}{14.2}\right) \times 2.5 \quad\left[\text { Here, } \frac{n _2}{n _1}=\frac{w _2}{w _1}\right] $$
$$ =1.48 atm $$
Also, pressure of gas outside the cylinder is $1.0 atm$.
$$ \begin{aligned} \Rightarrow \quad p V & =n R T \ \Rightarrow \quad V & =\frac{n R T}{p}=\frac{(14.2-8.4) \times 10^{3}}{58} \times \frac{0.082 \times 30}{1} L \ & =2460 L=2.46 m^{3} \end{aligned} $$
