States of Matter 1 Question 78
78. An LPG (liquefied petroleum gas) cylinder weighs $14.8 \mathrm{~kg}$ when empty. When full it weighs $29.0 \mathrm{~kg}$ and shows a pressure of $2.5 \mathrm{~atm}$. In the course of use at $27^{\circ} \mathrm{C}$, the weight of the full cylinder reduces to $23.2 \mathrm{~kg}$. 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 $0^{\circ} \mathrm{C}$.
(1994, 3M)
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Answer:
Correct Answer: 78. $\left(1020 \mathrm{~g} \mathrm{~mol}^{-1}\right)$ 83. $(5.23 \mathrm{~L}) \quad$ 85. (10)
Solution:
- Weight of butane gas in filled cylinder $=29-14.8 \mathrm{~kg}$
$$ =14.2 \mathrm{~kg} $$
$\Rightarrow$ During the course of use, weight of cylinder reduces to $23.2 \mathrm{~kg}$
$\Rightarrow$ Weight of butane gas remaining now
$$ =23.2-14.8=8.4 \mathrm{~kg} $$
Also, during use, $V$ (cylinder) and $T$ remains same.
Therefore, $\quad \frac{p_{1}}{p_{2}}=\frac{n_{1}}{n_{2}}$
$$ \Rightarrow \quad p_{2}=\left(\frac{n_{2}}{n_{1}}\right) p_{1}=\left(\frac{8.4}{14.2}\right) \times 2.5 \quad\left[\text { Here, } \frac{n_{2}}{n_{1}}=\frac{w_{2}}{w_{1}}\right] $$
$$ =1.48 \mathrm{~atm} $$
Also, pressure of gas outside the cylinder is $1.0 \mathrm{~atm}$.
$$ \begin{aligned} \Rightarrow \quad p V & =n R T \ \Rightarrow \quad V & =\frac{n R T}{p}=\frac{(14.2-8.4) \times 10^{3}}{58} \times \frac{0.082 \times 30}{1} \mathrm{~L} \ & =2460 \mathrm{~L}=2.46 \mathrm{~m}^{3} \end{aligned} $$
