Question

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 21$^{\circ}$ C, the weight of the Hill 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 $0^{\circ} C.$

Answer 1

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, $ \frac{p_{1}}{p_{2}}=\frac{n_{1}}{n_{2}}$
$\Rightarrow p_{2}=\left(\frac{n_{2}}{n_{1}}\right) p_{1}=\left(\frac{8.4}{14.2}\right) \times 2.5 \left[\right.$ Here, $\left.\frac{n_{2}}{n_{1}}=\frac{w_{2}}{w_{1}}\right]$
$=1.48 \,atm$
Also, pressure of gas outside the cylinder is $1.0\, atm$.
$\Rightarrow p V =n R T$
$\Rightarrow V =\frac{n R T}{p}=\frac{(14.2-8.4) \times 10^{3}}{58} \times \frac{0.082 \times 30}{1} L$
$=2460 \,L =2.46\, m ^{3}$