Question

An L.P.G cyclinder weighs $14.8\text{kg}$ when empty. When full,it weighs $29.0\text{kg}$ and shows a pressure of $3.36\text{atm}$. In the course of use at ${27}^o\text{C}$, the mass of the full cyclinder is reduced to $23.2\text{kg}$. The final pressure inside the cyclinder is found to be $x\text{atm}$. The volume of the gas in cubic meters used up at the normal usage conditions of ${27}^{\circ }\text{C}$ and $1\text{atm}$ is $y{\text{m}}^3$. Assume L.P.G to be n-Butane with normal boiling point of $0^o\text{C}$. Find $x+y$. (Round the answer upto two decimal places.)
(Given : $R=0.082{\text{L atm K}}^{-1}{\text{mol}}^{-1}$ )

Answer 1

Weight of L.P.G. present originally $=29-14.8=14.2\text{kg}$
Weight of L.P.G after use $=23.2-14.8=8.4\text{kg}$
Pressure $=2\text{atm}$.
Applying the solution, PV = nRT. As volume of the cylinder is constant, we write
$\frac{P_1}{P_2}=\frac{n_1}{n_2}=\frac{w_1/m}{w_2/m}\Rightarrow \frac{P_1}{P_2}=\frac{w_1}{w_2}$
Substituting the values, we get,
$\frac{2.84}{P_2}=\frac{14.2}{8.4}\Rightarrow P_2=\frac{2.84\times 8.4}{14.2}=1.68\text{atm}$
$\therefore x=1.68$
Weight of the used gas $=14.2-8.4=5.8$
No. of moles of gas = $\frac{5.8\times {10}^3}{58}=100$ moles.
Under normal condition, $P=1\text{atm}$
$T=27+273=300\text{K}$
Volume of $100$ moles of L.P.G at $1\text{atm}$ and $300\text{K}$ is
$V=\frac{nRT}{P}=\frac{100\times 0.082\times 300}{1}=2460\text{L}=2.46{\text{m}}^3$
$\therefore y=2.46$
$\therefore x+y=1.68+2.46=4.14$