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An empty LPG cylinder weighs $14.8 \mathrm{~kg}$. When full, it weighs $29.0 \mathrm{~kg}$ and shows a pressure of $3.47$ $\mathrm{atm}$. In the course of use at ambient temperature, the mass of the cylinder is reduced to $23.0 \mathrm{~kg}$. The final pressure inside of the cylinder is atm. (Nearest integer)
(Assume LPG of be an ideal gas)
Initial mass of gas $=29-14.8=14.2 \mathrm{Kg}$
mass of gas used $=29-23=6 \mathrm{Kg}$
gas left $=14.2-6=8.2 \mathrm{Kg}$
(1) $3.47 \times \mathrm{V}=\left(\frac{14.2 \times 10^{3}}{\mathrm{M}}\right) \times \mathrm{R} \times \mathrm{T}$
(2) $\mathrm{p} \times \mathrm{V}=\left(\frac{8.2 \times 10^{3}}{\mathrm{M}}\right) \times \mathrm{R} \times \mathrm{T}$
Divide :
$\frac{(1)}{(2)} \Rightarrow \frac{3.47}{\mathrm{P}}=\frac{14.2}{8.2}$
$P=2.003$
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