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 of ${27}^{o}C$, the mass of full cylinder is reduced to 23.2 kg. Find out the volume of the gas in cubic meters used up at the normal usage conditions and the final pressure inside the cylinder. Assume LPG to be n-butane with normal boiling point of $0^{o}C$ (as nearest integer).

Answer 1

Mass of butane in a cylinder

$=29.0-14.8=14.2=14.2\times {10}^{3}g$

$P=2.5atm,T=300K$, molar mass of butane $=58gmo{l}^{-1}$

$\therefore PV=\frac{w}{M}RT$

$2.5\times V=\frac{14.2\times {10}^3}{58}\times 0.0821\times 300$

$V=2.4120\times {10}^{3}litre=2.4120m^3$

This is a volume of cylinder or volume of gas.

Now, the mass of gas left after use $=23.2-14.8=8.4kg$

$=8.4\times {10}^{3}g$

The volume remains constant.

Again using, $PV=\left(w/M\right)RT$

$P\times 2.412\times {10}^3=\frac{8.4\times {10}^3}{58}\times 0.0821\times 300$

Pressure (P) of gas left in cylinder $=1.48atm$

Now, Pressure of gas given out $=1$

Mass of gas given out $=(29.0-23.2)kg$

$=5.8kg=5.8\times {10}^{3}g$

Thus, the volume of gas given out under these conditions are

$1\times V=\frac{5.8\times {10}^3}{58}\times 0.0821\times 300$

$V=2463litre=2.463m^3$

So, the nearest integer value is 2.