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° 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° C.

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° 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° C. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Weight of butane gas in filled cylinder

= 29 - 14.8 k g

= 14.2 k g

\Rightarrow

During the course of use, weight of cylinder reduces to

23.2 k g

\Rightarrow

Weight of butane gas remaining now

= 23.2 - 14.8 = 8.4 k g

Also, during use,

V

(cylinder) and

T

remains same.
Therefore,

\frac{p _{1}}{p _{2}} = \frac{n _{1}}{n _{2}}

\Rightarrow p_{2} = (\frac{n _{2}}{n _{1}}) p_{1} = (\frac{8.4}{14.2}) \times 2.5 [

Here,

\frac{n _{2}}{n _{1}} = \frac{w _{2}}{w _{1}}]

= 1.48 a t m

Also, pressure of gas outside the cylinder is

1.0 a t m

.

\Rightarrow p V = n RT

\Rightarrow V = \frac{n RT}{p} = \frac{( 14.2 - 8.4 ) \times 1 0 ^{3}}{58} \times \frac{0.082 \times 30}{1} L

= 2460 L = 2.46 m^{3}