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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 27^oC, the weight of the full cylinder reduces to 23.3 kg. Find out the volume of the gas in cubic meters used up at the normal usage conditions, find the final pressure inside the cylinder. Assume LPG to be n-butane with normal boiling point of 0^oC.

Amit Saxena

12 Years agoGrade upto college level

2 Answers

Navjyot Kalra

12 Years ago

Calculation of volume of gas :

Weight of cylinder with gas = 29.0 kg

Weight of empty cylinder = 14.8 kg

∴ Weight of gas in the cylinder = 14.2 kg

Pressure in cylinder = 2.5 atm

∴ No. of moles (n) in 14.2 kg (14.2 * 10³ g) of butane

N = Wt. of butane/Mol. wt. of butane = 14.2 * 10³/58 = 244.83 mol

Applying gas equation,

V = nRT/P = 244.83 * 0.0821 * 300/2.5 = 2412 litres

[27^oC = 273 + 27 = 300]

Calculation of pressure in cylinder after use.

Weight of cylinder after use = 23.2 kg

Weight of empty cylinder = 14.8 kg

∴ Wt. of unused gas = 8.4 kg = 8.4 *10₃^/58 moles of butane

Thus P = nRT/V = 8.4 * 10³ * 0.0821 * 300/58 * 2412 = 1.478 atm

[V = 2412 L]

Calculation of volume of used gas at 2.5 atm and 27^oC.

Weight f used gas = 14.2 – 8.4 = 5.8 kg

Pressure under normal usage conditions = 1 atm

V = nRT/P = 5.8 *10³/58 * 0.0821 *300/1 [∵ n = 5.8/58]

= 246 litres = 2.463 m³

Omkar Dubey

8 Years ago

The lpg is in liquid form in the cylinder and thus we have the vapour pressure and not the pressure of the gas. And the vapour pressure only depends on the temperature and not on the amount of the liquid.And since the temperature of the gas remains same, the vapour pressure also remains same. Thus the pressure in the cylinder remaining 2.5 atm as earlier.

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