Chemistry - Ideal Gas Question with Solution | TestHub
An LPG (liquefied petroleum gas) cylinder weighs 14.8 kg when empty. When full, it weighs 28.8 kg and shows a pressure of 35.5 atm. In the course of use at 300 K, the weight of the full cylinder is reduced to 23.2 kg. Assume LPG to be n-butane with normal boiling point of 273 K. If
ab = (1/100) × the volume (in litre) of the gas used up at the normal usage conditions, and
cd = final pressure (in atm) inside the cylinder
then the value of abcd is
Answer:
Solution:
Gas used = 28.8 - 23.2 = 5.6 kg = 5600/58 = 96.55 mol. At 300 K (above the 273 K boiling point, so all gaseous) and 1 atm: V = 96.55 × 0.0821 × 300 = 2379 L ⇒ ab = 23.79 = 24 (nearest integer).
Cylinder volume from the full state (14.0 kg = 241.4 mol at 35.5 atm, 300 K): Vc = 241.4 × 0.0821 × 300/35.5 = 167.5 L.
Gas left = 23.2 - 14.8 = 8.4 kg = 144.8 mol: P = 144.8 × 0.0821 × 300/167.5 = 21.3 atm ⇒ cd = 21.
abcd = 2421.
