Chemistry - Ideal Gas Question with Solution | TestHub
An LPG cylinder weighs 12.0 kg when empty. When full, it weighs 23.0 kg and shows a pressure of 30 atm. In the course of use at 300 K, the weight of the cylinder is reduced to 18.6 kg. Assume LPG to be propane (M = 44) with normal boiling point of 231 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 = 23.0 - 18.6 = 4.4 kg = 4400/44 = 100 mol. At 300 K (above the 231 K boiling point) and 1 atm: V = 100 × 0.08 × 300 = 2400 L ⇒ ab = 24.
Cylinder volume from the full state (11.0 kg = 250 mol at 30 atm, 300 K): Vc = 250 × 0.08 × 300/30 = 200 L.
Gas left = 18.6 - 12.0 = 6.6 kg = 150 mol: P = 150 × 0.08 × 300/200 = 18 atm ⇒ cd = 18.
abcd = 2418.
