Chemistry - Ideal Gas Question with Solution | TestHub
A volume of 6 l H2O is placed in a closed evacuated room of volume 827 l at the temperature 300 K. The density of liquid water at 300 K is 1.0 g/ml. The vapour pressure of water at 300 K is 22.8 mm Hg. Neglect the change in volume of liquid water by vaporization.
Column I | Column II |
|---|---|
A) Mass of water vapour formed (in g) | P) 6 |
B) Moles of water vapour formed | Q) 18 |
C) Approximately mass of liquid water left (in kg) | R) 3 |
D) Total moles of atoms in vapour form | S) 1 |
Options:
Answer:
Solution:
The vapour fills the room at P = 22.8/760 = 0.03 atm, so n = PV/RT = (0.03 × 827)/(0.0821 × 300) ≈ 1.0 mol (B→S). Mass of vapour = 1 × 18 = 18 g (A→Q). Moles of atoms in vapour = 3 × 1 = 3 mol (D→R). Liquid water initially = 6000 g; after losing 18 g about 5982 g ≈ 6 kg is left (C→P). Hence A→Q; B→S; C→P; D→R.
