Chemistry - Ideal Gas Question with Solution | TestHub
A volume of 4 l H₂O is placed in a closed evacuated room of volume 410.5 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 45.6 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) 4 |
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 = 45.6/760 = 0.06 atm, so n = PV/RT = (0.06 × 410.5)/(0.0821 × 300) ≈ 1.0 mol (BS). Mass of vapour = 1 × 18 = 18 g (AQ). Moles of atoms in vapour = 3 × 1 = 3 mol (DR). Liquid water initially = 4000 g; after losing 18 g about 3982 g ≈ 4 kg is left (CP). Hence AQ; BS; CP; DR.
