Question

A 1.4-L container of liquid nitrogen is kept in a closet measuring 1.1 m by 1.0 m by 1.9 m .

Assuming that the container is completely full, that the temperature is 27.6 ∘C, and that the atmospheric pressure is 1.0 atm , calculate the percent (by volume) of air that would be displaced if all of the liquid nitrogen evaporated. (Liquid nitrogen has a density of 0.807 g/mL.)

Answer 1

Ans- 1) the closet isn't sealed so that Pressure air initial = Pressure N₂ vaporized = Pressure mixture final
2) the volumes are additive.. ie volume air + volume N₂ vaporized = volume final
3) the composition of the final gas is uniform

volume air initially = (1.1m x 1.0m x 1.9m) x (1000L / m³) = 2090 L
volume N₂ vaporized = nRT/P = (1.4L x 1000mL/L x 0.807g/mL x 1mole / 28.0g) x (0.082106 Latm/moleK) x (295.25K) / (1.1 atm) = 889.23 L

total volume = volume air + volume N₂ vaporized = 2090L + 889.23 L = 2979.23 L

and that means only 2090L / 2979.23L = 70.15% of the air remains in the room and 29.85% of the air was displaced.
Assuming the mixture isn't uniform and that as a result, 1 L of N₂ displaces 1 L of air, then 889.23 L of air was displaced. 889.23L / 2090L = 42.5% was displaced.
Of course N₂ will diffuse about 7% faster than O₂ due to it's molecular weight,
and the position of the container relative to any openings in the closet will play a factor in which gas escapes and of course air is indeed 79% N₂. So you really can't distinguish between displacement of air molecules vs displacement of vaporized N₂ molecules.