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 N_{2} vaporized = Pressure
mixture final

2) the volumes are additive.. ie volume air + volume N_{2}
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 N_{2} 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 N_{2} 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
N_{2} displaces 1 L of air, then 889.23 L of air was
displaced. 889.23L / 2090L = 42.5% was displaced.

Of course N_{2} will diffuse about 7% faster than
O_{2} 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% N_{2}. So you really can't distinguish between
displacement of air molecules vs displacement of vaporized
N_{2} molecules.

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