Question

Assuming that the container is completely full, that the temperature is 22.0 ∘C, and that the atmospheric pressure is 1.2 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

Answer:

The % of displaced volume of nitrogen is 29.06%.

Explanation:

Volume of nitrogen = 1.2 L = 1200 mL

Density of nitrogen = 0.807 g/ml

Mass of nitrogen = $Density \times Volume= 0.807 \times 1200 = 1044 g$

Molar mass of nitrogen = 28 g/mol

$Number\,of\,moles\,nitrogen=\frac{Mass}{Molar\,mass}$

$= \frac{1044}{28}=37.28$

The ideal gas equation is as follows

$PV = nRT$

Rearrange the equation is as follows.

$V= \frac{nRT}{P}...............(1)$

n= Number of moles = 37.28

R = Gas constant = 0.0820

T = Temperature = 22+ 273 = 295

P = Pressure = 1.2 atm

Substitute the all values in equation (1)

$V= \frac{37.28 \times 0.0820 \times 295 }{1.2}= 751.5L= 0.751 \,m^{3}$

$0.751 \,m^{3}$ of nitrogen will displace same amount of air.

$Volume \,\,of\,\, closet= 1 \times 1.3 \times 2 = 2.6\,m^{3}$

$%\,displaced\,volume=\frac{0.751}{2.6}=28.8%$

Therefore, The % of displaced volume of nitrogen is 29.06%.