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

The % of displaced volume of nitrogen is 29.06%.

Step-by-step explanation:

Volume of nitrogen = 1.2 L = 1200 mL

Density of nitrogen = 0.807 g/ml

Mass of nitrogen =
`Density * Volume= 0.807 * 1200 = 1044 g`

Molar mass of nitrogen = 28 g/mol

`Number\,of\,moles\,nitrogen=(Mass)/(Molar\,mass)`

`= (1044)/(28)=37.28`

The ideal gas equation is as follows

`PV = nRT`

Rearrange the equation is as follows.

`V= (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= (37.28 * 0.0820 * 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 * 1.3 * 2 = 2.6\,m^(3)`

`%\,displaced\,volume=(0.751)/(2.6)=28.8%`

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