A1.2-l container of liquid nitrogen is kept in a closet measuring 1.0m by 1.3m by 2.0m . assume that the container is completely filled to the top with liquid nitrogen, that the temperature is 23.5? c, and that the atmospheric pressure is 1.2atm .calculate the percent (by volume) of air that would be displaced if all of the liquid nitrogen evaporated into the closet. the closet is ventilated such that the temperature and pressure remain constant through this process. (liquid nitrogen has a density of 0.807 g/ml.)

26.99 % of air that will be displaced

Explanation:

Step 1: Data given

A 1.2-L container of liquid nitrogen is kept in a closet measuring 1.0m by 1.3m by 2.0m

Temperature = 23.5 °C

Atmospheric pressure = 1.2 atm

Liquid nitrogen has a density of 0.807 g/mL

Molar mass of N2 = 28 g/mol

Step 2: Calculate mass of nitrogen

Mass of nitrogen = density * volume

Mass of nitrogen = 0.807 g/mL * 1200 mL

Mass of nitrogen = 968.4 grams

Step 3: Calculate moles of N2

Moles N2 = mass N2 / molar mass N2

Moles N2 = 968.4 grams /28 g/mol

Moles N2 = 34.586 moles

Step 4: Calculate volume

p*V = n*R*T

⇒ p = the the pressure = 1.2 atm

⇒ V = the volume of N2 = TO BE DETERMINED

⇒ n = the number of moles = 34.586 moles

⇒ R = the gas constant = 0.08206 L*atm/K*mol

⇒ T = the temperature = 23.5 °C = 296.65

V = (n*R*T)/p

V = (34.586 * 0.08206 * 296.65)/1.2

V = 701.61 L

Step 5: Calculate the total volume of the chamber

1.0 m * 1.3 m * 2 m = 2.6 m³ = 2600 L

Step 6: Calculate the percent volume displaced

(701.61 L  / 2600 L) * 100% = 26.99%

26.99 % of air that will be displaced

if the concentration of products is decreased, the equilibrium will shift toward the products). add an inert gas (one that is not involved in the reaction) to the constant-volume reaction mixture: this will increase the total pressure of the system, but will have no effect on the equilibrium condition.

explanation: