contestada

A 1.2-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.)

Respuesta :

Answer:

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

The percent (by volume) of air that would be displaced is 26.99 %.

Given information:

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

The calculation of the percent of air:

Mass of nitrogen = density  × volume

[tex]= 0.807 g/mL \times 1200 mL[/tex]

= 968.4 grams

Now

Moles N2 = mass N2 ÷  molar mass N2

[tex]= 968.4 grams \div 28 g/mol[/tex]

= 34.586 moles

Now volume

[tex]p\times V = n \times R \times T[/tex]

Here p = the the pressure = 1.2 atm

n = the number of moles = 34.586 moles

R = the gas constant = [tex]0.08206 L\times atm/K\times mol[/tex]

T = the temperature = 23.5 °C = 296.65

[tex]V = (n\times R\times T)\div p\\\\= (34.586 \times 0.08206 \times 296.65)\div 1.2[/tex]

= 701.61 L

Now the total volume of the chamber is

[tex]= 1.0 m \times 1.3 m \times 2 m \\\\= 2.6 m^3[/tex]

= 2600 L

Now finally the percent volume displaced

[tex]= (701.61 L \div 2600 L) \times 100\%[/tex]

= 26.99%

Find out more information about the  Temperature here:https://brainly.com/question/7510619?referrer=searchResults

Otras preguntas