Question

Assuming that the container is completely full, that the temperature is 22.1 ∘C, and that the atmospheric pressure is 1.1 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.) Express your answer using two significant figures.

Answer 1

Answer : The percent (by volume) of air displaced if all of the liquid nitrogen evaporated would be 29 %

Explanation :

Let us assume that 1.2 L container of liquid nitrogen is kept in a closet measuring 1.0m by 1.3m by 2.0m.

First we have to calculate the mass of nitrogen.

$Density=\frac{Mass}{Volume}$

Given:

Density of nitrogen = 0.807 g/mL

Volume of nitrogen = 1.2 L = 1200 mL    (1 L = 1000 mL)

$0.807g/mL=\frac{Mass}{1200mL}$

$Mass=0.807g/mL\times 1200mL=968.4g$

Now we have to calculate the moles of nitrogen.

$\text{Moles of }N_2=\frac{\text{Mass of }N_2}{\text{Molar mass of }N_2}$

Molar mass of nitrogen $(N_2)$ = 28 g/mole

$\text{Moles of }N_2=\frac{968.4g}{28g/mole}=34.58mole$

Now we have to calculate the volume of nitrogen.

As we know that,

PV = nRT

where,

P = pressure of nitrogen = 1.1 atm

V = volume of nitrogen = ?

n = number of moles of nitrogen = 34.58 mole

R = gas constant = 0.0821 L.atm/mol.K

T = temperature of nitrogen = $22.1^oC=273+22.1=295.1K$

Now put all the given values in this formula, we get:

$1.1atm\times V=34.58mole\times 0.0821L.atm/mol.K\times 295.1K$

$V=761.6L$

Now we have to calculate the volume of container.

The formula of volume of cuboid, we use the equation:

$V=lbh$

where,

V = volume of container = ?

l = length of container = 1.0 m

b = breadth of container = 1.3 m

h = height of container = 2.0 m

Putting values in above equation, we get:

$V=1.0m\times 1.3m\times 2.0m=2.6m^3=2600L$

Conversion used : $1m^3=1000L$

Now we have to calculate the percent (by volume) of air that would be displaced if all of the liquid nitrogen evaporated.

$\% \text{ of displaced volume}=\frac{761.6L}{2600L}\times 100=29.29\% \approx 29\%$

Therefore, the percent (by volume) of air displaced if all of the liquid nitrogen evaporated would be 29 %

Answer 2

The percent (by volume) of air that would be displaced if all the liquid nitrogen evaporated is 29%.

What is volume?

The volume of any substance is the space it occupies.

Given,

The temperature of the container is 22.1∘C

The density of the nitrogen is 0.807 g/mL

The atmospheric pressure is 1.1 atm.

Step 1: Calculating the mass

Let's assume the volume of nitrogen is 1.2 l

converting into ml 1200 ml

$\bold{mass = volume \times density}$

Mass = 1200 x 0.807 = 968.4 g

Step 2: Calculating the moles of nitrogen

$\bold{Number\;of \;moles= \dfrac{mass}{molar\;mass}}$

$\bold{Number\;of \;moles= \dfrac{968.4\;g}{28\;g\;mole}= 34.58\;moles}$

Step 3: Calculating the volume of nitrogen

By the formula of

PV = nRT

Put the values in the formula

$\bold{1.1 atm. \times V = 34.58\;mole\times0.0821\;L\;atm.\mol.K\times 295.1\;K = 761.6\;L}$

Step 4: Calculating the volume of the container

V = lbh

Putting the values

$\bold{V = 1.0\;m \times 1.3\;m \times 2.0\;m= 2.6m^3 = 2600\;L}$

Step 5: Calculating the percent by volume

$\bold{\dfrac{761.6}{2,600} \times 100 = 29.2\%}$

Thus, the percent (by volume) of air would be displaced if all the liquid nitrogen evaporated is 29%.

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