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Step 1: In the Earth’s atmosphere, the concentration of rare gas argon (Ar) is 0

ID: 894713 • Letter: S

Question

Step 1:

In the Earth’s atmosphere, the concentration of rare gas argon (Ar) is 0.934 % volume.

The volume of our classroom has been estimated to be about 600 m3.

Assuming argon behaves as a perfect gas, and assuming the pressure to be 1.00 atm, and the temperature to be 22 °C, calculate the mass of argon in our classroom. Atomic mass of argon is 39.95 amu ( 1 point)

What is the partial pressure of Ar in our classroom, expressed in Pa? (1 point)

Considering argon to behave as a real gas, calculate its partial pressure in our classroom, under the conditions stated in (a), using van der Waals’ equation.

(van der Waals coefficients for argon are : a = 1.337 atm dm6 mol-2 , and b= 3.20 x 10-2 dm3 mol-1 ) (1 point)

Among the assumptions made in the kinetic theory of gases, is the one stating that atoms and molecules are very small and thus occupy only a negligible portion of the volume, under normal conditions of temperature and pressure. This can be illustrated by calculating the volume physically occupied by the air molecules in our classroom. In order to do this, we can take all individual air molecules (both nitrogen and oxygen) as being the equivalent of spheres with a diameter of 0.71 nm.

Calculate the volume physically occupied by air molecules in our classroom, under the same conditions as stated in question 1 above.

(Reminder : Avogadro’s number is : 6.022 x 1023 molecules/mol)

Explanation / Answer

PV = nRT

n = PV / RT

V = 600 m3 x 1000 L / 1m3 = 600000 L

V = 600000 x 0.934/100 = 5604 L

n = 1 x 5604 L / 0.082 x (22+273)

n = 231.66 moles

m = 231.66 moles x 39.95 g/mol = 9255.05 g

The wan der waals equation:

(p + n2a/V2) (V - nb) = nRT

p + (231.662 x 1.337 / (56042) ) (5604 - 231.66x0.032) = 231.66 x 0.082 x 295

(p + 2.28x10-3) (5596.59) = 5603.85

5596.59p + 12.76 = 5603.85

p = 5603.85 - 12.76 / 5596.59

p = 0.999 atm

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