Question

1.How many air molecules are in a 14.5×12.0×10.0 ft room? Assume
atmospheric pressure of 1.00 atm, a room temperature of 20.0 ∘C,
and ideal behavior. **Volume conversion:**There are
28.2 liters in one cubic foot.

2. At an underwater depth of 255 ft , the pressure is 8.33 atm

What should the mole percent of oxygen be in the diving gas for the partial pressure of oxygen in the mixture to be 0.21 atm, the same as in air at 1 atm?

Answer #1

Q1.)

Solution:

Given

Volume = 14.5 ft x 12.0 ft x 10.0 ft

Pressure = 1.0 atm

T = 20.0 deg C + 273.15 = 293.15 K

Volume in L

= 14.5 ft x 12.0 ft x 10.0 ft

Volume in L

= ( 14.5 ft x 12.0 ft x 10.0 ft ) x 28.2 L / 1 cubic ft

=49068.0 L

Calculation of moles of air.

We use ideal gas law

pV = nRT

n = pV/ RT

=1.0 atm x 49068.0 L / ( 0.08206 L atm (Kmol)^{-1}x
293.15 K )

=2039.75 mol air molecule

Number of molecules of air

= 2039.75 mol air molecule x 6.02 E 23 Air molecules

=1.23 E27 air molecules

**Number of air molecules = 1.23 E 27**

Q.2)

Total pressure = 8.33 atm

Partial pressure = 0.21 atm

Partial pressure = mol fraction x total pressure

Mole fraction = partial pressure / total pressure

=0.21 atm / 9.33 atm

=0.0225

Mole percent = 0.0225 x 100 %

=2.25 %

**Mole percent of Oxygen = 2.25**

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