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?
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)-1x 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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