How many air molecules are in a 10.0×12.0×10.0 ft
room ( 28.2 L=1 ft3
)? Assume atmospheric pressure of 1.00 atm
, a room temperature of 20.0 ∘C
, and ideal behavior.
First, we need to convert the volume of the room from ft^3 to L:
10.0 ft x 12.0 ft x 10.0 ft = 1200 ft^3
1200 ft^3 x 28.2 L/1 ft^3 = 33,840 L
Next, we need to convert the room temperature to Kelvin:
20.0 °C + 273.15 = 293.15 K
Now we can use the Ideal Gas Law to calculate the number of air molecules in the room:
PV = nRT
(1.00 atm) x (33,840 L) = n x (0.0821 L atm / K mol) x (293.15 K)
n = (1.00 atm x 33,840 L) / (0.0821 L atm / K mol x 293.15 K)
n = 1415.5 mol
Finally, we can convert moles to molecules:
1 mol = 6.022 x 10^23 molecules
1415.5 mol x 6.022 x 10^23 molecules/mol = 8.52 x 10^26 molecules
Therefore, there are approximately 8.52 x 10^26 air molecules in the room.