How many pounds of air is in a 10 by 10 bedroom?

Answer 1

If the room is at sea level and the temperature is 68° F, then
the weight of the air in the room is

(7.623) x (height of the ceiling, in feet) pounds.

Below, a more patient contributor patiently explains how to calculate it.

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I assume you mean 10 feet by 10 feet. You also need to know the height of the
ceiling. Tall rooms contain more air than short rooms. A typical ceiling height is 8
feet, so we'll use that.

The WEIGHT (measured in pounds) of air is proportional to its VOLUME. Volume of a
rectangular room is length x width x height = 10*10*8 = 800 cubic feet.

Next, look up the density of air.
Depends on temperature & pressure, naturally, but let's say 70 degrees Fahrenheit
("room temperature") with barometer at one atmosphere (gauge pressure zero).
Then the chart says 0.075 pounds per cubic foot.
"Per" means "divided by"
Cubic feet can be written "ft^3" (feet to the third power)
So .075 pounds/ft^3
also written .075 pounds-ft^-3

Now you know volume and density. Density = mass/volume implies that
mass = volume*density. That's the formula you need for this problem.

So mass of air in room = (volume of room)(density of air)
Mass of air (pounds) = (800 ft^3)*(.075 pounds/ft^3)

= (800)(.075) (ft^3 * pounds/ft^3)
= 60 pounds
Note you multiply & divide units as you do numbers - after canceling out ft^3 because it's in numerator & denominator, it reduces to pounds. It all works out.

Btw I didn't use the common abbreviation for pounds "lbs" - or sometimes " # "