Question

The density of air at ordinary atmospheric pressure and 25 ∘C is 1.19 g/L. What is the mass of the air in a room that measures 14.5×16.5×6.0 ft?

Answer 1

40,827.34 g

Step-by-step explanation:

First find the volume of the room and then multiply it by the density of air at ordinary atmospheric pressure and 25 ∘C.

In this case, the room has a length of 14.5 ft, a width of 16.5 ft, and a height of 6.0 ft.

volume = 14.5 ft x 16.5 ft x 6.0 ft = 1209.25 ft^3

To convert ft^3 to L, we can use the conversion factor 1ft^3 = 28.316846592 L

1209.25 ft^3 = 1209.25 * 28.316846592 = 34,416.99 L

Calculate the mass of the air in the room by multiplying the volume of the room by the density of air:

mass = 34,416.99 L x 1.19 g/L = 40,827.34 g

So, the mass of the air in the room is 40,827.34 g