Question

An auditorium measures 40.0 m 3 20.0 m 3 12.0 m. The density of air is 1.20 kg/m3. What are (a) the volume of the room in cubic feet and (b) the weight of air in the room in pounds?

Answer 1

We use the formula,

m = V\rho

Here, m is the mass, V is the volume and  $\rho$ density

Also

$V = l w h$

Here l is length, w is width and h is height.

(a) The volume of the room,

$V=40.0 \ m \times 3 20.0 \ m \times 3 12.0\ m = 3.99 \times 10^{6} m^3$

The volume of the room in cubic feet,

$V = 3.99 \times 10^{6} m^3 \times(\frac{3.281 \ ft}{m} )^3 = 4.3 \times 10^7 \ ft^3$

(b) Now the mass of the air in room,

$m= (3.99 \times 10^{6} \ m^3) (1.20 \ kg/m^3) = 4.8 \times 10^6 kg$.

Therefore, the weight of the air in room,

$W = mg= 4.8 \times 10^6 kg \times 9.8 m/s^2 = 46.9 \times 10^6 \ N \\\\ W = 4.69 \times 10^7 \ N$.

The weight of air in the room in pounds,

$W = 4.69 \times 10^7 \ N ( \frac{1 \ lb}{4.448 \ N} ) = 1.1 \times 10^7 \ lb$