# What is the weight in pounds of the air in rectangular room of volume 3.39xx10^5 " ft"^3, given the density of air is rho_"air"=1.29 " kg"/"m"^3?

Mar 20, 2016

The weight is $2.74 \times {10}^{4} \text{ lbs}$, which is a little more than the typical large elephant!

#### Explanation:

First, convert the cubic feet of the room into cubic meters.

$3.39 \times {10}^{5} {\text{ ft"^3 = 3.39xx10^5 " ft"^3 * ((1 " m")/(3.281 " ft"))^3approx 9,600 " m}}^{3}$

Next, solve the formula for density for mass (in kg).

${\rho}_{\text{air"="mass"/"volume}}$

$\text{mass "=rho_"air"xx"volume}$
${\text{mass "=1.29 " kg"/"m"^3xx9,600 " m}}^{3}$
$\text{mass "=12,384" kg} = 1.24 \times {10}^{4}$

Now, using the fact that $1 \text{ kg"=2.21 " lbs}$, you can find the weight in pounds.

$\text{mass "=1.24xx10^4" kg"xx(2.21 " lbs")/(1 " kg")=2.74xx10^4 " lbs}$