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 $rho_"air"=1.29 " kg"/"m"^3$ ?

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Answer 1 · 2016-03-20T12:52:26

The weight is $2.74xx10^4 " lbs"$ , which is a little more than the typical large elephant!

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

$3.39xx10^5 " ft"^3 = 3.39xx10^5 " ft"^3 * ((1 " m")/(3.281 " ft"))^3approx 9,600 " m"^3$

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

$rho_"air"="mass"/"volume"$

$"mass "=rho_"air"xx"volume"$
$"mass "=1.29 " kg"/"m"^3xx9,600 " m"^3$
$"mass "=12,384" kg"=1.24xx10^4$

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

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

What is the weight in pounds of the air in rectangular room of volume 3.39xx10^5 " ft"^33.39xx10^5 " ft"^3, given the density of air is rho_"air"=1.29 " kg"/"m"^3rho_"air"=1.29 " kg"/"m"^3?