A living room has floor dimensions of $3.5 \mathrm{~m}$ and $4.2 \mathrm{~m}$ and a height of $2.4 \mathrm{~m}$.
What does the air in the room weigh when the air pressure is $1.0 \mathrm{~atm}$ ?
(1) The air's weight is equal to $m g$, where $m$ is its mass.
(2) Mass $m$ is related to the air density $\rho$ and the air volume $V$ by Eq.14-2 $(\rho=m / V)$.

Calculation: Putting the two ideas together and taking the density of air at 1.0 atm from Table 14-1, we find
$$
\begin{aligned}
m g & =(\rho V) g \\
& =\left(1.21 \mathrm{~kg} / \mathrm{m}^3\right)(3.5 \mathrm{~m} \times 4.2 \mathrm{~m} \times 2.4 \mathrm{~m})\left(9.8 \mathrm{~m} / \mathrm{s}^2\right) \\
& =418 \mathrm{~N} 
\end{aligned}
$$