Question

A room has a volume of 60 m^3 and is filled with air of an average molecular mass of 29 u. What is the mass of the air in the room at a pressure of 1.0 atm and temperature of 22°C? R = 0.082 L⋅atm/mol⋅K a. 2.4 kg b. 2 400 kg c. 72 kg d. 700 kg

Answer 1

Use a formula that relates molecular mass, density, pressure, temperature, and the constant R.

$\rho=\frac{M\cdot P}{RT}$

Using the given information, we have

$\rho=\frac{29\cdot\frac{gr}{\text{mol}}\cdot1\text{atm}}{0.082\cdot\frac{L\cdot\text{atm}}{\text{mol}\cdot K}\cdot(295.15K)}=\frac{29}{24.2023}=\frac{1.2gr}{L}$

The density is 1.2 grams per liter.

Then, use the density formula to find the mass.

$\begin{gathered} \rho=\frac{m}{V}\to m=\rho\cdot V \\ m=\frac{1.2gr}{L}\cdot60m^3 \end{gathered}$

But, 1 liter equals 0.001 m^3 and 1kg equals 1000gr.

$\begin{gathered} m=\frac{1.2gr\cdot\frac{1\operatorname{kg}}{1000gr}}{1L\cdot\frac{0.001m^3}{1L}}\cdot60m^3=\frac{0.072}{0.001}kg \\ m=72\operatorname{kg} \end{gathered}$

Therefore, the answer is c. 72 kg.