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=(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)}=(29)/(24.2023)=(1.2gr)/(L)`

The density is 1.2 grams per liter.

Then, use the density formula to find the mass.

`\begin{gathered} \rho=(m)/(V)\to m=\rho\cdot V \\ m=(1.2gr)/(L)\cdot60m^3 \end{gathered}`

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

Therefore, the answer is c. 72 kg.