Question 12.14: Calculating the Density of a Gas Using the Ideal Gas Law in ......
Calculating the Density of a Gas Using the Ideal Gas Law in Modified Form
In gas law calculations, air is often considered to be a single gas with a molar mass of 29 g/mole. On this basis, calculate the density of air, in grams per liter, on a hot summer day (41 °C) when the atmospheric pressure is 0.91 atm.
The ideal gas equation in the modified form
d = \frac{PM}{RT}is used to calculate the density of a gas.
All the quantities on the right side of this equation are known.
P = 0.91 atm, M = 29 g/mole
R = 0.08206\frac{\textrm{atm} \cdot \textrm{L}}{\textrm{mole} \cdot \textrm{K}} T = 41°C = 314 K
Substitution of these values into the equation gives
d = \frac{(0.91 \cancel{\textrm{atm}}) \Bigl( 29 \frac{\textrm{g}}{\cancel{\textrm{mole}}} \Bigr) }{ \Bigl( 0.08206 \frac{\cancel{\textrm{atm}} \cdot \textrm{L}}{\cancel{\textrm{mole}} \cdot \cancel{\textrm{K}}} \Bigr) (314 \cancel{\textrm{K}})}All units cancel except for the desired ones, grams per liter.
Doing the arithmetic, we obtain a value of 1.0 g/L for the density of air at the specified temperature and pressure.
d = \frac{0.91 \times 29}{0.08206 \times 314} \frac{\textrm{g}}{\textrm{L}} = 1.0241845 \frac{\textrm{g}}{\textrm{L}} (calculator answer)
= \textbf{1.0} \frac{\textbf{g}}{\textbf{L}} (correct answer)