Chapter 5
Q. 5.8
Q. 5.8
Calculate the density of carbon dioxide (CO_{2}) in grams per liter (g/L) at 0.990 atm and 55°C.
Strategy We need Equation (5.11) d=\frac{m}{V}=\frac{P.M}{RT}
to calculate gas density. Is sufficient information provided in the problem? What temperature unit should be used?

Step-by-Step
Verified Solution
To use Equation (5.11), we convert temperature to kelvins (T = 273 + 55 = 328 K) and use 44.01 g for the molar mass of CO_{2}; that is,
d=\frac{P.M}{RT}=\frac{(0.990 atm)(44.01 g/mol)}{(0.0821 L . atm/K . mol)(328 K)}=1.62 g/L
Alternatively, we can solve for the density by writing
density \frac{mass}{volume}Assuming that we have 1 mole of CO_{2}, the mass is 44.01 g. The volume of the gas can be obtained from the ideal gas equation
V=\frac{nRT}{P}=\frac{(1 mol)(0.0821 L . atm/K . mol)(328K)}{0.990 atm}
=27.2 L
Therefore, the density of CO_{2} is given by
\frac{44.01 g}{27.2 L}=1.62 g/LComment In units of grams per milliliter, the gas density is 1.62 \times 10^{-3} g/mL, which is a very small number. In comparison, the density of water is 1.0 g/mL and that of gold is 19.3 g/cm^{3}.