Rank the following gases from least dense to most dense at 1.00 atm and \(298 \mathrm{~K}: \mathrm{SO}_{2}, \mathrm{HBr}, \mathrm{CO}_{2} .\) Explain.

The ideal gas law is given by the formula: \(PV = nRT\) where P is the pressure of the gas, V is its volume, n is the number of moles of the gas, R is the ideal gas constant, and T is the temperature of the gas.

The density of the gas (\(\rho\)) can be calculated using the formula: \(\rho = \frac{m}{V}\) where m is the mass of the gas and V is its volume. We can rewrite the mass in terms of the number of moles (n) and molar mass (M) as follows: \(m = nM\) We can substitute this expression into the density formula: \(\rho = \frac{nM}{V}\) Now, we can use the ideal gas law to relate V and n: \(PV = nRT\) \(\frac{V}{n} = \frac{RT}{P}\) \(\frac{1}{V} = \frac{P}{RT}\) Substitute this expression for \(\frac{1}{V}\) into the density formula: \(\rho = \frac{nM}{\frac{RT}{P}}\) \(\rho = \frac{MP}{RT}\)

We are given that the pressure is 1.00 atm and the temperature is 298 K. We can now calculate the molar mass of each gas and then use the density formula to determine their densities. Molar mass of SO2: M_SO2 = 32.07 (S) + 2 * 16.00 (O) = 64.07 g/mol Molar mass of HBr: M_HBr = 1.01 (H) + 79.90 (Br) = 80.91 g/mol Molar mass of CO2: M_CO2 = 12.01 (C) + 2 * 16.00 (O) = 44.01 g/mol Density of SO2 (\(\rho_{SO2}\)): \(\rho_{SO2} = \frac{(64.07\times1.00)}{(0.0821\times298)} = 2.63 \frac{g}{L}\) Density of HBr (\(\rho_{HBr}\)): \(\rho_{HBr} = \frac{(80.91\times1.00)}{(0.0821\times298)} = 3.29 \frac{g}{L}\) Density of CO2 (\(\rho_{CO2}\)): \(\rho_{CO2} = \frac{(44.01\times1.00)}{(0.0821\times298)} = 1.82 \frac{g}{L}\)

From the calculated densities, we can rank the gases from least dense to most dense as follows: CO2 (1.82 g/L) < SO2 (2.63 g/L) < HBr (3.29 g/L)