Question

Sulfur dioxide \(\left(\mathrm{SO}_{2}\right)\) is one of the principal pollutants that the Environmental Protection Agency (EPA) has listed in the National Ambient Air Quality Standards (NAAQS), as required by the Clean Air Act (40 CFR part 50 ). The level of \(\mathrm{SO}_{2}\) in ambient air set by the NAAQS is \(0.5\) parts per million (ppm) by volume for an average time of $3 \mathrm{~h}$ to protect public welfare. Beyond that level the \(\mathrm{SO}_{2}\) becomes a harmful pollutant to the environment. If the mass fraction of \(\mathrm{SO}_{2}\) in an ambient air at \(20^{\circ} \mathrm{C}\) is \(0.5 \mathrm{mg} / \mathrm{kg}\), determine whether the level of \(\mathrm{SO}_{2}\) is above the limit set by the NAAQS.

Short answer

Answer: No, the level of SO2 in the given ambient air is not above the NAAQS limit, as the calculated SO2 level of 0.2355 ppm is less than the limit of 0.5 ppm.

Step-by-step solution

Convert mass fraction to mass per volume

First, we need to convert the mass fraction to the mass per unit volume of the air. To do this, we'll use the ideal gas law for the air, which states that \(PV=nRT\), where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature. Keep in mind that we need to work with SI units, so we must change the temperature from Celsius to Kelvin: \(T=20+273.15=293.15 \mathrm{K}\). Assuming that the air is under standard atmospheric pressure, i.e., \(P = 101325 \mathrm{Pa}\). We need to find the mass of air in a cubic meter at the given temperature and pressure. Convert the mass fraction of SO\(_2\) (\(0.5 \mathrm{mg}/\mathrm{kg}\)) to mass per unit volume (e.g., micrograms per cubic meter) using the density of air. For ideal gases, we know that \(\rho=\dfrac{PM}{RT}\), where \(\rho\) is the density, and \(M\) is the molecular weight of the gas. For air, the approximate molecular weight is \(28.97 \mathrm{g/mol}\), so the air's density at given temperature \(T\) and pressure \(P\) is: $$\rho_{air} = \dfrac{101325 \cdot 28.97}{8.314 \cdot 293.15}=1.205 \mathrm{kg/m^3}$$ Since the ambient air is \(0.5 \mathrm{mg}\) \(\mathrm{SO}_{2}\) for \(1 \mathrm{kg}\) of air, we can find how much \(\mathrm{SO}_{2}\) is there in \(1\mathrm{m^3}\) of air: $$\rho_{SO2} = \dfrac{0.5 \mathrm{mg}}{1 \mathrm{kg}}\cdot1.205 \mathrm{kg/m^3}=0.6025 \mathrm{mg/m^3}$$

Find the molar concentration of SO\(_2\) in the air

Next, we need to find the molar concentration of SO\(_2\) in the air. We do this by dividing the mass of SO\(_2\) in one cubic meter by its molar mass. The molar mass of SO\(_2\) is \(32+16+16 = 64 \mathrm{g/mol}\). $$C_{SO2}=\dfrac{0.6025 \mathrm{mg}}{64 \mathrm{g/mol}}$$ Converting from milligrams to grams, we get: $$C_{SO2}=\dfrac{0.0006025 \mathrm{g}}{64 \mathrm{g/mol}} = 9.414\times 10^{-6} \mathrm{mol/m^3}$$

Convert the molar concentration to volume fraction in ppm

Now, we can find the volume fraction of SO\(_2\) in the air by dividing the molar concentration of SO\(_2\) by the molar concentration of air. We can calculate the molar concentration of air using the ideal gas law: $$C_{air} = \dfrac{PV}{RT} = \dfrac{101325}{8.314 \cdot 293.15} = 39.962\times10^(3) \mathrm{mol/m^3}$$ Then, the volume fraction (in ppm) of \(\mathrm{SO}_{2}\) in air is: $$\dfrac{C_{SO2}}{C_{air}} \cdot 10^6 = \dfrac{9.414\times 10^{-6} \mathrm{mol/m^3}}{39.962\times10^{3} \mathrm{mol/m^3}}\cdot 10^6 \approx 0.2355 \mathrm{ppm}$$

Compare the calculated SO\(_2\) level with the NAAQS limit

Finally, we compare the calculated SO\(_2\) level (0.2355 ppm) to the limit set by the NAAQS (0.5 ppm) to determine if it is above that limit: $$0.2355 \mathrm{ppm} < 0.5 \mathrm{ppm}$$ Since the calculated SO\(_2\) level of 0.2355 ppm is less than the NAAQS limit of 0.5 ppm, we can conclude that the level of SO\(_2\) in the given ambient air is not above the limit set by the NAAQS.