After the large eruption of Mount St. Helens in 1980 , gas samples from the volcano were taken by sampling the downwind gas plume. The unfiltered gas samples were passed over a goldcoated wire coil to absorb mercury (Hg) present in the gas. The mercury was recovered from the coil by heating it and then analyzed. In one particular set of experiments scientists found a mercury vapor level of \(1800 \mathrm{ng}\) of Hg per cubic meter in the plume at a gas temperature of \(10^{\circ} \mathrm{C}\). Calculate (a) the partial pressure of Hg vapor in the plume, (b) the number of \(\mathrm{Hg}\) atoms per cubic meter in the gas, \((\mathrm{c})\) the total mass of Hg emitted per day by the volcano if the daily plume volume was \(1600 \mathrm{~km}^{3}\).

: To find the partial pressure of mercury vapor in the plume, we will use the Ideal Gas Law equation: \[PV = nRT\] We are given the mercury vapor concentration in the plume (1800 ng/m³) and the gas temperature (10°C). First, we need to convert the temperature to Kelvin: \(T(K) = T(°C) + 273.15\) \(T(K) = 10°C + 273.15 = 283.15 K\) Now, we need to find the number of moles of mercury per cubic meter. The molar mass of mercury is approximately 200.59 g/mol. We can convert the given concentration to moles per cubic meter: \(\frac{1800ng}{1\,m^3} × \frac{1g}{10^9ng} × \frac{1\,mol}{200.59 \, g} = 8.97 × 10^{-12}\, mol/m^3\) Now, we can use the Ideal Gas Law equation to find the partial pressure of mercury vapor: \(P = \frac{nRT}{V}\) where \(n\) is the number of moles per cubic meter, \(R\) is the universal gas constant (\(8.314 \, J \, mol^{-1} K^{-1}\)), \(T\) is the temperature in Kelvin, and \(V\) is the volume. \(P_{Hg} = \frac{(8.97 × 10^{-12}\, mol/m^3)(8.314\, J \, mol^{-1} K^{-1})(283.15 K)}{1\,m^3}\) \(P_{Hg} = 2.10 × 10^{-9}\, Pa\) Thus, the partial pressure of mercury vapor in the plume is approximately \(2.10 × 10^{-9}\, Pa\).

: To find the number of mercury atoms per cubic meter in the gas, we can use Avogadro's number, which tells us the number of atoms/molecules in a mole. Avogadro's number is approximately \(6.022 × 10^{23}\, atoms/mol\). We already calculated the number of moles of mercury per cubic meter in part (a), which is \(8.97 × 10^{-12}\, mol/m^3\). Now, we can use Avogadro's number to find the number of mercury atoms per cubic meter: \(N_{Hg} = (8.97 × 10^{-12}\, mol/m^3)(6.022 × 10^{23}\, atoms/mol)\) \(N_{Hg} = 5.40 × 10^{12}\, Hg\,atoms/m^3\) Therefore, there are approximately \(5.40 × 10^{12}\, Hg\) atoms per cubic meter in the gas.

: We are given the daily plume volume of 1600 km³, and we need to find the total mass of mercury emitted per day by the volcano. First, we need to convert the volume to cubic meters: \(1600\, km^3 × \frac{10^9 \, m^3}{1\, km^3} = 1.60 × 10^{12}\, m^3\) Next, we are going to multiply the volume by the given concentration of mercury vapor to find the total mass of mercury emitted per day: \(1.60 × 10^{12}\, m^3 × \frac{1800 \, ng}{1\, m^3} = 2.88 × 10^{15}\, ng\) Now, let's convert nanograms to grams: \(2.88 × 10^{15}\, ng × \frac{1\, g}{10^9\, ng} = 2880\, g\) Therefore, the total mass of mercury emitted per day by the volcano is approximately 2880 grams.