Question

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 \(1600 \mathrm{~km}^{3}\).

Short answer

The partial pressure of mercury vapor in the plume is approximately \(2.10 × 10^{-9}\, Pa\). There are approximately \(5.40 × 10^{12}\, Hg\) atoms per cubic meter in the gas. The total mass of mercury emitted per day by the volcano is approximately 2880 grams.

Step-by-step solution

(a) Find the partial pressure of Hg vapor

: 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 \(2.10 × 10^{-9}\, Pa\).

(b) Find the number of Hg atoms per cubic meter

: 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.

(c) Find the total mass of Hg emitted per day

: 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.

Key concepts

Understanding Partial Pressure

Partial pressure is an important concept in chemistry, especially when dealing with mixtures of gases. In simpler terms, it is the pressure that a single gas in a mixture would exert if it occupied the entire volume on its own. It's part of the Ideal Gas Law, which states that the physical properties of gases—pressure, volume, and temperature—are related.

In the context of the Ideal Gas Law, partial pressure is represented by the equation: \( P = \frac{nRT}{V} \), where \( n \) is the number of moles, \( R \) is the ideal gas constant, \( T \) is the temperature in Kelvin, and \( V \) is the volume.

• The partial pressure can help determine the behavior and characteristics of each component gas in a mixture.
• Remember, the total pressure of a mixture of gases is simply the sum of all individual partial pressures.

Partial pressures are often used in various fields like meteorology, respiratory physiology, and environmental science to measure and understand how gases interact.

The Role of Molar Mass

Molar mass is the mass of a given substance (chemical element or compound) divided by the amount of substance. It's expressed in grams per mole (g/mol). Molar mass serves as a bridge between the microscopic world of atoms and molecules and the macroscopic world of grams and liters.

For instance, in our exercise involving mercury (Hg), it is essential to know that:

• The molar mass of mercury is approximately 200.59 g/mol.
• This allows conversion between the mass of mercury and the number of moles, which is crucial for calculations utilizing the Ideal Gas Law.

Molar mass helps us:

• Understand how much of a substance we have in terms that are directly usable in chemical equations.
• Calculate the number of molecules or atoms present in a given volume, which is key when dealing with reactions and mixtures.

Avogadro's Number: Counting the Invisible

Avogadro's number is one of the fundamental constants in chemistry and is approximately \(6.022 \times 10^{23}\). It is the number of atoms, molecules, or particles in one mole of a substance. Avogadro's number is vital in understanding the scale of chemical reactions.

For example, to find the number of mercury atoms in the gas sample, we used Avogadro's number multiplied by the number of moles: \( N_{Hg} = (8.97 \times 10^{-12}\, \text{mol/m}^3)(6.022 \times 10^{23}\, \text{atoms/mol})\).

• This calculation provides a tangible number from what is otherwise an abstract mole concept.
• Helps visualize the size of particles in chemistry and biology for practices like determining concentrations in solutions and mixtures.

Avogadro's number acts like a bridge making it possible to "count" molecules by weighing them.