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 gold-coated wire coil to absorb mercury (Hg) present in the gas. The mercury was recovered from the coil by heating it, and then analyzed. \(\underline{\text { In one }}\) particular set of experiments scientists found a mercury vapor level of \(1800 \mathrm{ng}\) of \(\mathrm{Hg}\) per cubic meter in the plume, at a gas temperature of \(10^{\circ} \mathrm{C}\). Calculate (a) the partial pressure of \(\mathrm{Hg}\) vapor in the plume, \((\mathrm{b})\) 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 Hg vapor in the plume is \(2.08 \times 10^{-7}\) atm. The number of Hg atoms per cubic meter in the gas is \(5.405 \times 10^{15}\) Hg atoms/m³. The total mass of Hg emitted per day by the volcano is 2880 g Hg/day.

Step-by-step solution

Convert Temperature to Kelvin

Since the gas temperature is given in Celsius, we need to convert it to Kelvin. T(K) = T(°C) + 273.15 T(K) = 10°C + 273.15 = 283.15 K

Calculate the Moles of Mercury

Using the given mass (1800ng Hg) and the molar mass of Hg (200.59g/mol), we can calculate the number of moles of Hg. Number of moles = (mass Hg) / (molar mass Hg) First, convert nanograms (ng) to grams (g): 1800 ng Hg = 1800 * 10^{-9} g Hg = \(1.8 \times 10^{-6} \)g Hg Now, find the number of moles: \(n = \frac{1.8 \times 10^{-6} \text{g}}{200.59 \text{g/mol}} = 8.98 \times 10^{-9} \) moles Hg

Calculate the Partial Pressure of Mercury Vapor

According to the Ideal Gas Law, PV = nRT P = nRT / V We have n, R (the ideal gas constant, 0.0821 L atm/mol K), and T. We are given the concentration (in g/m³) and need to find P in atm. To convert concentration to pressure, we need to change the volume unit from liters to cubic meters. We have 1 mole of Hg per 1 m³. So we have: V = 1 m³ = 1000 L Then, we can find the partial pressure P: P = nRT / V = (8.98 x 10^{-9} moles)(0.0821 L atm/mol K)(283.15 K) / 1000 L = \(2.08 \times 10^{-7}\) atm (a) The partial pressure of Hg vapor in the plume is \(2.08 \times 10^{-7}\) atm.

Calculate the Number of Hg Atoms per Cubic Meter in the Gas

To find the number of Hg atoms per cubic meter, we can use Avogadro's number (6.022 x 1023 atoms/mol) and multiply it by the number of moles of Hg found in Step 2. Number of Hg atoms per m³ = (n) (Avogadro's number) = \(8.98 \times 10^{-9} \) moles * 6.022 x 10²³ atoms/mol = \(5.405 \times 10^{15}\) Hg atoms/m³ (b) The number of Hg atoms per cubic meter in the gas is \(5.405 \times 10^{15}\) Hg atoms/m³.

Calculate the Total Mass of Hg Emitted per Day by the Volcano

Given the daily plume volume (1600 km³) and Hg concentration (1800 ng Hg/m³), we can find the total mass of Hg emitted per day. First, convert the volume from km³ to m³: 1600 km³ = 1600 * 10^9 m³ Now, find the total mass of Hg emitted per day: Total mass Hg = daily plume volume * Hg concentration = 1600 * 10^9 m³ * 1800 * 10^{-9} g Hg/m³ = 2880 g Hg/day (c) The total mass of Hg emitted per day by the volcano is 2880 g Hg/day.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation that relates the pressure, volume, temperature, and number of moles of a gas. The equation is expressed as \( PV = nRT \), 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 in Kelvin. The ideal gas constant \( R \) is typically \( 0.0821 \text{ L atm/mol K}\).

In the context of volcanic emission calculations, the Ideal Gas Law helps us determine the pressure of gas samples based on their volume and temperature. By knowing the concentration of a specific gas (like mercury vapor), we can calculate its partial pressure in the gaseous mixture. This helps scientists understand the behavior of gases emitted during volcanic eruptions and assess their impact on the environment.

Mercury Vapor Pressure

Mercury vapor pressure refers to the partial pressure exerted by mercury vapor within a mixture of gases. When mercury present in a volcanic plume is analyzed, its vapor pressure is determined by the concentration and temperature according to the Ideal Gas Law.

The partial pressure gives an insight into how much mercury is present in gaseous form in the plume, which is crucial for calculating the spread and the potential health implications of the mercury emissions. Determining the vapor pressure aids in assessing the mercury's potential to disperse into the atmosphere and possibly affect regions far from the volcano.

In the exercise, the partial pressure of mercury vapor was calculated as a small fraction of atmospheric pressure, demonstrating how trace elements like mercury can exist in volcanic gases at low pressures.

Avogadro's Number

Avogadro's Number is a constant that describes the number of atoms, ions, or molecules in one mole of a substance. Its value is \( 6.022 \times 10^{23} \) entities per mole. This concept helps convert moles of a substance into the total number of particles present.

For volcanic emission calculations, after finding the amount of mercury in moles within a cubic meter, Avogadro's number allows us to calculate the exact count of mercury atoms in that space. This is useful for assessing the concentration of mercury and its potential environmental impact.

In the exercise, multiplying the number of mercury moles by Avogadro's number helps determine that there are approximately \( 5.405 \times 10^{15} \) mercury atoms per cubic meter of the volcanic gas.

Volume Conversion

Volume conversion is critical in calculations involving gases, especially when dealing with large volumes, like those emitted from volcanoes. Standard practice is to convert between different units such as liters (L) and cubic meters (m³) because many scientific formulas use liters, but environmental scales use cubic meters or even cubic kilometers.

In the given exercise, it was necessary to convert the volume from cubic kilometers (\( \text{km}^3 \)) to cubic meters (\( \text{m}^3 \)) to compute the total mercury emitted per day. This conversion ensures that all units are consistent, allowing accurate calculation of the total mass based on mercury concentration.

The conversion factor is \( 1 \, \text{km}^3 = 10^9 \, \text{m}^3 \), highlighting the essential nature of volume conversions in scientific calculations.

Mass Calculation

Mass calculation involves determining the total mass of a substance in a given system. With gases, this often requires understanding the concentration and volume of the gas. In the case of volcanic emissions, determining the total mass of mercury emitted requires knowledge of the volumetric flow rate of the volcanic plume and the concentration of mercury within that plume.

The exercise outlined the step to calculate the total mass of mercury by multiplying the plume's daily volume in cubic meters by the concentration of mercury in grams per cubic meter. This results in the total mass of mercury being emitted per day.

Such calculations are vital in quantifying emissions from natural sources, thereby allowing researchers to understand and predict the potential environmental and health impacts of volcanic activities. In this scenario, the volcano was found to emit 2880 grams of mercury daily, illustrating the significant quantity that can be released into the atmosphere.