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Hydrogen gas at \(750 \mathrm{kPa}\) and \(85^{\circ} \mathrm{C}\) is stored in a spherical nickel vessel. The vessel is situated in a surrounding of atmospheric air at \(1 \mathrm{~atm}\). Determine the molar and mass concentrations of hydrogen in the nickel at the inner and outer surfaces of the vessel.

Short Answer

Expert verified
Answer: According to Siegert's law, the solubility of hydrogen in a metal is directly proportional to the square root of the pressure of the surrounding hydrogen gas.

Step by step solution

01

Identifying the given information

We know the following: - Pressure of hydrogen gas inside the vessel: \(P_{H2} = 750 \mathrm{kPa}\) - Temperature of hydrogen gas inside the vessel: \(T = 85^{\circ} \mathrm{C}\) - Pressure of atmospheric air outside the vessel: \(P_{air} = 1 \mathrm{~atm}\) Converting the given temperature to Kelvin: \(T(K) = T(^\circ C) + 273.15\)
02

Convert the pressures to the same unit

Convert \(P_{air}\) to kPa, since \(P_{H2}\) is given in kPa. \(P_{air} = 1 \mathrm{~atm} \times \frac{101.3 \mathrm{kPa}}{1 \mathrm{~atm}} = 101.3 \mathrm{kPa}\) Now we have both pressures in kPa: - \(P_{H2} = 750 \mathrm{kPa}\) - \(P_{air} = 101.3 \mathrm{kPa}\)
03

Calculate the solubility of hydrogen in the nickel (Siegert's law)

Siegert's law states that the solubility of hydrogen in a metal is directly proportional to the square root of the pressure of the surrounding hydrogen gas. Mathematically, it is represented as: \(C = k \sqrt{P}\) where \(C\) is the concentration of hydrogen in the metal, \(k\) is a proportionality constant, and \(P\) is the pressure of the surrounding hydrogen gas. We will need to apply this equation at the inner and outer surfaces of the vessel.
04

Calculate the molar concentration of hydrogen at the inner and outer surfaces

Using Siegert's law, we can find the molar concentration of hydrogen at the inner and outer surfaces: At the inner surface: \(C_{inner} = k \sqrt{P_{H2}}\) At the outer surface: \(C_{outer} = k \sqrt{P_{air}}\) Note: The proportionality constant, \(k\), will be the same for both the inner and the outer surfaces.
05

Calculate the mass concentration of hydrogen at the inner and outer surfaces

To find the mass concentration of hydrogen, we need to multiply the molar concentration by the molar mass of hydrogen (\(M_{H2} = 2.016 \mathrm{g/mol}\)): At the inner surface: \(mass\_conc_{inner} = C_{inner} \times M_{H2} = k \sqrt{P_{H2}} \times M_{H2}\) At the outer surface: \(mass\_conc_{outer} = C_{outer} \times M_{H2} = k \sqrt{P_{air}} \times M_{H2}\) In conclusion, to determine the molar and mass concentrations of hydrogen at the inner and outer surfaces of the vessel, we need to know the proportionality constant \(k\). Once we have the value of \(k\), we can use Siegert's law and the given pressures to calculate the concentrations.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Siegert's Law
Siegert's law is a handy tool we use to find out how much hydrogen gas dissolves in metals like nickel. This is crucial when dealing with hydrogen storage in metal vessels. This law tells us that the amount of hydrogen absorbed is directly tied to the square root of the hydrogen gas's pressure around the metal. In simpler words:
  • The higher the pressure of hydrogen gas, the more hydrogen gets absorbed by the metal.
To put it mathematically: \[ C = k \sqrt{P} \] Here, \( C \) is the concentration of hydrogen in the metal, \( k \) is a constant representing the "ease" of absorption (which depends on the metal and temperature), and \( P \) is the pressure of the hydrogen gas. This equation is straightforward but very valuable for understanding hydrogen storage. Once you know the pressure, you can quickly find the concentration of hydrogen using this law. Add in the constant \( k \), and you're ready to calculate the specifics of hydrogen solubility.
Concentration Calculation
Calculating the concentration of hydrogen is important because it tells us how much hydrogen is present in a certain quantity of nickel. In general, the formula used, thanks to Siegert's law, is:
  • Molar Concentration at inner surface: \( C_{inner} = k \sqrt{P_{H2}} \)
  • Molar Concentration at outer surface: \( C_{outer} = k \sqrt{P_{air}} \)
These equations allow us to determine the molar concentration at both the inside and outside of the vessel. Understanding molar concentration is quite simple; it's just about knowing how many moles of hydrogen exist per unit volume of nickel. This can in turn be converted into mass concentration. This is essential because industries and science need to know how much hydrogen is in the metal for predicting behavior, safety, and performance in processes.
Pressure Conversion
Pressure conversion might sound technical, but it's straightforward. When dealing with gases, pressure can be reported in different units, which can lead to confusion. In the solved exercise case, we needed the pressures in kilopascals (kPa) for both indoor and atmospheric conditions. Originally, the hydrogen gas was given in kPa, but the atmospheric air was provided in atmospheres (atm). To have a coherent system:
  • You convert the atmospheric pressure from atm to kPa.
This involves using the conversion factor: \[ 1 \, \text{atm} = 101.3 \, \text{kPa} \]So for air pressure: \[ P_{air} = 1 \, \text{atm} \times \frac{101.3 \, \text{kPa}}{1 \, \text{atm}} = 101.3 \, \text{kPa} \]This simple conversion is essential for calculations that follow, ensuring consistency and reliability of results when using formulas like Siegert's law.
Molar and Mass Concentration
Understanding molar and mass concentration is key to comprehend how much gas is present and what it might weigh. The concept makes it easy to transition from the number of molecules to their weight within the metal. First, molar concentration represents how much hydrogen (moles) is in the nickel per volume. You derive this from Siegert's law:
  • For the inner surface, it's \( C_{inner} = k \sqrt{P_{H2}} \).
  • For the outer surface, it's \( C_{outer} = k \sqrt{P_{air}} \).
To get from molar to mass concentration:
  • Simply multiply the molar concentration by the molar mass of hydrogen, which is 2.016 g/mol.
This gives you:
  • Mass Concentration at inner surface: \( \text{mass\_conc}_{inner} = C_{inner} \times 2.016 \)
  • Mass Concentration at outer surface: \( \text{mass\_conc}_{outer} = C_{outer} \times 2.016 \)
Knowing both molar and mass concentrations provides a full picture of hydrogen's presence in the nickel, valuable for applications requiring specific hydrogen content or density calculations.

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Most popular questions from this chapter

During a hot summer day, a \(2-L\) bottle drink is to be cooled by wrapping it in a cloth kept wet continually and blowing air to it with a fan. If the environment conditions are \(1 \mathrm{~atm}, 80^{\circ} \mathrm{F}\), and 30 percent relative humidity, determine the temperature of the drink when steady conditions are reached.

Air at \(52^{\circ} \mathrm{C}, 101.3 \mathrm{kPa}\), and 10 percent relative humidity enters a 5 -cm-diameter tube with an average velocity of \(5 \mathrm{~m} / \mathrm{s}\). The tube inner surface is wetted uniformly with water, whose vapor pressure at \(52^{\circ} \mathrm{C}\) is \(13.6 \mathrm{kPa}\). While the temperature and pressure of air remain constant, the partial pressure of vapor in the outlet air is increased to \(10 \mathrm{kPa}\). Detemine \((a)\) the average mass transfer coefficient in \(\mathrm{m} / \mathrm{s},(b)\) the log-mean driving force for mass transfer in molar concentration units, \((c)\) the water evaporation rate in \(\mathrm{kg} / \mathrm{h}\), and \((d)\) the length of the tube.

Determine the mole fraction of the water vapor at the surface of a lake whose temperature is \(15^{\circ} \mathrm{C}\) and compare it to the mole fraction of water in the lake. Take the atmospheric pressure at lake level to be \(92 \mathrm{kPa}\).

Benzene-free air at \(25^{\circ} \mathrm{C}\) and \(101.3 \mathrm{kPa}\) enters a 5 -cm-diameter tube at an average velocity of \(5 \mathrm{~m} / \mathrm{s}\). The inner surface of the \(6-m\)-long tube is coated with a thin film of pure benzene at \(25^{\circ} \mathrm{C}\). The vapor pressure of benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) at \(25^{\circ} \mathrm{C}\) is \(13 \mathrm{kPa}\), and the solubility of air in benezene is assumed to be negligible. Calculate \((a)\) the average mass transfer coefficient in \(\mathrm{m} / \mathrm{s},(b)\) the molar concentration of benzene in the outlet air, and \((c)\) the evaporation rate of benzene in \(\mathrm{kg} / \mathrm{h}\).

Consider a brick house that is maintained at \(20^{\circ} \mathrm{C}\) and 60 percent relative humidity at a location where the atmospheric pressure is \(85 \mathrm{kPa}\). The walls of the house are made of 20 -cm-thick brick whose permeance is \(23 \times 10^{-12} \mathrm{~kg} / \mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{Pa}\). Taking the vapor pressure at the outer side of the wallboard to be zero, determine the maximum amount of water vapor that will diffuse through a \(3-\mathrm{m} \times 5-\mathrm{m}\) section of a wall during a 24-h period.

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