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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

You probably have noticed that balloons inflated with helium gas rise in the air the first day during a party but they fall down the next day and act like ordinary balloons filled with air. This is because the helium in the balloon slowly leaks out through the wall while air leaks in by diffusion. Consider a balloon that is made of \(0.1\)-mm-thick soft rubber and has a diameter of \(15 \mathrm{~cm}\) when inflated. The pressure and temperature inside the balloon are initially \(110 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\). The permeability of rubber to helium, oxygen, and nitrogen at \(25^{\circ} \mathrm{C}\) are \(9.4 \times 10^{-13}, 7.05 \times 10^{-13}\), and \(2.6 \times 10^{-13} \mathrm{kmol} / \mathrm{m} \cdot \mathrm{s} \cdot\) bar, respectively. Determine the initial rates of diffusion of helium, oxygen, and nitrogen through the balloon wall and the mass fraction of helium that escapes the balloon during the first \(5 \mathrm{~h}\) assuming the helium pressure inside the balloon remains nearly constant. Assume air to be 21 percent oxygen and 79 percent nitrogen by mole numbers and take the room conditions to be \(100 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\).

Hydrogen can cause fire hazards, and hydrogen gas leaking into surrounding air can lead to spontaneous ignition with extremely hot flames. Even at very low leakage rate, hydrogen can sustain combustion causing extended fire damages. Hydrogen gas is lighter than air, so if a leakage occurs it accumulates under roofs and forms explosive hazards. To prevent such hazards, buildings containing source of hydrogen must have adequate ventilation system and hydrogen sensors. Consider a metal spherical vessel, with an inner diameter of \(5 \mathrm{~m}\) and a thickness of \(3 \mathrm{~mm}\), containing hydrogen gas at \(2000 \mathrm{kPa}\). The vessel is situated in a room with atmospheric air at \(1 \mathrm{~atm}\). The ventilation system for the room is capable of keeping the air fresh, provided that the rate of hydrogen leakage is below \(5 \mu \mathrm{g} / \mathrm{s}\). If the diffusion coefficient and solubility of hydrogen \(\mathrm{gas}\) in the metal vessel are \(1.5 \times 10^{-12} \mathrm{~m}^{2} / \mathrm{s}\) and \(0.005 \mathrm{kmol} / \mathrm{m}^{3}\).bar, respectively, determine whether or not the vessel is safely containing the hydrogen gas.

A rubber object is in contact with nitrogen \(\left(\mathrm{N}_{2}\right)\) at \(298 \mathrm{~K}\) and \(250 \mathrm{kPa}\). The solubility of nitrogen gas in rubber is \(0.00156 \mathrm{kmol} / \mathrm{m}^{3}\).bar. The mass density of nitrogen at the interface is (a) \(0.049 \mathrm{~kg} / \mathrm{m}^{3}\) (b) \(0.064 \mathrm{~kg} / \mathrm{m}^{3}\) (c) \(0.077 \mathrm{~kg} / \mathrm{m}^{3}\) (d) \(0.092 \mathrm{~kg} / \mathrm{m}^{3}\) (e) \(0.109 \mathrm{~kg} / \mathrm{m}^{3}\)

Consider a tank that contains moist air at \(3 \mathrm{~atm}\) and whose walls are permeable to water vapor. The surrounding air at \(1 \mathrm{~atm}\) pressure also contains some moisture. Is it possible for the water vapor to flow into the tank from surroundings? Explain.

When the ___ is unity, one can expect the momentum and mass transfer by diffusion to be the same. (a) Grashof (b) Reynolds (c) Lewis (d) Schmidt (e) Sherwood

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