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A 2-mm-thick 5-L vessel made of nickel is used to store hydrogen gas at \(358 \mathrm{~K}\) and \(300 \mathrm{kPa}\). If the total inner surface area of the vessel is \(1600 \mathrm{~cm}^{2}\), determine the rate of gas loss from the nickel vessel via mass diffusion. Also, determine the fraction of the hydrogen lost by mass diffusion after one year of storage.

Short Answer

Expert verified
Answer: 26.5% of the hydrogen is lost via mass diffusion after one year.

Step by step solution

01

Use Fick's First Law of Diffusion

Fick's First Law of Diffusion states that the flux of a species (mass per area per time) is proportional to the negative concentration gradient in the medium. The law can be represented as: $$J = -D\frac{dC}{dx}$$ where \(J\) is the diffusion flux, \(D\) is the diffusion coefficient, and \(\frac{dC}{dx}\) is the concentration gradient. We will use this formula to determine the rate of gas loss (flux) through the nickel vessel.
02

Calculate the Diffusion Coefficient for Hydrogen in Nickel

Before we start, let's convert the temperature to Kelvin so that we can use it in our calculations. Given: Temperature, \(T = 358 K\) Now, we will use the given temperature and material to find the diffusion coefficient (\(D\)) for hydrogen in nickel. From literature or a table, we find that the diffusion coefficient for hydrogen in nickel is: $$D = 7.0 \times 10^{-11} \frac{m^2}{s}$$ This diffusion coefficient will help us in finding the rate of gas loss.
03

Calculate Concentration Gradient

To calculate the concentration gradient, we need to determine the difference in concentration between the hydrogen inside the vessel and the hydrogen outside the vessel. Assuming ideal gas behavior and using the Ideal Gas Law, we can get the hydrogen molar concentration at 300 kPa: $$C_{inside} = \frac{P}{RT}$$ where \(C_{inside}\) is the molar concentration inside the vessel, \(P\) is the pressure, \(R\) is the universal gas constant, and \(T\) is the temperature. Given: Pressure, \(P = 300 kPa = 300000 Pa\) Since R (gas constant) = 8.314 J/mol K, we can solve for \(C_{inside}\): $$C_{inside} = \frac{300000 Pa}{8.314 \frac{J}{mol K} \times 358 K} = 9.42 \times 10^3 \frac{mol}{m^3}$$ Assuming no hydrogen outside the vessel and that the vessel thickness is very small, we can approximate the concentration gradient as: $$\frac{dC}{dx} = \frac{C_{inside} - C_{outside}}{X}$$ Given: Vessel thickness, \(X = 2 mm = 2 \times 10^{-3} m\) $$\frac{dC}{dx} = \frac{9.42 \times 10^3 \frac{mol}{m^3}}{2 \times 10^{-3} m} = 4.71 \times 10^6 \frac{mol}{m^4}$$
04

Calculate the Rate of Gas Loss

Now, we can use Fick's Law to calculate the rate of hydrogen loss through the nickel: $$J = -D \times \frac{dC}{dx} = -7.0 \times 10^{-11} \frac{m^2}{s} \times 4.71 \times 10^6 \frac{mol}{m^4} = -3.3 \times 10^{-4} \frac{mol}{m^2 s}$$ Since the rate of loss is negative, it means there is a net flux from the inside to the outside of the vessel. To find the total rate of gas loss, we multiply the flux by the total inner surface area: Given: Total inner surface area, \(A = 1600 cm^2 = 1.6 \times 10^{-1} m^2\) $$Rate \ of \ Gas \ Loss = -J \times A = -(-3.3 \times 10^{-4} \frac{mol}{m^2 s})(1.6 \times 10^{-1} m^2) = 5.28 \times 10^{-5} \frac{mol}{s}$$ So, the rate of gas loss of hydrogen via diffusion is \(5.28 \times 10^{-5}\) mol/s.
05

Calculate the Fraction of Hydrogen Loss After One Year

To find the fraction of hydrogen lost after one year, we will first determine the total amount of hydrogen in the vessel initially. Using the ideal gas law again, we can get the initial amount of hydrogen: Given: Vessel volume, \(V = 5 L = 5 \times 10^{-3} m^3\) $$n_{initial} = PV/RT = \frac{300000 Pa \times 5 \times 10^{-3} m^3}{8.314 \frac{J}{mol K} \times 358 K} = 6.31 mol$$ Now, multiply the rate of gas loss by the total time to find the hydrogen loss: Time in 1 year, \(t = 1 \times 365 \ days \times 24 \ hours \times 60 \ min \times 60 \ sec = 3.1536 \times 10^7 s\) $$n_{loss} = Rate \ of \ Gas \ Loss \times t = 5.28 \times 10^{-5} \frac{mol}{s} \times 3.1536 \times 10^7 s = 1.67 mol$$ Finally, calculate the fraction of hydrogen loss after one year: $$Fraction \ of \ Hydrogen \ Loss = \frac{n_{loss}}{n_{initial}} = \frac{1.67 mol}{6.31 mol} = 0.265$$ Therefore, 26.5% of the hydrogen stored in the vessel is lost via mass diffusion after one year.

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Key Concepts

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

Understanding Fick's First Law of Diffusion
Fick's First Law of Diffusion is a principle that plays a fundamental role in explaining how substances disperse from regions of high concentration to areas of low concentration. It is akin to thinking about how the smell of a flower spreads throughout a room. The law mathematically defines the flux, denoted as \( J \), which is the amount of substance that will flow through a unit area over time. It correlates directly with the concentration gradient, \( \frac{dC}{dx} \), which is the rate at which the concentration changes spatially. Essentially, the law tells us that the flux is proportional to this gradient, with the proportionality constant being the diffusion coefficient \( D \). That is, \( J = -D\frac{dC}{dx} \).

Why include the negative sign? It's to convey that the diffusion direction is from high to low concentration, meaning that diffusion effectively ‘descends’ the concentration gradient. For the problem involving the nickel vessel storing hydrogen gas, we use Fick's First Law to determine the rate of gas loss due to mass diffusion. By doing so, students can gain a deeper understanding of how molecular movement is quantified and its implications in real-world applications. Such an understanding is essential for tackling challenges in fields like material science, environmental engineering, and even pharmacology, where diffusion processes are critical.
The Role of the Diffusion Coefficient in Mass Diffusion
The diffusion coefficient, symbolized by \( D \), is a property that measures the ease with which particles spread out due to thermal energy, relevant in various disciplines from material science to biophysics. It is scale-dependent, varying not just between different materials but also according to temperature, and sometimes pressure.

In our example, the diffusion coefficient describes how readily hydrogen atoms penetrate the nickel walls of the vessel. A higher coefficient implies faster diffusion and shorter characteristic times for gas to disseminate. It's essential to collect accurate values of the diffusion coefficient from reliable references or experimental data when solving problems like the one at hand. Understanding the diffusion coefficient's significance helps students appreciate that the kinetics of a diffusion process are contingent on the material properties and external conditions, a concept with vast implications on things like controlling the release of medication from delivery systems or the design of membranes for gas separation.
Concentration Gradient: Driving Force of Diffusion
A concentration gradient refers to the change in the concentration of particles over a distance, representing the 'slope' that particles move down during diffusion. It is the driving force that propels particles from where they are more densely packed toward regions of lower density, similar to how a ball rolls down a hill.

The steeper the gradient, much like a steeper hill, the faster the particles will move. This concept is instrumental in problems of diffusion, as seen with the hydrogen gas in the vessel. By determining the gradient, which is the difference in concentration from one side of the vessel wall to the other, one can predict the rate at which hydrogen will escape. In cases where it is safe to assume that the particle concentration on one side of a barrier is zero (outside the vessel, for example), the concentration gradient is simply the internal concentration divided by the thickness of the barrier. In the classroom and beyond, mastering the concept of concentration gradients empowers students to solve and understand a myriad of processes, from how a sugar cube dissolves in water to how gases exchange in our lungs.
Applying the Ideal Gas Law to Diffusion Problems
The Ideal Gas Law is a cornerstone equation in chemistry and physics, stating that the pressure, volume, and temperature of an ideal gas are related through the equation \( PV = nRT \). In our hydrogen storage problem, it's used to calculate the initial concentration of hydrogen gas inside the vessel.

The constant \( R \) in the equation is the universal gas constant, and \( T \) stands for temperature. As ideal gases are theoretical constructs, the law assumes no intermolecular forces and that the particles occupy no space, which isn't entirely true in real life. However, it often provides a reasonable approximation for gases at high temperatures and low pressures, which is valuable for students to recognize as they apply this law to a myriad of gas-related problems. When used in conjunction with concepts like diffusion and mass loss over time, the Ideal Gas Law allows for impressive and powerful predictions about how a gas will behave in confined spaces, such as the nickel vessel, broadening students' capabilities in problem-solving and critical thinking in the scientific arena.

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Most popular questions from this chapter

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