(a) Certain alloys such as LaNi5 can store hydrogen at room temperature. A plate of LaNi5 containing no hydrogen is placed in a chamber filled with pure hydrogen and maintained at a constant pressure. At what depth from the surface will the concentration of hydrogen be 1/2 the surface concentration after 1 hour? Assume the diffusivity of hydrogen in the alloy is 3.091×10−6 cm2/s. Use the approximation erf(x)≅x for 0<x<0.6 if appropriate.

Question

(a) Certain alloys such as LaNi5 can store hydrogen at room temperature. A plate of LaNi5 containing no hydrogen is placed in a chamber filled with pure hydrogen and maintained at a constant pressure. At what depth from the surface will the concentration of hydrogen be 1/2 the surface concentration after 1 hour? Assume the diffusivity of hydrogen in the alloy is 3.091×10−6 cm2/s. Use the approximation erf(x)≅x for 0<x<0.6 if appropriate.

Give your answer in unit centimeters with up to three decimal spaces

Answer 1

HYDROGEN STORAGE IN ALLOYS - THE FUTURE? (10/10 points)

Certain alloys such as LaNi5 can store hydrogen at room temperature. A plate of LaNi5 containing no hydrogen is placed in a chamber filled with pure hydrogen and maintained at a constant pressure. At what depth from the surface will the concentration of hydrogen be half the surface concentration after 1 hour? (Express your answer in centimeters). Assume the diffusivity of hydrogen in the alloy is 3.091x10-6 cm2/sec. Use the approximation erf(x) = x for x with values between 0 and 0.6, if appropriate. Use the Erf Table on the Course Info page for values greater than 0.6.

0.105 - correct

The activation energy for H diffusion in LaNi5 is 0.25 eV. How far will the H diffuse (where will it reach half the surface concentration) if the experiment above is repeated at 35°C? (Express your answer in centimeters).

0.124 - correct

Answer 2

To solve this problem, we can use Fick's second law of diffusion, which relates the concentration of a diffusing species to its position and time. Fick's second law is given by:

∂C/∂t = D * ∂^2C/∂x^2

Where:
- C is the concentration of hydrogen at a given position x and time t.
- D is the diffusivity of hydrogen in the alloy.

We can solve this differential equation by separation of variables and then apply boundary conditions. However, since the problem asks for an approximate solution, we will use an approximation for the error function, as given in the problem, to simplify the calculations.

Let's denote the surface concentration of hydrogen as Cs and the concentration at a depth x as C(x), and let t = 1 hour.

We want to find the depth x where C(x) = Cs/2 after 1 hour.

First, we need to determine the value of the constant D. It is given as 3.091×10^(-6) cm^2/s in the problem.

Next, we can use the approximation of the error function (erf(x) ≈ x) to approximate the solution.

The concentration profile can be written as:

C(x, t) = Cs * (1 - erf(x / (2 * sqrt(D * t)) ))

Plugging in the values into the equation, we have:

C(x, 1 hour) = Cs * (1 - erf(x / (2 * sqrt(3.091×10^(-6) * 3600))))

Now, we want to find the depth x where C(x, 1 hour) = Cs/2.

Setting C(x, 1 hour) = Cs/2, we can solve for x:

Cs/2 = Cs * (1 - erf(x / (2 * sqrt(3.091×10^(-6) * 3600))))

Simplifying the equation, we have:

1/2 = 1 - erf(x / (2 * sqrt(3.091×10^(-6) * 3600)))

erf(x / (2 * sqrt(3.091×10^(-6) * 3600))) = 1/2

Since we have been given an approximation for the error function, we can assume that the argument of the error function (x / (2 * sqrt(3.091×10^(-6) * 3600))) is small, and thus the error function can be approximated as:

erf(x / (2 * sqrt(3.091×10^(-6) * 3600))) ≈ x / (2 * sqrt(3.091×10^(-6) * 3600))

Now we can solve for x:

x / (2 * sqrt(3.091×10^(-6) * 3600)) ≈ 1/2

Simplifying further:

x ≈ sqrt(3.091×10^(-6) * 3600)

Evaluating this expression:

x ≈ 0.030 cm

Therefore, after 1 hour, the concentration of hydrogen will be half the surface concentration at a depth of approximately 0.030 cm from the surface.