When \(\alpha\) -iron is subjected to an atmosphere of nitrogen gas, the concentration of nitrogen in the iron, \(C_{\mathrm{N}}\) (in weight percent), is a function of hydrogen pressure, \(p_{\mathrm{N}_{2}}(\text { in } \mathrm{MPa}),\) and absolute temperature \((T)\) according to $$C_{\mathrm{N}}=4.90 \times 10^{-3} \sqrt{p_{\mathrm{N}_{2}}} \exp \left(-\frac{37.6 \mathrm{kJ} / \mathrm{mol}}{R T}\right)$$.Furthermore, the values of \(D_{0}\) and \(Q_{d}\) for this diffusion system are \(3.0 \times 10^{-7} \mathrm{m}^{2} / \mathrm{s}\) and \(76,150 \mathrm{J} / \mathrm{mol}\), respectively. Consider a thin iron membrane \(1.5 \mathrm{mm}\) thick that is at \(300^{\circ} \mathrm{C}\) Compute the diffusion flux through this membrane if the nitrogen pressure on one side of the membrane is \(0.10 \mathrm{MPa}(0.99 \mathrm{atm}),\) and on the other side \(5.0 \mathrm{MPa}(49.3 \mathrm{atm})\).

Step by step solution

01

Calculate \(C_N\) for both sides of the membrane

Using the equation provided: $$C_{\mathrm{N}}=4.90 \times 10^{-3} \sqrt{p_{\mathrm{N}_{2}}} \exp\left(-\frac{37.6 \mathrm{kJ} / \mathrm{mol}}{R T}\right)$$ We calculate the nitrogen concentration for both sides of the membrane. First, let's convert the temperature to absolute scale (Kelvin): $$T_{\text{Kelvin}} = T_{\text{Celsius}} + 273.15 = 300 + 273.15 = 573.15\,\text{K}$$ Now, the concentrations for both sides are: $$C_{\mathrm{N_1}}=4.90 \times 10^{-3} \sqrt{0.10} \exp\left(-\frac{37.6 \times 10^3}{8.314 \time 573.15}\right)$$ $$C_{\mathrm{N_2}}=4.90 \times 10^{-3} \sqrt{5.0} \exp\left(-\frac{37.6 \times 10^3}{8.314 \times 573.15}\right)$$

02

Calculate the concentration gradient

The concentration gradient is the difference in concentrations divided by the distance between both sides: $$\frac{dC_N}{dx} = \frac{C_{\mathrm{N_2}} - C_{\mathrm{N_1}}}{1.5 \times 10^{-3}}$$ Here, \(x\) is the iron membrane thickness in meters.

03

Calculate the diffusion coefficient \(D\)

We can calculate the diffusion coefficient \(D\) using the provided values of \(D_0\) and \(Q_d\), and the temperature in Kelvin: $$D = D_{0} \exp\left(-\frac{Q_d}{RT}\right) = 3.0 \times 10^{-7} \exp\left(-\frac{76,150}{8.314 \times 573.15}\right)$$

04

Calculate the diffusion flux using Fick's first law

Finally, we can calculate the diffusion flux \(J\) using Fick's first law: $$J = -D \frac{dC_N}{dx} = -D \frac{C_{\mathrm{N_2}} - C_{\mathrm{N_1}}}{1.5 \times 10^{-3}}$$ Compute all the values above, and the obtained diffusion flux \(J\) will be the answer.

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