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Briefly explain the concept of steady state as it applies to diffusion. Briefly

ID: 976174 • Letter: B

Question

Briefly explain the concept of steady state as it applies to diffusion. Briefly explain the concept of a driving force. What is the driving force for steady-state difussion? The purification of hydrogen gas by diffusion through a palladium sheet was discussed in Section 5.3. Compute the number of kilograms of hydrogen that pass per hour through a 5-mm-thick sheet of palladium having an area of 0.25 m^2 500degree C. Assume a diffusion coefficient of 1.0 times 10^-8 m^2/s, that the concentrations at the high- and low-pressure sides of the plate are 2.4 and 0.6 kg of hydrogen per cubic meter of palladium, and that steady-state conditions have been attained. A sheet of steel 1.8 mm thick has nitrogen atmospheres on both sides at 1200degree C and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient for nitrogen in steel at this temperature is 6 times 10^-11 m^2/s, and the diffusion flux is found to be 1.2 times 10^-7 kg/m^2 middot s. Also, it is known that the concentration of nitrogen in the steel at the high-pressure surface is 4 kg/m^3. How far into the sheet from this high-pressure side will the concentration be 2.0 kg/m^3? Assume a linear concentration profile.

Explanation / Answer

Q.5.4: A steady state in diffusion is defined as the situation in which the rate of diffusion into a given system is exactly equal to the rate of diffusion out of the system, such that there is neither deposition nor depletion of the diffusing species inside the system.

Hence in case of steady state diffusion, the flux is independent of time.

Q.5.5: (a):The force that brings about the reaction to occur is called driving force. Hence a driving force initiates a reaction.

(b): Concentration gradient is the driving force for steady state diffusion.

Given the diffusion coefficient, Dc = 1.0x10-8 m2/s

Area, A = 0.25 m2

time, t = 3600 s/hour

dC = (0.6 - 2.4) Kg/m3 =

dx = 5 mm = 5.0x10-3 m

Kg of hydrogen passed per hour can be calculated as

M = - DcAxtx(dD / dx)

=> M = - 1.0x10-8 m2/s x 0.25 m2 x (3600 s/hour) x ( - 1.8 Kg/m3) /  5.0x10-3 m

=> M = 3.24x10-3 Kg/m3 (answer)

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