Question

Calculate the value of D at 575°C for the diffusion of some species in a metal for which the values of D0 and Qd are 4.5 × 10-5 m2/s and 160 kJ/mol, respectively.

Answer 1

Answer:

The value of D = 4.4 × $10^{-5}$

Explanation:

Temperature T = 575 °c = 848 K

$D_{0}$ = 4.5 × $10^{-5}$ $\frac{m^{2} }{sec}$

$Q_{d}$ = 160 $\frac{KJ}{mol}$

The value of $D_{0}$ is given by the formula

$D_{0}$ = D exp   $( \frac{Q_{d} }{RT} )$

Put all the values in above formula we get,

4.5 × $10^{-5}$ = D exp $\frac{160}{7050.272}$

4.5 × $10^{-5}$ = D × 1.023

D = 4.4 × $10^{-5}$

This is the value of D.