Kinetics of Materials

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Eq. 3.100

$\displaystyle c_1^A(T_2) = c_1^A(T_1)  
\exp \left[ \frac{-84000 \left(T_1 - T_2 \right)}{N_\circ k T_1 T_2} \right]
$

Eq. 3.101

$\displaystyle \Delta c_1 = c_1(T_1)   \left\{ \exp \left[ \frac{-84000 \left(T_1 - T_2
\right)}{N_\circ k T_1 T_2} \right] - 1 \right\}
$

Eq. 3.104

$\displaystyle J_1 = -\frac{D_1c_1}{kT^2} \left( \Delta H + Q^\mathrm{trans} \right)
\FD {T}{x}
$

Text after Eq. 3.104

Because $ \left( \Delta H + Q^\mathrm{trans} \right)$ is positive, the C atoms will be swept toward the cold end, as observed.

Eq. 3.105

$\displaystyle \vec F_1 = - \Omega_1   A   \nabla \left[ \frac{y'}{x'^2 + y'^2}
\right]
$

Eq. 3.106

$\displaystyle \vec F_1 = - \Omega_1   A   \nabla \left[
\frac{y}{x^2 + 2 x R + R^2 + y^2} \right]
$




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Kinetics of Materials 2006-01-04