Thermodynamics, 5th Edition

(7.3.9c)

To satisfy the inequality (7.3.8), the two integrals on the right must be so formulated that the integrands remain positive. These two integrals then are required to have a form that may be evaluated by standard methods to yield

$I_1=(D/(V_g-V_c)){\int }_{V_c}^{V_g}{(V-V_c)}^{\delta }dV=D{(V_g-V_c)}^{\delta }/(1+\delta )$ (7.3.10a)

(7.3.10a)

and

$I_2=(D/(V_c-V_l)){\int }_{V_l}^{V_c}{(V_c-V)}^{\delta }dV=D{(V_c-V_l)}^{\delta }/(1+\delta ).$ (7.3.10b)

(7.3.10b)

Then,

$I_1+I_2=\frac{D}{1+\delta }[{(V_g-V_c)}^{\delta }+{(V_c-V_l)}^{\delta }]=\frac{2D{B}^{\delta }{T}_c^{\beta \delta }}{1+\delta }{\left(1-\frac{T}{T_c}\right)}^{\beta \delta }.$ (7.3.11)

(7.3.11) ...