Chapter 11.13, Problem 104AAP

To determine

The value for the activation energy for viscous flow.

Answer to Problem 104AAP

The value for the activation energy for viscous flow is $457.3\text{kJ}/\text{mol}$.

Explanation of Solution

Write the equation involving the viscosity to temperature for a viscous flow in the glass.

  $\eta ={\eta }_0{e}^{Q/RT}$                                                                                                            (I)

Here, viscosity of the glass is $\eta$, pre-exponential constant is ${\eta }_0$,molar activation energy for viscous flow is $Q$, universal gas constant is $R$ and absolute temperature is $T$.

For the annealing point, write the equation involving the viscosity to temperature for a viscous flow in the glass.

  ${\eta }_{ap}={\eta }_0{e}^{Q/R{T}_{ap}}$                                                                                                       (II)

Here, viscosity of the glass at strain point is ${\eta }_{ap}$ and annealing point temperature is $T_{ap}$.

For the softening point, write the equation involving the viscosity to temperature for a viscous flow in the glass.

  ${\eta }_{sp}={\eta }_0{e}^{Q/R{T}_{sp}}$                                                                                                      (III)

Here, viscosity of the glass at softening point is ${\eta }_{sp}$ and softening point temperature is $T_{sp}$.

Conclusion:

Convert the unit of annealing point temperature to absolute value.

 $\begin{array}{c}{T}_{ap}=600°\text{C}\\ =600+273\text{K}\\ =873\text{K}\end{array}$

Convert the unit of softening point temperature to absolute value.

 $\begin{array}{c}{T}_{sp}=800°\text{C}\\ =800+273\text{K}\\ =1073\text{K}\end{array}$

Divide Equation (II) by Equation (III).

  $\begin{array}{c}\frac{{\eta }_{ap}}{{\eta }_{sp}}=\frac{{\eta }_0{e}^{Q/R{T}_{ap}}}{{\eta }_0{e}^{Q/R{T}_{sp}}}\\ =\frac{e^{Q/R{T}_{ap}}}{e^{Q/R{T}_{sp}}}\\ =\exp \left[\frac{Q}{R}\left(\frac{1}{T_{ap}}-\frac{1}{T_{sp}}\right)\right]\end{array}$                                                                                  (IV)

Substitute ${10}^{12.5}\text{P}$ for ${\eta }_{ap}$, ${10}^{7.4}\text{P}$ for ${\eta }_{sp}$, $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$, $873\text{K}$ for $T_{ap}$ and $1073\text{K}$ for $T_{sp}$ in Equation (IV).

 $\begin{array}{l}\frac{{10}^{12.5}\text{P}}{{10}^{7.4}\text{P}}=\exp \left[\frac{Q}{8.314\text{J}/\text{mol}\cdot \text{K}}\left(\frac{1}{873\text{K}}-\frac{1}{1073\text{K}}\right)\right]\\ {10}^{5.1}=\exp \left(\left(2.568\times {10}^{-5}\right)Q\right)\\ \ln {10}^{5.1}=\left(2.568\times {10}^{-5}\right)Q\end{array}$

 $\begin{array}{c}Q=4.573\times {10}^5\text{J}/\text{mol}\\ =457.3\text{kJ}/\text{mol}\end{array}$

Thus, the value for the activation energy for viscous flow is $457.3\text{kJ}/\text{mol}$.