Chapter 11.13, Problem 102AAP

To determine

The viscosity of the soda-lime glass at the temperature of $675°\text{C}$.

Answer to Problem 102AAP

The viscosity of the soda-lime glass at the temperature of $675°\text{C}$ is ${10}^{11.33}\text{P}$.

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 $T$.

Conclusion:

At the temperature of $560°\text{C}$, write the equation involving the viscosity to temperature for a viscous flow in the glass.

  ${\eta }_1={\eta }_0{e}^{Q/R{T}_1}$                                                                                                          (II)

Here, viscosity of the soda-lime glass at the temperature of $560°\text{C}$ is ${\eta }_1$ and initial temperature of the soda-lime glass is $T_1$.

At the temperature of $675°\text{C}$, write the equation involving the viscosity to temperature for a viscous flow in the glass.

  ${\eta }_2={\eta }_0{e}^{Q/R{T}_2}$                                                                                                        (III)

Here, viscosity of the soda-lime glass at the temperature of $675°\text{C}$ is ${\eta }_2$ and final temperature of the soda-lime glass is $T_2$.

Divide Equation (II) by Equation (III).

  $\begin{array}{c}\frac{{\eta }_1}{{\eta }_2}=\frac{{\eta }_0{e}^{Q/R{T}_1}}{{\eta }_0{e}^{Q/R{T}_2}}\\ =\frac{e^{Q/R{T}_1}}{e^{Q/R{T}_2}}\\ =\exp \left[\frac{Q}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)\right]\end{array}$                                                                                  (IV)

Substitute ${10}^{14.6}\text{P}$ for ${\eta }_1$, $430\text{kJ}/\text{mol}$ for $Q$, $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$, $560°\text{C}$ for $T_1$ and $675°\text{C}$ for $T_2$ in Equation (IV).

 $\begin{array}{l}\frac{{10}^{14.6}\text{P}}{{\eta }_2}=\exp \left[\frac{430\text{kJ}/\text{mol}}{8.314\text{J}/\text{mol}\cdot \text{K}}\left(\frac{1}{560°\text{C}}-\frac{1}{675°\text{C}}\right)\right]\\ \frac{{10}^{14.6}\text{P}}{{\eta }_2}=\exp \left[\frac{430\times {10}^3\text{J}/\text{mol}}{8.314\text{J}/\text{mol}\cdot \text{K}}\left(\frac{1}{833\text{K}}-\frac{1}{948\text{K}}\right)\right]\\ \frac{{10}^{14.6}\text{P}}{{\eta }_2}=\exp \left(7.532\right)\\ \frac{{10}^{14.6}\text{P}}{{\eta }_2}=1866.6\end{array}$

 $\begin{array}{c}{\eta }_2=\frac{{10}^{14.6}\text{P}}{1866.6}\\ =2.133\times {10}^{11}\text{P}\\ ={10}^{11.329}\text{P}\\ ={10}^{11.33}\text{P}\end{array}$

Thus, the viscosity of the soda-lime glass at the temperature of $675°\text{C}$ is ${10}^{11.33}\text{P}$.