Chapter 9, Problem 9.1P

To determine

The enthalpy of activation for viscous flow in soda lime glass in the temperature range 700K to 800K.

Answer to Problem 9.1P

The enthalpy of activation for viscous flow in soda lime glass in the temperature range 700K to 800K is $4.19\text{}\text{ eV/atom}$ .

Explanation of Solution

Given:

Temperature range is $700\text{ K}$ to $800\text{ K}$ .

Concept used:

Write the expression for viscosity as function of temperature.

  $\eta \left(T\right)={\eta }_0\exp \left(\frac{\Delta H_p}{kT}\right)$ …… (1)

Here, $\eta \left(T\right)$ is the viscosity as function of temperature, ${\eta }_0$ is the coefficient of viscosity, $\Delta H_p$ is the activation enthalpy, $k$ is the Boltzmann constant and $T$ is the temperature.

Calculation:

Refer to Figure 9.3 “The temperature dependence of the viscosity of silica glass $\left({\text{SiO}}_{\text{2}}\right)$ soda-lime glass, and boron oxide $\left({\text{B}}_{\text{2}}{\text{O}}_{\text{3}}\right)$ ” to obtain the viscosity of soda lime glass at $700\text{ K}$ as ${10}^{12.9}\text{ Pa}\cdot \text{s}$ and at $800\text{ K}$ as ${10}^{9.1}\text{ Pa}\cdot \text{s}$ .

Substitute $700\text{ K}$ for $T$ and ${10}^{12.9}\text{ Pa}\cdot \text{s}$ for $\eta \left(T\right)$ in equation (1).

  ${10}^{12.9}\text{ Pa}\cdot \text{s}={\eta }_0\exp \left(\frac{\Delta H_p}{k\left(700\text{K}\right)}\right)$ …… (2)

Substitute $800\text{ K}$ for $T$ and ${10}^{9.1}\text{ Pa}\cdot \text{s}$ for $\eta \left(T\right)$ in equation (1).

  ${10}^{9.1}\text{ Pa}\cdot \text{s}={\eta }_0\exp \left(\frac{\Delta H_p}{k\left(800\text{K}\right)}\right)$ …… (3)

Divide equation (2) by equation (3).

  $\begin{array}{c}\frac{{10}^{12.9}\text{ Pa}\cdot \text{s}}{{10}^{9.1}\text{ Pa}\cdot \text{s}}=\frac{{\eta }_0\exp \left(\frac{\Delta H_p}{k\left(700\text{K}\right)}\right)}{{\eta }_0\exp \left(\frac{\Delta H_p}{k\left(800\text{K}\right)}\right)}\\ 6.3\times {10}^3=\frac{\exp \left(\frac{\Delta H_p}{k}\left(1.43\times {10}^{-3}\right)\right)}{\exp \left(\frac{\Delta H_p}{k}\left(1.25\times {10}^{-3}\right)\right)}\\ 6.3\times {10}^3=\exp \left(\frac{\Delta H_p}{k}\left(1.43\times {10}^{-3}-1.25\times {10}^{-3}\right)\right)\\ 6.3\times {10}^3=\exp \left(\frac{\Delta H_p}{k}\left(0.18\times {10}^{-3}\right)\right)\end{array}$

Take logarithm on both sides.

  $\ln \left(6.3\times {10}^3\right)=\frac{\Delta H_p}{k}\left(0.18\times {10}^{-3}\right)$

Substitute $8.62\times {10}^{-5}\text{ eV/atom}\cdot \text{K}$ for $k$ in above expression.

  $\begin{array}{c}8.74=\frac{\Delta H_p}{8.62\times {10}^{-5}}\left(0.18\times {10}^{-3}\right)\\ \Delta H_p=\frac{8.74\left(8.62\times {10}^{-5}\right)}{\left(0.18\times {10}^{-3}\right)}\\ \Delta H_p=4.19\text{ eV/atom}\end{array}$

Conclusion:

Thus, the enthalpy of activation for viscous flow in soda lime glass in the temperature range 700K to 800K is $4.19\text{}\text{ eV/atom}$ .