Following are data on the viscosity of liquid carbon tetrachloride:

T/°C 0 15 20 30 40 50 60

η/cP 1.329 1.038 0.969 0.843 0.739 0.651 0.585

The viscosity coefficient values are given in centipoise (cP). The poise is an older unit of viscosity, equal to $1g cm^{−1 } s^{−1}$ , so that $1 poise =0.1 kgm^{−1} s^{−1}$ . Find the values of $η_{0} and E_{aη}.$

Question

Following are data on the viscosity of liquid carbon tetrachloride:

T/°C	0	15	20	30	40	50	60
η/cP	1.329	1.038	0.969	0.843	0.739	0.651	0.585

The viscosity coefficient values are given in centipoise (cP). The poise is an older unit of viscosity, equal to [latex]1g cm^{−1 } s^{−1}[/latex], so that [latex]1 poise =0.1 kgm^{−1} s^{−1}[/latex]. Find the values of [/latex]η_{0} and E_{aη}.[/latex]

Accepted Answer

We linearize [latex]Eq. (10.4-6): η=η_{0}e^{Ea,η/RT}[/latex] by taking logarithms:

[latex]\ln (η/η_{0})=\frac{E_{aη}}{RT} [/latex]

A linear least squares fit of ln (η/1cP) to 1/T gives an intercept of −4.279 and a slope of 1245 K, with a correlation coefficient of 0.9998 (a good fit).

[latex]E_{aη}=(slope)×R=(1245K) \bigl(8.3145 J K^{−1} mol^{−1}\bigr)[/latex]

[latex]=1.035×10^{4} J mol^{−1}=10.4 kJ mol^{−1}[/latex]

[latex]η_{0}=(1 cP)e^{−4.279}=1.39×10^{-2} cP=1.39×10^{−5} kgm^{−1} s^{−1}[/latex]

The value of this energy of activation is typical of liquids with small molecules. As might be expected, it is somewhat smaller than the energy of vaporization of [latex]CCl_{4}, 33.9 kJ mol^{−1}[/latex], which corresponds to removing a molecule from all of its nearest neighbors.

Question 10.17: Following are data on the viscosity of liquid carbon tetrach...

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