HardChemistryHigh School (11–12)States of MatterNumericalJEE Advanced AP A-Level IB

The viscosity of water is 1.002 mPa s at 20 °C and 0.355 mPa s at 80 °C…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

The viscosity of water is $1.002\ \text{mPa s}$ at $20\ ^\circ\text{C}$ and $0.355\ \text{mPa s}$ at $80\ ^\circ\text{C}$ . Assume $\eta = Ae^{E/RT}$ . (a) Find the activation energy $E$ of viscous flow. (b) Predict the viscosity at $50\ ^\circ\text{C}$ and at $100\ ^\circ\text{C}$ . (c) The measured values are $0.547$ and $0.282\ \text{mPa s}$ . By what percentage is each prediction off?

Need a reminder? The key ideaShow

Poiseuille: $Q = \dfrac{\pi\Delta Pr^4}{8\eta L}$ . Stokes: a steady speed $v = \dfrac{2r^2(\rho_s - \rho)g}{9\eta}$ , valid when $Re = \dfrac{\rho vd}{\eta}$ is below about $1$ . Ostwald viscometer: $\dfrac{\eta_1}{\eta_2} = \dfrac{\rho_1t_1}{\rho_2t_2}$ . For a liquid, $\eta = Ae^{E/RT}$ , which falls as $T$ rises.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Given. $T_1 = 293.15\ \text{K}$ with $\eta_1 = 1.002\ \text{mPa s}$ , and $T_2 = 353.15\ \text{K}$ with $\eta_2 = 0.355\ \text{mPa s}$ .

Step 1. The activation energy. $E = \dfrac{R\ln(\eta_1/\eta_2)}{1/T_1 - 1/T_2} = \dfrac{8.314 \times \ln(1.002/0.355)}{3.4112 \times 10^{-3} - 2.8316 \times 10^{-3}} = 14\,890\ \text{J/mol} = 14.9\ \text{kJ/mol}$ .

Step 2. At $50\ ^\circ\text{C}$ . $T = 323.15\ \text{K}$ , and $\dfrac{E}{R} = \dfrac{14\,890}{8.314} = 1791\ \text{K}$ .

\displaystyle \ln\dfrac{\eta}{\eta_1} = 1791 \times (0.0030945 - 0.0034112) = -0.567

So $\eta = 1.002 \times e^{-0.567} = 0.568\ \text{mPa s}$ .

Step 3. At $100\ ^\circ\text{C}$ . $T = 373.15\ \text{K}$ , and the same formula gives $0.271\ \text{mPa s}$ .

Step 4. The errors. At $50\ ^\circ\text{C}$ : $\dfrac{0.568 - 0.547}{0.547} = +3.8\%$ . At $100\ ^\circ\text{C}$ : $\dfrac{0.271 - 0.282}{0.282} = -3.9\%$ .

Check. Predictions that are within about $4\%$ over a range of $80\ \text{K}$ show that the Arrhenius form is a good model for the viscosity of water.

Answer: $14.9\ \text{kJ/mol}$ , $0.568\ \text{mPa s}$ at $50\ ^\circ\text{C}$ and $0.271\ \text{mPa s}$ at $100\ ^\circ\text{C}$ .

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