Chapter 11: Problem 65
In terms of the arrangement and freedom of motion of the molecules, how are the nematic liquid crystalline phase and an ordinary liquid phase similar? How are they different?
Chapter 11: Problem 65
In terms of the arrangement and freedom of motion of the molecules, how are the nematic liquid crystalline phase and an ordinary liquid phase similar? How are they different?
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Get started for freeIndicate whether each statement is true or false: (a) The critical pressure of a substance is the pressure at which it turns into a solid at room temperature. \((\mathbf{b})\) The critical temperature of a substance is the highest temperature at which the liquid phase can form. (c) Generally speaking, the higher the critical temperature of a substance, the lower its critical pressure. (d) In general, the more intermolecular forces there are in a substance, the higher its critical temperature and pressure.
(a) Would you expect the viscosity of isopropanol, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHOH},\) to be larger or smaller than the viscosity of ethanol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) ? (b) Would you expect the viscosity of isopropanol to be smaller or larger than the viscosity of 1-propanol, $\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{2} \mathrm{OH}$ ?
Which type of intermolecular force accounts for each of these differences? (a) Acetone, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO},\) boils at $56^{\circ} \mathrm{C} ;\( dimethyl sulfoxide or DMSO, (CH \)\left._{3}\right)_{2}$ SO, boils at \(189^{\circ}\) C. (b) \(\mathrm{CCl}_{4}\) is a liquid at atmospheric pressure and room temperature, whereas \(\mathrm{CH}_{4}\) is a gas under the same conditions. \((\mathbf{c})\) \(\mathrm{H}_{2} \mathrm{O}\) boils at \(100^{\circ} \mathrm{C}\) but \(\mathrm{H}_{2} \mathrm{~S}\) boils at \(-60^{\circ} \mathrm{C}\). (d) 1 -propanol boils at \(97^{\circ} \mathrm{C}\), whereas 2 -propanol boils at \(82.6^{\circ} \mathrm{C}\).
For a given substance, the liquid crystalline phase tends to be more viscous than the liquid phase. Why?
Suppose the vapor pressure of a substance is measured at two different temperatures. (a) By using the Clausius-Clapeyron equation (Equation 11.1) derive the following relationship between the vapor pressures, \(P_{1}\) and \(P_{2}\), and the absolute temperatures at which they were measured, \(T_{1}\) and \(T_{2}\) : $$ \ln \frac{P_{1}}{P_{2}}=-\frac{\Delta H_{\mathrm{vap}}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right) $$ (b) Gasoline is a mixture of hydrocarbons, a component of which is octane $\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\right)$. Octane has a vapor pressure of \(1.86 \mathrm{kPa}\) at \(25^{\circ} \mathrm{C}\) and a vapor pressure of \(19.3 \mathrm{kPa}\) at \(75^{\circ} \mathrm{C}\). Use these data and the equation in part (a) to calculate the heat of vaporization of octane. \((\mathbf{c})\) By using the equation in part (a) and the data given in part (b), calculate the normal boiling point of octane. Compare your answer to the one you obtained from Exercise 11.81 . (d) Calculate the vapor pressure of octane at \(-30^{\circ} \mathrm{C}\).
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