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?

Short Answer

Expert verified
In terms of the arrangement and freedom of motion of molecules, nematic liquid crystalline phase and ordinary liquid phase share the similarity of molecular mobility and fluidity. Both phases exhibit motion and fluidity as molecules slide past each other. However, they differ in molecular arrangement, shape, and degree of freedom. In nematic liquid crystals, molecules are partially ordered with a preferred orientation and are usually elongated in shape, resulting in restricted motion. In ordinary liquids, there is no specific order in arrangement, and molecules have various shapes and sizes, which leads to a higher degree of freedom in their motion.

Step by step solution

01

Describe the nematic liquid crystalline phase

The nematic liquid crystalline phase is a unique phase of matter that exists between the solid and liquid phases. In this phase, molecules are partially ordered, meaning that they have a preferred orientation (or direction) but still exhibit fluid-like properties. The molecules in nematic liquid crystals are usually elongated and rod-like in shape, which allows them to align with one another.
02

Describe the ordinary liquid phase

In an ordinary liquid phase, molecules have little to no order in their arrangement, and they are free to move around and slide past one another. The forces of attraction between the molecules are weak, which allows for a significant degree of molecular motion and fluidity.
03

Compare the arrangement and freedom of motion in both phases

Similarities: 1. Mobility: In both nematic liquid crystalline phases and ordinary liquid phases, the molecules retain a significant degree of motion and fluidity. They can flow and slide past each other, which gives both phases their distinct liquid-like properties. Differences: 1. Order: The arrangement of molecules in nematic liquid crystals is partially ordered, as they have a preferred orientation and tend to align with one another. In ordinary liquids, there is little to no order in the arrangement of molecules. 2. Molecular shape: The molecules in nematic liquid crystals are usually elongated and rod-like in shape, which contributes to their alignment and partial order. In ordinary liquids, the molecules can be of various shapes and sizes, which leads to their random arrangement. 3. Degree of motion: The motion of molecules in nematic liquid crystals is more restricted compared to the motion in ordinary liquids. In nematic phases, the elongated molecules can rotate and slide past each other while maintaining their preferred orientation. While in ordinary liquids, the molecules have a greater range of motion since they are not restricted to any specific alignment. In conclusion, the nematic liquid crystalline phase and an ordinary liquid phase have some similarities in terms of molecular mobility and fluidity. However, they differ in their molecular arrangement, shape, and the degree of freedom that the molecules have in their motion.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Indicate 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 \(-30^{\circ} \mathrm{C}\).

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free