Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

It often happens that a substance possessing a smectic liquid crystalline phase just above the melting point passes into a nematic liquid crystalline phase at a higher temperature. Account for this type of behavior.

Short Answer

Expert verified
In summary, a substance transitions from a smectic liquid crystalline phase to a nematic liquid crystalline phase as the temperature increases because the added thermal energy disrupts the positional order within the layers present in the smectic phase. As the order breaks down, the substance loses its layered structure and transitions to the nematic phase, where molecules maintain a preferred orientation without the well-defined layers or positional order.

Step by step solution

01

1. Understand the smectic liquid crystalline phase

A smectic liquid crystal phase is a phase in which the molecules of a substance align themselves into layers (with a well-defined average spacing between them), while maintaining a certain degree of positional order within the layers. This means that the molecules are organized in a more ordered fashion compared to the isotropic liquid phase, but they still possess some liquid-like behavior, which allows for some degree of translation and rotation.
02

2. Understand the nematic liquid crystalline phase

A nematic liquid crystal phase is another type of liquid crystal phase in which the molecules of a substance align themselves with the same average orientation but without any layering or a well-defined positional order as in the smectic phase. This means that the nematic phase is more disordered than the smectic phase in terms of positional order, but it still maintains a preferred orientation for the molecules.
03

3. Effects of temperature on liquid crystalline phases

The behavior and properties of liquid crystalline phases can be affected by temperature changes. Ab increasing temperature adds thermal energy to the system, which can disrupt the order present in the smectic and nematic phases. This additional energy can cause the molecules in the system to become more disordered and exhibit more liquid-like behavior.
04

4. Transition from smectic to nematic phase as temperature increases

When the temperature of a substance with a smectic liquid crystalline phase increases, the thermal energy introduced into the system can disrupt the positional order of the layers present in the smectic phase. As the positional order within the layers breaks down, the substance starts losing its layered structure and might transition to a nematic phase, in which the molecules still maintain a preferred orientation, but without the positional order associated with the layers. This marks the transition from the smectic to the nematic phase as the temperature increases.
05

5. Conclusion

In summary, the behavior of a substance transitioning from a smectic liquid crystalline phase to a nematic liquid crystalline phase as the temperature increases can be explained by the disruption of the positional order within the layers present in the smectic phase due to an increase in thermal energy. As this positional order breaks down, the substance loses its layered structure and transitions to a nematic phase, where the molecules only maintain an average orientation without the well-defined layers or positional order.

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

Acetone \(\left[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}\right]\) is widely used as an industrial solvent. (a) Draw the Lewis structure for the acetone molecule and predict the geometry around each carbon atom. (b) Is the acetone molecule polar or nonpolar? (c) What kinds of intermolecular attractive forces exist between acetone molecules? (d) 1-Propanol \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\right)\) has a molecular weight that is very similar to that of acetone, yet acetone boils at \(56.5^{\circ} \mathrm{C}\) and 1 -propanol boils at \(97.2^{\circ} \mathrm{C}\). Explain the difference.

The table shown here lists the molar heats of vaporization for several organic compounds. Use specific examples from this list to illustrate how the heat of vaporization varies with (a) molar mass, (b) molecular shape, (c) molecular polarity, (d) hydrogen-bonding interactions. Explain these comparisons in terms of the nature of the intermolecular forces at work. (You may find it helpful to draw out the structural formula for each compound.)

One of the attractive features of ionic liquids is their low vapor pressure, which in turn tends to make them nonflammable. Why do you think ionic liquids have lower vapor pressures than most room-temperature molecular liquids?

True or false: (a) \(\mathrm{CBr}_{4}\) is more volatile than \(\mathrm{CCl}_{4}\). (b) \(\mathrm{CBr}_{4}\) has a higher boiling point than \(\mathrm{CCl}_{4}\) - (c) \(\mathrm{CBr}_{4}\) has weaker intermolecular forces than \(\mathrm{CCl}_{4}\) - (d) \(\mathrm{CBr}_{4}\) has a higher vapor pressure at the same temperature than \(\mathrm{CCl}_{4}\).

Explain the following observations: (a) Water evaporates more quickly on a hot, dry day than on a hot, humid day. (b) It takes longer to cook an egg in boiling water at high altitudes than it does at lower altitudes.

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