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

For each of the following pairs, choose the substance with the higher entropy per mole at a given temperature: (a) \(\operatorname{Ar}(l)\) or \(\mathrm{Ar}(g),\) (b) \(\mathrm{He}(g)\) at 3 atm pressure or \(\mathrm{He}(g)\) at 1.5 atm pressure, (c) \(1 \mathrm{~mol}\) of \(\mathrm{Ne}(g)\) in \(15.0 \mathrm{~L}\) or \(1 \mathrm{~mol}\) of \(\mathrm{Ne}(g)\) in \(1.50 \mathrm{~L}\), (d) \(\mathrm{CO}_{2}(g)\) or \(\mathrm{CO}_{2}(s)\).

Short Answer

Expert verified
(a) Ar(g) is higher, (b) He(g) at 1.5 atm is higher, (c) 1 mole of Ne(g) in 15 L is higher, and (d) CO2(g) is higher.

Step by step solution

01

Pair (a) Comparison

In pair (a), we compare Ar in liquid state and Ar in gaseous state. Gases are generally in a more disorderly state than liquids due to their molecules' higher amount of kinetic energy and freedom of motion. Therefore, Ar(g) has a higher entropy than Ar(l) per mole at the same temperature.
02

Pair (b) Comparison

For pair (b), both substances are gaseous helium, but at different pressures. Higher pressure means that the molecules are packed more tightly, so the randomness decreases. Therefore, He(g) at 1.5 atm has a higher entropy per mole at the same temperature than He(g) at 3 atm.
03

Pair (c) Comparison

In pair (c), we compare 1 mole of Ne gas in two different volumes. Gas in a larger volume has more space to move around, so its randomness or disorder increases. Therefore, 1 mole of Ne(g) in 15 L has a higher entropy per mole at the same temperature than 1 mole of Ne(g) in 1.50 L.
04

Pair (d) Comparison

Finally, in pair (d), we compare the entropy of CO2 in gaseous and solid states at the same temperature. Gases, due to their higher kinetic energy and freedom of motion, are more disorderly than solids. Therefore, CO2(g) has a higher entropy per mole at the same temperature than CO2(s). In conclusion, based on the concept of entropy as a measure of randomness or disorder, we can compare different substances and conditions: (a) Ar(g) has a higher entropy per mole at the same temperature than Ar(l). (b) He(g) at 1.5 atm has a higher entropy per mole at the same temperature than He(g) at 3 atm. (c) 1 mole of Ne(g) in 15 L has a higher entropy per mole at the same temperature than 1 mole of Ne(g) in 1.50 L. (d) CO2(g) has a higher entropy per mole at the same temperature than CO2(s).

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Entropy and Temperature
Entropy, often symbolized by the letter 'S', is a fundamental concept in chemistry that measures the level of disorder or randomness in a system. Understanding the relationship between entropy and temperature can give insight into the behavior of substances under different thermal conditions.

As temperature increases, the kinetic energy of the particles within a substance also increases, promoting greater movement and spreading out of particles. This enhanced movement results in a higher level of disorder, leading to an increase in entropy. For example, when a liquid like argon (Ar(l)) transitions to its gaseous form (Ar(g)), this change occurs at a temperature where the added thermal energy allows the argon atoms to move around more freely, hence increasing its entropy.

When comparing substances at the same temperature, one must consider their phase. Gases typically exhibit higher entropy than liquids and solids because their particles can move freely and spread out, creating a higher degree of randomness. This is why, at the same temperature, Ar(g) possesses a higher entropy than Ar(l).
Gas Entropy and Pressure
The entropy of a gas is also greatly influenced by pressure. Since gases are compressible, changing the pressure can alter their volume and therefore their entropy. When a gas is compressed to a higher pressure, the molecules are forced closer together, reducing the randomness of their movement and hence decreasing the entropy.

Conversely, if the pressure is decreased, the gas expands, the molecules have more space to move randomly, and thus, the entropy increases. This explains why helium gas (He(g)) at a lower pressure of 1.5 atm has a higher entropy than at a higher pressure of 3 atm, at the same temperature. The molecules at lower pressure are less confined and have more available microstates, increasing the system's disorder.
Molar Entropy and Volume
The concept of molar entropy is deeply connected to the volume occupied by a substance, particularly gases. Molar entropy represents the entropy of one mole of a substance, and for gases, an increase in volume provides more accessible microstates for the particles to occupy, driving up the entropy.

For instance, when comparing neon gas (Ne(g)) in a 15.0 L volume to the same amount in a 1.50 L volume, the larger volume allows for more particle positions and random movements. The Neon gas has more space in which to distribute itself, which leads to a greater number of possible microstates and therefore a higher molar entropy in the 15.0 L volume.
States of Matter and Entropy
The state of matter—solid, liquid, or gas—affects the entropy of a substance because it determines the degree of particle movement and arrangement. Solids, with their structured lattice and limited particle movement, exhibit the lowest entropy. Liquids, with some degree of fluidity, have higher entropy than solids. Gases, which are not constrained by a lattice and have the most freedom to move, have the highest entropy among the three states.

Thus, when comparing carbon dioxide as a solid (CO2(s)) and as a gas (CO2(g)), the gaseous form manifests significantly higher entropy due to the greater level of particle disorder. In a solid state, the structured arrangement of particles constrains movement and limits the number of accessible microstates, leading to lower entropy.

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

The following data compare the standard enthalpies and free energies of formation of some crystalline ionic substances and aqueous solutions of the substances: $$ \begin{array}{lrr} \text { Substance } & \Delta \boldsymbol{H}_{f}^{\circ}(\mathbf{k} \mathbf{J} / \mathbf{m o l}) & \Delta \mathbf{G}_{f}^{\circ}(\mathbf{k J} / \mathbf{m o l}) \\ \hline \mathrm{AgNO}_{3}(s) & -124.4 & -33.4 \\ \mathrm{AgNO}_{3}(a q) & -101.7 & -34.2 \\ \mathrm{MgSO}_{4}(s) & -1283.7 & -1169.6 \\ \mathrm{MgSO}_{4}(a q) & -1374.8 & -1198.4 \end{array} $$ (a) Write the formation reaction for \(\mathrm{AgNO}_{3}(s) .\) Based on this reaction, do you expect the entropy of the system to increase or decrease upon the formation of \(\mathrm{AgNO}_{3}(s) ?\) (b) Use \(\Delta H_{f}^{\circ}\) and \(\Delta G_{f}^{\circ}\) of \(\mathrm{AgNO}_{3}(s)\) to determine the entropy change upon formation of the substance. Is your answer consistent with your reasoning in part (a)? (c) Is dissolving \(\mathrm{AgNO}_{3}\) in water an exothermic or endothermic process? What about dissolving \(\mathrm{MgSO}_{4}\) in water? (d) For both \(\mathrm{AgNO}_{3}\) and \(\mathrm{MgSO}_{4},\) use the data to calculate the entropy change when the solid is dissolved in water. (e) Discuss the results from part (d) with reference to material presented in this chapter and in the "A Closer Look" box on page 540 .

Predict the sign of \(\Delta S_{\text {sys }}\) for each of the following processes: (a) Molten gold solidifies. (b) Gaseous \(\mathrm{Cl}_{2}\) dissociates in the stratosphere to form gaseous \(\mathrm{Cl}\) atoms. (c) Gaseous CO reacts with gaseous \(\mathrm{H}_{2}\) to form liquid methanol, \(\mathrm{CH}_{3} \mathrm{OH}\). (d) Calcium phosphate precipitates upon mixing \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(a q)\) and \(\left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4}(a q)\)

Suppose we vaporize a mole of liquid water at \(25^{\circ} \mathrm{C}\) and another mole of water at \(100{ }^{\circ} \mathrm{C}\). (a) Assuming that the enthalpy of vaporization of water does not change much between \(25^{\circ} \mathrm{C}\) and \(100^{\circ} \mathrm{C},\) which process involves the larger change in entropy? (b) Does the entropy change in either process depend on whether we carry out the process reversibly or not? Explain.

(a) What is meant by calling a process irreversible? (b) After a particular irreversible process, the system is restored to its original state. What can be said about the condition of the surroundings after the system is restored to its original state? (c) Under what conditions will the condensation of a liquid be an irreversible process?

The standard entropies at \(298 \mathrm{~K}\) for certain of the group \(4 \mathrm{~A}\) elements are as follows: \(\mathrm{C}(s,\) diamond \()=2.43 \mathrm{~J} / \mathrm{mol}-\mathrm{K},\) \(\mathrm{Si}(s)=18.81 \mathrm{~J} / \mathrm{mol}-\mathrm{K}, \mathrm{Ge}(s)=31.09 \mathrm{~J} / \mathrm{mol}-\mathrm{K},\) and \(\operatorname{Sn}(s)=\) \(51.818 \mathrm{~J} / \mathrm{mol}-\mathrm{K} .\) All but \(\mathrm{Sn}\) have the diamond structure. How do you account for the trend in the \(S^{\circ}\) values?

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