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).

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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.

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Most popular questions from this chapter

Propanol \(\left(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}\right)\) melts at \(-126.5^{\circ} \mathrm{C}\) and boils at \(97.4{ }^{\circ} \mathrm{C}\). Draw a qualitative sketch of how the entropy changes as propanol vapor at \(150^{\circ} \mathrm{C}\) and 1 atm is cooled to solid propanol at \(-150^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\).

Trouton's rule states that for many liquids at their normal boiling points, the standard molar entropy of vaporization is about $88 \mathrm{~J} / \mathrm{mol}-\mathrm{K}$. (a) Estimate the normal boiling point of bromine, \(\mathrm{Br}_{2}\), by determining \(\Delta H_{\text {vap }}^{t}\) for \(\mathrm{Br}_{2}\) using data from Appendix C. Assume that $\Delta H_{\text {ap }}^{a}$ remains constant with temperature and that Trouton's rule holds. (b) Look hal boiling point of Br\(r_{2}\) in a chemistry handbook or at the WebElements Web site (www,webelements. com) and compare it to your calculation. What are the possible sources of error, or incorrect assumptions, in the calculation?

(a) What is the meaning of the standard free-energy change, \(\Delta G^{\circ},\) as compared with \(\Delta G\) ? (b) For any process that occurs at constant temperature and pressure, what is the significance of \(\Delta G=0 ?(c)\) For a certain process, \(\Delta G\) is large and negative. Does this mean that the process necessarily occurs rapidly?

A system goes from state 1 to state 2 and back to state 1 . (a) What is the relationship between the value of \(\Delta E\) for going from state 1 to state 2 to that for going from state 2 back to state \(1 ?\) (b) Without further information, can you conclude anything about the amount of heat transferred to the system as it goes from state 1 to state 2 as compared to that upon going from state 2 back to state \(1 ?(\mathrm{c})\) Suppose the changes in state are reversible processes. Can you conclude anything about the work done by the system upon going from state 1 to state 2 as compared to that upon going from state 2 back to state \(1 ?\)

The relationship between the temperature of a reaction, its standard enthalpy change, and the equilibrium constant at that temperature can be expressed as the following linear equation: $$ \ln K=\frac{-\Delta H^{\circ}}{R T}+\text { constant } $$ (a) Explain how this equation can be used to determine \(\Delta H^{\circ}\) experimentally from the equilibrium constants at several different temperatures. (b) Derive the preceding equation using relationships given in this chapter. To what is the constant equal?

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