(a) What sign for \(\Delta S\) do you expect when the volume of 0.200 mol of an ideal gas at \(27^{\circ} \mathrm{C}\) is increased isothermally from an initial volume of \(10.0 \mathrm{~L} ?(\mathbf{b})\) If the final volume is 18.5 L, calculate the entropy change for the process. (c) Do you need to specify the temperature to calculate the entropy change? Explain.

Short Answer

Expert verified
The expected sign of \(\Delta S\) is positive since the gas is expanding at a constant temperature, increasing its randomness. The entropy change for the process can be calculated using the formula \(\Delta S = nR \ln \frac{V_f}{V_i}\), giving \(\Delta S \approx 0.516 \,\text{J/K}\). While the temperature does not directly appear in the formula for the entropy change, it is necessary to specify that the process is isothermal (constant temperature) for the formula's validity. However, the actual value of the temperature doesn't directly affect the result in this case.

Step by step solution

01

Determine the expected sign of \(\Delta S\)

Entropy is a measure of the dispersal of energy. When a process takes place spontaneously, the entropy of the system tends to increase. In this case, we have an isothermal expansion, which means the gas is expanding at a constant temperature. As the gas expands, it occupies a larger volume and, as a consequence, has more possible microstates or ways to arrange its particles. This increases the randomness of the system, and therefore, the entropy should also increase. So, we expect a positive sign for \(\Delta S\).
02

Calculate the entropy change

To calculate the entropy change for an ideal gas undergoing an isothermal process, we can use the following formula: \[\Delta S = nR \ln \frac{V_f}{V_i}\] Where \(n\) is the number of moles of the gas, \(R\) is the gas constant (8.314 J/molK or, ideal gas constant 0.0821 atm L/mol K), \(V_f\) is the final volume, and \(V_i\) is the initial volume. We are given \(n = 0.200\) mol, \(V_i = 10.0\) L, and \(V_f = 18.5\) L. Substituting these values into the formula, we have: \[\Delta S = (0.200 \,\text{mol}) (8.314 \,\text{J/mol K}) \ln \frac{18.5 \,\text{L}}{10.0 \,\text{L}}\]
03

Calculate the result

Now, we can calculate the entropy change: \[\Delta S = (0.200 \,\text{mol}) (8.314 \,\text{J/mol K}) \ln 1.85\] \[\Delta S \approx 0.5156 \,\text{J/K}\] So, the entropy change for the process is approximately 0.516 J/K.
04

Discuss the necessity of specifying the temperature

The question asks whether it's necessary to specify the temperature when calculating the entropy change. In our calculations, we used the formula \(\Delta S = nR \ln \frac{V_f}{V_i}\) to find the entropy change, and this formula does not directly involve the temperature. However, it's important to note that this formula is derived from the assumption that the process is isothermal, meaning the temperature remains constant while the gas undergoes the volume change. So, even though the temperature does not explicitly appear in the formula for the entropy change, its constancy is an essential condition for the validity of the formula. In summary, it is necessary to specify that the process is isothermal (constant temperature) to apply the given formula for entropy change. However, the actual value of the temperature doesn't directly affect the result in this case.

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Key Concepts

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

Isothermal Process
An isothermal process is characterized by its constant temperature throughout the entire operation. This temperature stability is key in certain thermodynamic evaluations, most notably those involving ideal gases. When a gas undergoes such a process, it can either expand or compress isothermally. In both cases, since the temperature remains unchanged, the internal energy of an ideal gas also remains constant.

During an isothermal expansion, a gas does work on its surroundings because it increases in volume. This applies to our example where the volume of an ideal gas expands from 10.0 L to 18.5 L. Since this process is done without a change in temperature, we can deduce that the heat energy must have flowed into the gas, allowing the expansion to be an isothermal one. Understanding this concept is vital when calculating the entropy change for the process.
Ideal Gas Law
The ideal gas law is a fundamental relation in chemistry and physics, represented by the equation \( PV=nRT \), where \( P \) stands for pressure, \( V \) is the volume, \( n \) signifies the amount of substance (moles), \( R \) is the gas constant, and \( T \) denotes the temperature. This equation describes the state of an ideal gas—a theoretical gas in which there are no intermolecular attractions and whose particles occupy negligible space.

The utility of the ideal gas law is extensive as it can predict the behavior of a gas under different conditions, as long as it behaves ideally, which most gases approximately do under normal conditions. It is crucial in deriving other relationships, such as the one used for entropy change during isothermal processes. The equality between the internal energy variations by isothermal expansion or compression and heat exchange with the surroundings guides us to assess entropy without the need to know the specific temperature, assuming ideal gas behavior.
Entropy and Spontaneity
The concept of entropy is central to the second law of thermodynamics and provides essential insight into the spontaneity of processes. Entropy, represented by \( S \), is a measure of a system's disorder or randomness. Spontaneous processes tend to lead to an increase in the total entropy of the universe, which includes the entropies of the system and its surroundings.

In our textbook example, the expansion of a gas without energy input (isothermal process) results in an increase in entropy since the gas particles have more space to disperse, hence more ways to arrange themselves. This means there's an increase in the number of possible microstates for the system, which corresponds to a larger entropy value. Calculating the entropy change can thus predict the spontaneity of a process, where a positive entropy change suggests a spontaneous process under constant temperature and pressure conditions. The direct connection between entropy change and spontaneity is a fundamental principle in thermodynamics that governs a multitude of natural and industrial processes.

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Most popular questions from this chapter

Explain qualitatively how \(\Delta G\) changes for each of the following reactions as the partial pressure of \(\mathrm{O}_{2}\) is increased: (a) \(2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)\) (b) \(2 \mathrm{H}_{2} \mathrm{O}_{2}(l) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)\) (c) \(2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\)

Consider the vaporization of liquid water to steam at a pressure of 1 atm. (a) Is this process endothermic or exothermic? (b) In what temperature range is it a spontaneous process? (c) In what temperature range is it a nonspontaneous process? (d) At what temperature are the two phases in equilibrium?

Consider the following reaction between oxides of nitrogen: $$ \mathrm{NO}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 3 \mathrm{NO}(g) $$ (a) Use data in Appendix \(\mathrm{C}\) to predict how \(\Delta \mathrm{G}^{\circ}\) for the reaction varies with increasing temperature. (b) Calculate \(\Delta G^{\circ}\) at \(800 \mathrm{~K}\), assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not change with temperature. Under standard conditions is the reaction spontaneous at \(800 \mathrm{~K} ?\) (c) Calculate \(\Delta G^{\circ}\) at \(1000 \mathrm{~K}\). Is the reaction spontaneous under standard conditions at this temperature?

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For the isothermal expansion of a gas into a vacuum, \(\Delta E=0, q=0\), and \(w=0\). (a) Is this a spontaneous process? (b) Explain why no work is done by the system during this process. (c) In thermodynamics, what is the "driving force" for the expansion of the gas?

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