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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?

Short Answer

Expert verified
The isothermal expansion of a gas into a vacuum is a spontaneous process because the Gibbs Free Energy (ΔG) is negative. No work is done during the process because there is no external pressure against the gas expansion. The driving force for the expansion of the gas is the increase in entropy (ΔS > 0), as nature tends to favor an increase in disorder.

Step by step solution

01

(a) Determining Spontaneity of the Process

To determine if the process is spontaneous, we need to find the change in Gibbs Free Energy (ΔG). The equation for ΔG is as follows: ΔG = ΔH - TΔS If ΔG < 0, then the process is spontaneous. However, we only know that ΔE = q + w = 0, and we cannot directly find ΔG using this information. Since the process is isothermal, ΔH = 0 (no change in the internal energy of the gas). Also, as the gas is expanding into a vacuum, the entropy (S) will increase due to an increase in volume. Therefore, ΔS > 0. ΔH = ΔE = 0, ΔS > 0 Putting these values in the ΔG equation: ΔG = 0 - T(ΔS) Since T > 0 (assuming the system isn't at absolute zero), and ΔS > 0, this implies that: ΔG < 0 So, the isothermal expansion of a gas into a vacuum is a spontaneous process.
02

(b) Explaining Why No Work is Done during the Process

During an isothermal expansion, there is no change in temperature. Work is done by the system when there is external pressure acting against the expansion of the gas. In this case, however, the gas is expanding into a vacuum, which means there is no external pressure. The formula for work done during an expansion is: w = -P_ext(ΔV) Since the pressure, P_ext = 0 (vacuum), the work (w) done by the system is also zero: w = -0(ΔV) = 0 This explains why no work is done by the system during the isothermal expansion of a gas into a vacuum.
03

(c) Identifying the "Driving Force" for Gas Expansion

In thermodynamics, the "driving force" for a process is directly related to the change in entropy (ΔS). Entropy represents the degree of disorder in a system. An increase in entropy generally leads to a spontaneous process, as nature tends to favor an increase in disorder. For the case of the isothermal expansion of a gas into a vacuum, the gas expands and occupies a larger volume. As a result, the number of microstates (possible arrangements of gas particles) increases, leading to an increase in entropy (ΔS). Thus, the "driving force" for the expansion of the gas in this scenario is the increase in entropy (ΔS > 0).

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Key Concepts

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

Gibbs Free Energy
Gibbs Free Energy (G) is a fundamental concept in thermodynamics that helps determine whether a process will occur spontaneously. It’s an energy measure that combines the system's enthalpy (H), temperature (T), and entropy (S) to predict the favorability of a chemical reaction or physical change. A spontaneous process is characterized by a decrease in Gibbs Free Energy ()
In the case of isothermal expansion of a gas into a vacuum, where
Since the change in enthalpy (
A negative value for )
Entropy
Entropy is a measure of the disorder or randomness in a system. In thermodynamics, it is widely accepted that processes tend to move towards a state of increased entropy. During the isothermal expansion of a gas, as the volume of gas increases, so does the entropy, because the gas molecules have more space to occupy and can therefore arrange themselves in a greater number of ways. This increase in entropy (
In our discussed exercise, the expansion into a vacuum presents no opposition to this spreading out of molecules, causing an unequivocal rise in entropy. This aligns with the second law of thermodynamics, which basically states that for any spontaneous process, the entropy of the universe increases. Therefore, )
Thermodynamics
Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. It involves studying systems as they move towards thermal equilibrium, where no net transfer of energy occurs. The first law of thermodynamics, also known as the law of conservation of energy, states that energy cannot be created or destroyed, only transformed from one form to another. This is why, in our isothermal expansion example, even though the internal energy change (
Thermodynamics also covers the concepts of heat transfer, work done by or on the system, the thermodynamic properties of materials, and the laws that describe the transfer and transformation of energy. The phenomenon of gas expanding into a vacuum without doing work and without heat exchange is a unique scenario perfectly encapsulated by thermodynamic principles.
Spontaneous Process
A spontaneous process is one that occurs without the need for external energy. It’s driven by a decrease in Gibbs Free Energy, an increase in entropy, or both. For a process to be spontaneous at all temperatures, its entropy must increase, as indicated by a positive change in entropy (
In the scenario of an isothermal expansion of a gas into a vacuum, we have a clear example of a spontaneous process. There's no energy input, yet the gas expands due to the natural predisposition towards a state of higher entropy. This expansion is spontaneous despite no work being performed and no heat exchange, again confirming to the basic premise that systems will spontaneously evolve towards configurations that maximize their entropy.
Internal Energy
The concept of internal energy (
When examining isothermal processes, such as the isothermal expansion into a vacuum, the temperature of the system remains constant. Since temperature is a measure of the average kinetic energy of the molecules in a gas, and )

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Most popular questions from this chapter

The reaction $$ \mathrm{SO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons 3 \mathrm{~S}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ is the basis of a suggested method for removal of \(\mathrm{SO}_{2}\) from power-plant stack gases. The standard free energy of each substance is given in Appendix \(\mathrm{C}\). (a) What is the equilibrium constant for the reaction at \(298 \mathrm{~K}\) ? (b) In principle, is this reaction a feasible method of removing \(\mathrm{SO}_{2} ?\) (c) If \(P_{\mathrm{SO}_{2}}=P_{\mathrm{H}_{2} \mathrm{~S}}\) and the vapor pressure of water is 25 torr, calculate the equilibrium \(\mathrm{SO}_{2}\) pressure in the system at \(298 \mathrm{~K}\). (d) Would you expect the process to be more or less effective at higher temperatures?

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 \(\mathrm{CO}_{2}(s)\).

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?

A particular constant-pressure reaction is spontaneous at \(390 \mathrm{~K}\). The enthalpy change for the reaction is \(+23.7 \mathrm{~kJ}\). What can you conclude about the sign and magnitude of \(\Delta S\) for the reaction?

Using \(S^{\circ}\) values from Appendix C, calculate \(\Delta S^{\circ}\) values for the following reactions. In each case account for the sign of \(\Delta S^{\circ} .\) (a) \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)\) (b) \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) (c) \(\mathrm{Be}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{BeO}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) (d) \(2 \mathrm{CH}_{3} \mathrm{OH}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\)

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