The Second and Third Laws of Thermodynamics Quiz
The question sheet
Reveal any answer as you study-
The second law of thermodynamics states that all spontaneous changes cause:
- No change in universe entropy
- An increase in universe entropy
- A decrease in universe entropy
- A decrease in system energy
Reveal answer
Answer: An increase in universe entropy
Source evidence
PDF page 882: univ These results lead to a profound statement regarding the relation between entropy and spontaneity known as the second law of thermodynamics: all spontaneous changes cause an increase in the entropy of the universe. A summary of these three relations is provided in Table 16.1. The Second Law of Thermodynamics spontaneous ΔSuniv > 0 ΔSuniv < 0 nonspontaneous (spontaneous in opposite direction) at equilibrium ΔSuniv = 0
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According to thermodynamic models, the system and surroundings together comprise:
- Only the surroundings
- Only the system
- The atmosphere
- The universe
Reveal answer
Answer: The universe
Source evidence
PDF page 881: In the quest to identify a property that may reliably predict the spontaneity of a process, a promising candidate has been identified: entropy. Processes that involve an increase in entropy of the system (ΔS > 0) are very often spontaneous; however, examples to the contrary are plentiful. By expanding consideration of entropy changes to include the surroundings, we may reach a significant conclusion regarding the relation between this property and spontaneity. In thermodynamic models, the system and surroundings comprise everything, that is, the universe, and so the following is true: ΔS = ΔSsys + ΔSsurr univ To illustrate this relation, consider again the process of heat flow between two objects, one identified as the system and the other as the surroundings. There are three possibilities for such a process:
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Which equation expresses the entropy change of the universe?
- ΔSuniv = ΔSsys − ΔSsurr
- ΔSuniv = ΔSsys × ΔSsurr
- ΔSuniv = ΔSsys + ΔSsurr
- ΔSuniv = ΔSsurr / ΔSsys
Reveal answer
Answer: ΔSuniv = ΔSsys + ΔSsurr
Source evidence
PDF page 881: In the quest to identify a property that may reliably predict the spontaneity of a process, a promising candidate has been identified: entropy. Processes that involve an increase in entropy of the system (ΔS > 0) are very often spontaneous; however, examples to the contrary are plentiful. By expanding consideration of entropy changes to include the surroundings, we may reach a significant conclusion regarding the relation between this property and spontaneity. In thermodynamic models, the system and surroundings comprise everything, that is, the universe, and so the following is true: ΔS = ΔSsys + ΔSsurr univ To illustrate this relation, consider again the process of heat flow between two objects, one identified as the system and the other as the surroundings. There are three possibilities for such a process:
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A process is spontaneous when the entropy change of the universe is:
- Less than zero
- Undefined
- Greater than zero
- Equal to zero
Reveal answer
Answer: Greater than zero
Source evidence
PDF page 882: univ These results lead to a profound statement regarding the relation between entropy and spontaneity known as the second law of thermodynamics: all spontaneous changes cause an increase in the entropy of the universe. A summary of these three relations is provided in Table 16.1. The Second Law of Thermodynamics spontaneous ΔSuniv > 0 ΔSuniv < 0 nonspontaneous (spontaneous in opposite direction) at equilibrium ΔSuniv = 0
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When ΔSuniv equals zero, the system is:
- At absolute zero
- Spontaneous
- Nonspontaneous
- At equilibrium
Reveal answer
Answer: At equilibrium
Source evidence
PDF page 882: are essentially the same for both the system and the surroundings. In this case, the entropy change of the universe is zero, and the system is at equilibrium. Ssys Ssurr| |Δ | ≈ |Δ
PDF page 882: univ These results lead to a profound statement regarding the relation between entropy and spontaneity known as the second law of thermodynamics: all spontaneous changes cause an increase in the entropy of the universe. A summary of these three relations is provided in Table 16.1. The Second Law of Thermodynamics spontaneous ΔSuniv > 0 ΔSuniv < 0 nonspontaneous (spontaneous in opposite direction) at equilibrium ΔSuniv = 0
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When ΔSuniv is negative, the process is:
- At absolute zero
- Spontaneous
- At equilibrium
- Nonspontaneous
Reveal answer
Answer: Nonspontaneous
Source evidence
PDF page 882: univ These results lead to a profound statement regarding the relation between entropy and spontaneity known as the second law of thermodynamics: all spontaneous changes cause an increase in the entropy of the universe. A summary of these three relations is provided in Table 16.1. The Second Law of Thermodynamics spontaneous ΔSuniv > 0 ΔSuniv < 0 nonspontaneous (spontaneous in opposite direction) at equilibrium ΔSuniv = 0
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When heat flows from hotter system to cooler surroundings, the universe entropy:
- Increases
- Becomes negative
- Stays zero
- Decreases
Reveal answer
Answer: Increases
Source evidence
PDF page 882: Ssys Ssurr| |Δ | < |Δ ΔS = ΔSsys + ΔSsurr > 0 univ
PDF page 881: observed to occur spontaneously. Designating the hotter object as the system and invoking the definition of entropy yields the following: −qrev qrev and ΔSsys = ΔSsurr = Tsys Tsurr The magnitudes of −qrev and qrev are equal, their opposite arithmetic signs denoting loss of heat by the system and gain of heat by the surroundings. Since Tsys > Tsurr in this scenario, the entropy decrease of the system will be less than the entropy increase of the surroundings, and so the entropy of the universe will increase:
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For a vast surroundings, qsurr is a good approximation of:
- qrev
- the temperature
- W
- ΔSsys
Reveal answer
Answer: qrev
Source evidence
PDF page 882: For many realistic applications, the surroundings are vast in comparison to the system. In such cases, the heat gained or lost by the surroundings as a result of some process represents a very small, nearly infinitesimal, fraction of its total thermal energy. For example, combustion of a fuel in air involves transfer of heat from a system (the fuel and oxygen molecules undergoing reaction) to surroundings that are infinitely more massive (the earth’s atmosphere). As a result, qsurr is a good approximation of qrev, and the second law may be stated as the following: qsurr
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The third law of thermodynamics states the entropy of a pure perfect crystal at 0 K is:
- Negative
- Infinite
- One
- Zero
Reveal answer
Answer: Zero
Source evidence
PDF page 883: This limiting condition for a system’s entropy represents the third law of thermodynamics: the entropy of a pure, perfect crystalline substance at 0 K is zero. Careful calorimetric measurements can be made to determine the temperature dependence of a substance’s entropy and to derive absolute entropy values under specific conditions. Standard entropies (S°) are for one mole of substance under standard conditions (a pressure of 1 bar and a temperature of 298.15 K; see details regarding standard conditions in the thermochemistry chapter of this text). The standard entropy change (ΔS°) for a reaction may be computed using standard entropies as shown below:
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A pure, perfectly crystalline solid at 0 K is described by how many microstates?
- Infinite
- One
- Two
- Zero
Reveal answer
Answer: One
Source evidence
PDF page 883: The previous section described the various contributions of matter and energy dispersal that contribute to the entropy of a system. With these contributions in mind, consider the entropy of a pure, perfectly crystalline solid possessing no kinetic energy (that is, at a temperature of absolute zero, 0 K). This system may be described by a single microstate, as its purity, perfect crystallinity and complete lack of motion means there is but one possible location for each identical atom or molecule comprising the crystal (W = 1). According to the Boltzmann equation, the entropy of this system is zero.
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According to the Boltzmann equation, S = k ln(1) equals:
- ln k
- 0
- k
- 1
Reveal answer
Answer: 0
Source evidence
PDF page 883: S = k ln W = k ln(1) = 0
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Standard entropies (S°) are given for one mole of substance under conditions of:
- 2 bar and 273 K
- 0 bar and 0 K
- 1 atm and 0 K
- 1 bar and 298.15 K
Reveal answer
Answer: 1 bar and 298.15 K
Source evidence
PDF page 883: This limiting condition for a system’s entropy represents the third law of thermodynamics: the entropy of a pure, perfect crystalline substance at 0 K is zero. Careful calorimetric measurements can be made to determine the temperature dependence of a substance’s entropy and to derive absolute entropy values under specific conditions. Standard entropies (S°) are for one mole of substance under standard conditions (a pressure of 1 bar and a temperature of 298.15 K; see details regarding standard conditions in the thermochemistry chapter of this text). The standard entropy change (ΔS°) for a reaction may be computed using standard entropies as shown below:
Chemistry 2e
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