Entropy and Second Law of Thermodynamics
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AP Physics 2 › Entropy and Second Law of Thermodynamics
A sealed, insulated container is divided into two equal volumes by a valve. Side A contains an ideal gas at $2.0,\text{atm}$; side B contains the same gas at $1.0,\text{atm}$, both at $300,\text{K}$. The valve is opened and the gases mix until pressure equalizes; the process is spontaneous. Which statement best describes the entropy change of the gases?
The entropy is zero because no heat crosses the container boundary.
The entropy increases because the combined system evolves toward a state with more accessible microstates.
The entropy decreases because the final state has a single uniform pressure.
The entropy increases only if the gases have different chemical identities.
Explanation
This question tests understanding of entropy and the second law of thermodynamics. When gases at different pressures mix spontaneously in an isolated container, the system evolves toward equilibrium with uniform pressure. The second law requires entropy to increase for this spontaneous process. As molecules redistribute throughout the available volume, the number of accessible microstates increases dramatically, causing entropy to rise even without heat transfer. Choice A incorrectly assumes uniform conditions mean lower entropy, confusing equilibrium with the number of microstates. The key insight is: spontaneous mixing and expansion processes increase entropy by increasing the number of accessible molecular configurations.
A 400 K object transfers 200 J of heat to a 300 K reservoir; the process is spontaneous. Which statement best compares the entropy changes of object and reservoir?
Both entropy changes are equal in magnitude because the same heat is transferred.
The object’s entropy decrease is larger in magnitude than the reservoir’s entropy increase.
The reservoir’s entropy increase is larger in magnitude than the object’s entropy decrease.
Both entropy changes are zero because the transfer is spontaneous and requires no work.
Explanation
This question tests understanding of entropy and the second law of thermodynamics. When 200 J of heat flows spontaneously from a 400 K object to a 300 K reservoir, both objects experience entropy changes. The hot object loses entropy: ΔS_object = -200/400 = -0.5 J/K, while the cold reservoir gains entropy: ΔS_reservoir = +200/300 = +0.67 J/K. The reservoir's entropy increase (0.67 J/K) is larger in magnitude than the object's entropy decrease (0.5 J/K), resulting in a net positive entropy change of +0.17 J/K for the universe. This positive total ensures the process is spontaneous according to the second law. Choice C incorrectly assumes equal heat transfer means equal entropy changes—but entropy change depends on both heat and temperature (ΔS = Q/T). For spontaneous heat transfer, always verify that the cold object's entropy gain exceeds the hot object's entropy loss.
A 300 K metal block is placed in contact with a 500 K reservoir and absorbs 600 J of heat; the process is spontaneous. Which conclusion is consistent with the second law?
The total entropy change is positive only if the block’s temperature stays constant.
The total entropy change is negative because the block becomes more ordered.
The total entropy change is positive because heat flows from hot to cold.
The total entropy change is zero because energy is conserved.
Explanation
This question tests understanding of entropy and the second law of thermodynamics. When heat flows spontaneously from a hot reservoir (500 K) to a cooler block (300 K), the total entropy of the universe must increase according to the second law. The entropy decrease of the hot reservoir (ΔS = -Q/T_hot = -600/500 = -1.2 J/K) is smaller in magnitude than the entropy increase of the cold block (ΔS = +Q/T_cold = +600/300 = +2.0 J/K), resulting in a net positive entropy change of +0.8 J/K. Choice B incorrectly confuses energy conservation with entropy conservation—while energy is conserved, entropy increases in spontaneous processes. For any spontaneous heat transfer, always check that the total entropy change is positive by calculating ΔS_total = Q(1/T_cold - 1/T_hot).
A perfectly insulated box contains two equal copper blocks at 350 K and 250 K that are placed in thermal contact; the process is spontaneous. Which statement best describes the entropy change?
The total entropy of the two-block system increases until thermal equilibrium is reached.
The total entropy of the two-block system becomes negative because temperature differences create order.
The total entropy of the two-block system remains constant because it is isolated.
The total entropy of the two-block system decreases because energy flows out of the hot block.
Explanation
This question tests understanding of entropy and the second law of thermodynamics. When two blocks at different temperatures (350 K and 250 K) are placed in thermal contact within an isolated system, heat flows spontaneously from hot to cold until thermal equilibrium is reached. During this process, the hot block loses entropy (ΔS_hot < 0) while the cold block gains entropy (ΔS_cold > 0), but the magnitude of the cold block's entropy gain exceeds the hot block's entropy loss because |ΔS| = |Q|/T is larger for lower temperatures. The total entropy increases throughout the process until equilibrium is reached at 300 K. Choice C incorrectly assumes that isolation means constant entropy—isolation prevents energy exchange with surroundings but doesn't prevent internal entropy increases. For any spontaneous thermal equilibration in an isolated system, total entropy always increases.
A heat engine operates between reservoirs at $T_h=600,\text{K}$ and $T_c=300,\text{K}$ in a complete cycle. Which conclusion is consistent with the second law?
The efficiency can exceed 1 if the engine’s working gas is ideal.
The efficiency can exceed 0.50 if the engine is well insulated from the cold reservoir.
The efficiency cannot exceed $1-\frac{T_c}{T_h}=0.50$ for any engine between these reservoirs.
The efficiency must equal 1 because the engine returns to its initial state each cycle.
Explanation
This question tests understanding of entropy and the second law of thermodynamics. The second law places a fundamental limit on heat engine efficiency through the Carnot efficiency: η_max = 1 - T_c/T_h, which represents the maximum possible efficiency for any heat engine operating between two temperature reservoirs. For reservoirs at 600 K and 300 K, this maximum efficiency is 1 - 300/600 = 0.50 or 50%, regardless of the engine's design or working substance. No real engine can exceed this limit because doing so would require decreasing the total entropy of the universe, violating the second law. Choice B incorrectly suggests that using an ideal gas could allow efficiency above the Carnot limit—but the Carnot limit applies to all engines, regardless of working substance. When analyzing heat engines, always apply the Carnot efficiency formula to find the theoretical maximum.
A proposed cyclic device absorbs 400 J from a single 350 K reservoir each cycle and converts all of it to work; it is claimed to operate spontaneously. Which conclusion is consistent with the second law?
The device is possible if the reservoir is large enough to keep its temperature constant.
The device is possible because entropy in an isolated system can decrease during a cycle.
The device is impossible because a cycle cannot convert heat from one reservoir entirely into work.
The device is possible because energy conservation allows $W=Q$ in a cycle.
Explanation
This question tests understanding of entropy and the second law of thermodynamics. The proposed device claims to convert all heat from a single reservoir into work in a complete cycle, which would be a perfect heat engine with 100% efficiency. This violates the Kelvin-Planck statement of the second law: no cyclic process can convert heat entirely into work while operating with a single heat reservoir. Such a device would decrease the entropy of the universe (reservoir loses entropy, nothing gains it), making it impossible. All real heat engines must reject some heat to a cold reservoir, limiting their efficiency to less than 100%. Choice B incorrectly suggests energy conservation allows this—while energy is conserved, the second law imposes additional restrictions beyond conservation. Remember that any cyclic heat engine requires at least two reservoirs at different temperatures.
An ideal gas undergoes a reversible isothermal expansion at 350 K while absorbing 500 J of heat from a reservoir; the process can be made quasi-static. Which statement is consistent with the second law?
The gas’s entropy change is negative because expansion makes the gas more disordered.
The gas’s entropy change is $\Delta S=\frac{T}{Q_{rev}}=\frac{350}{500},\text{J/K}$.
The gas’s entropy change is zero because the internal energy of an ideal gas is constant at constant $T$.
The gas’s entropy change is $\Delta S=\frac{Q_{rev}}{T}=\frac{500}{350},\text{J/K}$.
Explanation
This question tests understanding of entropy and the second law of thermodynamics. For a reversible isothermal process, the entropy change is calculated using ΔS = Q_rev/T, where Q_rev is the heat absorbed reversibly and T is the constant temperature. When the ideal gas expands isothermally at 350 K while absorbing 500 J of heat, its entropy increases by ΔS = 500/350 = 10/7 ≈ 1.43 J/K. This positive entropy change reflects the increased volume and thus increased number of accessible microstates for the gas molecules. Choice D incorrectly inverts the formula as T/Q_rev—entropy has units of J/K, which requires Q/T not T/Q. For any reversible isothermal process, use ΔS = Q_rev/T to calculate entropy change, with positive Q for heat absorbed.
A refrigerator removes 900 J of heat from a 270 K compartment and exhausts heat to a 300 K room; it operates non-spontaneously using electrical work. Which statement best explains why work is required?
Work is required because the room’s entropy must always decrease during cooling.
Moving heat from cold to hot would decrease total entropy unless work is added.
Moving heat from cold to hot conserves energy but violates the first law unless work is added.
Work is required because entropy must decrease in any cyclic device.
Explanation
This question tests understanding of entropy and the second law of thermodynamics. A refrigerator moves heat from a cold space (270 K) to a warmer room (300 K), which is the opposite of spontaneous heat flow. This process would decrease total entropy if done alone: removing heat from the cold compartment decreases its entropy more than adding the same heat to the warm room increases the room's entropy (since |ΔS| = |Q|/T is larger at lower T). To make this process possible while still increasing total entropy, work must be added to the system, which ultimately gets converted to additional heat expelled to the room. Choice B incorrectly claims this violates the first law—energy is conserved, but the second law requires work input. Remember that moving heat from cold to hot requires work input to ensure total entropy increases.
A gas in a cylinder is compressed rapidly with significant friction, and 300 J of heat is transferred to the surroundings; the process is spontaneous and irreversible. Which statement best describes the total entropy change?
The total entropy of gas plus surroundings decreases because friction creates order.
The total entropy of gas plus surroundings increases only if the gas temperature remains constant.
The total entropy of gas plus surroundings is zero because the gas loses heat.
The total entropy of gas plus surroundings increases because irreversibility produces entropy.
Explanation
This question tests understanding of entropy and the second law of thermodynamics. During rapid compression with friction, work is done on the gas in an irreversible manner, and 300 J of heat flows to the surroundings. Irreversible processes always generate entropy, so even though the gas might lose entropy (due to compression and heat loss), the surroundings gain more entropy than the gas loses. The friction converts organized mechanical work into disorganized thermal energy, creating additional entropy beyond what reversible compression would produce. The total entropy of the gas plus surroundings must increase for any spontaneous, irreversible process. Choice C incorrectly claims friction creates order—friction actually increases disorder by converting mechanical energy to heat. For any irreversible process, total entropy of the universe always increases, regardless of local decreases.
A refrigerator removes $Q_c=200,\text{J}$ from the cold space each cycle while consuming $W=50,\text{J}$ of electrical work; the process is non-spontaneous and requires input work. Which statement is consistent with the second law for this device?
It violates the second law because it moves heat from cold to hot, which can never occur.
It violates the second law because the coefficient of performance is greater than $1$.
It is consistent only if it converts more than $100%$ of the work into extracted heat.
It is consistent because external work allows heat to be transferred from cold to hot in a cycle.
Explanation
This question tests understanding of entropy and the second law of thermodynamics. A refrigerator moves heat from cold to hot, which doesn't occur spontaneously, but the second law allows this when external work is supplied. The device must reject Qh = Qc + W = 250 J to the hot side while consuming 50 J of work, maintaining positive entropy production overall. The coefficient of performance (COP = Qc/W = 4) being greater than 1 is perfectly allowed and doesn't violate any laws. Choice B incorrectly claims heat can never move from cold to hot, misunderstanding that work input makes this possible. Remember: non-spontaneous processes can occur when external work compensates for entropy decrease.