Thermodynamics — Set 2

Correct Answer: C. It is zero

• **It is zero** = In free expansion into vacuum the external pressure is zero, so W = ∫ P_ext dV = 0; the gas does no work against any opposing force. • **Joule's experiment** — this result, confirmed by Joule, shows that for an ideal gas ΔU = 0 and ΔT = 0 in free expansion since Q = 0 and W = 0. • 💡 Wrong-option analysis: It is maximum: maximum work is obtained in a reversible process against maximum opposing pressure, not in free expansion into vacuum; It is negative: negative work means work is done on the gas, which requires an external agent compressing it; It equals nRT: nRT is the work done in an isothermal reversible expansion, not in free expansion into vacuum.

Correct Answer: B. PV^γ = constant

• **PV^γ = constant** = In a reversible adiabatic process, combining Q = 0 with the first law and ideal gas equation yields PV^γ = constant, where γ = Cp/Cv. • **Steeper than isothermal** — the adiabatic curve on a P-V diagram is steeper than the isothermal curve because γ > 1, meaning pressure drops faster for the same volume increase. • 💡 Wrong-option analysis: P + V = constant: pressure and volume have different units so their sum is not physically meaningful; PV = constant: this is Boyle's law for an isothermal process, not an adiabatic one; P/V = constant: this ratio has no standard thermodynamic significance in adiabatic processes.

Correct Answer: B. η = 1 - Tc/Th

• **η = 1 − Tc/Th** = The Carnot efficiency formula shows that the fraction of heat converted to work equals one minus the ratio of cold to hot reservoir temperatures in kelvin. • **Maximum possible efficiency** — no real engine operating between Th and Tc can exceed this value; as Tc → 0 K, η → 1, but absolute zero is unattainable. • 💡 Wrong-option analysis: η = 1 − Th/Tc: inverting the ratio gives a negative efficiency when Th > Tc, which is physically impossible; η = Th/Tc: this exceeds 1 when Th > Tc, violating the second law; η = Tc/Th: this is the fraction of heat wasted to the cold reservoir, not the efficiency.

Correct Answer: D. Qc/W

• **Qc/W** = The COP of a refrigerator is the heat extracted from the cold space divided by the work input; a higher COP means more cooling per unit of work. • **COP = Tc/(Th − Tc)** — for a reversible refrigerator, the COP expressed in terms of reservoir temperatures shows it increases as the temperature difference decreases. • 💡 Wrong-option analysis: W/Qh: this is the inverse of the heat pump COP and does not represent refrigerator performance; W/Qc: this is the reciprocal of the refrigerator COP, not the COP itself; Qh/W: this is the COP of a heat pump, which delivers heat to the hot space, not removes heat from the cold space.

Correct Answer: B. No cyclic engine can convert all heat from a single reservoir into work

• **No cyclic engine can convert all heat from a single reservoir into work** = The Kelvin-Planck statement forbids a 100% efficient cyclic heat engine operating from a single thermal reservoir. • **Second reservoir required** — some heat must always be rejected to a cold reservoir; this is why real power plants need cooling towers or cold water bodies. • 💡 Wrong-option analysis: All processes are reversible in practice: real processes involve friction, heat leaks, and turbulence making them irreversible; Heat cannot flow from cold to hot without work: this is the Clausius statement, a different but equivalent form of the second law; Entropy of an isolated system always decreases: the second law says entropy cannot decrease in an isolated system.

Correct Answer: B. Heat cannot spontaneously flow from a colder body to a hotter body without external work

• **Heat cannot spontaneously flow from a colder body to a hotter body without external work** = The Clausius statement says heat moves spontaneously only from hot to cold; moving it the other way requires work input, as in a refrigerator. • **Refrigerator example** — a refrigerator uses electrical work to pump heat from the cold interior to the hot surroundings, satisfying Clausius by requiring external work. • 💡 Wrong-option analysis: Work cannot be converted to heat: work can and does convert to heat through friction and other dissipative processes; Internal energy depends on volume only: for an ideal gas internal energy depends only on temperature, not volume; Heat always flows from hot to cold only in reversible paths: heat flows from hot to cold in both reversible and irreversible paths spontaneously.

Correct Answer: C. dS = δQrev/T

• **dS = δQrev/T** = Entropy change for a reversible process is defined as the reversible heat transfer divided by the absolute temperature at which it occurs. • **State function** — integrating dS along any reversible path between the same two states always gives the same ΔS, confirming entropy is a state function. • 💡 Wrong-option analysis: dS = δWrev/T: using work instead of heat mixes up the two distinct energy transfers; dS = T/δQrev: inverting the fraction makes entropy decrease when heat is added, which contradicts the definition; dS = δQrev × T: multiplying by T gives units of J·K, which is not the unit of entropy (J/K).

Correct Answer: C. It is zero

• **It is zero** = In a reversible process, the entropy lost by the hot body exactly equals the entropy gained by the cold body, so the total entropy change of the universe is zero. • **Ideal limit** — a reversible process is an idealization where no entropy is generated; all real processes are irreversible and produce a net positive entropy change. • 💡 Wrong-option analysis: It depends only on pressure: entropy change depends on heat exchange and temperature, not pressure alone; It is always positive: positive total entropy change characterizes irreversible processes, not reversible ones; It is always negative: entropy of the universe can never decrease according to the second law.

Correct Answer: C. Entropy cannot decrease

• **Entropy cannot decrease** = An isolated system exchanges neither heat nor work with surroundings; the second law states its total entropy can only remain constant (reversible) or increase (irreversible). • **ΔS_universe ≥ 0** — equality holds only for reversible processes; all spontaneous irreversible processes inside an isolated system increase its entropy. • 💡 Wrong-option analysis: Entropy must decrease: this directly contradicts the second law; spontaneous processes increase or maintain entropy; Entropy equals heat absorbed: entropy has units J/K while heat absorbed has units J, so they cannot be equal; Entropy is always zero: entropy is zero only at absolute zero (third law) and is generally nonzero.

Correct Answer: D. A process that can be reversed leaving no net change in system and surroundings

• **A process that can be reversed leaving no net change in system and surroundings** = A reversible process is an idealized quasi-static process with no dissipation; reversing it restores both system and surroundings exactly to their initial states. • **Zero entropy generation** — reversibility requires no friction, no finite temperature gradients, and no turbulence; such processes generate no entropy and have maximum work output. • 💡 Wrong-option analysis: A process with large friction: friction is a dissipative effect that makes processes irreversible, the opposite of reversible; A process that finishes quickly: rapid processes are far from equilibrium and highly irreversible; A process with no temperature change: this describes an isothermal process, which can be either reversible or irreversible.