Questions: Entropy Changes in Thermodynamic Processes

Question 1

An ideal gas undergoes free expansion into a vacuum: no heat is exchanged, no work is done, and temperature is unchanged. What is the entropy change of the gas?

• A. Zero, because Q = 0 and the temperature didn't change
• B. Negative, because the gas expanded spontaneously without doing useful work
• C. Positive, equal to nR ln(V₂/V₁), computed via an equivalent reversible isothermal path
• D. Impossible to determine because the process is irreversible

Answer: (C)

Free expansion is irreversible — you cannot use ΔS = Q/T because Q = 0 applies to the actual (irreversible) process, not the reversible path. Since entropy is a state function, ΔS depends only on the initial and final states (same T, larger V). The reversible path connecting those states is an isothermal expansion, giving ΔS = nR ln(V₂/V₁) > 0. Entropy increased even though no heat flowed — this illustrates that irreversibility generates entropy independently of heat transfer.

Question 2

For an irreversible process at constant temperature T where actual heat Q_irrev flows in, how is ΔS correctly calculated?

• A. ΔS = Q_irrev / T, since the temperature is constant throughout
• B. ΔS = 0, because irreversible processes don't change thermodynamic state functions
• C. Identify the initial and final states, construct any reversible path between them, and integrate dQ_rev/T
• D. ΔS = Q_irrev / T only if the surroundings are at the same temperature

Answer: (C)

The Clausius inequality states dS ≥ dQ/T, with equality only for reversible processes. For an irreversible process, dS > dQ_irrev/T — using Q_irrev/T gives the wrong (too small) answer. The correct procedure exploits the fact that S is a state function: its change depends only on endpoints, not path. Construct any convenient reversible path between the same endpoints and integrate dQ_rev/T along it. The actual heat of the irreversible process is irrelevant to ΔS.

Question 3

For a reversible adiabatic process, the entropy change of the system is zero.

• T. True
• F. False

Answer: True

True. A reversible adiabatic process has dQ_rev = 0 at every infinitesimal step by definition (adiabatic means no heat exchange). Since ΔS = ∫dQ_rev/T and dQ_rev = 0 throughout, ΔS = 0. These are 'isentropic' processes — they move along curves of constant entropy in any thermodynamic diagram. Note the crucial qualifier 'reversible': an irreversible adiabatic process still generates entropy (ΔS > 0) even though Q = 0, because entropy is also produced by internal irreversibilities.

Question 4

In any thermodynamic process, the entropy change of the universe is zero because energy is conserved.

• T. True
• F. False

Answer: False

False — this conflates energy conservation (the first law) with entropy. Energy is always conserved, but entropy is not. For irreversible processes, the total entropy of the universe (system + surroundings) strictly increases. Only in reversible processes does universe entropy remain constant. The second law says entropy is generated by irreversibility, not destroyed. The universe's entropy increases monotonically over time — this is why processes are directional.

Question 5

Why must you construct a reversible path to calculate the entropy change of an irreversible process, even though the actual process followed a different path?

Model answer: Because entropy is a state function — its change depends only on the initial and final states, not on the path taken. The entropy change between two states is uniquely defined regardless of how the system got there. The formula ΔS = ∫dQ_rev/T requires a reversible path because only on a reversible path does the equality dS = dQ/T hold exactly. For an irreversible path, dS > dQ_irrev/T, so using the actual heat of the irreversible process gives too small a value. The reversible path is just a computational device — it gives the correct ΔS for any pair of endpoints.

This is the central skill: recognizing that computing ΔS is a two-step process. First, identify the initial and final equilibrium states. Second, choose any reversible path between them (pick the easiest to integrate). The choice of reversible path doesn't affect the answer because S is a state function. Common reversible paths: isothermal for ideal gas volume changes, isobaric heating for constant-pressure processes. The actual irreversible process is physically interesting but thermodynamically irrelevant to computing ΔS.