An ideal gas undergoes free expansion into a vacuum: it doubles in volume, but no heat is transferred and no work is done (Q = 0, W = 0). What is the entropy change ΔS of the gas?
AZero, because no heat was transferred (Q = 0)
BZero, because internal energy didn't change (ΔU = 0)
CnR ln(2), positive — calculated via an equivalent reversible path
DNegative, because the gas expanded spontaneously into disorder
Entropy is a state function, so ΔS depends only on the initial and final states, not on the actual (irreversible) path. Even though Q_actual = 0, you must compute ΔS by imagining a reversible isothermal expansion between the same two states, giving ΔS = nR ln(V_f/V_i) = nR ln(2) > 0. Option A is the most tempting mistake: using Q_actual in dS = dQ/T. The subscript 'rev' in dS = dQ_rev/T is essential — it demands the reversible path, not the actual one.
Question 2 Multiple Choice
A Carnot engine absorbs Q_H = 1000 J from a reservoir at T_H = 600 K and exhausts Q_C to a reservoir at T_C = 300 K. What is the total entropy change of the universe per cycle?
AZero — a Carnot engine is reversible, so the universe's entropy doesn't change
BPositive — heat flows from hot to cold, so entropy must increase
CNegative — work is extracted, reducing the universe's disorder
D+Q_H/T_H = 1000/600 ≈ 1.67 J/K
For a Carnot engine with efficiency η = 1 − T_C/T_H = 0.5, the work output is 500 J and the heat rejected is Q_C = 500 J. The entropy change of the hot reservoir is −Q_H/T_H = −1000/600, and of the cold reservoir is +Q_C/T_C = +500/300. These cancel exactly: −5/3 + 5/3 = 0. This is the defining property of a reversible engine — zero net entropy change. Any irreversible engine would produce Q_C/T_C > Q_H/T_H, giving a positive total ΔS.
Question 3 True / False
Entropy is a state function, which means its change between two states is the same regardless of which path — reversible or irreversible — connects them.
TTrue
FFalse
Answer: True
This is the foundational property that makes entropy calculable. Just as the change in gravitational potential energy between two heights is path-independent, ΔS between two thermodynamic states is fixed. This is why you are allowed — and required — to use a reversible path to compute ΔS even when the actual process was irreversible. The actual process might involve ΔQ_actual ≠ Q_rev, but the entropy change is the same.
Question 4 True / False
For any real process involving heat transfer, you can calculate the entropy change using ΔS = Q_actual/T, where Q_actual is the heat exchanged during the process.
TTrue
FFalse
Answer: False
This formula is only valid for reversible processes. For irreversible processes, Q_actual ≠ Q_rev, and using Q_actual gives the wrong answer — as the free-expansion example shows starkly (Q_actual = 0, yet ΔS > 0). The correct definition is dS = dQ_rev/T, where the heat integral must be taken along a reversible path connecting the same endpoints. For irreversible processes, ΔS_universe > 0, which is encoded in the Clausius inequality: ∮ dQ/T ≤ 0, with equality only for reversible cycles.
Question 5 Short Answer
Why must you use a reversible path to calculate entropy change, even when the actual process was irreversible?
Think about your answer, then reveal below.
Model answer: Entropy is a state function — it has a definite value at each equilibrium state, and the change between two states is path-independent. This means you can choose any convenient path between the same initial and final states to calculate ΔS. Because the definition dS = dQ/T is only rigorously valid for reversible processes (where the system is always near equilibrium), you must construct a reversible path. The actual irreversible path may involve different heat exchanges, but since ΔS is path-independent, the result calculated via the reversible path applies to any process between those states.
The subtlety is that dQ/T along an irreversible path does not give the entropy change — it gives a lower bound (Clausius inequality). Only along a reversible path does dQ_rev/T exactly equal dS. The state-function property is what licenses this substitution: you compute using the convenient reversible path, and the answer is valid for the actual path too.