Entropy & Second Law Problem Patterns (IPhO Thermo)
IPhO thermodynamics lesson on entropy bookkeeping: computing entropy changes fast and spotting second-law constraints in common setups.
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Second-law marks usually come from two moves: compute Δ S using a reversible surrogate path, then use Δ Sᵤₙᵢᵥ to test feasibility or bound work/heat.
- Entropy change of a system: Δ S = ∫ (δ Qᵣₑᵥ)/T (choose any reversible path between the same endpoints).
- Reservoir at fixed temperature T₀: Δ Sᵣₑₛ = -Q_sys/T₀.
- Second law: Δ Sᵤₙᵢᵥ = Δ S_sys + Δ Sₛᵤᵣᵣ ≥ 0 with equality only for a reversible overall process.
1. Definitions (Must Know)
- Entropy S: a state function measuring how many microstates are compatible with a macrostate. In problems: it is a bookkeeping tool for irreversibility.
- Reversible process: an idealized process that can be reversed with no net change to system plus surroundings. In entropy terms: Δ Sᵤₙᵢᵥ = 0.
- Irreversible process: real process with dissipation (friction, finite temperature difference heat flow, free expansion). It satisfies Δ Sᵤₙᵢᵥ ≥ 0 with strict inequality.
- Clausius inequality: for a cycle, ∮ (δ Q)/T ≤ 0.
- Entropy generation S_gen: the nonnegative extra entropy produced by irreversibility. Commonly: Δ Sᵤₙᵢᵥ = S_gen if the universe is otherwise closed.
- Heat reservoir: a body with (effectively) constant T even when it exchanges heat. This makes Δ S easy.
- Carnot engine limit: maximum efficiency between reservoirs Tₕ and T_c is ηₘₐₓ = 1-T_c/Tₕ.
2. Key Ideas (What Earns Marks)
- Compute Δ S_sys with a reversible path: even if the real process is irreversible, entropy is a state function.
- Compute reservoir entropy directly: for a reservoir, do not integrate. Use Δ Sᵣₑₛ = -Q/T₀.
- Always decide what is the “universe”: isolated composite? system plus reservoirs? This decides which entropy sum is constrained.
- Second law as a feasibility test: if your computed Δ Sᵤₙᵢᵥ is negative, the proposed process cannot occur as stated.
- Second law as an optimization bound: “maximum work”, “minimum heat input”, “best possible COP” nearly always means “assume reversible and use equality”.
3. Detailed Explanations
A. The fastest entropy balance template
Write Δ Sᵤₙᵢᵥ = Δ S_(system(s)) + ∑ᵢ Δ S_(reservoir i).
- If the only surroundings are reservoirs at fixed temperatures: Δ Sᵤₙᵢᵥ ≥ 0 gives an inequality in Qᵢ and the system state change.
- If the “universe” is an isolated composite that comes to equilibrium: Δ Sᵤₙᵢᵥ is just the total entropy change of the composite, and it must be nonnegative.
B. High-yield formulas for ideal gases
For an ideal gas with constant heat capacities: Δ S = nC_V ln(T_f/Tᵢ) + nR ln(V_f/Vᵢ) or equivalently Δ S = nC_P ln(T_f/Tᵢ)-nR ln(P_f/Pᵢ).
These are reversible-path results, but Δ S is the same for any real path connecting the same endpoints.
Micro-pattern: adiabatic does not mean zero entropy change
- Reversible adiabatic: Q = 0 and Δ S = 0.
- Irreversible adiabatic (friction, free expansion): Q = 0 but Δ S > 0 for the system, because S_gen is produced internally.
C. Heat exchange between finite bodies (common IPhO composite)
If two bodies exchange heat in isolation, the final equilibrium temperature is found from energy conservation, but the direction and feasibility are guaranteed by the second law.
For constant heat capacities C₁ and C₂:
- Energy: C₁(T_f-T₁) + C₂(T_f-T₂) = 0.
- Entropy change: Δ S = C₁ ln(T_f/T₁) + C₂ ln(T_f/T₂).
D. “One reservoir + a system” problems
If a system changes state while in thermal contact with a single reservoir at T₀: Δ Sᵤₙᵢᵥ = Δ S_sys-Q/T₀ ≥ 0.
This is a clean way to bound the heat Q (and through the first law, the work).
E. Engine and refrigerator constraints (entropy is the gatekeeper)
Between reservoirs Tₕ and T_c:
- Reversible engine: Qₕ/Tₕ = Q_c/T_c and W = Qₕ-Q_c.
- Any real engine: Qₕ/Tₕ-Q_c/T_c = Δ Sᵤₙᵢᵥ ≥ 0.
So given Qₕ and Tₕ,T_c, you get an immediate bound on Q_c and therefore on the maximum possible work.
4. Common Mistakes
- Using Δ S = ∫ δ Q/T along an irreversible path (you must use δ Qᵣₑᵥ or a reversible surrogate).
- Forgetting the reservoir sign: if the system absorbs heat Q, the reservoir loses that heat and its entropy decreases by Q/T₀.
- Treating any adiabatic process as Δ S = 0 (only true if the process is reversible).
- Mixing Celsius and kelvin inside logarithms.
- Computing Δ S_sys correctly but forgetting to include the surroundings, then making a wrong second-law conclusion.
- Dropping the degeneracy-style effect in mixing problems: entropy can increase even when Δ U = 0.
5. Exam Tips
- For any proposed process: do a one-line check of Δ Sᵤₙᵢᵥ.
- When you see “maximum” or “minimum”, assume the reversible limit first and use equality (Δ Sᵤₙᵢᵥ = 0) to get the bound.
- For ideal gases, memorize Δ S = nC_V ln(T_f/Tᵢ) + nR ln(V_f/Vᵢ) and use it aggressively.
- If you do not know the real path, pick a convenient reversible path (two-step isothermal plus isochoric often works).
- If algebra gets messy, take logs late, and keep temperatures symbolic until the last step.
6. Worked Examples
Example 1: Two bodies come to equilibrium (isolation)
Two identical blocks each have constant heat capacity C. One starts at T₁ and the other at T₂ with T₂ hotter. They are placed in thermal contact in an insulated box. Find T_f and Δ Sᵤₙᵢᵥ.
Worked solution (energy first, then entropy)
Energy conservation in the isolated box gives C(T_f-T₁) + C(T_f-T₂) = 0 ⇒ T_f = (T₁ + T₂)/2.
Total entropy change is the sum:
Since T_f is the arithmetic mean, T_f² ≥ T₁T₂ (AM-GM), so Δ S ≥ 0 with equality only if T₁ = T₂.
Example 2: Free expansion of an ideal gas (adiabatic but irreversible)
One mole of ideal gas in an insulated container expands freely into vacuum, doubling its volume from V to 2V. Find Δ S of the gas and Δ Sᵤₙᵢᵥ.
Worked solution (use a reversible surrogate path)
Free expansion into vacuum does no work and exchanges no heat: Q = 0 and W = 0, so Δ U = 0.
For an ideal gas, U depends only on T, so Δ U = 0 implies Δ T = 0. The endpoints are therefore (T,V) to (T,2V).
Compute entropy using a reversible isothermal expansion between the same endpoints: Δ S = nR ln(V_f/Vᵢ) = R ln 2.
The surroundings are vacuum and insulated walls, so Δ Sᵤₙᵢᵥ = Δ S_gas = R ln 2.
Example 3: Second-law bound on work from two reservoirs
An engine takes heat Qₕ from a hot reservoir at Tₕ and rejects heat Q_c to a cold reservoir at T_c, producing work W = Qₕ-Q_c. What is the maximum possible work for given Qₕ, Tₕ, and T_c?
Worked solution (reversible limit gives the bound)
The universe entropy change is Δ Sᵤₙᵢᵥ = -Qₕ/Tₕ + Q_c/T_c ≥ 0.
Therefore Q_c ≥ Qₕ T_c/Tₕ.
So the work satisfies W = Qₕ-Q_c ≤ Qₕ(1-T_c/Tₕ).
Equality holds in the reversible (Carnot) limit.
7. Mind Stretchers
- A system at temperature T is brought into contact with a reservoir at T₀ and allowed to equilibrate. What is the maximum possible work you can extract during this relaxation if you are allowed to run any engine you like in between?
- A piston expands with kinetic energy and then dissipates that energy as internal heating. Track where the entropy is produced and which term(s) in Δ Sᵤₙᵢᵥ “see” it.
- Entropy decreases in the system: give a concrete example where Δ S_sys < 0 but Δ Sᵤₙᵢᵥ ≥ 0.
Mind-stretcher (solution idea): maximum work from a finite hot body to a cold reservoir
Suppose a body with heat capacity C(T) cools reversibly from initial temperature Tᵢ to the reservoir temperature T₀.
At each step, you can run an infinitesimal reversible engine between the body (as the hot side) at temperature T and the reservoir at T₀. For a reversible engine, (δ Q_c)/T₀ = (δ Qₕ)/T, δ W = δ Qₕ-δ Q_c = δ Qₕ(1-T₀/T).
The body supplies heat δ Qₕ = -C(T) dT (positive when T decreases), so Wₘₐₓ = ∫_T₀^Tᵢ C(T)(1-T₀/T)dT.
Any irreversibility increases Δ Sᵤₙᵢᵥ and reduces the extractable work below this bound.
8. Practice
- For each past problem, force yourself to write Δ Sᵤₙᵢᵥ as a sum of system plus reservoirs.
- Drill the three templates: two bodies equilibrating, one reservoir plus a system, and two-reservoir engine bounds.
Syllabus and review details
No official syllabus alignment is listed for this lesson.