Systems, State & the Four Laws
Control mass vs control volume, the state postulate, zeroth through second laws with the Clausius inequality, and theKelvin–Planck statement.
- Classify systems (open/closed/isolated) and apply the state postulate
- State the zeroth, first, second and third laws and what each forbids
- Use Clausius inequality to decide if a cycle is possible, reversible or impossible
A heat engine that converted all heat to work would end civilization's energy crisis overnight — the Second Law says it is not just hard, it is mathematically forbidden. Every power plant, refrigerator and living cell obeys the same boundary.
System vocabulary:
- Open (control volume): mass crosses boundary — turbines, nozzles, compressors.
- Closed (control mass): fixed mass, energy crosses as heat/work — piston-cylinder.
- Isolated: neither mass nor energy — universe as a whole (theoretical).
- State postulate: two independent intensive properties fix the state of a simple compressible system — the foundation of every property table.
Zeroth law: if A equilibrates with B and B with C, then A with C — defines temperature and makes thermometers possible.
First law (energy conservation): for a closed system, δQ − δW = dU; path-independent U, path-dependent Q and W. Rate form: Q̇ − Ẇ = dU/dt. For a control volume use the SFEE (steady-flow energy equation): Q̇ − Ẇ = ṁ(h₂ − h₁ + Δke + Δpe).
Second law (directionality): heat does not spontaneously flow cold→hot; work does not convert to heat with 100% efficiency in a cycle. Equivalent statements:
- Clausius: ∮δQ/T ≤ 0 (equality reversible).
- Kelvin–Planck: no cycle can convert heat entirely to work; some must reject to a low-T sink.
- Entropy (Clausius definition): dS = δQ_rev/T; for any process ΔS ≥ ∫δQ/T.
Third law: entropy of a perfect crystal at 0 K is zero — absolute zero is unreachable in finite steps.
| One-line statement | |
|---|---|
| Zeroth | Thermal equilibrium is transitive — temperature exists |
| First | Energy is conserved: ΔU = Q − W (closed) |
| Second | Entropy of an isolated system never decreases |
| Third | S = 0 at 0 K for a perfect crystal; absolute 0 unattainable |
Q > 0 into system, W > 0 done BY system — pick this everywhere and never flip mid-problem.
Energy bookkeeping alone; the Second Law rejects the second kind (100%-efficient cycle).
- 1
Compute ∮δQ/T
1000/800 − 600/300 = 1.25 − 2.00 = −0.75 kJ/K.
- 2
Compare to zero
Negative, so Clausius inequality is satisfied — the cycle is possible.
- 3
Reversible?
Equality holds only for reversible cycles. Here ∮δQ/T = −0.75 ≠ 0, so irreversible.
- 4
Cross-check efficiency
η = 1 − 600/1000 = 0.40. Carnot bound at same temps: 1 − 300/800 = 0.625. 0.40 < 0.625 — consistent with irreversibility.
- 5
Verdict
Possible and irreversible; if ∮δQ/T > 0 it would be impossible.
Solution
∮δQ/T = −0.75 kJ/K < 0 → possible but irreversible (not a Carnot cycle).
A proposed cycle has ∮δQ/T = +0.2 kJ/K. What is the verdict, and why is the sign decisive?
Hint: Clausius allows ≤ 0 only.
Treating Q and W as state functions
Only U, H, S, T, P are state functions; Q and W are path-dependent — never write dQ, write δQ.
Applying the closed-system First law to a turbine
Turbines are control volumes — use SFEE with enthalpy h, not internal energy U.
State the Kelvin–Planck statement of the Second Law and its equivalence to Clausius.
No cycle can have as its sole result the conversion of heat from a single reservoir entirely into work. Equivalence proof: assume a Clausius violator (net heat cold→hot with no work) and couple it to a Kelvin–Planck engine to build a forbidden engine.
Source: Standard proof question in ME exams.
- State postulate: two independent intensive properties fix a simple compressible state.
- Zeroth defines T; First conserves energy; Second sets direction; Third anchors S.
- Clausius inequality ∮δQ/T ≤ 0 is the universal cycle feasibility test.
