Gas processes, entropy and reversible cycles
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| coefficient of performance | 性能系数 | xìng néng xì shù |
| entropy production | 熵产生 | shāng chǎn shēng |
A decision before an answer
- A refrigerator can move more heat than the work supplied to it. That ratio is a coefficient of performance, not an efficiency that must be below one.
- Your goal: Apply first-law work and heat signs to gas processes.
Read the relationship
- Use ΔU=Q−W, where Q is heat entering the system and W is work done by it. Quasistatic boundary work is ∫P dV, positive during expansion. For a fixed amount of ideal gas, internal energy depends only on temperature. At constant volume W=0, so ΔU=Q; for an isothermal ideal-gas process ΔU=0, so Q=W. Adiabatic means Q=0, not constant temperature. Over a cycle, the state returns and ΔU=0, so net Q equals net W even though individual legs have different heat transfers.
- Compare reversible entropy balances and P–V cycle areas.
In a reversible process of a system and its environment, which must hold?
Reversibility produces no total entropy; individual system quantities may change.
Use the defining rule
- For reversible isothermal expansion of an ideal gas, PV is constant. With constant heat capacities, a reversible adiabatic path obeys PV^γ=constant, with γ=Cp/Cv>1; through the same initial state its pressure falls faster as volume increases. An isobaric path is horizontal on a P–V plot. A clockwise closed cycle has positive ∮P dV; reversing direction changes the sign. For a rectangular loop the magnitude is ΔPΔV, and a triangular loop has half the corresponding bounding-rectangle area. Use Pa and m³ for joules, not an unconverted litre value.
- Distinguish engine efficiency from refrigerator coefficient of performance.
A reversible refrigerator has TC=270 K and TH=300 K. Its COP is:
QC/W=270/(300−270)=9.
Check the conditions
- For reversible heat transfer, dS=δQ_rev/T; entropy is a state function while heat is path dependent. A reversible system-plus-environment process has zero total entropy change, but the system’s entropy alone can increase or decrease. Irreversible spontaneous processes produce nonnegative total entropy. A reversible isothermal expansion increases the gas entropy by nR ln(V2/V1); the reservoir loses the same amount. Reversibility does not require constant system temperature, zero work or zero internal-energy change.
- Distinguish engine efficiency from refrigerator coefficient of performance.
A clockwise P–V rectangle spans 1×10⁵ to 3×10⁵ Pa and 0.001 to 0.004 m³. Work and net entering heat are 600 J. Reversing it gives −600 J. A Carnot refrigerator operates at TC=280 K and TH=300 K with entropy transfer 2 J/K: QC=560 J, QH=600 J, work input=40 J and COP=14.
A clockwise rectangular P–V cycle has ΔP=4000 Pa and ΔV=0.002 m³. Net work is ____ J.
Positive clockwise work is ΔPΔV=8 J.
Apply the task format
- A reversible engine between TH and TC has efficiency W/QH=1−TC/TH. A reversible refrigerator instead has COP=QC/W=TC/(TH−TC), and a heat pump has QH/W=TH/(TH−TC). Use absolute kelvin temperatures. On a reversible T–S diagram, heat magnitude along an isotherm is T times the entropy change; the Carnot rectangle’s area is the work magnitude. Refrigeration moves heat from cold to hot by consuming work, reversing the engine cycle. COP can exceed one without violating conservation because the moved heat is not supplied solely by the work.
- Distinguish engine efficiency from refrigerator coefficient of performance.
Zero total entropy production does not mean zero system entropy change. A refrigerator uses TC/(TH−TC), not the engine efficiency formula.
Which answer fits this case?
Apply first-law work and heat signs to gas processes
For any complete gas cycle, net entering heat equals net work done by the gas.
ΔU=0 because internal energy is a state function and the initial state returns.
Keep the distinctions
- entropy production 熵产生 — Nonnegative total entropy generated by irreversibility.
- coefficient of performance 性能系数 — Useful heat transferred divided by work input for a refrigerator or heat pump.
- Apply first-law work and heat signs to gas processes.
- Compare reversible entropy balances and P–V cycle areas.
- Distinguish engine efficiency from refrigerator coefficient of performance.
Match each term with its precise meaning in this lesson.
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
The order follows the stated process; check each stage before the next.