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TS.1 · Gas processes, entropy and reversible cycles

GRE · GRE Subject Test · GRE Physics · Topic 19

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19.1

Gas processes, entropy and reversible cycles

A refrigerator 制冷机 can move more heat than the work supplied to it. That ratio is a coefficient of performance 性能系数, not an efficiency that must be below one.

Prerequisites: 4.

  • 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
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English
coefficient of performance/ˌkəʊɪˈfɪʃənt ɒv pəˈfɔːməns/
refrigerator
19.2

Choose the system and model

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.

19.3

Use the governing relation

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.

19.4

Apply 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.

19.5

Check the conclusion

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.

19.6

Worked method

A reversible refrigerator transfers entropy Delta S from Tc to Th.

$$Q_c=T_c\Delta S=(270\,\mathrm K)(3.0\,\mathrm{J/K})=810\,\mathrm J.$$
$$Q_h=T_h\Delta S=(300\,\mathrm K)(3.0\,\mathrm{J/K})=900\,\mathrm J.$$
$$W_{in}=Q_h-Q_c=900\,\mathrm J-810\,\mathrm J=90\,\mathrm J.$$
$$\mathrm{COP}=Q_c/W_{in}=(810\,\mathrm J)/(90\,\mathrm J)=9.$$
COP above one is possible because heat is moved, rather than all produced from work.

Gas processes, entropy and reversible cycles: GRE original diagram
Gas processes, entropy and reversible cycles: original GRE teaching diagram.
19.7

Check conditions and vocabulary

Zero total entropy production 熵产生 does not mean zero system entropy change. A refrigerator uses TC/(TH−TC), not the engine efficiency formula.

entropy production: Nonnegative total entropy generated by irreversibility.

coefficient of performance: Useful heat transferred divided by work input for a refrigerator or heat pump.

Vocabulary Train
English
entropy production/ˈentrəpi prəˈdʌkʃn/

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