Quá trình khí, entropy và chu trình thuận nghịch

Moving more heat than the work supplied

• A refrigerator pumping 560 J out of a 280 K interior with only 40 J of work has a coefficient of performance 性能系数 (COP) of 14. That is not an efficiency above 100% — the heat is moved, not created.
• GRE thermodynamics tests first-law signs, process paths on $P$–$V$ diagrams, entropy and the Carnot limits for engines and refrigerators.

First law along a path

• $\Delta U = Q - W$, $W = \int P\,dV$ (by the gas). Ideal gas: $U$ depends only on $T$.
• Constant volume: $W = 0$, $\Delta U = Q$. Isothermal 等温的: $\Delta U = 0$, $Q = W$. Adiabatic 绝热的: $Q = 0$ (not constant $T$). Cycle: $\Delta U = 0$, net $Q$ = net $W$.

Reading $P$–$V$ diagrams

• Isotherm $PV =$ const; adiabat $PV^\gamma =$ const, steeper. Clockwise cycle: positive net work equal to the enclosed area.
• Rectangle from $1$ to $3\times10^5$ Pa and $0.001$ to $0.004$ m³: $W = 2\times10^5 \times 0.003 = 600$ J. Use Pa and m³, not litres.

Entropy

• $dS = \delta Q_{\text{rev}}/T$; $S$ is a state function. Reversible isothermal expansion: $\Delta S_{\text{gas}} = nR\ln(V_2/V_1)$; the reservoir loses the same.
• A reversible process has zero total entropy change; the system's entropy can still change.

Engines, refrigerators, heat pumps

• Engine $\eta = 1 - T_C/T_H$; refrigerator $\text{COP} = \dfrac{T_C}{T_H - T_C}$; heat pump $\dfrac{T_H}{T_H - T_C}$ (all Carnot limits, kelvin).

Recap