Unit B.4: Thermodynamics单元 B.4：热力学 HL only

Internal energy $U$ is a property of the system's current state: a gas at a given $p$, $V$, $T$ has one definite $U$, however it got there. So $\oint dU = 0$ around any closed cycle, and $\Delta U$ depends only on the endpoints.

内能 $U$ 是系统当前状态的属性：给定 $p$、$V$、$T$ 的气体只有一个确定的 $U$，与到达方式无关。故任意闭合循环 $\oint dU = 0$，$\Delta U$ 只取决于始末点。

Heat $Q$ and work $W$ are path functions: they measure energy in transit and depend on the route taken in the $p$-$V$ plane. Two different paths between the same two states transfer different $Q$ and $W$, yet always give the same $\Delta U = Q - W$. This is the experimental content of the first law.

热量 $Q$ 与功 $W$ 是路径函数：它们度量传递中的能量，取决于在 $p$-$V$ 平面上走的路线。同样两个状态间的不同路径传递的 $Q$、$W$ 不同，但 $\Delta U = Q - W$ 始终相同。这正是第一定律的实验内涵。

Take a gas from state X to state Y two ways: (1) expand at high pressure, then drop pressure; (2) drop pressure first, then expand at low pressure. Both reach the same Y, but the areas under the two $p$-$V$ paths differ, so they do different amounts of work.

把气体由状态 X 带到状态 Y 有两种走法：(1) 先高压膨胀再降压；(2) 先降压再低压膨胀。两者都到达同一 Y，但两条 $p$-$V$ 路径下方面积不同，故做的功不同。

This is exactly why $W$ is a path function. The difference in work between the two routes is matched by an equal difference in heat $Q$, so that $\Delta U = Q - W$ comes out the same for both — consistent with $U$ being a state function (B4.1).

这正是 $W$ 为路径函数的原因。两条路线做功之差由热量 $Q$ 的等量之差补偿，使 $\Delta U = Q - W$ 两路相同——与 $U$ 为态函数（B4.1）一致。

Along an isotherm, $pV = $ constant, so $p \propto 1/V$. Along an adiabat, $pV^{\gamma} = $ constant with $\gamma = C_{p}/C_{V} > 1$, so $p \propto V^{-\gamma}$. Because $\gamma > 1$, the adiabatic curve falls off more rapidly with $V$: at any point it is steeper than the isotherm through the same point.

沿等温线 $pV = $ 常数，故 $p \propto 1/V$。沿绝热线 $pV^{\gamma} = $ 常数，$\gamma = C_{p}/C_{V} > 1$，故 $p \propto V^{-\gamma}$。因 $\gamma > 1$，绝热曲线随 $V$ 下降得更快：在任一点处都比过该点的等温线更陡。

Physically: during an adiabatic expansion no heat enters, so the gas does work entirely at the expense of its internal energy, cooling as it expands. The extra temperature drop pulls the pressure down faster than the isothermal case where temperature is held fixed. (The IB does not require the $pV^{\gamma}$ relation itself, only the qualitative "steeper" shape.)

物理上：绝热膨胀中无热量进入，气体完全以内能为代价做功，膨胀时降温。额外的降温使压强下降得比保持温度不变的等温情形更快。（IB 不要求 $pV^{\gamma}$ 关系本身，只要求定性的"更陡"形状。）

Microscopically, entropy measures the number of microstates $\Omega$ consistent with a system's macrostate, via Boltzmann's relation $S = k_{B} \ln \Omega$. A spread-out, disordered configuration corresponds to vastly more microstates than an ordered one.

微观上，熵通过玻尔兹曼关系 $S = k_{B} \ln \Omega$ 度量与某宏观态对应的微观态数目 $\Omega$。分散、无序的组态对应的微观态数目远多于有序组态。

Spontaneous change moves toward the macrostate with the most microstates simply because it is overwhelmingly more probable. The second law is therefore statistical: a decrease in total entropy is not strictly impossible, just astronomically unlikely for any macroscopic system. This statistical asymmetry is what gives time a forward direction. (IB requires the macroscopic $\Delta S = \Delta Q / T$ statement; the Boltzmann picture is enrichment.)

自发变化趋向微观态最多的宏观态，仅仅因为它的概率压倒性地大。第二定律因此是统计性的：总熵减小并非严格不可能，只是对任何宏观系统而言概率小到天文级。正是这种统计不对称赋予时间以前进方向。（IB 要求宏观的 $\Delta S = \Delta Q / T$ 陈述；玻尔兹曼图景为拓展。）

The Kelvin form of the second law states: no cyclic engine can take heat from a single reservoir and convert it entirely into work. If $Q_{\mathrm{out}} = 0$, then $\eta = 1$, which this statement forbids.

第二定律的开尔文表述：没有任何循环热机能从单一热源吸热并将其完全转化为功。若 $Q_{\mathrm{out}} = 0$ 则 $\eta = 1$，正是该表述所禁止的。

Entropy makes the reason explicit. Drawing heat $Q_{\mathrm{in}}$ from the hot reservoir lowers its entropy by $Q_{\mathrm{in}}/T_{H}$. The engine itself returns to its start each cycle ($\Delta S_{\text{engine}} = 0$). To keep $\Delta S_{\text{total}} \ge 0$, some heat $Q_{\mathrm{out}}$ must be dumped to the cold reservoir, raising its entropy by $Q_{\mathrm{out}}/T_{C}$. A nonzero $Q_{\mathrm{out}}$ is therefore mandatory, so $\eta < 1$ for every real engine.

熵把原因说清楚了。从热源取热 $Q_{\mathrm{in}}$ 使其熵下降 $Q_{\mathrm{in}}/T_{H}$。热机每循环回到起点（$\Delta S_{\text{engine}} = 0$）。为保持 $\Delta S_{\text{total}} \ge 0$，必须向冷源排出一些热量 $Q_{\mathrm{out}}$，使其熵升高 $Q_{\mathrm{out}}/T_{C}$。因此非零的 $Q_{\mathrm{out}}$ 是必需的，故任何真实热机 $\eta < 1$。

Carnot's theorem says every reversible engine between two reservoirs has the same efficiency $\eta_{\text{Carnot}}$, and no engine can beat it. The reason is entropy. A reversible cycle produces zero total entropy: the entropy drawn from the hot reservoir, $Q_{\mathrm{in}}/T_{H}$, exactly equals the entropy delivered to the cold one, $Q_{\mathrm{out}}/T_{C}$.

卡诺定理指出：两热源间的每台可逆热机效率都相同，等于 $\eta_{\text{Carnot}}$，且无机可超越。原因在于熵。可逆循环产生零总熵：从热源取走的熵 $Q_{\mathrm{in}}/T_{H}$ 恰好等于排向冷源的熵 $Q_{\mathrm{out}}/T_{C}$。

Setting $Q_{\mathrm{in}}/T_{H} = Q_{\mathrm{out}}/T_{C}$ gives $Q_{\mathrm{out}}/Q_{\mathrm{in}} = T_{C}/T_{H}$. Substituting into $\eta = 1 - Q_{\mathrm{out}}/Q_{\mathrm{in}}$ yields $\eta_{\text{Carnot}} = 1 - T_{C}/T_{H}$ directly. Any real (irreversible) engine generates extra entropy, which forces $Q_{\mathrm{out}}/Q_{\mathrm{in}} > T_{C}/T_{H}$ and hence $\eta < \eta_{\text{Carnot}}$.

令 $Q_{\mathrm{in}}/T_{H} = Q_{\mathrm{out}}/T_{C}$，得 $Q_{\mathrm{out}}/Q_{\mathrm{in}} = T_{C}/T_{H}$。代入 $\eta = 1 - Q_{\mathrm{out}}/Q_{\mathrm{in}}$ 直接得 $\eta_{\text{Carnot}} = 1 - T_{C}/T_{H}$。任何真实（不可逆）热机产生额外熵，迫使 $Q_{\mathrm{out}}/Q_{\mathrm{in}} > T_{C}/T_{H}$，故 $\eta < \eta_{\text{Carnot}}$。