IB Physics: Thermodynamics – Key Concepts & Exam Tips | IB物理热力学考点精讲

📚 IB Physics: Thermodynamics – Key Concepts & Exam Tips | IB物理热力学考点精讲

Thermodynamics is a cornerstone of IB Physics, bridging microscopic particle behaviour with macroscopic energy transfers. This guide covers the essential SL and HL topics—from the zeroth law through entropy and Carnot cycles—with clear definitions, key equations, and common pitfalls to help you ace both the theory and calculation questions on your exam.

热力学是IB物理的核心板块之一，它将微观粒子行为与宏观能量传递紧密连接。本篇考点精讲梳理了SL与HL的必考内容—从第零定律到熵和卡诺循环—提供清晰的定义、关键公式和常见易错点，助你顺利攻克理论与计算题。

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1. Temperature, Heat, and Internal Energy | 温度、热量与内能

s or type of material. The SI base unit is the kelvin (K). A temperature in degrees Celsius can be converted by T(K) = θ(°C) + 273.15.

温度是度量物质内粒子无规则运动平均动能的宏观物理量，与物质的质量和种类无关。其SI基本单位是开尔文(K)，摄氏温度换算关系为 T(K) = θ(°C) + 273.15。

Heat is the energy transferred from one body to another solely because of a temperature difference. It is not a property stored in an object; instead, an object possesses thermal energy. The unit of heat is the joule (J).

热量是纯粹由于温差而从一物体传递到另一物体的能量，它不是物体本身储存的性质；物体拥有的是热能。热量的单位是焦耳(J)。

Internal energy U is the sum of the total random kinetic energy and the total potential energy of all the particles in a system. Kinetic energy changes with temperature, while potential energy changes with the state (e.g., during phase changes).

内能U是系统内所有粒子的无规则运动总动能与总势能之和。动能随温度变化，势能则随物态变化（例如相变期间）。

The zeroth law of thermodynamics states that if two systems A and B are each in thermal equilibrium with a third system C, then A and B are in thermal equilibrium with each other. This law underpins the use of thermometers.

热力学第零定律指出：若两个系统A和B分别与第三个系统C处于热平衡，则A与B彼此也处于热平衡。该定律是使用温度计测量温度的基础。

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2. Specific Heat Capacity and Latent Heat | 比热容与潜热

s given by Q = mcΔT. The unit of c is J·kg⁻¹·K⁻¹. The thermal capacity of an object is C = mc, in J·K⁻¹.

比热容c是指单位质量物质温度升高1 K所需吸收的能量，且不发生物态变化。传递的热量由 Q = mcΔT 计算。c的单位是 J·kg⁻¹·K⁻¹。物体的热容为 C = mc，单位 J·K⁻¹。

change the state of 1 kg of a substance at constant temperature. The formula is Q = mL. L_f is for fusion (melting/freezing) and L_v for vaporisation (boiling/condensing). During phase changes, temperature remains constant while internal potential energy changes.

比潜热L是指单位质量物质在恒温下发生物态变化所需吸收的能量。公式为 Q = mL。L_f 为熔解潜热，L_v 为汽化潜热。相变过程中温度保持不变，但内能中的势能发生变化。

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3. The First Law of Thermodynamics | 热力学第一定律

ms, you will encounter the convention where W is positive for compression (work done on gas) and negative for expansion. An alternative form is ΔU = Q − W_by, where W_by eck the sign convention used in your syllabus; the standard IB expectation uses ΔU = Q + W_on.

热力学第一定律是能量守恒的表述：ΔU = Q + W。式中Q为系统吸收的热量，W为外界对系统所做的功。在IB的题目中，通常规定气体被压缩时外界做功为正，气体膨胀时外界做功为负。另一种等价形式为 ΔU = Q − W_by（W_by是系统对外做功）。务必确认考纲所用的符号约定；IB常用的是 ΔU = Q + W_on。

−pΔV. More generally, the area under a p-V curve represents the work done BY the gas, so W_on = − (area under p-V curve). Special cases: isochoric (V constant) ⇒ W = 0 ⇒ ΔU = Q. Adiabatic (Q = 0) ⇒ ΔU = W. Isothermal (ΔU = 0 for ideal gas) ⇒ Q = −W.

对于恒压下的气体，外界对气体做功为 W = −pΔV。普遍地，p-V图上曲线下的面积代表气体对外做功，因此外界对气体做功 W_on = −(p-V曲线下的面积)。特殊情况：等容（V恒定）⇒ W = 0 ⇒ ΔU = Q；绝热 (Q = 0) ⇒ ΔU = W；等温（理想气体ΔU = 0）⇒ Q = −W。

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4. Thermodynamic Processes | 热力学过程

as. Isothermal: T constant, ΔU = 0, pV = constant. Adiabatic: no heat exchange (Q = 0), pV^γ = constant, where γ = C_p/C_v = 0. An adiabatic curve on a p-V diagram is steeper than an isothermal at the same point.

IB常考理想气体的四种理想化过程。等温：T恒定，ΔU = 0，pV = 常数。绝热：无热量交换 (Q = 0)，pV^γ = 常数，其中 γ = C_p/C_v。等压：p恒定，W = −pΔV。等容：V恒定，W = 0。在p-V图中，绝热线比等温线更陡峭。

Process	ΔU	Q	W (on gas)
Isothermal	0	Q = −W	W = −nRT ln(V2/V1)
Adiabatic	ΔU = W	0	W = ΔU
Isobaric	ΔU = Q + W	Q = nCpΔT	W = −pΔV
Isochoric	ΔU = Q	Q = nCvΔT	0

理解这些过程与p-V图的对应关系是解题的基础，尤其需区分等温与绝热膨胀/压缩时的p、V、T变化。

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5. Ideal Gas Equation and Kinetic Model | 理想气体状态方程与分子动理论模型

s, and k = 1.38×10⁻²³ J·K⁻¹. This equation combines Boyle’s law, Charles’s law, and the pressure law.

理想气体的宏观行为由状态方程 pV = nRT = NkT 描述，其中n为摩尔数，R = 8.31 J·mol⁻¹·K⁻¹，N为分子数，k = 1.38×10⁻²³ J·K⁻¹。该方程统一了玻意耳定律、查理定律和压力定律。

ible compared with the gas volume; there are no intermolecular forces except during collisions; and the number of particles is large enough for statistical treatment. From these assumptions, one can derive pV = (1/3)Nm⟨c²⟩, where ⟨c²⟩ is the mean square speed.

分子动理论的基本假设为：分子视为质点且完全相同；做无规则运动，彼此及与器壁的碰撞均为弹性碰撞；分子本身体积与气体所占体积相比可忽略；除碰撞瞬间外分子间无相互作用力；分子数足够多以进行统计处理。基于这些假设可推导出 pV = (1/3)Nm⟨c²⟩，⟨c²⟩为均方速率。

The average translational kinetic energy of a molecule is directly proportional to the absolute temperature: ⟨E_k⟩ microscopic world to the macroscopic thermometer reading. The root-mean-square speed is c_rms = √⟨c²⟩ = √3kT/m = √3RT/M.

分子平均平动动能与绝对温度成正比：⟨E_k⟩ = (3/2)kT，这一关系将微观运动与宏观温度联系起来。方均根速率为 c_rms = √⟨c²⟩ = √3kT/m = √3RT/M。

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6. p-V Diagrams and Work Done | p-V图与做功

p-V diagrams are central to thermal physics questions. The work done BY a gas when it expands from volume V₁ to V₂ is equal to the area under the p-V curve: W_by = ∫ p dV. The work done ON the gas is the negative of this area, W_on = −W_by. For a cyclic process, the net work done by the gas is the area enclosed by the cycle. If the cycle is clockwise, it represents a heat engine; if counterclockwise, it represents a refrigerator or heat pump.

p-V图是热学题目的核心工具。气体从体积V₁膨胀到V₂时，对外做功等于p-V曲线下的面积：W_by = ∫ p dV。外界对气体做功则为该面积的负值，W_on = −W_by。在循环过程中，气体对外做的净功等于循环所围成的面积。顺时针循环代表热机；逆时针循环代表制冷机或热泵。

V, and T. Pay attention to sign conventions when applying the first law. The temperature at any point can be determined from pV = nRT.

分析p-V图时，先判断过程属于等温、绝热、等压还是等容，利用状态方程将p、V、T关联起来。应用第一定律时需格外留意符号规定。任一点的气体温度都可以通过 pV = nRT 求得。

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7. The Second Law of Thermodynamics | 热力学第二定律

The second law has two classic statements. The Clausius statement: Heat cannot pass spontaneously from a colder body to a hotter body. The Kelvin-Planck statement: It is impossible to construct a heat engine that, operating in a cycle, converts all the heat from a single heat reservoir into work without any other effect. Both statements emphasise the directional nature of thermal processes and the necessity of a cold reservoir.

热力学第二定律有两种经典表述。克劳修斯表述：热量不可能自发地从低温物体传向高温物体。开尔文-普朗克表述：不可能制造出一种循环工作的热机，它只从单一热源吸热并将之全部转化为功，而不产生其他影响。两种表述均强调了热过程的方向性以及需要低温热源这一事实。

In practice, this means every heat engine must reject some waste heat to a cold reservoir. The second law also gives rise to the concept of entropy and explains why certain processes (like the expansion of a gas into a vacuum) are irreversible.

在实践中，这意味着任何热机都必然要向低温热源排放一部分废热。第二定律还衍生出熵的概念，并解释了为何某些过程（如气体向真空自由膨胀）是不可逆的。

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8. Entropy | 熵

the change in entropy of a system is ΔS = Q/T, where Q is the heat transferred at constant temperature T (in kelvin). The unit of entropy is J·K⁻¹.

熵S是系统无序程度或热力学状态可实现的微观状态数目的量度。对于可逆等温过程，系统熵的变化为 ΔS = Q/T，其中Q为在恒定温度T(K)下传递的热量。熵的单位是 J·K⁻¹。

The second law in entropy terms: The total entropy of an isolated system never decreases. It remains constant in a reversible process and increases in an irreversible process. Thus, for the universe, ΔS_total ≥ 0. When calculating entropy changes, consider both the system and the surroundings.

用熵表述的第二定律为：孤立系统的总熵永不减小。可逆过程中总熵不变，不可逆过程中总熵增加。因此，对整个宇宙而言 ΔS_total ≥ 0。计算熵变时需同时考虑系统与环境。

A common HL calculation: the entropy change when ice melts at 0 °C, ΔS = mL_f/T. Another is heating water: ΔS = mc ln(T₂/T₁) for a constant-pressure heating process if c is constant.

常考的HL计算包括：0 °C时冰融化的熵变 ΔS = mL_f/T；以及水在恒压下加热的熵变，当比热容恒定时 ΔS = mc ln(T₂/T₁)。

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9. Heat Engines and Efficiency | 热机与效率

A heat engine operates in a cycle between a hot reservoir at temperature T_H and a cold reservoir at T_C. It absorbs heat Q_H, does net work W, and rejects Q_C to the cold reservoir. By energy conservation, W = Q_H − Q_C. The thermal efficiency is defined as η = W/Q_H = 1 − Q_C/Q_H. Efficiency is always less than 1 (and less than the Carnot efficiency for real engines).

热机在高温热源T_H与低温热源T_C之间循环工作，从高温热源吸热Q_H，做净功W，并向低温热源排放Q_C。由能量守恒得 W = Q_H − Q_C。热效率定义为 η = W/Q_H = 1 − Q_C/Q_H，效率恒小于1（对于真实热机也小于对应的卡诺效率）。

Many exam questions ask you to calculate efficiency from a p-V cycle by finding net work (enclosed area) and heat input (often using the first law on individual legs of the cycle). Ensure you only include Q_H that is actually absorbed from the hot reservoir, not the net heat transfer.

许多考题要求从p-V循环图中计算效率，方法是先求净功（图形所围面积），再求实际从高温热源吸收的热量Q_H（通常需对各分过程使用第一定律）。注意只计入真正从高温热源吸收的热量，而非净传热量。

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10. Carnot Cycle and Carnot Efficiency (HL) | 卡诺循环与卡诺效率（HL）

The Carnot cycle is an idealised, reversible cycle consisting of two isothermal and two adiabatic processes. It operates between T_H and T_C and achieves the maximum possible efficiency for any heat engine working between these two temperatures. The Carnot efficiency is η_Carnot = 1 − T_C/T_H, with temperatures in kelvin. No real engine can exceed this efficiency.

卡诺循环是由两条等温过程和两条绝热过程组成的理想可逆循环，工作在T_H与T_C之间，可实现这两温度之间任何热机的最高效率。卡诺效率为 η_Carnot = 1 − T_C/T_H，温度需用开尔文。任何真实热机的效率都不可能超过这一值。

For refrigerators and heat pumps, the coefficient of performance (COP) is used instead of efficiency. For a refrigerator, COP_R = Q_C/W = Q_C/(Q_H − Q_C). For a heat pump, COP_HP = Q_H/W. The ideal (Carnot) limits are COP_{R, Carnot} = T_C/(T_H − T_C) and COP_{HP, Carnot} = T_H/(T_H − T_C).

对于制冷机和热泵，常用性能系数(COP)代替效率。制冷机：COP_R = Q_C/W = Q_C/(Q_H − Q_C)；热泵：COP_HP = Q_H/W。理想（卡诺）极限值分别为 COP_{R, Carnot} = T_C/(T_H − T_C) 与 COP_{HP, Carnot} = T_H/(T_H − T_C)。

Understanding the Carnot cycle reinforces the idea that efficiency is determined only by reservoir temperatures for reversible engines, and that to improve efficiency one must either raise T_H or lower T_C.

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