📚 A-Level Physics: Key Formula Derivations for Thermodynamics and Kinetics (A2 Physical Unit 3) | A-Level 物理：热力学与动力学关键公式推导（A2 物理单元3）
A-Level Physics demands not just the ability to use equations, but a deep understanding of where they come from. Thermodynamics and kinetics are central topics in the A2 specification, bridging microscopic behaviour with macroscopic measurements. This article walks through the step-by-step derivations of the most important formulas you will encounter in your Oxford AQA International A-Level Physical Unit 3 content, helping you master both the ‘how’ and the ‘why’.
A-Level 物理不仅要求会使用方程，更需要深入理解它们的来源。热力学和动力学是 A2 课程的核心主题，连接着微观行为与宏观测量。本文将逐步推导你在 Oxford AQA International A-Level 物理单元 3 中将遇到的最重要公式，帮助你掌握’如何做’以及’为什么这么做’。
1. Introduction to Thermodynamic Derivation | 热力学推导简介
Thermodynamic derivations link the properties of individual particles to the bulk properties of a system. Starting from simple models such as the kinetic theory of gases, we can obtain equations like pV = nRT and the Maxwell-Boltzmann distribution. Each derivation reinforces fundamental principles of conservation of energy, momentum, and statistical mechanics.
热力学推导将单个粒子的性质与系统的宏观性质联系起来。从气体分子运动论等简单模型出发，我们可以得到 pV = nRT 和麦克斯韦-玻尔兹曼分布等方程。每一个推导都强化了能量守恒、动量守恒和统计力学的基本原理。
2. Kinetic Theory Assumptions & Derivation of Pressure | 分子运动论假设与压强公式推导
To derive the pressure exerted by an ideal gas, imagine a cubic box of side L containing N molecules of mass m, moving with random velocities. The key assumptions are: (i) molecules are point particles, (ii) collisions with walls are perfectly elastic, (iii) no intermolecular forces, and (iv) large N so statistical treatment is valid.
为推导理想气体的压强，想象一个边长为 L 的立方体容器，内有 N 个质量为 m 的分子随机运动。关键假设包括：(i) 分子视为质点，(ii) 与器壁的碰撞是完全弹性的，(iii) 分子间无作用力，(iv) N 很大，统计处理有效。
Consider a molecule with velocity v (components vx, vy, vz). When it hits a wall perpendicular to the x‑axis, its momentum change is –2mvx. The time between collisions with the same wall is 2L/vx, so the average force on that wall is F = (2mvx) / (2L/vx) = mvx²/L. Summing over all molecules, the total force on the wall is F = (m/L) ∑ vxi². Pressure p = F / L² = (m/L³) ∑ vxi² = (m/V) ∑ vxi². Since the motion is random, ⟨vx²⟩ = (1/3)⟨v²⟩, giving ∑ vxi² = (N/3)⟨v²⟩. Hence:
考虑一个速度为 v（分量 vx, vy, vz）的分子。当它撞击垂直于 x 轴的器壁时，动量变化为 –2mvx。两次撞击同一器壁的时间间隔为 2L/vx，因此对该壁的平均力为 F = (2mvx) / (2L/vx) = mvx²/L。对所有分子求和，墙上的总力 F = (m/L) ∑ vxi²。压强 p = F / L² = (m/L³) ∑ vxi² = (m/V) ∑ vxi²。由于运动是随机的，⟨vx²⟩ = (1/3)⟨v²⟩，从而 ∑ vxi² = (N/3)⟨v²⟩。因此：
p = ⅓ (N m / V) ⟨v²⟩ or pV = ⅓ N m ⟨v²⟩
This is the fundamental kinetic equation linking pressure to the mean square speed of molecules.
这就是将压强与分子方均速率联系起来的基本动力学方程。
3. Ideal Gas Equation pV = nRT | 理想气体方程 pV = nRT 的推导
The kinetic model links the microscopic ⟨v²⟩ to the thermodynamic temperature. The average translational kinetic energy per molecule is (1/2)m⟨v²⟩. For an ideal gas, this is directly proportional to the absolute temperature: (1/2)m⟨v²⟩ = (3/2)kT, where k = 1.38×10⁻²³ J K⁻¹ is the Boltzmann constant. Substituting into pV = (1/3)N m ⟨v²⟩ gives:
分子运动模型将微观的 ⟨v²⟩ 与热力学温度联系起来。每个分子的平均平动动能为 (1/2)m⟨v²⟩。对于理想气体，这直接正比于绝对温度：(1/2)m⟨v²⟩ = (3/2)kT，其中 k = 1.38×10⁻²³ J K⁻¹ 为玻尔兹曼常数。代入 pV = (1/3)N m ⟨v²⟩ 得到：
pV = N k T
Introducing the amount of substance n (in moles) and the Avogadro constant Nₐ = 6.02×10²³ mol⁻¹, we have N = n Nₐ. The universal gas constant R = Nₐ k = 8.31 J mol⁻¹ K⁻¹. Thus:
引入物质的量 n（以摩尔为单位）和阿伏伽德罗常数 Nₐ = 6.02×10²³ mol⁻¹，有 N = n Nₐ。普适气体常数 R = Nₐ k = 8.31 J mol⁻¹ K⁻¹。于是：
pV = n R T
This derivation unites the microscopic world with the measurable pressure, volume, and temperature of a gas.
这一推导将微观世界与气体的可测压力、体积和温度统一起来。
4. Maxwell-Boltzmann Speed Distribution (Simplified Derivation) | 麦克斯韦-玻尔兹曼速率分布（简化推导）
The distribution of molecular speeds in a gas at temperature T is not single-valued but follows a probability density function. Starting from the Boltzmann factor, the probability that a molecule has a particular velocity vector v is proportional to e–m(vx²+vy²+vz²)/2kT. To find the speed distribution, we consider velocities in the range v to v + dv, which correspond to a spherical shell of volume 4πv² dv in velocity space. Integrating over all directions and normalising yields:
温度为 T 的气体中分子速率的分布并非单一值，而是遵循一个概率密度函数。从玻尔兹曼因子出发，分子具有特定速度矢量 v 的概率正比于 e–m(vx²+vy²+vz²)/2kT。为找到速率分布，我们考虑速率在 v 到 v + dv 范围内的速度，这对应于速度空间中体积为 4πv² dv 的球壳。对所有方向积分并归一化后得到：
f(v) = 4π (m/2πkT)3/2 v² e–mv²/2kT
From this distribution, we can derive the most probable speed vmp = √(2kT/m), the average speed ⟨v⟩ = √(8kT/πm), and the root-mean-square speed vrms = √(3kT/m). Note vrms² = ⟨v²⟩, consistent with the earlier kinetic derivation.
从这个分布函数，我们可以推导出最概然速率 vmp = √(2kT/m)，平均速率 ⟨v⟩ = √(8kT/πm)，以及方均根速率 vrms = √(3kT/m)。注意 vrms² = ⟨v²⟩，与之前的动力学推导一致。
5. First Law of Thermodynamics & Enthalpy | 热力学第一定律与焓
The first law states that the change in internal energy ΔU of a closed system equals the heat added to the system Q plus the work done on the system W: ΔU = Q + W. For pressure‑volume work, a reversible expansion does work –p dV on the system, so dU = dQ – p dV.
热力学第一定律指出，封闭系统内能的变化 ΔU 等于加入系统的热量 Q 加上对系统做的功 W：ΔU = Q + W。对于压力-体积功，可逆膨胀对系统做功 –p dV，因此 dU = dQ – p dV。
Enthalpy H is defined as H = U + pV. Under constant pressure, the change in enthalpy is dH = dU + p dV + V dp = (dQ – p dV) + p dV + V dp = dQ + V dp. At constant pressure dp = 0, so dH = dQp. Thus the heat absorbed or released at constant pressure equals the enthalpy change, which is directly measurable in a calorimeter.
焓 H 定义为 H = U + pV。在恒压条件下，焓的变化为 dH = dU + p dV + V dp = (dQ – p dV) + p dV + V dp = dQ + V dp。恒压下 dp = 0，故 dH = dQp。因此恒压下吸收或放出的热量等于焓变，这可以直接用量热计测量。
6. Arrhenius Equation from Collision Theory | 从碰撞理论推导阿伦尼乌斯方程
The rate of a bimolecular gas‑phase reaction A + B → products depends on the frequency of collisions and the fraction of those collisions that have sufficient energy and correct orientation. Collision frequency ZAB for molecules in a container is proportional to the number densities and the average relative speed: ZAB = σ √(8kT/πμ) × NA² [A][B], where σ is the collision cross‑section, μ is the reduced mass. The rate constant k can be expressed as:
双分子气相反应 A + B → 产物的速率取决于碰撞频率以及具有足够能量和正确取向的碰撞分数。容器中分子的碰撞频率 ZAB 正比于数密度和平均相对速率：ZAB = σ √(8kT/πμ) × NA² [A][B]，其中 σ 为碰撞截面，μ 为约化质量。速率常数 k 可表示为：
k = ZAB p e–Eₐ/RT
Here p is the steric factor (probability of correct orientation) and e–Eₐ/RT is the fraction of collisions with energy ≥ Eₐ. Absorbing the pre‑exponential factors into a single constant A (the pre‑exponential factor) yields the Arrhenius equation:
这里 p 是方位因子（正确取向的概率），e–Eₐ/RT 是具有能量 ≥ Eₐ 的碰撞分数。将所有指数前因子合并为一个常数 A（指前因子），即得阿伦尼乌斯方程：
k = A e–Eₐ/RT
The linear form ln k = ln A – Eₐ/RT is used to determine activation energy from a graph of ln k against 1/T.
线性形式 ln k = ln A – Eₐ/RT 用于从 ln k 对 1/T 的图中确定活化能。
7. Relationship between ΔG° and Equilibrium Constant | 标准自由能变与平衡常数的关系
For a reaction aA + bB ⇌ cC + dD, the Gibbs free energy change under non‑standard conditions is given by the reaction quotient Q:
对于反应 aA + bB ⇌ cC + dD，非标准条件下的吉布斯自由能变由反应商 Q 给出：
ΔG = ΔG° + RT ln Q
where Q = [C]c[D]d / [A]a[B]b. At equilibrium, ΔG = 0 and Q = K (the equilibrium constant). Substituting gives 0 = ΔG° + RT ln K, hence:
其中 Q = [C]c[D]d / [A]a[B]b。在平衡时，ΔG = 0 且 Q = K（平衡常数）。代入得到 0 = ΔG° + RT ln K，因此：
ΔG° = –RT ln K
This powerful equation allows calculation of equilibrium constants from thermodynamic data, and explains why a negative ΔG° corresponds to K > 1 (product‑favoured).
这个强有力的方程允许从热力学数据计算平衡常数，并解释了为何负的 ΔG° 对应于 K > 1（有利于产物）。
8. van der Waals Equation Derivation | 范德瓦尔斯方程推导
Real gases deviate from ideal behaviour due to (i) finite molecular volume and (ii) intermolecular attractive forces. The van der Waals equation corrects the ideal gas law by introducing two gas‑specific constants a and b. The available volume for molecules is reduced because each molecule occupies space, so V is replaced by V – n b, where b is the excluded volume per mole. Attractive forces reduce the pressure exerted on the walls; the internal pressure correction is proportional to the square of the density, i.e. a n²/V². Thus:
实际气体因 (i) 分子占有体积和 (ii) 分子间引力而偏离理想行为。范德瓦尔斯方程通过引入两个气体特定常数 a 和 b 来修正理想气体定律。由于每个分子占有空间，可供分子运动的体积减小，因此 V 被替换为 V – n b，其中 b 是每摩尔排除体积。引力减小了作用于器壁的压强；内压力修正项正比于密度的平方，即 a n²/V²。于是：
(p + a n²/V²)(V – n b) = n R T
For one mole, (p + a/V²)(V – b) = RT. The constants a and b are related to the critical constants Tc, pc and Vc.
对于 1 摩尔，(p + a/V²)(V – b) = RT。常数 a 和 b 与临界常数 Tc、pc 和 Vc 有关。
9. Summary of Key Derivations | 关键推导总结
The derivations covered in this article form the backbone of physical unit 3. They show how macroscopic observables arise from microscopic models, and they appear repeatedly in A‑level exam questions. Mastering these logical progressions will not only improve your problem‑solving skills but also deepen your appreciation for the unity of physical laws.
本文涵盖的推导构成了物理单元 3 的支柱。它们展示了宏观可观测量如何从微观模型产生，并在 A‑level 考试中反复出现。掌握这些逻辑进程不仅能提高解题能力，还能加深你对物理定律统一性的理解。
- Kinetic theory → pV = ⅓ N m ⟨v²⟩ → pV = nRT
- Boltzmann statistics → Maxwell‑Boltzmann speed distribution
- First law → Enthalpy H = U + pV
- Collision theory → Arrhenius equation k = A e–Eₐ/RT
- Free energy → ΔG° = –RT ln K
- Real gas corrections → van der Waals equation
Re‑derive these yourself with pen and paper, and you will be well‑prepared for any topic test on thermodynamics and kinetics.
用笔和纸亲自再推导一遍，你将完全准备好应对任何热力学和动力学的单元测试。
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