📚 Edexcel IAL Physics Unit 5: Core Topics from Questions 231-295 | Edexcel IAL物理第五单元231-295题核心考点
Edexcel IAL Physics Unit 5 brings together thermal physics, radioactivity, oscillations and astrophysics. Questions 231-295 test the key definitions, equations and graph skills that regularly decide grade boundaries. This revision guide works through the most important ideas point by point, with English and Chinese paired explanations, so you can check both your conceptual understanding and your ability to use the equations confidently.
Edexcel IAL物理第五单元综合了热物理、放射性、振动与天体物理。第231-295题考察的核心定义、方程和图像技能，常常决定等级线的高低。这篇复习指南逐点梳理最重要的考点，每个要点都给出英文与中文配对讲解，帮助你同时检查概念理解与公式运用的熟练度。
1. Internal Energy and Kinetic Theory | 内能与分子动理论
Internal energy is the sum of the random kinetic energy and the random potential energy of all the particles in a system. In an ideal gas the molecules are assumed to have no intermolecular forces, so the potential energy term is zero and the internal energy depends only on the temperature.
内能是系统内所有粒子随机动能与随机势能的总和。理想气体模型假设分子之间没有相互作用力，因此势能项为零，内能只由温度决定。
le is given by the equation below, where k is the Boltzmann constant (1.38 × 10⁻²³ J K⁻¹).
开尔文温度衡量分子平均随机动能的大小。对单原子气体，每个分子的平均动能由下式给出，其中 k 是玻耳兹曼常数（1.38 × 10⁻²³ J K⁻¹）。
Eₖ = (3/2)kT
Doubling the thermodynamic temperature of a fixed mass of monatomic gas doubles the mean kinetic energy per molecule, so its internal energy also doubles. Adding energy at constant temperature, as happens during melting, increases potential energy instead of kinetic energy.
对固定质量的单原子气体，热力学温度加倍时分子平均动能加倍，内能也随之加倍。而在温度不变的熔化过程中输入能量，增加的是势能而不是动能。
For diatomic and polyatomic gases the molecules can also rotate and vibrate, so extra degrees of freedom contribute to the internal energy and the (3/2)kT result no longer applies. Unit 5 questions normally restrict this calculation to a monatomic gas.
对于双原子和多原子气体，分子还可以转动和振动，额外的自由度也会贡献内能，此时 (3/2)kT 的结果不再适用。第五单元的题目通常只要求对单原子气体做这种计算。
Exam phrasing matters: internal energy is a property of the particles in a system, while heat is energy in transit because of a temperature difference. Marks are often lost by mixing these two terms in definitions.
答题措辞很关键：内能是系统内粒子自身的属性，而热量是由于温差而传递的能量。把这两个概念混淆是最常见的失分点。
2. The Ideal Gas Equation | 理想气体方程
the molar gas constant and k is the Boltzmann constant, and the two are linked by R = Nₐk.
理想气体方程把压强 p、体积 V 和热力学温度 T 联系起来。对 n mol 气体，pV = nRT；对 N 个分子，pV = NkT。其中 R 是摩尔气体常数，k 是玻耳兹曼常数，二者由 R = Nₐk 联系。
pV = nRT = NkT
ol⁻¹. The number of molecules is N = nNₐ, so a mole of gas contains 6.02 × 10²³ particles.
需要记住的常数：R = 8.31 J mol⁻¹ K⁻¹，k = 1.38 × 10⁻²³ J K⁻¹，Nₐ = 6.02 × 10²³ mol⁻¹。分子数 N = nNₐ，即 1 mol 气体含有 6.02 × 10²³ 个粒子。
For a fixed mass of gas, pV/T is constant, which gives the combined gas law below. This is the fastest route through before-and-after problems, especially when one of the three quantities is held constant.
对固定质量的气体，pV/T 为常量，由此得到下面的联合气体定律。处理变化前后问题时这是最快的思路，尤其是当压强、体积或温度中某一个量保持不变时。
p₁V₁ / T₁ = p₂V₂ / T₂
Graph questions in this topic are very predictable. At constant temperature, p against 1/V is a straight line through the origin; pV against p is a horizontal line; p/T against T is a straight line through the origin, and its gradient equals nR/V.
本考点的图像题非常有规律。温度不变时，p 对 1/V 的图像是过原点的直线；pV 对 p 的图像是水平直线；p/T 对 T 的图像是过原点的直线，斜率等于 nR/V。
× 10³ Pa). Substituting °C or cm³ directly is one of the most common errors at this level.
单位陷阱：温度必须用开尔文（摄氏度加 273.15），体积用 m³（1 cm³ = 1 × 10⁻⁶ m³），压强用 Pa（1 kPa = 1 × 10³ Pa）。直接代入摄氏度或立方厘米是本阶段最常见的错误之一。
3. First Law of Thermodynamics | 热力学第一定律
The first law is a statement of energy conservation for a gas: the increase in internal energy equals the energy added by heating minus the work done by the gas. ΔU is the change in internal energy, Q is the energy transferred to the system by heating and W is the work done by the system.
热力学第一定律是气体系统中的能量守恒表述：内能的增加量等于通过加热输入的能量减去气体对外做的功。ΔU 是内能变化，Q 是传递给系统的热量，W 是系统对外做的功。
ΔU = Q − W
s is compressed, ΔV is negative and W is negative, meaning work is done on the gas and its internal energy rises.
在等压过程中，气体膨胀对外做功 W = pΔV，因此第一定律也可写成 ΔU = Q − pΔV。若气体被压缩，ΔV 为负，W 为负，表示外界对气体做功，其内能升高。
iabatic change Q = 0, so ΔU = −W, which is why a rapid compression makes a gas hotter and a rapid expansion cools it.
题目中最常出现两种特殊情况。等温过程中 ΔU = 0，因此 Q = W，加热输入的能量全部转化为对外做功。绝热过程中 Q = 0，因此 ΔU = −W，这正是快速压缩使气体升温、快速膨胀使气体降温的原因。
s the work done on the gas. The physics is identical, but the numerical sign of W flips, and mixed conventions cost easy marks.
代入前先看清题目给出的符号约定。有些教材写作 ΔU = Q + W，把 W 定义为外界对气体做的功。物理本质相同，但 W 的数值符号相反，混用约定会白丢分。
4. Specific Heat Capacity and Latent Heat | 比热容与潜热
quired to change the state of 1 kg of a substance without any change in temperature, so Q = mL.
比热容 c 是使 1 kg 物质温度升高 1 K 所需的能量，即 Q = mcΔT。比潜热 L 是在温度不变的情况下使 1 kg 物质发生物态变化所需的能量，即 Q = mL。
Q = mcΔT and Q = mL
water = 2.26 × 10⁶ J kg⁻¹. The large value for vaporisation explains why steam causes worse burns than boiling water.
常用参考值：水的 c = 4200 J kg⁻¹ K⁻¹，冰的熔化比潜热 = 3.34 × 10⁵ J kg⁻¹，水的汽化比潜热 = 2.26 × 10⁶ J kg⁻¹。汽化潜热数值很大，这正是水蒸气烫伤比沸水更严重的原因。
The standard experiment uses an electric heater of known power, a joulemeter or ammeter-voltmeter pair, an insulated container and a thermometer. Plotting energy supplied against temperature rise gives a straight line whose gradient is mc, which reduces the effect of random reading errors.
标准实验使用已知功率的电热器、焦耳计或电流表与电压表、保温容器和温度计。作出输入能量对温升的图像，得到的直线斜率即为 mc，这样可以减小随机读数误差的影响。
Key systematic errors and corrections: heat loss to the surroundings makes the measured c too large, so plot a cooling correction or extrapolate the graph; the thermometer and heater also absorb energy; and the heater must be fully immersed and switched on only when the readings are steady.
主要系统误差与修正：向环境散热会使测得的 c 偏大，因此需要作冷却修正或对图像外推；温度计和加热器本身也吸收热量；加热器必须完全浸没，且待读数稳定后再开始计时。
A heating curve of temperature against energy supplied shows flat plateaus at the melting and boiling points. During a plateau the energy supplied breaks intermolecular bonds rather than increasing kinetic energy, so the temperature stays constant even though energy is still being transferred.
温度对输入能量的加热曲线在熔点和沸点处出现平台。平台期间输入的能量用于破坏分子间作用力而不是增加动能，所以尽管能量持续输入，温度仍保持不变。
5. Radioactivity: Decay Modes and Nuclear Equations | 放射性：衰变方式与核反应方程
Radioactive decay is random and spontaneous: you cannot predict which nucleus decays next or when, and the process is not affected by temperature, pressure or chemistry. Each decay changes the nucleus towards a more stable configuration.
放射性衰变是随机且自发的：无法预测下一个衰变的是哪个核、何时发生，且不受温度、压强或化学状态影响。每一次衰变都使原子核朝更稳定的状态靠近。
| Radiation | Nature | Change in A / Z | Ionising power |
| Alpha α | 2 protons + 2 neutrons | A − 4, Z − 2 | Strongest, range a few cm in air |
| Beta-minus β⁻ | Fast electron + antineutrino | A unchanged, Z + 1 | Medium, absorbed by few mm of aluminium |
| Beta-plus β⁺ | Positron + neutrino | A unchanged, Z − 1 | Medium, annihilates with an electron |
| Gamma γ | High-energy photon | A and Z unchanged | Weakest, needs thick lead |
Nuclear equations must balance both nucleon number A and charge Z. For example, alpha decay of uranium-238 is written below; the totals are
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