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物质的粒子模型

AQA · GCSE · 物理 · 知识点 3

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3.1

The particle model: matter from the inside

Why does a metal spoon sink while a huge ship floats? Why does sweat cool you down? Both answers live in the particle model. This reference covers AQA GCSE Physics 8463, topic 4.3 Particle model of matter.

How the exam treats this topic:

  • Paper 1 (4.1–4.4) carries this topic. The equations $\rho = m/V$, $\Delta E = mc\Delta\theta$, $E = mL$ and $pV = \text{constant}$ are all printed on the enclosed sheet.
  • Pressure in gases and doing work on a gas are physics only (work on a gas also Higher Tier).
  • You must interpret heating and cooling graphs that include changes of state.
  • You must distinguish specific heat capacity from specific latent heat — a favourite one-mark check.
3.1

材料的密度
教学大纲

Density of materials (AQA 8463 statement 4.3.1.1).

  1. Use density = mass / volume with the units kg/m3 and g/cm3, converting between them.
  2. Use the particle model to explain the different states of matter and the differences in density between them.
  3. Recognise and draw simple diagrams that model solids, liquids and gases.
  4. Required practical 5: determine the densities of regular and irregular solid objects and liquids, using dimensions, a balance and a displacement technique.

来源：Cambridge International 教学大纲

$$\rho = \frac{m}{V}$$
  • $\rho$ density 密度 in kg/m³; $m$ mass in kg; $V$ volume in m³.
  • Common trap: g/cm³ must become kg/m³ before substituting. $1\ \text{g/cm}^3 = 1000\ \text{kg/m}^3$ (multiply by 1000: a cm³ is a millionth of a m³ and a gram is a thousandth of a kg).

The particle model explains the states of matter:

The particle arrangement in a solid, a liquid and a gas.
Pattern, contact, spacing.
State Particle arrangement Particle motion Density trend
solid packed in a fixed pattern, touching vibrate about fixed positions highest
liquid touching, but free to slide past each other random motion, no fixed shape slightly lower than solid
gas far apart, random fast, random, straight lines between collisions much lower
  • Solids and liquids have similar densities because their particles are similarly packed; a gas is mostly empty space.
  • Ice is unusual: water expands on freezing, so ice is slightly less dense than water.

Worked example. A ring has mass 9.46 g and volume 0.44 cm³. Find its density in kg/m³.

  • Convert both first: $m = 9.46\ \text{g} = 9.46\times10^{-3}\ \text{kg}$; $V = 0.44\ \text{cm}^3 = 4.4\times10^{-7}\ \text{m}^3$.
    $$\rho = \frac{m}{V} = \frac{9.46\times10^{-3}\ \text{kg}}{4.4\times10^{-7}\ \text{m}^3} = 21\,500\ \text{kg/m}^3$$

Required practical 5: density

Measuring density: a rectangular block measured with a ruler, and an irregular object lowered into a displacement (eureka) can.
Regular shapes from dimensions; irregular shapes by displacement.
  • Regular solid: measure length, width and thickness with a ruler (or micrometer/Vernier callipers), multiply for $V$; find $m$ on a balance; $\rho = m/V$.
  • Irregular solid: fill a displacement (eureka) can to the spout, wait for dripping to stop, lower the object in on thin string; the volume of water collected in a measuring cylinder equals the object's volume.
  • Liquid: find the mass of an empty measuring cylinder, then the mass with a known volume inside; subtract for $m$.
  • Accuracy points: read the measuring cylinder at eye level on a flat surface (avoid parallax); use thin string so it displaces almost no water; repeat and average.
词汇 训练
English 中文 拼音
density/ˈdensɪti/ 密度 mì dù
3.2

物态变化与内能
教学大纲

物态变化与内能（AQA 8463 陈述 4.3.1.2-4.3.2.1）。

  1. 描述熔化、凝固、沸腾、蒸发、液化和升华过程，并指出质量守恒。
  2. 解释物态变化属于物理变化，其逆过程可恢复原有性质。
  3. 定义内能为系统内所有粒子的动能与势能之和。
  4. 解释加热要么导致温度升高，要么引起物态变化。

来源：Cambridge International 教学大纲

When a substance melts, freezes, boils, evaporates, condenses or sublimates 升华:

  • Mass is conserved — the number of particles does not change.
  • Changes of state are physical changes 物理变化: reverse the change and the material recovers its original properties. (A chemical change makes new substances; melting does not.)

Internal energy 内能 is the total kinetic and potential energy of all the particles that make up a system. Heating a system increases the particles' energy, and that energy goes one of two ways:

  1. it raises the temperature — the particles' kinetic energy grows;
  2. it produces a change of state — the particles' potential energy grows as bonds break; the temperature stays constant.
词汇 训练
English 中文 拼音
internal energy/ɪnˈtɜːnl ˈenədʒi/ 内能 nèi néng
physical changes/ˈfɪzɪkl ˈtʃeɪndʒɪz/ 物理变化 wù lǐ biàn huà
sublimates/ˈsʌblɪmeɪts/ 升华 shēng huá
3.3

比热容与温度变化
教学大纲

比热容与温度变化（AQA 8463 陈述 4.3.2.2）。

  1. 在温度变化中使用公式 dE = m c dθ，理解 c 的单位为每千克每摄氏度。
  2. 从粒子角度解释比热容的含义。
  3. 结合单位换算，求解能量、质量、比热容或温度变化量中的未知项。

来源：Cambridge International 教学大纲

While the temperature changes, the energy needed follows (also met in topic 1):

$$\Delta E = m\,c\,\Delta\theta$$

Specific heat capacity 比热容 $c$ (J/kg °C) is the energy needed to raise the temperature of one kilogram by one degree Celsius. It measures how hard it is to warm the substance — from the particle view, how much energy its particles store per degree of kinetic energy rise.

Worked example. 0.030 kg of olive oil ($c = 1800$ J/kg °C) warms from 21 °C to 96 °C.

  • Temperature change first: $\Delta\theta = 96 - 21 = 75$ °C.
    $$\Delta E = mc\Delta\theta = 0.030 \times 1800 \times 75 = 4050\ \text{J}$$

The RP1 method, error analysis and percentage-difference work are covered on sheet 1.3 — the same equation, the same practical.

词汇 训练
English 中文 拼音
specific heat capacity/spəˈsɪfɪk hiːt kəˈpæsɪti/ 比热容 bǐ rè róng
3.4

比潜热
教学大纲

比潜热与加热曲线（AQA 8463 陈述 4.3.2.3）。

  1. 使用物态变化所需能量 = 质量 × 比潜热（E = mL）进行计算。
  2. 定义比潜热，并区分熔解潜热与汽化潜热。
  3. 解读包含物态变化的加热与冷却曲线图。
  4. 区分比热容与比潜热。

来源：Cambridge International 教学大纲

While a substance changes state, its temperature stops rising even though energy keeps flowing in. The energy needed is called latent heat 潜热:

$$E = mL$$
  • $E$ energy for the change of state in J; $m$ mass that changes state in kg; $L$ specific latent heat 比潜热 in J/kg.
  • Specific latent heat is the energy needed to change the state of one kilogram of a substance with no change of temperature.
  • Fusion 熔化: solid to liquid. Vaporisation 汽化: liquid to vapour. Vaporisation is much larger than fusion — breaking free of the liquid completely takes more energy than loosening a solid.
A heating graph: temperature rises, plateaus at the melting point, rises again, plateaus at the boiling point, rises as steam.
Read the plateaus as changes of state.

Reading the graph:

  • Rising sections: energy goes into kinetic energy — the temperature climbs ($\Delta E = mc\Delta\theta$).
  • Flat sections: energy goes into potential energy — the state is changing ($E = mL$). The longer the plateau, the more mass changed state.
  • Cooling graphs are the mirror image: flat while the substance freezes or condenses, releasing latent heat.

Distinguishing the two (a credited 1–2 mark check): specific heat capacity involves a temperature change; specific latent heat involves a change of state at constant temperature.

Worked example. A heater supplies 0.0075 kW... (keep units honest) — a 30 W heater runs for 11 minutes and boils off $6.6\times10^{-3}$ kg of water. Find $L$.

  • Convert: $E = Pt = 30\ \text{W} \times 660\ \text{s} = 19\,800$ J.
    $$L = \frac{E}{m} = \frac{19\,800\ \text{J}}{6.6\times10^{-3}\ \text{kg}} = 3.0\times10^6\ \text{J/kg}$$
词汇 训练
English 中文 拼音
latent heat/ˈleɪtənt hiːt/ 潜热 qián rè
specific latent heat/spəˈsɪfɪk ˈleɪtənt hiːt/ 比潜热 bǐ qián rè
Fusion/ˈfjuːʒn/ 熔化 róng huà
Vaporisation/ˌveɪpəraɪˈzeɪʃn/ 汽化 qì huà
3.5

气体中的粒子运动
教学大纲

气体中的粒子运动（AQA 8463 陈述 4.3.3.1）。

  1. 描述气体分子处于永不停息的无规则运动中。
  2. 阐述气体的温度与其分子平均动能之间的关系。
  3. 从分子与容器壁碰撞的角度解释气体压强。
  4. 定性说明一定体积的气体压强如何随温度变化。

来源：Cambridge International 教学大纲

The molecules of a gas are in constant random motion. Its temperature is related to the average kinetic energy of the molecules: hotter gas, faster particles.

Gaseous pressure from the particle model (each clause earns credit):

Gas molecules colliding with the container walls make pressure; compressing the gas raises it.
Collisions at right angles to the wall.
  1. the moving molecules collide with the container walls;
  2. each collision exerts a force at right angles to the wall;
  3. pressure is force per unit area — the total of many tiny collisions spread over the wall.

Temperature up (constant volume) → pressure up: the molecules move faster on average, so they hit the walls more often and harder (larger force each impact), so the force per unit area rises.

3.6

气体压强与对气体做功（仅物理）
教学大纲

气体压强及对气体做功，仅限物理学科（AQA 8463 陈述 4.3.3.2-4.3.3.3）。

  1. 对于质量固定的气体，在恒定温度下应用压强×体积=常数。
  2. 当压强或体积发生变化时，计算新的压强或体积值。
  3. 利用粒子模型解释增大气体体积为何会降低其压强。
  4. （仅高阶）解释对气体做功如何增加其内能并可能升高温度，例如在自行车气筒中。

来源：Cambridge International 教学大纲

A gas can be compressed or expanded by pressure changes. For a fixed mass of gas at constant temperature:

$$pV = \text{constant}$$
  • $p$ pressure in pascals, Pa; $V$ volume in m³.
  • Before/after form: $p_1V_1 = p_2V_2$.

The particle explanation of each direction:

  • Volume up → pressure down (constant temperature): the same number of molecules spread over a larger wall area collide less often, so force per unit area falls.
  • Volume down → pressure up: molecules hit a smaller area more often.

Worked example. A syringe holds 50 cm³ of air at 100 kPa. It is compressed to 20 cm³ at constant temperature.

  • Convert or keep consistent: volumes in cm³ cancel; pressures must be consistent.
    $$p_1V_1 = p_2V_2 \quad\Rightarrow\quad p_2 = \frac{100\ \text{kPa} \times 50}{20} = 250\ \text{kPa}$$
3.6

气体压强与对气体做功（仅物理）
教学大纲

气体压强及对气体做功，仅限物理学科（AQA 8463 陈述 4.3.3.2-4.3.3.3）。

  1. 对于质量固定的气体，在恒定温度下应用压强×体积=常数。
  2. 当压强或体积发生变化时，计算新的压强或体积值。
  3. 利用粒子模型解释增大气体体积为何会降低其压强。
  4. （仅高阶）解释对气体做功如何增加其内能并可能升高温度，例如在自行车气筒中。

来源：Cambridge International 教学大纲

Work is the transfer of energy by a force. Compressing a gas — doing work on it — increases the gas's internal energy, which can raise its temperature.

The credited chain (bicycle pump): pushing the pump's handle does work on the trapped gas → energy is transferred to the gas's particles → their average kinetic energy rises → the temperature of the gas increases (the pump feels warm).

The reverse is true too: a gas expanding does work on its surroundings and cools.

3.6

Checklist before you call this topic done
  • Convert g/cm³ to kg/m³, and cm³ to m³, before using $\rho = m/V$.
  • Describe RP5 for regular solids, displacement and liquids, with accuracy points.
  • State that mass is conserved in changes of state and that they are physical changes.
  • Define internal energy as total kinetic plus potential energy of the particles.
  • Choose between $\Delta E = mc\Delta\theta$ (temperature changes) and $E = mL$ (state changes).
  • Read heating graphs: rising = kinetic energy, plateau = latent heat.
  • Explain gas pressure from wall collisions; use $pV =$ constant with consistent units.
  • (physics only, HT) Explain why doing work on a gas raises its temperature.

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