A-Level Physics: Pressure — Key Concepts and Exam Question Patterns | A-Level 物理：压强的考点与题型归纳

📚 A-Level Physics: Pressure — Key Concepts and Exam Question Patterns | A-Level 物理：压强的考点与题型归纳

Pressure is a fundamental concept in A-Level Physics that bridges mechanics, fluids, and thermal physics. In this article, we systematically review the key definitions, derivations, and common exam question patterns related to pressure, tailored to the CIE A-Level syllabus.

压强是 A-Level 物理中连接力学、流体与热学的基础概念。本文将根据 CIE A-Level 考纲，系统梳理与压强相关的核心定义、公式推导以及常见题型。

1. Definition of Pressure | 压强的定义

Pressure is defined as the normal force exerted per unit area. The SI unit of pressure is the pascal (Pa), where 1 Pa = 1 N m⁻².

压强定义为单位面积上所受到的法向力。压强的国际单位是帕斯卡（Pa），其中 1 Pa = 1 N m⁻²。

p = F / A

Here, F is the magnitude of the normal force acting perpendicular to the surface, and A is the area over which the force acts.

其中 F 是垂直于表面作用力的大小，A 是力作用的面积。

Pressure is a scalar quantity, even though force is a vector.

压强是标量，尽管力是矢量。
For a fluid at rest, pressure acts equally in all directions.

对于静止流体，压强在各个方向上大小相等。
Pressure increases with depth in a fluid due to the weight of the fluid above.

在流体中，由于上方流体的重力，压强随深度增加而增大。

2. Pressure in a Liquid | 液体中的压强

For a liquid of density ρ at a depth h below the surface, the pressure due to the liquid column is given by:

对于密度为 ρ 的液体，在液面下方深度 h 处，由液柱产生的压强为：

p = ρ g h

This equation assumes that the density is constant and that g is the gravitational field strength (approximately 9.81 N kg⁻¹ on Earth).

该公式假设液体密度恒定，g 为重力场强度（地球上约为 9.81 N kg⁻¹）。

The pressure depends only on the vertical depth, not on the shape of the container.

压强只取决于垂直深度，与容器的形状无关。
The total pressure at depth h is the sum of atmospheric pressure and the liquid pressure: p_total = p_atm + ρgh.

深度 h 处的总压强等于大气压加上液体压强：p_total = p_atm + ρgh。
This principle is used in hydraulic systems and barometers.

这一原理应用于液压系统和气压计中。

3. Atmospheric Pressure and Barometers | 大气压强与气压计

Atmospheric pressure is the pressure exerted by the weight of the Earth’s atmosphere. At sea level, standard atmospheric pressure is approximately 1.01 × 10⁵ Pa, often called 1 atmosphere (atm).

大气压强是地球大气层重力所产生的压强。在海平面，标准大气压约为 1.01 × 10⁵ Pa，通常称为 1 个标准大气压（atm）。

A mercury barometer measures atmospheric pressure by balancing the weight of a mercury column against the atmospheric pressure. At standard pressure, the mercury column height is approximately 760 mm.

水银气压计通过使水银柱的重力与大气压平衡来测量大气压。在标准大气压下，水银柱高度约为 760 mm。

p_atm = ρ_mercury × g × h

For mercury, ρ ≈ 13,600 kg m⁻³, g = 9.81 N kg⁻¹, and h = 0.760 m, giving p_atm ≈ 1.01 × 10⁵ Pa.

对于水银，ρ ≈ 13,600 kg m⁻³，g = 9.81 N kg⁻¹，h = 0.760 m，因此 p_atm ≈ 1.01 × 10⁵ Pa。

4. Gauge Pressure and Absolute Pressure | 表压与绝对压强

Gauge pressure is the pressure measured relative to atmospheric pressure. Absolute pressure is the total pressure measured relative to a perfect vacuum.

表压是相对于大气压测量的压强。绝对压强是相对于绝对真空测量的总压强。

p_absolute = p_gauge + p_atm

In exam questions, always check whether the given pressure is absolute or gauge. For example, a tyre pressure gauge reading of 2.2 bar is a gauge pressure; the absolute pressure inside the tyre is 2.2 bar + 1.0 bar = 3.2 bar.

在考试题目中，务必检查给定的压强是绝对压强还是表压。例如，轮胎气压表读数为 2.2 bar 是表压；轮胎内部的绝对压强为 2.2 bar + 1.0 bar = 3.2 bar。

5. Boyle’s Law and Isothermal Changes | 波义耳定律与等温变化

For a fixed mass of gas at constant temperature, the pressure and volume are inversely proportional. This is Boyle’s law.

对于一定质量的理想气体，在温度恒定时，压强与体积成反比。这就是波义耳定律。

p₁V₁ = p₂V₂

On a pressure-volume (p-V) graph, an isothermal change appears as a hyperbolic curve. Each curve corresponds to a different temperature, with higher temperatures lying further from the origin.

在压强-体积（p-V）图上，等温变化表现为双曲线。每条曲线对应不同的温度，温度越高，曲线离原点越远。

Boyle’s law applies only to ideal gases at constant temperature.

波义耳定律仅适用于恒温条件下的理想气体。
Common exam questions involve compressing a gas slowly so that heat can be exchanged with the surroundings.

常见考题涉及缓慢压缩气体，以便气体与外界交换热量。
Units of p and V must be consistent on both sides of the equation.

方程两侧的 p 和 V 单位必须一致。

6. Pressure and Kinetic Theory of Gases | 压强与气体动理论

The kinetic theory of gases explains gas pressure as the result of collisions between gas molecules and the container walls. The pressure exerted by an ideal gas is related to the mean square speed of its molecules.

气体动理论将气体压强解释为气体分子与容器壁碰撞的结果。理想气体产生的压强与其分子的均方根速度有关。

p = (1/3) × (N m ⟨c²⟩) / V

Here, N is the number of molecules, m is the mass of one molecule, ⟨c²⟩ is the mean square speed, and V is the volume of the gas.

其中 N 是分子数量，m 是单个分子的质量，⟨c²⟩ 是分子速度平方的平均值，V 是气体的体积。

From this equation, the root-mean-square (r.m.s.) speed can be derived: c_rms = √(3p / ρ), where ρ is the gas density.

由此方程可以推导出均方根速度：c_rms = √(3p / ρ)，其中 ρ 是气体的密度。

7. Pressure in an Ideal Gas Equation | 理想气体方程中的压强

The ideal gas equation combines Boyle’s law, Charles’s law, and the pressure law into one expression:

理想气体方程将波义耳定律、查理定律和压强定律结合为一个表达式：

pV = nRT

In this equation, p is the absolute pressure in pascals, V is the volume in cubic metres, n is the number of moles, R is the molar gas constant (8.31 J K⁻¹ mol⁻¹), and T is the absolute temperature in kelvin.

在此方程中，p 为绝对压强（单位 Pa），V 为体积（单位 m³），n 为物质的量（单位 mol），R 为摩尔气体常数（8.31 J K⁻¹ mol⁻¹），T 为绝对温度（单位 K）。

When using pV = nRT, always convert temperature to kelvin.

使用 pV = nRT 时，务必把温度转换为开尔文。
If the mass of gas is given, use n = m / M, where M is the molar mass.

若给出气体质量，则用 n = m / M 计算物质的量，其中 M 为摩尔质量。
For a fixed mass of gas, pV/T = constant.

对于固定质量的气体，pV/T 为常数。

8. Exam Question Pattern: Manometer Calculations | 题型归纳：U 形管压强计计算

A U-tube manometer is a common device used to measure gas pressure. In exam questions, you are often given the height difference between two liquid columns and asked to find the pressure of a gas.

U 形管压强计是测量气体压强的常用装置。在考题中，通常会给出两液柱的高度差，要求计算气体的压强。

For a U-tube open to the atmosphere, the gas pressure is:

对于一端开口于大气的 U 形管，气体压强为：

p_gas = p_atm + ρ g h (if the gas side is lower)

p_gas = p_atm − ρ g h (if the gas side is higher)

When the manometer is connected to a gas supply on one side and open to the atmosphere on the other, the side with the lower liquid level has the higher gas pressure.

当 U 形管一端连接气体源、另一端开口于大气时，液面较低的一侧气体压强较大。

Condition	Pressure Relationship
Gas side liquid level lower	p_gas = p_atm + ρgh
Gas side liquid level higher	p_gas = p_atm − ρgh
Both levels equal	p_gas = p_atm

9. Exam Question Pattern: Hydraulic Systems | 题型归纳：液压系统

Hydraulic systems use an incompressible liquid to transmit pressure. According to Pascal’s principle, a pressure applied to an enclosed fluid is transmitted undiminished to every point in the fluid and to the walls of the container.

液压系统利用不可压缩的液体传递压强。根据帕斯卡原理，施加在封闭流体上的压强会毫无衰减地传递到流体的每一个点和容器壁上。

F₁ / A₁ = F₂ / A₂

In a hydraulic lift, a small force applied to a small piston produces a large force on a larger piston. The work done is conserved, so the smaller piston moves a larger distance.

在液压升降机中，施加在小活塞上的小力可以在大活塞上产生较大的力。由于做功守恒，小活塞移动的距离更大。

Always identify which piston is the input and which is the output.

务必分清哪个活塞是输入端，哪个是输出端。
Use consistent units for area and force.

面积和力的单位要保持一致。
Remember that pressure throughout the fluid is the same if the fluid is at rest and gravity is neglected.

记住：在忽略重力且流体静止的情况下，整个流体的压强相同。

10. Exam Question Pattern: Boyle’s Law Problems | 题型归纳：波义耳定律问题

Boyle’s law problems typically involve a gas trapped in a cylinder by a piston, or a gas bubble rising in a liquid. When the volume changes, the pressure changes inversely at constant temperature.

波义耳定律问题通常涉及被活塞封闭在气缸中的气体，或者在液体中上升的气泡。在温度不变时，体积变化会引起压强反向变化。

Example: A gas occupies 200 cm³ at a pressure of 1.5 × 10⁵ Pa. What volume will it occupy at a pressure of 3.0 × 10⁵ Pa, assuming constant temperature?

例题：一定质量的气体在压强 1.5 × 10⁵ Pa 时体积为 200 cm³。若温度不变，当压强变为 3.0 × 10⁵ Pa 时，体积为多少？

p₁V₁ = p₂V₂ → V₂ = p₁V₁ / p₂ = (1.5 × 10⁵ × 200) / (3.0 × 10⁵) = 100 cm³

Note that the pressure doubled, so the volume halved. This inverse relationship is a quick check for your answer.

注意压强变为原来的两倍，因此体积变为原来的一半。这一反比关系可快速检验答案。

11. Exam Question Pattern: Kinetic Theory Calculations | 题型归纳：气体动理论计算

Kinetic theory questions often ask you to calculate the r.m.s. speed of gas molecules or the number of molecules in a container. These questions require careful unit conversion.

气体动理论问题常要求计算气体分子的均方根速度或容器中的分子数。这类问题需要仔细进行单位换算。

For example, calculate the r.m.s. speed of oxygen molecules at a pressure of 1.0 × 10⁵ Pa and a density of 1.43 kg m⁻³.

例如，在压强为 1.0 × 10⁵ Pa、密度为 1.43 kg m⁻³ 时，计算氧分子的均方根速度。

c_rms = √(3p / ρ) = √(3 × 1.0 × 10⁵ / 1.43) ≈ 458 m s⁻¹

Always check that the final units are m s⁻¹. If the density is not given, use ρ = m / V and the ideal gas equation to find it.

始终检查最终单位是否为 m s⁻¹。若题目未给出密度，可用 ρ = m / V，结合理想气体方程求出密度。

12. Common Pitfalls and Final Tips | 常见错误与复习建议

Many students lose marks in pressure questions due to avoidable mistakes. Here are the most common pitfalls and how to avoid them.

许多学生在压强题目中因可避免的错误而失分。以下是常见错误及其规避方法。

Forgetting to convert temperatures to kelvin before using pV = nRT. Absolute zero is −273 °C, so T(K) = T(°C) + 273.15.

使用 pV = nRT 前忘记将温度转换为开尔文。绝对零度为 −273 °C，因此 T(K) = T(°C) + 273.15。
Confusing gauge pressure with absolute pressure. Always add atmospheric pressure when required.

混淆表压与绝对压强。需要时务必加上大气压。
Using inconsistent units, such as mixing cm³ with m³ in the same equation.

使用不一致的单位，例如在同一方程中混用 cm³ 与 m³。
Ignoring the density of the liquid in a manometer when calculating pressure differences.

在计算 U 形管压强计的压强差时忽略液体的密度。
Forgetting that pressure is a scalar, so direction does not matter when summing pressures in a fluid.

忘记压强是标量，因此在流体中叠加压强时无需考虑方向。

To master pressure topics, practice past paper questions involving manometers, hydraulic lifts, and ideal gas calculations. Draw a clear diagram for every problem and label all known quantities before starting the algebra.

要掌握压强相关内容，请多练习涉及 U 形管、液压升降机和理想气体计算的历年真题。每道题先画示意图，标出所有已知量，再进行代数运算。

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A-Level Physics: Pressure — Key Concepts and Exam Question Patterns | A-Level 物理：压强的考点与题型归纳