📚 Pressure | 压强
s that describes how a force is distributed over an area. In the Cambridge International A‑ ret pressure measurements in a variety of practical contexts. This article builds your knowledge step by step, from the basic definition p = F/A through to the kinetic‑theory explanation of gas pressure, linking each idea to typical CIE exam requirements.
压强是物理学中的一个基本概念，它描述力在面积上的分布情况。在剑桥国际A‑Level物理课程中，你需要理解固体、液体和气体中的压强，准确应用相关方程，并在各种实际情境中解释压强的测量结果。本文从基本定义 p = F/A 出发，逐步深入到气体压强的分子动理论解释，并将每个知识点与典型的CIE考试要求联系起来。
1. Defining Pressure | 压强的定义
force in newtons (N) and A is the area in square metres (m²). The SI unit of pressure is the pascal (Pa), which is equivalent to one newton per square metre: 1 Pa = 1 N m⁻². Although pressure itself is a scalar quantity, the force responsible for it has a direction, and the area is always taken as the surface upon which the force acts normally.
压强定义为垂直作用于单位面积上的力的大小。数学上，压强 p 表示为 p = F / A，其中 F 是以牛顿（N）为单位的力，A 是以平方米（m²）为单位的面积。压强的国际单位是帕斯卡（Pa），相当于每平方米一牛顿：1 Pa = 1 N m⁻²。虽然压强本身是标量，但产生压强的力具有方向，而面积始终取为力垂直作用的那个面。
a). Atmospheric pressure at sea level is approximately 101 kPa, which is a useful figure to remember for order‑of‑magnitude estimates.
物理学中常用的帕斯卡倍数有千帕（1 kPa = 10³ Pa）和兆帕（1 MPa = 10⁶ Pa）。海平面的标准大气压约为 101 kPa，这是在估算数量级时很有用的参考值。
2. Pressure in Fluids at Rest | 静止流体中的压强
ssure difference Δp between two points separated by a vertical height h is given by Δp = ρgh, where g is the acceleration of free fall. If the pressure at the top surface of the fluid is p₀ (for example, atmospheric pressure), the absolute pressure at a depth h below the surface is p = p₀ + ρgh.
在静止流体中，由于上方流体的重量，压强随深度而增加。对于密度 ρ 恒定的不可压缩流体，垂直高度差为 h 的两点之间的压强差 Δp 由公式 Δp = ρgh 给出，其中 g 是自由落体加速度。如果流体上表面的压强为 p₀（例如大气压），则表面以下深度 h 处的绝对压强为 p = p₀ + ρgh。
Δp = ρgh
This relationship explains why dams are thicker at the bottom and why divers experience greater pressures at greater depths. It also tells us that the pressure at points at the same horizontal level in a continuous, static fluid is the same, irrespective of the shape of the container.
这个关系式解释了为什么水坝底部较厚，以及为什么潜水员在较深的地方会承受更大的压强。它也告诉我们，在连续的静止流体中，同一水平面上的压强相同，与容器的形状无关。
3. Measuring Pressure: The Manometer | 测量压强：U型管压力计
A manometer is a U‑tube partially filled with a liquid, often mercury or water, that is used to measure the pressure difference between two points, or the pressure of a gas relative to the atmosphere. When one limb is connected to a gas supply and the other limb is open to the air, the difference in liquid column heights, Δh, is related to the gauge pressure pgauge . If the gas pressure is less than atmospheric pressure, the liquid rises on the gas side.
U型管压力计是一种部分装有液体（通常为汞或水）的U型管，用来测量两点间的压强差，或者某个气体相对于大气的压强。当一端连接到气源而另一端敞开通大气时，液柱高度差 Δh 与表压 pgauge = ρgΔh 相对应。如果气体压强小于大气压，则气体一侧的液面会上升。
For precise work, the manometer can be used to determine absolute gas pressure by adding the atmospheric pressure exerted on the open limb: pgas = patm + ρgΔh (if the gas side is lower) or pgas = patm − ρgΔh (if the gas side is higher). In exam diagrams, you will often see a manometer with one limb sealed or open, and you must interpret the heights correctly.
在精确测量中，可利用U型管压力计求得气体的绝对压强，方法是加上作用在开口端的大气压强：pgas = patm + ρgΔh（当气体一侧液面较低时）或 pgas = patm − ρgΔh（当气体一侧液面较高时）。在考试示意图中，你常会看到一端密封或开口的 U 型管，必须正确地解读液柱高度。
4. Barometers and Atmospheric Pressure | 气压计与大气压
A mercury barometer provides a direct measurement of atmospheric pressure. It consists of a long glass tube, closed at one end, filled with mercury, and then inverted into a mercury reservoir. The space above the mercury column is a vacuum (Torricellian vacuum), so the height h of the column is supported entirely by the atmospheric pressure pushing on the reservoir surface. Hence patm = ρgHgh.
水银气压计可直接测量大气压。它由一根一端封闭的长玻璃管组成，管内充满汞，然后倒置插入汞槽中。汞柱上方的空间为真空（托里拆利真空），因此汞柱的高度 h 完全由作用在槽面的大气压强来支撑，故有 patm = ρgHgh。
0⁵ Pa. Water barometers are possible in principle, but the column would be about 10.3 m high, making them impractical. The aneroid barometer, which uses a partially evacuated metal capsule, is a more compact alternative and is often examined in CIE papers as a calibration exercise.
标准大气压定义为 0°C 下 760 毫米汞柱（760 mmHg），相当于 1.013 × 10⁵ Pa。水气压计在原理上可行，但其液柱高度约为 10.3 m，不实用。无液气压计使用部分抽空的金属膜盒，是一种更加紧凑的替代方案，在 CIE 试卷中常作为校准题出现。
5. Pascal’s Principle | 帕斯卡原理
Pascal’s principle states that if an external pressure is applied to an enclosed incompressible fluid at rest, the pressure is transmitted undiminished to every part of the fluid and to the walls of the containing vessel. This principle underpins the operation of all hydraulic machines, where a force applied at one piston creates an equal pressure increase throughout the fluid, enabling a larger force to be generated at a wider piston.
帕斯卡原理指出，如果对密闭静止的不可压缩流体施加外部压强，该压强会大小不变地传递到流体的各个部分以及容器壁。这一原理是所有液压机械工作的基础：在一个活塞上施加力，会在整个流体中产生相等的压强增量，从而能够在较宽的活塞上产生更大的力。
It is crucial to note that Pascal’s principle applies to changes in pressure, not to the total existing pressure, and the fluid must be confined and incompressible. In CIE questions, you will often be asked to calculate the force multiplication factor using the ratio of piston areas, or to identify the work done in a hydraulic press.
务必注意，帕斯卡原理适用于压强的变化，而不是已有的总压强，而且流体必须被限制且不可压缩。在 CIE 题目中，常会要求你利用活塞面积比计算力的放大倍数，或者求液压机中所做的功。
6. Hydraulic Systems and Force Multiplication | 液压系统与力放大
A simple hydraulic press consists of two connected cylinders of different cross‑ F₁/A₁ in the fluid. By Pascal’s principle, the same pressure increase acts on the large piston, so the output force is F₂ = Δp × A₂ = (A₂/A₁) F₁. Thus the force is multiplied by the area ratio A₂/A₁, which can be very large.
一个简单的液压机由两个横截面积分别为 A₁ 和 A₂ 的连通缸体组成，内部充满不可压缩的油液。当对小活塞施加较小的力 F₁ 时，它使流体产生压强增量 Δp = F₁/A₁。根据帕斯卡原理，同样的压强增量作用在大活塞上，因此输出力为 F₂ = Δp × A₂ = (A₂/A₁) F₁。这样力就被放大了面积比 A₂/A₁ 倍，这个比值可以非常大。
F₂ / F₁ = A₂ / A₁
ans the distance moved by the large piston is smaller: d₂ = (A₁/A₂) d₁. Hydraulic brakes, jacks, and presses all exploit this principle.
然而，根据能量守恒，忽略摩擦时，对小活塞做的功等于大活塞输出的功。这意味着大活塞移动的距离更小：d₂ = (A₁/A₂) d₁。液压制动器、千斤顶和压力机都利用了该原理。
7. Force Due to Pressure on Submerged Surfaces | 压强对浸没表面的作用力
ce is horizontal and at a constant depth. For a vertical or inclined surface, the pressure varies linearly with depth, so the average pressure acting on the wall of a container or a dam is ρg(h⁄2), where h is the maximum depth. The resultant force acts through the centre of pressure, which is lower than the centroid of the area.
当一个表面浸没在流体中时，流体施加的总作用力并非简单地是 p × 面积，除非该表面是水平的且处于恒定深度。对于竖直或倾斜的表面，压强随深度线性变化，因此作用在容器壁或水坝侧壁上的平均压强为 ρg(h⁄2)，其中 h 为最大深度。合力的作用点（压力中心）低于该面积的形心。
depth to the centroid of the submerged area and multiply the average pressure by that area. For a rectangular wall of width w and height h, the force is F = (½ ρgh) × (wh) = ½ ρgw h².
CIE 试题常会给出一个浸没的矩形闸门或鱼缸，要求计算水的作用力。你需要确定浸没面积形心的深度，然后将平均压强乘以该面积。对于宽为 w、高为 h 的矩形壁面，力为 F = (½ ρgh) × (wh) = ½ ρgw h²。
8. Pressure in Gases: The Ideal Gas Equation | 气体压强：理想气体状态方程
K⁻¹ is the molar gas constant, and T is the absolute temperature in kelvin. An alternative form is pV = NkT, with N being the number of molecules and k = 1.38 × 10⁻²³ J K⁻¹. This equation summarises the macroscopic behaviour of a gas under changing conditions.
对于理想气体，压强、体积、温度和气体的量由方程 pV = nRT 联系起来，其中 n 为摩尔数，R = 8.31 J mol⁻¹ K⁻¹ 为摩尔气体常数，T 是以开尔文为单位的绝对温度。另一种形式为 pV = NkT，其中 N 为分子数，k = 1.38 × 10⁻²³ J K⁻¹。该方程归纳了气体在变化条件下的宏观行为。
pV = nRT
; at constant pressure, V/T = constant (Charles’s law); and at constant volume, p/T = constant (Gay‑Lussac’s law). The absolute temperature must always be used.
由理想气体定律可以推导出 CIE 所要求的几种特殊情况：对于一定质量的气体，在温度不变时，pV = 常数（玻意耳定律）；在压强不变时，V/T = 常数（查理定律）；在体积不变时，p/T = 常数（盖‑吕萨克定律）。计算时必须始终使用绝对温度。
9. Kinetic Theory Model of Gas Pressure | 气体压强的分子动理论模型
and negligible intermolecular forces, the pressure exerted by an ideal gas can be expressed as p = ⅓ ρ ⟨c²⟩, where ρ is the density of the gas and ⟨c²⟩ is the mean square speed of the molecules. Equivalently, pV = ⅓ N m ⟨c²⟩, where m is the mass of a single molecule.
分子动理论将气体压强解释为大量分子与容器壁碰撞的结果。在假设完全弹性碰撞和分子间作用力可忽略的前提下，理想气体产生的压强可表示为 p = ⅓ ρ ⟨c²⟩，其中 ρ 为气体密度，⟨c²⟩ 为分子方均速率。等效地，pV = ⅓ N m ⟨c²⟩，其中 m 是单个分子的质量。
p = ⅓ ρ ⟨c²⟩
Combining pV = ⅓ N m ⟨c²⟩ elationship between the average kinetic energy of a molecule and absolute temperature: ½ m ⟨c²⟩ nal kinetic energy of its molecules, a link frequently tested in CIE structured questions.
将 pV = ⅓ N m ⟨c²⟩ 与 pV = N k T 结合，可得出分子平均动能与绝对温度之间的关系：½ m ⟨c²⟩ = (3/2) kT。这表明理想气体的温度与其分子的平均平动动能成正比，该联系在 CIE 结构化试题中经常被考查。
10. Applications, Pitfalls, and Exam Strategy | 实际应用、常见误区与应试策略
Pressure concepts appear across many CIE Physics topics, from fluid statics to thermal physics. Key applications include hydraulic car lifts, liquid‑column pressure gauges, scuba diving depth warnings, and the design of pressure vessels. In exams, you should be comfortable converting between pressure units (Pa, atm, mmHg, bar), applying ρgh to manometers and barometers correctly, and interpreting p–V diagrams for gases.
压强概念出现在 CIE 物理的许多主题中，从流体静力学到热学。主要应用包括液压举升机、液柱压力计、潜水深度警示以及压力容器的设计。在考试中，你应该能熟练地进行压强单位换算（Pa、atm、mmHg、bar），正确地对 U 型管压力计和气压计应用 ρgh，并能解读气体的 p–V 图线。
- In manometer problems, always decide whether the gas pressure is greater or less than atmospheric pressure before writing the equation.
- 当处理 U 型管压力计题目时，先在写方程前判断气体压强是大于还是小于大气压。
- Remember that the pressure given by ρgh is a pressure difference, not an absolute pressure unless you add the external pressure at the open surface.
- 记住由 ρgh 得到的压强是压强差，而不是绝对压强，除非加上开口端的外部压强。
- For kinetic theory, be ready to explain why pressure increases when volume decreases at constant temperature (more frequent collisions) or when temperature increases at constant volume (higher molecular speeds and harder collisions).
- 在分子动理论部分，要准备好解释为什么恒温下体积减小压强会增大（碰撞更频繁），以及恒容下温度升高压强会增大（分子速率更大且碰撞更剧烈）。
Always draw a clear diagram when a question describes a manometer or a hydraulic system. Labelling heights, densities, and reference levels reduces careless mistakes. When energy or work appears alongside Pascal’s principle, equate the work done on the input and output pistons, accounting for any efficiency if specified.
当题目描述 U 型管压力计或液压系统时，一定要画一个清晰的示意图。标出高度、密度和参考水平可以减少粗心错。当能量或功与帕斯卡原理一同出现时，应使输入活塞和输出活塞的功相等，如果给出了效率则需将其考虑在内。
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