📚 Edexcel A-Level Physics: Combined Gas Law and Ideal Gas Equation | 组合气体定律与理想气体方程
The combined gas law links pressure, volume and absolute temperature for a fixed mass of an ideal gas. It appears across Edexcel A-Level Physics in thermal physics, kinetic theory and ideal gas calculations. Understanding how to manipulate P, V and T, and how the law leads to the ideal gas equation, will help you answer both structured and multiple-choice questions with confidence.
组合气体定律将一定质量理想气体的压强、体积和绝对温度联系起来。它在 Edexcel A-Level 物理的热学、气体动理论以及理想气体计算中反复出现。掌握如何变换 P、V、T，以及该定律如何导出理想气体方程，将帮助你从容应对结构题和选择题。
1. Statement and Formula of the Combined Gas Law | 组合气体定律的表述与公式
For a fixed mass of an ideal gas, pressure P, volume V and absolute temperature T are related by the condition PV/T = constant. If a gas changes from state 1 to state 2, the combined gas law is written as:
对于一定质量的理想气体，压强 P、体积 V 和绝对温度 T 满足 PV/T = 常数。如果气体从状态 1 变化到状态 2，组合气体定律可以写成：
P₁V₁ / T₁ = P₂V₂ / T₂
All temperatures must be expressed in kelvin, not degrees Celsius. The law is only linear on the absolute temperature scale because 0 K represents zero thermal motion.
所有温度必须使用开尔文，而不是摄氏度。该定律只有在绝对温标下才是线性关系，因为 0 K 代表热运动为零。
A common mistake is to substitute 25 °C directly into the formula. You should convert to kelvin by adding 273.15: T(K) = θ(°C) + 273.15. For most exam calculations, T = θ + 273 is acceptable unless high precision is required.
一个常见错误是把 25 °C 直接代入公式。你应当通过加 273.15 转换为开尔文：T(K) = θ(°C) + 273.15。在大多数考试计算中，使用 T = θ + 273 即可，除非题目要求高精度。
2. Deriving the Combined Gas Law from the Individual Gas Laws | 由各气体定律推导组合气体定律
The combined gas law is obtained by combining Boyle’s law, Charles’s law and the pressure law. Boyle’s law states P ∝ 1/V at constant temperature, Charles’s law states V ∝ T at constant pressure, and the pressure law states P ∝ T at constant volume.
组合气体定律由玻意耳定律、查理定律和压强定律组合得到。玻意耳定律指出在恒温下 P ∝ 1/V，查理定律指出在恒压下 V ∝ T，压强定律指出在恒容下 P ∝ T。
If we combine the three proportionalities for a fixed amount of gas, we get PV ∝ T, so PV/T = constant. This is why the combined gas law applies whenever the amount of gas and the number of particles remain unchanged.
如果对一定量的气体合并三个比例关系，就得到 PV ∝ T，因此 PV/T = 常数。这就是为什么只要气体的量和粒子数保持不变，组合气体定律就适用。
You should be able to derive the relationship qualitatively from the kinetic model: raising temperature while keeping volume fixed increases collision frequency and mean kinetic energy, so pressure rises.
你应当能够从气体动理论模型定性推导该关系：在体积不变时升高温度，会提高碰撞频率和平均动能，因此压强增大。
3. The Ideal Gas Equation: pV = nRT | 理想气体方程：pV = nRT
The ideal gas equation extends the combined gas law by including the number of moles n. For an ideal gas, the state variables are related by:
理想气体方程通过引入摩尔数 n 扩展了组合气体定律。对于理想气体，状态参数之间的关系为：
pV = nRT
Here p is pressure in pascals, V is volume in cubic metres, n is amount in moles, T is absolute temperature in kelvin, and R is the molar gas constant, R = 8.31 J mol⁻¹ K⁻¹.
其中 p 是压强（单位帕斯卡），V 是体积（单位立方米），n 是物质的量（单位摩尔），T 是绝对温度（单位开尔文），R 是摩尔气体常数，R = 8.31 J mol⁻¹ K⁻¹。
When working with the number of molecules N rather than moles, the equation becomes pV = NkT, where k = R/NA = 1.38 × 10⁻²³ J K⁻¹ and NA is the Avogadro constant, 6.02 × 10²³ mol⁻¹.
当使用分子数 N 而不是摩尔数时，方程变为 pV = NkT，其中 k = R/NA = 1.38 × 10⁻²³ J K⁻¹，NA 是阿伏伽德罗常数，6.02 × 10²³ mol⁻¹。
The ideal gas equation assumes that the gas obeys the kinetic theory assumptions. It is most accurate for low pressure and high temperature, where real gases behave more ideally.
理想气体方程假设气体满足气体动理论的基本假定。它在低压和高温下最准确，此时真实气体表现得更加理想。
4. Key Assumptions of the Kinetic Theory of Gases | 气体动理论的关键假设
To use the ideal gas model, you need to recall the assumptions made about gas particles. These assumptions explain why pV/T is constant for a fixed mass of gas.
要使用理想气体模型，你需要记住关于气体粒子的假设。这些假设解释了为什么对一定质量的气体，pV/T 为常数。
First, gas particles are point masses with negligible volume compared with the container. Second, collisions between particles and with the container walls are perfectly elastic, so total kinetic energy is conserved. Third, there are no intermolecular forces between particles except during collisions.
首先，气体粒子是点质量，与容器相比其体积可以忽略。其次，粒子之间以及粒子与容器壁之间的碰撞是完全弹性的，因此总动能守恒。第三，除碰撞瞬间外，粒子之间没有分子间作用力。
Fourth, particles are in constant, random motion. Fifth, the average kinetic energy of the particles is proportional to the absolute temperature of the gas.
第四，粒子处于持续无规则运动状态。第五，粒子的平均动能与气体的绝对温度成正比。
Exam questions often ask you to state one assumption that breaks down at high pressure or low temperature. At high pressure, particle volume is no longer negligible; at low temperature, intermolecular attractions become significant.
考试题经常要求你写出一个在高压或低温下失效的假设。在高压下，粒子体积不再可以忽略；在低温下，分子间吸引力变得显著。
5. Molar Gas Volume and Standard Conditions | 摩尔气体体积与标准条件
The molar gas volume is the volume occupied by one mole of an ideal gas under specified conditions. At standard temperature and pressure, STP, taken as 273.15 K and 101 325 Pa, one mole occupies about 22.4 dm³.
摩尔气体体积是指一摩尔理想气体在特定条件下所占的体积。在标准温度与压强 STP 下，即 273.15 K 和 101 325 Pa，一摩尔气体约占 22.4 dm³。
In many Edexcel questions, room temperature and pressure, RTP, is taken as 298 K and 101 kPa. At RTP, one mole of an ideal gas occupies approximately 24.0 dm³, or 0.0240 m³.
在许多 Edexcel 题目中，常温常压 RTP 取为 298 K 和 101 kPa。在 RTP 下，一摩尔理想气体约占 24.0 dm³，即 0.0240 m³。
You can verify this using the ideal gas equation: V = nRT/p. With n = 1 mol, T = 298 K and p = 101 000 Pa, V ≈ 0.0245 m³, which is close to 24 dm³.
你可以用理想气体方程验证：V = nRT/p。当 n = 1 mol，T = 298 K，p = 101 000 Pa 时，V ≈ 0.0245 m³，接近 24 dm³。
6. Rearranging and Using Gas Law Calculations | 气体定律计算的重排与应用
You should be able to rearrange the combined gas law to find any unknown variable. For example, if you need the final pressure P₂, use:
你应当会重排组合气体定律来求任意未知量。例如，如果需要求末态压强 P₂，可以使用：
P₂ = P₁V₁T₂ / (V₂T₁)
Before substituting values, write down the known quantities and convert them into SI units. Volume must be in m³, pressure in Pa and temperature in K.
在代入数值之前，先写出已知量并将其转换为国际单位制。体积必须是 m³，压强必须是 Pa，温度必须是 K。
For ideal gas equation questions, you may be asked to find n, p, V or T. Rearrangements include n = pV/RT, T = pV/nR and p = nRT/V.
对于理想气体方程题，可能要求你求 n、p、V 或 T。重排形式包括 n = pV/RT、T = pV/nR 和 p = nRT/V。
Always show your working clearly. Edexcel examiners award marks for correct substitution and conversion, even if the final arithmetic contains a slip.
务必清晰展示解题步骤。即使最后计算出现小错误，Edexcel 考官也会对正确的代入和单位转换给予步骤分。
7. Units and Conversions in Gas Calculations | 气体计算中的单位与换算
Gas law calculations in A-Level Physics require strict SI units. The table below summarises the most important conversions.
A-Level 物理中的气体定律计算要求严格使用国际单位制。下表总结了最重要的换算关系。
| Quantity 物理量 | SI Unit 国际单位 | Common Conversion 常见换算 |
| Pressure 压强 | Pa | 1 atm = 101 325 Pa; 1 bar = 100 000 Pa |
| Volume 体积 | m³ | 1 dm³ = 10⁻³ m³; 1 cm³ = 10⁻⁶ m³ |
| Temperature 温度 | K | T(K) = θ(°C) + 273.15 |
When a question gives volume in cm³ or dm³, convert before using pV = nRT or the combined gas law. A ratio calculation may allow the same volume unit on both sides, but it is safer to convert to m³.
当题目给出的体积单位是 cm³ 或 dm³ 时，应先转换再使用 pV = nRT 或组合气体定律。比例计算可能允许两侧使用相同体积单位，但转换为 m³ 更保险。
Similarly, pressure given in kPa should be multiplied by 1000 to obtain pascals. For example, 101 kPa = 101 000 Pa.
同样，以 kPa 给出的压强应乘以 1000 得到帕斯卡。例如，101 kPa = 101 000 Pa。
8. Worked Example: Combined Gas Law | 例题：组合气体定律
A fixed mass of gas has a pressure of 2.0 × 10⁵ Pa, a volume of 3.0 × 10⁻³ m³ and a temperature of 300 K. It is compressed to 1.5 × 10⁻³ m³ and heated to 360 K. Calculate the new pressure.
一定质量气体在压强 2.0 × 10⁵ Pa、体积 3.0 × 10⁻³ m³、温度 300 K 下。它被压缩至 1.5 × 10⁻³ m³ 并加热至 360 K。求新的压强。
Use P₂ = P₁V₁T₂ / (V₂T₁). Substitute the values:
使用 P₂ = P₁V₁T₂ / (V₂T₁)。代入数值：
P₂ = (2.0 × 10⁵ × 3.0 × 10⁻³ × 360) / (1.5 × 10⁻³ × 300)
The volume units cancel because they are both in m³, so the final pressure is 4.8 × 10⁵ Pa. Notice that the pressure increases because the gas is compressed and heated.
体积单位由于都是 m³ 而相消，因此最终压强为 4.8 × 10⁵ Pa。注意压强增大是因为气体被压缩并加热。
9. Worked Example: Ideal Gas Equation | 例题：理想气体方程
Calculate the amount of gas, in mol, in a 2.0 × 10⁻² m³ container at a pressure of 101 000 Pa and a temperature of 298 K.
计算在 2.0 × 10⁻² m³ 的容器中，压强为 101 000 Pa、温度为 298 K 时气体的物质的量（单位 mol）。
Rearrange pV = nRT to give n = pV / RT. Substitute the values:
重排 pV = nRT 得到 n = pV / RT。代入数值：
n = (101 000 × 2.0 × 10⁻²) / (8.31 × 298)
This gives n ≈ 0.816 mol. Always check that your final answer has the correct unit and a sensible order of magnitude.
得到 n ≈ 0.816 mol。务必检查最终答案的单位是否正确，数量级是否合理。
10. Real Gases and Deviations from
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