A-Level Physics Unit 5 Insert (Jan 2020) Formula Derivations | A-Level 物理单元5 2020年1月公式表推导

📚 A-Level Physics Unit 5 Insert (Jan 2020) Formula Derivations | A-Level 物理单元5 2020年1月公式表推导

In the January 2020 A-Level Physics Unit 5 examination, students are provided with a formula insert that summarises key equations from thermodynamics, nuclear physics and oscillations. This article presents step-by-step derivations of those core formulas, linking macroscopic laws with microscopic models and mathematical principles. By working through these derivations, you gain a deeper understanding of the underlying physics and will be better prepared to apply the equations correctly in exams.

在2020年1月A-Level物理单元5的考试中，考生会收到一份公式表，归纳了热力学、核物理和振动中的关键方程。本文将这些核心公式进行逐步推导，将宏观定律与微观模型和数学原理联系起来。通过梳理这些推导过程，你可以更深刻地理解背后的物理，并在考试中更准确地运用这些方程。

1. Ideal Gas Equation and Microscopic Pressure | 理想气体方程与微观压强

The ideal gas equation, pV = nRT, combines the experimental gas laws (Boyle’s law, Charles’ law and Avogadro’s hypothesis). On a microscopic scale, the pressure exerted by a gas arises from the rate of change of momentum of molecules colliding with the container walls. For a single molecule of mass m travelling with speed u_x perpendicular to a wall of a cubic box of side L, the momentum change per collision is 2m u_x, and the time between collisions with that wall is 2L/u_x. The average force on the wall due to one molecule is therefore F = (2m u_x) / (2L/u_x) = m u_x²/L.

理想气体方程pV = nRT综合了实验气体定律（玻意耳定律、查理定律和阿伏伽德罗假说）。在微观尺度上，气体压强源于分子与容器壁碰撞时的动量变化率。对于质量为m、以速度u_x垂直撞向边长为L的立方体箱壁的单个分子，每次碰撞的动量变化为2m u_x，与同一器壁相邻两次碰撞的时间间隔为2L/u_x。因此，一个分子对器壁的平均作用力为F = (2m u_x) / (2L/u_x) = m u_x²/L。

Summing over all N molecules and taking the average of u_x², the total force on the wall is F = (m/L) Σ u_xi² = (N m / L) <u_x²>, where <u_x²> denotes the mean square x-component of speed. Because the motion is random, the mean square speed <c²> = <u_x²> + <u_y²> + <u_z²> = 3 <u_x²>. Hence

对所有N个分子求和，并取u_x²的平均值，墙上的总力为F = (m/L) Σ u_xi² = (N m / L) <u_x²>，其中<u_x²>表示x方向速度分量平方的平均值。由于运动无规，均方速率<c²> = <u_x²> + <u_y²> + <u_z²> = 3 <u_x²>。因此

p = F / L² = (N m / L³) <u_x²> = (1/3) (N m / V) <c²> = (1/3) ρ <c²>

where ρ = N m / V is the gas density. For an ideal gas, the macroscopic equation can be written as pV = NkT, where k is the Boltzmann constant and N is the number of molecules. Equating (1/3) N m <c²> with NkT gives a direct link between the micro- and macro-worlds.

式中ρ = N m / V为气体密度。对于理想气体，宏观方程可写成pV = NkT，其中k为玻尔兹曼常数，N为分子数。令(1/3) N m <c²>等于NkT，便可直接联系微观与宏观世界。

2. Temperature and Average Kinetic Energy | 温度与平均动能的关系

From the equivalence pV = (1/3) N m <c²> and the ideal gas equation pV = NkT, we can cancel N and rearrange:

由等式pV = (1/3) N m <c²>和理想气体状态方程pV = NkT，可消去N并整理：

(1/3) m <c²> = kT

The average translational kinetic energy of a single molecule is KE_avg = ½ m <c²>. Substituting gives KE_avg = (3/2) kT. This result is crucial: it shows that the absolute temperature of an ideal gas is directly proportional to the average random kinetic energy of its particles. For one mole of gas (N = N_A), the total internal kinetic energy is U = (3/2) nRT, where R = N_A k.

单个分子的平均平动动能为KE_avg = ½ m <c²>。代入得KE_avg = (3/2) kT。这一结果至关重要：它表明理想气体的绝对温度与其粒子平均无规动能成正比。对一摩尔气体（N = N_A），总的内动能U = (3/2) nRT，其中R = N_Ak。

3. Adiabatic Process Equation pV^γ = constant | 绝热过程方程pV^γ = 常数

In an adiabatic process, no heat enters or leaves the system (Q = 0). The first law of thermodynamics states ΔU = Q – W, so for an adiabatic change ΔU = -W = -p ΔV. For an ideal gas, the change in internal energy can also be written as ΔU = n C_v ΔT, where C_v is the molar heat capacity at constant volume. Therefore, n C_v ΔT = -p ΔV.

在绝热过程中，系统与外界无热量交换（Q = 0）。热力学第一定律ΔU = Q – W，因此绝热变化有ΔU = -W = -p ΔV。对于理想气体，内能变化也可写成ΔU = n C_v ΔT，其中C_v为等容摩尔热容。于是n C_v ΔT = -p ΔV。

For an ideal gas, pV = nRT. Differentiating both sides gives p dV + V dp = nR dT. Substituting dT from the energy relation, we have p dV + V dp = – (nR / n C_v) p dV = – (R / C_v) p dV. Using the relationship C_p – C_v = R, where C_p is the molar heat capacity at constant pressure, and introducing the adiabatic index γ = C_p / C_v, we write R/C_v = γ – 1. The differential equation becomes p dV + V dp = -(γ – 1) p dV, or V dp = -γ p dV.

对理想气体pV = nRT，两边微分得p dV + V dp = nR dT。利用能量关系代入dT，得到p dV + V dp = – (nR / n C_v) p dV = – (R / C_v) p dV。利用迈耶关系C_p – C_v = R，其中C_p为等压摩尔热容，并引入绝热指数γ = C_p / C_v，有R/C_v = γ – 1。上述微分方程化为p dV + V dp = -(γ – 1) p dV，即V dp = -γ p dV。

Separating variables and integrating:

分离变量并积分：

∫ dp/p = -γ ∫ dV/V ⇒ ln p = -γ ln V + constant

This yields ln(p V^γ) = constant, or

由此可得ln(p V^γ) = 常数，即

p V^γ = constant

which is the well-known adiabatic equation. Using pV = nRT, it can also be expressed as T V^{γ – 1} = constant or T p^{(1 – γ)/γ} = constant.

这就是著名的绝热过程方程。利用pV = nRT，还可表达为T V^{γ – 1} = 常数或T p^{(1 – γ)/γ} = 常数。

4. Efficiency of a Carnot Engine | 卡诺热机效率

A Carnot cycle consists of two reversible isothermal and two reversible adiabatic processes. For the high-temperature isothermal expansion at T_h, the heat absorbed from the hot reservoir is Q_h = nR T_h ln(V₂/V₁). For the low-temperature isothermal compression at T_c, the heat rejected to the cold reservoir is Q_c = nR T_c ln(V₃/V₄), with positive magnitude.

卡诺循环由两个可逆等温过程和两个可逆绝热过程组成。在高温T_h下的等温膨胀中，从高温热源吸收的热量为Q_h = nR T_h ln(V₂/V₁)。在低温T_c下的等温压缩中，向低温热源放出的热量为Q_c = nR T_c ln(V₃/V₄)（取正值）。

The two adiabatic stages link the volumes: T_h V₂^{γ – 1} = T_c V₃^{γ – 1} and T_h V₁^{γ – 1} = T_c V₄^{γ – 1}. Dividing these two equations gives (V₂/V₁)^{γ – 1} = (V₃/V₄)^{γ – 1}, hence V₂/V₁ = V₃/V₄. Therefore, the magnitudes of the heats satisfy Q_c/Q_h = T_c/T_h.

两个绝热阶段将体积关联起来：T_h V₂<

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