Topic 3 热学 / Thermal Physics
温度与热量 · 比热容 · 潜热 · 理想气体 · 气体动理论
IB SL / HLTopic 3热现象是能量传递的表现
热物理学研究热量（heat）的传递以及温度（temperature）变化对物质的影响。热是能量传递的一种形式，温度则是分子平均动能的量度。本章将学习热传递的计算、理想气体的行为以及气体动理论的基本概念——这是理解热力学和能量守恒的基础。
Thermal physics studies the transfer of heat and the effect of temperature changes on matter. Heat is a form of energy transfer, while temperature measures average molecular kinetic energy. This chapter covers heat transfer calculations, ideal gas behavior, and kinetic theory — the foundation for understanding thermodynamics and energy conservation.
3.1 温度与热量 / Temperature & Heat
IB 3.1核心基础
温度 / Temperature
$$T(\text{K}) = \theta(^{\circ}\text{C}) + 273.15$$
温度是物体冷热程度的量度，微观上是分子平均平动动能的量度。SI 单位是开尔文（K），是七个基本单位之一。摄氏温标（°C）以水的冰点（0°C）和沸点（100°C）为基准，热力学温标（K）以绝对零度（0 K = -273.15°C）为起点。温度差在两种温标中相同：$\Delta T(\text{K}) = \Delta\theta(^{\circ}\text{C})$。
Temperature measures how hot or cold an object is; microscopically, it measures the average translational kinetic energy of molecules. The SI unit is the kelvin (K), one of the seven base units. The Celsius scale (°C) uses water's freezing (0°C) and boiling (100°C) points; the thermodynamic scale (K) starts at absolute zero (0 K = -273.15°C). Temperature differences are the same on both scales: $\Delta T(\text{K}) = \Delta\theta(^{\circ}\text{C})$.
📐 详细说明 / Detailed Explanation
温度的本质：
① 宏观定义：温度决定热传递的方向——热量从高温物体自发传向低温物体。
② 微观定义：温度与分子的平均平动动能成正比：$\bar{E}_k = \frac32 k_B T$，其中 $k_B = 1.38 \times 10^{-23}$ J/K 是玻尔兹曼常数。
③ 绝对零度：0 K（-273.15°C）是所有分子热运动停止的理论温度，实际上无法达到。
④ 热平衡：两个物体接触足够长时间后达到相同温度，称为热平衡。
⑤ 温标换算：$T(\text{K}) = \theta(^{\circ}\text{C}) + 273.15$，温差 $\Delta T = \Delta\theta$。
常见温度参考：
• 室温：约 293 K (20°C)
• 人体温度：约 310 K (37°C)
• 水沸点：373 K (100°C)
• 液氮：77 K (-196°C)
• 太阳表面：约 5800 K
① 宏观定义：温度决定热传递的方向——热量从高温物体自发传向低温物体。
② 微观定义：温度与分子的平均平动动能成正比：$\bar{E}_k = \frac32 k_B T$，其中 $k_B = 1.38 \times 10^{-23}$ J/K 是玻尔兹曼常数。
③ 绝对零度：0 K（-273.15°C）是所有分子热运动停止的理论温度，实际上无法达到。
④ 热平衡：两个物体接触足够长时间后达到相同温度，称为热平衡。
⑤ 温标换算：$T(\text{K}) = \theta(^{\circ}\text{C}) + 273.15$，温差 $\Delta T = \Delta\theta$。
常见温度参考：
• 室温：约 293 K (20°C)
• 人体温度：约 310 K (37°C)
• 水沸点：373 K (100°C)
• 液氮：77 K (-196°C)
• 太阳表面：约 5800 K
Nature of temperature:
① Macroscopic: Temperature determines the direction of heat flow — heat spontaneously flows from hot to cold.
② Microscopic: Temperature is proportional to average translational kinetic energy: $\bar{E}_k = \frac32 k_B T$, where $k_B = 1.38 \times 10^{-23}$ J/K is Boltzmann's constant.
③ Absolute zero: 0 K (-273.15°C) — theoretical point where all molecular motion stops, unreachable in practice.
④ Thermal equilibrium: Two objects in contact reach the same temperature after sufficient time.
⑤ Scale conversion: $T(\text{K}) = \theta(^{\circ}\text{C}) + 273.15$, difference $\Delta T = \Delta\theta$.
Common temperature references:
• Room temperature: ~293 K (20°C)
• Human body: ~310 K (37°C)
• Water boiling: 373 K (100°C)
• Liquid nitrogen: 77 K (-196°C)
• Sun surface: ~5800 K
① Macroscopic: Temperature determines the direction of heat flow — heat spontaneously flows from hot to cold.
② Microscopic: Temperature is proportional to average translational kinetic energy: $\bar{E}_k = \frac32 k_B T$, where $k_B = 1.38 \times 10^{-23}$ J/K is Boltzmann's constant.
③ Absolute zero: 0 K (-273.15°C) — theoretical point where all molecular motion stops, unreachable in practice.
④ Thermal equilibrium: Two objects in contact reach the same temperature after sufficient time.
⑤ Scale conversion: $T(\text{K}) = \theta(^{\circ}\text{C}) + 273.15$, difference $\Delta T = \Delta\theta$.
Common temperature references:
• Room temperature: ~293 K (20°C)
• Human body: ~310 K (37°C)
• Water boiling: 373 K (100°C)
• Liquid nitrogen: 77 K (-196°C)
• Sun surface: ~5800 K
热量 / Heat
$$Q = mc\Delta T \quad \text{（无相变）}$$
热量（Q）是能量传递的一种形式，单位是焦耳（J）。当物体因温度差而吸收或释放能量时，我们称之为热传递。热传递有三种方式：传导（conduction）、对流（convection）和辐射（radiation）。
Heat (Q) is a form of energy transfer, measured in joules (J). When energy is transferred due to a temperature difference, we call it heat transfer. There are three mechanisms: conduction, convection, and radiation.
📐 热传递的三种方式 / Three Modes of Heat Transfer
① 传导 (Conduction)：
热量通过分子振动和自由电子在物质内部传递。固体中传导最快，气体中最慢。
• 导热系数 $k$（W/m·K）描述材料的导热能力
• 导热率公式：$\displaystyle \frac{Q}{t} = kA\frac{\Delta T}{d}$
• 良导体：金属（铜 $k \approx 400$ W/m·K）
• 不良导体：木材、塑料、空气
② 对流 (Convection)：
流体（液体或气体）因温度差导致密度差而产生的宏观流动。
• 自然对流：暖气片使热空气上升
• 强制对流：风扇、水泵
③ 辐射 (Radiation)：
物体通过电磁波（红外线）发射能量，不需要介质。
• 斯特藩-玻尔兹曼定律：$P = e\sigma A T^4$
• $\sigma = 5.67 \times 10^{-8}$ W/m·K⁴，$e$ 为发射率（0-1）
• 太阳辐射加热地球是辐射传热的最好例子
热量通过分子振动和自由电子在物质内部传递。固体中传导最快，气体中最慢。
• 导热系数 $k$（W/m·K）描述材料的导热能力
• 导热率公式：$\displaystyle \frac{Q}{t} = kA\frac{\Delta T}{d}$
• 良导体：金属（铜 $k \approx 400$ W/m·K）
• 不良导体：木材、塑料、空气
② 对流 (Convection)：
流体（液体或气体）因温度差导致密度差而产生的宏观流动。
• 自然对流：暖气片使热空气上升
• 强制对流：风扇、水泵
③ 辐射 (Radiation)：
物体通过电磁波（红外线）发射能量，不需要介质。
• 斯特藩-玻尔兹曼定律：$P = e\sigma A T^4$
• $\sigma = 5.67 \times 10^{-8}$ W/m·K⁴，$e$ 为发射率（0-1）
• 太阳辐射加热地球是辐射传热的最好例子
① Conduction:
Heat transfer through molecular vibrations and free electrons within a substance. Fastest in solids, slowest in gases.
• Thermal conductivity $k$ (W/m·K) describes a material's ability to conduct
• Conduction rate: $\displaystyle \frac{Q}{t} = kA\frac{\Delta T}{d}$
• Good conductors: metals (copper $k \approx 400$ W/m·K)
• Insulators: wood, plastic, air
② Convection:
Bulk movement of fluids (liquids/gases) due to density differences from temperature changes.
• Natural: warm air rising from a radiator
• Forced: fans, pumps
③ Radiation:
Emission of electromagnetic waves (infrared) — no medium needed.
• Stefan-Boltzmann law: $P = e\sigma A T^4$
• $\sigma = 5.67 \times 10^{-8}$ W/m·K⁴, $e$ is emissivity (0-1)
• Solar heating of Earth is the best example
Heat transfer through molecular vibrations and free electrons within a substance. Fastest in solids, slowest in gases.
• Thermal conductivity $k$ (W/m·K) describes a material's ability to conduct
• Conduction rate: $\displaystyle \frac{Q}{t} = kA\frac{\Delta T}{d}$
• Good conductors: metals (copper $k \approx 400$ W/m·K)
• Insulators: wood, plastic, air
② Convection:
Bulk movement of fluids (liquids/gases) due to density differences from temperature changes.
• Natural: warm air rising from a radiator
• Forced: fans, pumps
③ Radiation:
Emission of electromagnetic waves (infrared) — no medium needed.
• Stefan-Boltzmann law: $P = e\sigma A T^4$
• $\sigma = 5.67 \times 10^{-8}$ W/m·K⁴, $e$ is emissivity (0-1)
• Solar heating of Earth is the best example
🎯 IB 考试要点 / Exam Tips：① IB 中热量的符号是 $Q$，单位始终是 J；② 温度用开尔文还是摄氏取决于公式——温差公式中两者都可（因为 $\Delta T = \Delta\theta$），但涉及绝对温度的公式（如理想气体定律）必须用 K；③ 热平衡是 IB 常见考点——两个物体接触最终温度相同；④ 注意区分"热"和"温度"——一杯 50°C 的水和半杯 50°C 的水温度相同但含热量不同。
🎯 IB Exam Tips: ① IB uses $Q$ for heat, always in J; ② Use K or °C depends on the formula — for temperature differences either works ($\Delta T = \Delta\theta$), but formulas with absolute temperature (like ideal gas law) require K; ③ Thermal equilibrium is common — objects in contact reach the same final temperature; ④ Distinguish heat vs temperature — a cup and half-cup of water at 50°C have the same temperature but different heat content.
例 3.1 / Example 3.1
将以下温度转换为开尔文：(a) 25°C (b) -50°C (c) 100°C。
Convert to kelvin: (a) 25°C (b) -50°C (c) 100°C.
解题过程 / Solution
$T = \theta + 273.15$
(a) $T = 25 + 273.15 = 298.15$ K
(b) $T = -50 + 273.15 = 223.15$ K
(c) $T = 100 + 273.15 = 373.15$ K
(a) $T = 25 + 273.15 = 298.15$ K
(b) $T = -50 + 273.15 = 223.15$ K
(c) $T = 100 + 273.15 = 373.15$ K
$T = \theta + 273.15$
(a) $T = 25 + 273.15 = 298.15$ K
(b) $T = -50 + 273.15 = 223.15$ K
(c) $T = 100 + 273.15 = 373.15$ K
(a) $T = 25 + 273.15 = 298.15$ K
(b) $T = -50 + 273.15 = 223.15$ K
(c) $T = 100 + 273.15 = 373.15$ K
例 3.2 / Example 3.2
室内温度 20°C，室外 -5°C。墙壁面积 20 m²，厚度 0.25 m，导热系数 $k = 0.8$ W/m·K。求每小时通过墙壁的热量损失。
Indoor 20°C, outdoor -5°C. Wall area 20 m², thickness 0.25 m, $k = 0.8$ W/m·K. Find heat loss per hour.
解题过程 / Solution
$\Delta T = 20 - (-5) = 25$°C = 25 K
$\displaystyle \frac{Q}{t} = kA\frac{\Delta T}{d} = 0.8 \times 20 \times \frac{25}{0.25} = 1600$ W = 1600 J/s
$Q = 1600 \times 3600 = 5.76 \times 10^6$ J = 5.76 MJ
$\displaystyle \frac{Q}{t} = kA\frac{\Delta T}{d} = 0.8 \times 20 \times \frac{25}{0.25} = 1600$ W = 1600 J/s
$Q = 1600 \times 3600 = 5.76 \times 10^6$ J = 5.76 MJ
$\Delta T = 20 - (-5) = 25$°C = 25 K
$\displaystyle \frac{Q}{t} = kA\frac{\Delta T}{d} = 0.8 \times 20 \times \frac{25}{0.25} = 1600$ W = 1600 J/s
$Q = 1600 \times 3600 = 5.76 \times 10^6$ J = 5.76 MJ
$\displaystyle \frac{Q}{t} = kA\frac{\Delta T}{d} = 0.8 \times 20 \times \frac{25}{0.25} = 1600$ W = 1600 J/s
$Q = 1600 \times 3600 = 5.76 \times 10^6$ J = 5.76 MJ
3.2 比热与潜热 / Specific Heat & Latent Heat
IB 3.2计算核心
比热容 / Specific Heat Capacity
$$Q = mc\Delta T \quad \text{或} \quad c = \frac{Q}{m\Delta T}$$
比热容（$c$）是单位质量的物质温度升高（或降低）1 K 时所吸收（或释放）的热量。单位：J/(kg·K)。不同的物质比热容不同——水的比热容很大（$c = 4200$ J/kg·K），这是沿海地区气候温和的原因之一。
Specific heat capacity ($c$) is the heat energy required to raise the temperature of 1 kg of a substance by 1 K. Unit: J/(kg·K). Different substances have different $c$ values — water's high specific heat ($c = 4200$ J/kg·K) moderates coastal climates.
📐 详细推导 / Detailed Derivation
比热容的定义与推导：
由实验发现，物体吸收的热量 $Q$ 与质量 $m$ 和温度变化 $\Delta T$ 成正比。引入比例常数 $c$：
$Q \propto m\Delta T$
$Q = mc\Delta T$
其中 $c$ 就是比热容。
能量守恒（热平衡）：
当两个不同温度的物体接触达到热平衡时，放出的热量等于吸收的热量：
$Q_{\text{放}} = Q_{\text{吸}}$
$m_1 c_1 (T_1 - T_f) = m_2 c_2 (T_f - T_2)$
其中 $T_f$ 是最终平衡温度，$T_1 > T_2$。
常见物质的比热容：
• 水：4200 J/kg·K
• 冰：2100 J/kg·K
• 铝：900 J/kg·K
• 铜：390 J/kg·K
• 铁：450 J/kg·K
• 乙醇：2440 J/kg·K
由实验发现，物体吸收的热量 $Q$ 与质量 $m$ 和温度变化 $\Delta T$ 成正比。引入比例常数 $c$：
$Q \propto m\Delta T$
$Q = mc\Delta T$
其中 $c$ 就是比热容。
能量守恒（热平衡）：
当两个不同温度的物体接触达到热平衡时，放出的热量等于吸收的热量：
$Q_{\text{放}} = Q_{\text{吸}}$
$m_1 c_1 (T_1 - T_f) = m_2 c_2 (T_f - T_2)$
其中 $T_f$ 是最终平衡温度，$T_1 > T_2$。
常见物质的比热容：
• 水：4200 J/kg·K
• 冰：2100 J/kg·K
• 铝：900 J/kg·K
• 铜：390 J/kg·K
• 铁：450 J/kg·K
• 乙醇：2440 J/kg·K
Definition and derivation of specific heat capacity:
Experiments show that heat absorbed $Q$ is proportional to mass $m$ and temperature change $\Delta T$. The proportionality constant is $c$:
$Q \propto m\Delta T$
$Q = mc\Delta T$
Energy conservation (thermal equilibrium):
When two objects at different temperatures reach thermal equilibrium, heat lost equals heat gained:
$Q_{\text{lost}} = Q_{\text{gained}}$
$m_1 c_1 (T_1 - T_f) = m_2 c_2 (T_f - T_2)$
where $T_f$ is the final equilibrium temperature, $T_1 > T_2$.
Specific heat capacities of common substances:
• Water: 4200 J/kg·K
• Ice: 2100 J/kg·K
• Aluminium: 900 J/kg·K
• Copper: 390 J/kg·K
• Iron: 450 J/kg·K
• Ethanol: 2440 J/kg·K
Experiments show that heat absorbed $Q$ is proportional to mass $m$ and temperature change $\Delta T$. The proportionality constant is $c$:
$Q \propto m\Delta T$
$Q = mc\Delta T$
Energy conservation (thermal equilibrium):
When two objects at different temperatures reach thermal equilibrium, heat lost equals heat gained:
$Q_{\text{lost}} = Q_{\text{gained}}$
$m_1 c_1 (T_1 - T_f) = m_2 c_2 (T_f - T_2)$
where $T_f$ is the final equilibrium temperature, $T_1 > T_2$.
Specific heat capacities of common substances:
• Water: 4200 J/kg·K
• Ice: 2100 J/kg·K
• Aluminium: 900 J/kg·K
• Copper: 390 J/kg·K
• Iron: 450 J/kg·K
• Ethanol: 2440 J/kg·K
相变与潜热 / Phase Change & Latent Heat
$$Q = mL \quad \text{（相变时，温度不变）}$$
物质在相变（如熔化、凝固、汽化、液化）过程中吸收或释放热量，但温度保持不变。这部分热量称为潜热（latent heat）。$L$ 是比潜热（J/kg），$Q$ 是吸收或释放的热量。
During a phase change (melting, freezing, boiling, condensing), a substance absorbs or releases heat at constant temperature. This is latent heat. $L$ is the specific latent heat (J/kg), and $Q$ is the heat absorbed or released.
📐 详细说明 / Detailed Explanation
相变过程：
① 熔化（固体→液体）：吸收潜热 $L_f$（熔解潜热）
② 凝固（液体→固体）：释放潜热 $L_f$
③ 汽化（液体→气体）：吸收潜热 $L_v$（汽化潜热）
④ 液化（气体→液体）：释放潜热 $L_v$
加热曲线：
加热冰（-20°C→120°C 的水蒸气）的典型曲线：
• A→B：冰升温（$Q = m c_{\text{冰}} \Delta T$）
• B→C：冰在 0°C 熔化（$Q = m L_f$，温度不变）
• C→D：水升温（$Q = m c_{\text{水}} \Delta T$）
• D→E：水在 100°C 沸腾（$Q = m L_v$，温度不变）
• E→F：水蒸气升温（$Q = m c_{\text{汽}} \Delta T$）
常见潜热值：
• 水的熔解潜热 $L_f = 3.34 \times 10^5$ J/kg
• 水的汽化潜热 $L_v = 2.26 \times 10^6$ J/kg
• 汽化潜热远大于熔解潜热——汽化需要破坏所有分子间作用力
① 熔化（固体→液体）：吸收潜热 $L_f$（熔解潜热）
② 凝固（液体→固体）：释放潜热 $L_f$
③ 汽化（液体→气体）：吸收潜热 $L_v$（汽化潜热）
④ 液化（气体→液体）：释放潜热 $L_v$
加热曲线：
加热冰（-20°C→120°C 的水蒸气）的典型曲线：
• A→B：冰升温（$Q = m c_{\text{冰}} \Delta T$）
• B→C：冰在 0°C 熔化（$Q = m L_f$，温度不变）
• C→D：水升温（$Q = m c_{\text{水}} \Delta T$）
• D→E：水在 100°C 沸腾（$Q = m L_v$，温度不变）
• E→F：水蒸气升温（$Q = m c_{\text{汽}} \Delta T$）
常见潜热值：
• 水的熔解潜热 $L_f = 3.34 \times 10^5$ J/kg
• 水的汽化潜热 $L_v = 2.26 \times 10^6$ J/kg
• 汽化潜热远大于熔解潜热——汽化需要破坏所有分子间作用力
Phase change processes:
① Melting (solid → liquid): absorbs latent heat $L_f$ (fusion)
② Freezing (liquid → solid): releases latent heat $L_f$
③ Boiling (liquid → gas): absorbs latent heat $L_v$ (vaporization)
④ Condensing (gas → liquid): releases latent heat $L_v$
Heating curve:
Heating ice from -20°C to 120°C steam:
• A→B: Ice heats up ($Q = m c_{\text{ice}} \Delta T$)
• B→C: Ice melts at 0°C ($Q = m L_f$, constant T)
• C→D: Water heats up ($Q = m c_{\text{water}} \Delta T$)
• D→E: Water boils at 100°C ($Q = m L_v$, constant T)
• E→F: Steam heats up ($Q = m c_{\text{steam}} \Delta T$)
Common latent heats:
• Water's latent heat of fusion $L_f = 3.34 \times 10^5$ J/kg
• Water's latent heat of vaporization $L_v = 2.26 \times 10^6$ J/kg
• $L_v \gg L_f$ — vaporization breaks all intermolecular forces
① Melting (solid → liquid): absorbs latent heat $L_f$ (fusion)
② Freezing (liquid → solid): releases latent heat $L_f$
③ Boiling (liquid → gas): absorbs latent heat $L_v$ (vaporization)
④ Condensing (gas → liquid): releases latent heat $L_v$
Heating curve:
Heating ice from -20°C to 120°C steam:
• A→B: Ice heats up ($Q = m c_{\text{ice}} \Delta T$)
• B→C: Ice melts at 0°C ($Q = m L_f$, constant T)
• C→D: Water heats up ($Q = m c_{\text{water}} \Delta T$)
• D→E: Water boils at 100°C ($Q = m L_v$, constant T)
• E→F: Steam heats up ($Q = m c_{\text{steam}} \Delta T$)
Common latent heats:
• Water's latent heat of fusion $L_f = 3.34 \times 10^5$ J/kg
• Water's latent heat of vaporization $L_v = 2.26 \times 10^6$ J/kg
• $L_v \gg L_f$ — vaporization breaks all intermolecular forces
🎯 IB 考试要点 / Exam Tips：① 相变时温度不变——这是常见考点；② $Q = mL$ 中，注意是吸热还是放热；③ 热平衡计算中，相变和升温可能同时出现，需要分段计算；④ 汽化潜热远大于熔解潜热（水的 $L_v \approx 7 L_f$），因为汽化需要完全克服分子间作用力；⑤ 比热容和潜热都依赖于物质种类。
🎯 IB Exam Tips: ① Temperature is constant during phase change — common exam question; ② In $Q = mL$, note whether heat is absorbed or released; ③ In thermal equilibrium problems, phase changes and heating may occur together — calculate in stages; ④ Latent heat of vaporization is much larger than fusion (water's $L_v \approx 7 L_f$) because vaporization completely overcomes intermolecular forces; ⑤ Both specific heat and latent heat depend on the substance.
例 3.3 / Example 3.3
2 kg 的水从 20°C 加热到 80°C，需要吸收多少热量？水的比热容 $c = 4200$ J/kg·K。
Heat required to raise 2 kg of water from 20°C to 80°C. Water $c = 4200$ J/kg·K.
解题过程 / Solution
$\Delta T = 80 - 20 = 60$ K
$Q = mc\Delta T = 2 \times 4200 \times 60 = 504000$ J = 504 kJ
$Q = mc\Delta T = 2 \times 4200 \times 60 = 504000$ J = 504 kJ
$\Delta T = 80 - 20 = 60$ K
$Q = mc\Delta T = 2 \times 4200 \times 60 = 504000$ J = 504 kJ
$Q = mc\Delta T = 2 \times 4200 \times 60 = 504000$ J = 504 kJ
例 3.4 / Example 3.4
将 0.5 kg 的 0°C 冰块完全转化为 100°C 的水蒸气，需要吸收多少热量？$c_{\text{水}} = 4200$ J/kg·K，$L_f = 3.34 \times 10^5$ J/kg，$L_v = 2.26 \times 10^6$ J/kg。
Convert 0.5 kg of ice at 0°C to steam at 100°C. $c_{\text{water}} = 4200$ J/kg·K, $L_f = 3.34 \times 10^5$ J/kg, $L_v = 2.26 \times 10^6$ J/kg.
解题过程 / Solution
三步计算：
① 熔化：$Q_1 = mL_f = 0.5 \times 3.34 \times 10^5 = 1.67 \times 10^5$ J
② 水升温：$Q_2 = mc\Delta T = 0.5 \times 4200 \times 100 = 2.10 \times 10^5$ J
③ 汽化：$Q_3 = mL_v = 0.5 \times 2.26 \times 10^6 = 1.13 \times 10^6$ J
$Q_{\text{总}} = Q_1 + Q_2 + Q_3 = 1.67 \times 10^5 + 2.10 \times 10^5 + 1.13 \times 10^6 = 1.507 \times 10^6$ J ≈ $1.51 \times 10^6$ J
① 熔化：$Q_1 = mL_f = 0.5 \times 3.34 \times 10^5 = 1.67 \times 10^5$ J
② 水升温：$Q_2 = mc\Delta T = 0.5 \times 4200 \times 100 = 2.10 \times 10^5$ J
③ 汽化：$Q_3 = mL_v = 0.5 \times 2.26 \times 10^6 = 1.13 \times 10^6$ J
$Q_{\text{总}} = Q_1 + Q_2 + Q_3 = 1.67 \times 10^5 + 2.10 \times 10^5 + 1.13 \times 10^6 = 1.507 \times 10^6$ J ≈ $1.51 \times 10^6$ J
Three stages:
① Melting: $Q_1 = mL_f = 0.5 \times 3.34 \times 10^5 = 1.67 \times 10^5$ J
② Heating water: $Q_2 = mc\Delta T = 0.5 \times 4200 \times 100 = 2.10 \times 10^5$ J
③ Vaporization: $Q_3 = mL_v = 0.5 \times 2.26 \times 10^6 = 1.13 \times 10^6$ J
$Q_{\text{total}} = Q_1 + Q_2 + Q_3 = 1.507 \times 10^6$ J ≈ $1.51 \times 10^6$ J
① Melting: $Q_1 = mL_f = 0.5 \times 3.34 \times 10^5 = 1.67 \times 10^5$ J
② Heating water: $Q_2 = mc\Delta T = 0.5 \times 4200 \times 100 = 2.10 \times 10^5$ J
③ Vaporization: $Q_3 = mL_v = 0.5 \times 2.26 \times 10^6 = 1.13 \times 10^6$ J
$Q_{\text{total}} = Q_1 + Q_2 + Q_3 = 1.507 \times 10^6$ J ≈ $1.51 \times 10^6$ J
例 3.5 / Example 3.5
将 0.2 kg 的 90°C 热水倒入 0.3 kg 的 20°C 冷水中。忽略容器吸热，求平衡温度。水的 $c = 4200$ J/kg·K。
Pour 0.2 kg of 90°C water into 0.3 kg of 20°C water. Neglect container, find equilibrium temperature. Water $c = 4200$ J/kg·K.
解题过程 / Solution
设平衡温度为 $T_f$：
$Q_{\text{放}} = Q_{\text{吸}}$
$m_1 c (T_1 - T_f) = m_2 c (T_f - T_2)$
$0.2 \times 4200 \times (90 - T_f) = 0.3 \times 4200 \times (T_f - 20)$
$0.2(90 - T_f) = 0.3(T_f - 20)$
$18 - 0.2T_f = 0.3T_f - 6$
$24 = 0.5T_f$
$T_f = 48$°C
$Q_{\text{放}} = Q_{\text{吸}}$
$m_1 c (T_1 - T_f) = m_2 c (T_f - T_2)$
$0.2 \times 4200 \times (90 - T_f) = 0.3 \times 4200 \times (T_f - 20)$
$0.2(90 - T_f) = 0.3(T_f - 20)$
$18 - 0.2T_f = 0.3T_f - 6$
$24 = 0.5T_f$
$T_f = 48$°C
Let final temperature be $T_f$:
$Q_{\text{lost}} = Q_{\text{gained}}$
$m_1 c (T_1 - T_f) = m_2 c (T_f - T_2)$
$0.2 \times 4200 \times (90 - T_f) = 0.3 \times 4200 \times (T_f - 20)$
$0.2(90 - T_f) = 0.3(T_f - 20)$
$18 - 0.2T_f = 0.3T_f - 6$
$24 = 0.5T_f$
$T_f = 48$°C
$Q_{\text{lost}} = Q_{\text{gained}}$
$m_1 c (T_1 - T_f) = m_2 c (T_f - T_2)$
$0.2 \times 4200 \times (90 - T_f) = 0.3 \times 4200 \times (T_f - 20)$
$0.2(90 - T_f) = 0.3(T_f - 20)$
$18 - 0.2T_f = 0.3T_f - 6$
$24 = 0.5T_f$
$T_f = 48$°C
3.3 理想气体定律 / Ideal Gas Laws
IB 3.2SL & HL 核心
理想气体状态方程 / Ideal Gas Equation
$$PV = nRT$$
理想气体状态方程将压强 $P$、体积 $V$、物质的量 $n$ 和热力学温度 $T$ 联系起来。$R = 8.31$ J/(mol·K) 是摩尔气体常数。理想气体是一种简化模型：分子间无相互作用力，分子体积可忽略，分子碰撞是完全弹性的。
The ideal gas equation relates pressure $P$, volume $V$, amount of substance $n$, and thermodynamic temperature $T$. $R = 8.31$ J/(mol·K) is the molar gas constant. An ideal gas is a simplified model: no intermolecular forces, negligible molecular volume, perfectly elastic collisions.
📐 推导说明 / Derivation Note
理想气体定律的历史推导：
① 玻意耳定律（Boyle's Law, 1662）：
温度不变时，$P \propto 1/V$ 或 $PV = \text{常数}$
② 查理定律（Charles' Law, 1787）：
压强不变时，$V \propto T$ 或 $V/T = \text{常数}$
③ 盖-吕萨克定律（Gay-Lussac's Law, 1802）：
体积不变时，$P \propto T$ 或 $P/T = \text{常数}$
④ 阿伏伽德罗定律（Avogadro's Law, 1811）：
同温同压下，相同体积的气体含有相同数量的分子。
$V \propto n$
⑤ 将以上定律合并：$V \propto nT/P$，引入常数 $R$：
$\displaystyle PV = nRT$
① 玻意耳定律（Boyle's Law, 1662）：
温度不变时，$P \propto 1/V$ 或 $PV = \text{常数}$
② 查理定律（Charles' Law, 1787）：
压强不变时，$V \propto T$ 或 $V/T = \text{常数}$
③ 盖-吕萨克定律（Gay-Lussac's Law, 1802）：
体积不变时，$P \propto T$ 或 $P/T = \text{常数}$
④ 阿伏伽德罗定律（Avogadro's Law, 1811）：
同温同压下，相同体积的气体含有相同数量的分子。
$V \propto n$
⑤ 将以上定律合并：$V \propto nT/P$，引入常数 $R$：
$\displaystyle PV = nRT$
Historical derivation of ideal gas laws:
① Boyle's Law (1662):
At constant $T$, $P \propto 1/V$ or $PV = \text{constant}$
② Charles' Law (1787):
At constant $P$, $V \propto T$ or $V/T = \text{constant}$
③ Gay-Lussac's Law (1802):
At constant $V$, $P \propto T$ or $P/T = \text{constant}$
④ Avogadro's Law (1811):
Equal volumes of gases at the same $T$ and $P$ contain equal numbers of molecules. $V \propto n$
⑤ Combining all: $V \propto nT/P$, introducing constant $R$:
$\displaystyle PV = nRT$
① Boyle's Law (1662):
At constant $T$, $P \propto 1/V$ or $PV = \text{constant}$
② Charles' Law (1787):
At constant $P$, $V \propto T$ or $V/T = \text{constant}$
③ Gay-Lussac's Law (1802):
At constant $V$, $P \propto T$ or $P/T = \text{constant}$
④ Avogadro's Law (1811):
Equal volumes of gases at the same $T$ and $P$ contain equal numbers of molecules. $V \propto n$
⑤ Combining all: $V \propto nT/P$, introducing constant $R$:
$\displaystyle PV = nRT$
气体定律的多种形式 / Alternative Forms
$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \quad \text{（气体量不变时）}$$
当气体质量（物质的量）不变时，可以使用组合气体定律。注意所有温度必须使用开尔文。
When the amount of gas is constant, use the combined gas law. All temperatures must be in kelvin.
$$PV = Nk_B T \quad \text{（$N$ 为分子数）}$$
用分子数 $N$ 表示的形式：$PV = Nk_B T$，其中 $k_B = 1.38 \times 10^{-23}$ J/K 是玻尔兹曼常数。与 $PV = nRT$ 等价，因为 $n = N/N_A$，$R = N_A k_B$。
In terms of number of molecules $N$: $PV = Nk_B T$, where $k_B = 1.38 \times 10^{-23}$ J/K is Boltzmann's constant. Equivalent to $PV = nRT$ because $n = N/N_A$ and $R = N_A k_B$.
| 定律 / Law | 条件 / Condition | 公式 / Formula |
|---|---|---|
| 玻意耳定律 / Boyle's Law | $T$ 不变，$n$ 不变 | $P_1 V_1 = P_2 V_2$ |
| 查理定律 / Charles' Law | $P$ 不变，$n$ 不变 | $\displaystyle \frac{V_1}{T_1} = \frac{V_2}{T_2}$ |
| 盖-吕萨克定律 / Gay-Lussac's Law | $V$ 不变，$n$ 不变 | $\displaystyle \frac{P_1}{T_1} = \frac{P_2}{T_2}$ |
| 组合气体定律 / Combined Gas Law | $n$ 不变 | $\displaystyle \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$ |
| 理想气体状态方程 / Ideal Gas Eq. | 一般情况 | $PV = nRT$ |
🎯 IB 考试要点 / Exam Tips：① 所有温度必须转换为开尔文（K），否则结果错误；② $R = 8.31$ J/(mol·K) 是 IB 数据手册给出的值，考试中直接使用；③ 标准摩尔体积：STP（标准温度和压强，0°C = 273 K, 1 atm = 1.01 $\times 10^5$ Pa）下，1 mol 理想气体体积为 22.4 L = 0.0224 m³；④ 注意单位一致性——$V$ 用 m³，$P$ 用 Pa，$T$ 用 K。
🎯 IB Exam Tips: ① Always convert temperatures to kelvin (K) — using °C will give wrong results; ② $R = 8.31$ J/(mol·K) is in the IB data booklet; ③ Molar volume at STP (0°C = 273 K, 1 atm = 1.01 $\times 10^5$ Pa): 1 mol ideal gas occupies 22.4 L = 0.0224 m³; ④ Check unit consistency — $V$ in m³, $P$ in Pa, $T$ in K.
例 3.6 / Example 3.6
3 mol 的理想气体在 27°C、2.0 $\times 10^5$ Pa 下，体积是多少？
3 mol of ideal gas at 27°C and 2.0 $\times 10^5$ Pa. Find the volume.
解题过程 / Solution
$T = 27 + 273 = 300$ K
$PV = nRT$
$V = nRT/P = 3 \times 8.31 \times 300 / (2.0 \times 10^5)$
$V = 7479 / (2.0 \times 10^5) = 0.0374$ m³ = 37.4 L
$PV = nRT$
$V = nRT/P = 3 \times 8.31 \times 300 / (2.0 \times 10^5)$
$V = 7479 / (2.0 \times 10^5) = 0.0374$ m³ = 37.4 L
$T = 27 + 273 = 300$ K
$PV = nRT$
$V = nRT/P = 3 \times 8.31 \times 300 / (2.0 \times 10^5)$
$V = 7479 / (2.0 \times 10^5) = 0.0374$ m³ = 37.4 L
$PV = nRT$
$V = nRT/P = 3 \times 8.31 \times 300 / (2.0 \times 10^5)$
$V = 7479 / (2.0 \times 10^5) = 0.0374$ m³ = 37.4 L
例 3.7 / Example 3.7
气缸中气体初始压强 $2.5 \times 10^5$ Pa，体积 0.030 m³，温度 47°C。压缩后体积 0.010 m³，温度升到 127°C。求最终压强。
Gas in cylinder: initial $P = 2.5 \times 10^5$ Pa, $V = 0.030$ m³, $T = 47$°C. Compressed to $V = 0.010$ m³, $T = 127$°C. Find final pressure.
解题过程 / Solution
$T_1 = 47 + 273 = 320$ K，$T_2 = 127 + 273 = 400$ K
$\displaystyle \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$
$\displaystyle P_2 = \frac{P_1 V_1 T_2}{T_1 V_2} = \frac{2.5 \times 10^5 \times 0.030 \times 400}{320 \times 0.010}$
$P_2 = \frac{3.0 \times 10^6}{3.2} = 9.375 \times 10^5$ Pa ≈ $9.4 \times 10^5$ Pa
$\displaystyle \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$
$\displaystyle P_2 = \frac{P_1 V_1 T_2}{T_1 V_2} = \frac{2.5 \times 10^5 \times 0.030 \times 400}{320 \times 0.010}$
$P_2 = \frac{3.0 \times 10^6}{3.2} = 9.375 \times 10^5$ Pa ≈ $9.4 \times 10^5$ Pa
$T_1 = 47 + 273 = 320$ K, $T_2 = 127 + 273 = 400$ K
$\displaystyle \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$
$\displaystyle P_2 = \frac{P_1 V_1 T_2}{T_1 V_2} = \frac{2.5 \times 10^5 \times 0.030 \times 400}{320 \times 0.010}$
$P_2 = \frac{3.0 \times 10^6}{3.2} = 9.375 \times 10^5$ Pa ≈ $9.4 \times 10^5$ Pa
$\displaystyle \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$
$\displaystyle P_2 = \frac{P_1 V_1 T_2}{T_1 V_2} = \frac{2.5 \times 10^5 \times 0.030 \times 400}{320 \times 0.010}$
$P_2 = \frac{3.0 \times 10^6}{3.2} = 9.375 \times 10^5$ Pa ≈ $9.4 \times 10^5$ Pa
3.4 气体动理论 / Kinetic Theory
IB 3.2HL 重点 / SL 考点
分子平均动能 / Average Molecular Kinetic Energy
$$\bar{E}_k = \frac{3}{2}k_B T$$
气体分子平均平动动能只与温度有关，与气体种类无关。$k_B = 1.38 \times 10^{-23}$ J/K 是玻尔兹曼常数。这是温度微观定义的核心——温度越高，分子运动越剧烈。
The average translational kinetic energy of gas molecules depends only on temperature, not on the type of gas. $k_B = 1.38 \times 10^{-23}$ J/K is Boltzmann's constant. This is the core of the microscopic definition of temperature — higher temperature means more vigorous molecular motion.
压强与分子速度 / Pressure and Molecular Speed
$$P = \frac{1}{3}\rho \overline{v^2} \quad \text{或} \quad PV = \frac{1}{3} N m \overline{v^2}$$
压强来源于大量气体分子对容器壁的碰撞。$\rho$ 是气体密度，$\overline{v^2}$ 是分子速度平方的平均值。$N$ 是分子总数，$m$ 是单个分子的质量。重要推论：$P \propto \overline{v^2}$。
Pressure arises from countless collisions of gas molecules with container walls. $\rho$ is gas density, $\overline{v^2}$ is the mean square speed. $N$ is the number of molecules, $m$ is the mass of a single molecule. Key result: $P \propto \overline{v^2}$.
📐 推导过程 / Derivation
气体压强公式推导：
假设一个立方体容器边长 $L$，一个质量为 $m$ 的分子以速度 $v$ 沿 $x$ 方向运动。
① 碰撞一次器壁的动量变化：$\Delta p = 2mv_x$
② 两次碰撞之间的时间：$\Delta t = 2L/v_x$
③ 该分子对壁的平均力：$F = \Delta p/\Delta t = mv_x^2/L$
④ $N$ 个分子对壁的总力：$F_{\text{总}} = \sum mv_{x,i}^2/L = (m/L)\sum v_{x,i}^2$
⑤ 各向同性：$\overline{v_x^2} = \overline{v_y^2} = \overline{v_z^2} = \frac13 \overline{v^2}$
⑥ $\sum v_{x,i}^2 = N \cdot \frac13 \overline{v^2}$
⑦ 压强 $P = F/A = F/L^2 = (m/L^3)(N/3)\overline{v^2} = \frac13 \rho \overline{v^2}$
推论：
• $PV = \frac13 N m \overline{v^2}$
• $\overline{v^2} = 3k_B T / m$（结合 $PV = Nk_B T$）
• 方均根速度：$v_{\text{rms}} = \sqrt{\overline{v^2}} = \sqrt{3k_B T / m} = \sqrt{3RT / M}$
其中 $M$ 是摩尔质量（kg/mol）
假设一个立方体容器边长 $L$，一个质量为 $m$ 的分子以速度 $v$ 沿 $x$ 方向运动。
① 碰撞一次器壁的动量变化：$\Delta p = 2mv_x$
② 两次碰撞之间的时间：$\Delta t = 2L/v_x$
③ 该分子对壁的平均力：$F = \Delta p/\Delta t = mv_x^2/L$
④ $N$ 个分子对壁的总力：$F_{\text{总}} = \sum mv_{x,i}^2/L = (m/L)\sum v_{x,i}^2$
⑤ 各向同性：$\overline{v_x^2} = \overline{v_y^2} = \overline{v_z^2} = \frac13 \overline{v^2}$
⑥ $\sum v_{x,i}^2 = N \cdot \frac13 \overline{v^2}$
⑦ 压强 $P = F/A = F/L^2 = (m/L^3)(N/3)\overline{v^2} = \frac13 \rho \overline{v^2}$
推论：
• $PV = \frac13 N m \overline{v^2}$
• $\overline{v^2} = 3k_B T / m$（结合 $PV = Nk_B T$）
• 方均根速度：$v_{\text{rms}} = \sqrt{\overline{v^2}} = \sqrt{3k_B T / m} = \sqrt{3RT / M}$
其中 $M$ 是摩尔质量（kg/mol）
Derivation of gas pressure formula:
Consider a cubic container of side $L$, a molecule of mass $m$ moving with velocity $v$ along $x$.
① Momentum change per collision: $\Delta p = 2mv_x$
② Time between collisions: $\Delta t = 2L/v_x$
③ Average force from this molecule: $F = \Delta p/\Delta t = mv_x^2/L$
④ Total force from $N$ molecules: $F_{\text{total}} = \sum mv_{x,i}^2/L = (m/L)\sum v_{x,i}^2$
⑤ Isotropy: $\overline{v_x^2} = \overline{v_y^2} = \overline{v_z^2} = \frac13 \overline{v^2}$
⑥ $\sum v_{x,i}^2 = N \cdot \frac13 \overline{v^2}$
⑦ Pressure $P = F/A = F/L^2 = (m/L^3)(N/3)\overline{v^2} = \frac13 \rho \overline{v^2}$
Results:
• $PV = \frac13 N m \overline{v^2}$
• $\overline{v^2} = 3k_B T / m$ (combine with $PV = Nk_B T$)
• Root-mean-square speed: $v_{\text{rms}} = \sqrt{\overline{v^2}} = \sqrt{3k_B T / m} = \sqrt{3RT / M}$
where $M$ is molar mass (kg/mol)
Consider a cubic container of side $L$, a molecule of mass $m$ moving with velocity $v$ along $x$.
① Momentum change per collision: $\Delta p = 2mv_x$
② Time between collisions: $\Delta t = 2L/v_x$
③ Average force from this molecule: $F = \Delta p/\Delta t = mv_x^2/L$
④ Total force from $N$ molecules: $F_{\text{total}} = \sum mv_{x,i}^2/L = (m/L)\sum v_{x,i}^2$
⑤ Isotropy: $\overline{v_x^2} = \overline{v_y^2} = \overline{v_z^2} = \frac13 \overline{v^2}$
⑥ $\sum v_{x,i}^2 = N \cdot \frac13 \overline{v^2}$
⑦ Pressure $P = F/A = F/L^2 = (m/L^3)(N/3)\overline{v^2} = \frac13 \rho \overline{v^2}$
Results:
• $PV = \frac13 N m \overline{v^2}$
• $\overline{v^2} = 3k_B T / m$ (combine with $PV = Nk_B T$)
• Root-mean-square speed: $v_{\text{rms}} = \sqrt{\overline{v^2}} = \sqrt{3k_B T / m} = \sqrt{3RT / M}$
where $M$ is molar mass (kg/mol)
方均根速度 / Root-Mean-Square Speed
$$v_{\text{rms}} = \sqrt{\frac{3k_B T}{m}} = \sqrt{\frac{3RT}{M}}$$
方均根速度 $v_{\text{rms}}$ 是分子速度的一种统计度量。注意它不是简单的平均速度——先平方、再平均、再开方。$M$ 是摩尔质量（kg/mol）。温度越高，$v_{\text{rms}}$ 越大；分子越轻，$v_{\text{rms}}$ 越大。
The root-mean-square speed $v_{\text{rms}}$ is a statistical measure of molecular speed. Note it's not the simple average — square first, then average, then square root. $M$ is molar mass (kg/mol). Higher temperature gives larger $v_{\text{rms}}$; lighter molecules give larger $v_{\text{rms}}$.
🎯 IB 考试要点 / Exam Tips：① 注意 $v_{\text{rms}}$ 不是平均速度：对于同一组分子，$v_{\text{rms}}$ 总是大于算术平均速度；② 气体动理论的假设是IB常考点：分子大小可忽略、分子间无作用力、碰撞完全弹性、分子运动随机且各向同性；③ 温度相同时，不同气体的平均动能相同，但 $v_{\text{rms}}$ 不同（因为质量不同）；④ 公式 $\bar{E}_k = \frac32 k_B T$ 中的 $\bar{E}_k$ 是每个分子的平动动能。
🎯 IB Exam Tips: ① $v_{\text{rms}}$ is not the mean speed — for the same set of molecules, $v_{\text{rms}}$ is always larger than the arithmetic mean; ② Kinetic theory assumptions are common IB questions: negligible molecular size, no intermolecular forces, perfectly elastic collisions, random and isotropic motion; ③ At the same temperature, different gases have the same average kinetic energy but different $v_{\text{rms}}$ (different mass); ④ $\bar{E}_k = \frac32 k_B T$ gives the translational kinetic energy per molecule.
例 3.8 / Example 3.8
氮气（$N_2$）分子在 27°C 时的方均根速度是多少？氮的摩尔质量 $M = 28.0$ g/mol。
Find the rms speed of nitrogen ($N_2$) molecules at 27°C. Molar mass $M = 28.0$ g/mol.
解题过程 / Solution
$T = 27 + 273 = 300$ K
$M = 28.0 \times 10^{-3}$ kg/mol
$v_{\text{rms}} = \sqrt{3RT / M} = \sqrt{3 \times 8.31 \times 300 / 0.028}$
$v_{\text{rms}} = \sqrt{7479 / 0.028} = \sqrt{267107} = 517$ m/s
$M = 28.0 \times 10^{-3}$ kg/mol
$v_{\text{rms}} = \sqrt{3RT / M} = \sqrt{3 \times 8.31 \times 300 / 0.028}$
$v_{\text{rms}} = \sqrt{7479 / 0.028} = \sqrt{267107} = 517$ m/s
$T = 27 + 273 = 300$ K
$M = 28.0 \times 10^{-3}$ kg/mol
$v_{\text{rms}} = \sqrt{3RT / M} = \sqrt{3 \times 8.31 \times 300 / 0.028}$
$v_{\text{rms}} = \sqrt{7479 / 0.028} = \sqrt{267107} = 517$ m/s
$M = 28.0 \times 10^{-3}$ kg/mol
$v_{\text{rms}} = \sqrt{3RT / M} = \sqrt{3 \times 8.31 \times 300 / 0.028}$
$v_{\text{rms}} = \sqrt{7479 / 0.028} = \sqrt{267107} = 517$ m/s
例 3.9 / Example 3.9
在相同温度下，氢气（$H_2$，$M = 2.0$ g/mol）分子的方均根速度是氧气（$O_2$，$M = 32.0$ g/mol）的多少倍？
At the same temperature, how many times larger is the rms speed of hydrogen ($H_2$, $M = 2.0$ g/mol) compared to oxygen ($O_2$, $M = 32.0$ g/mol)?
解题过程 / Solution
$v_{\text{rms}} \propto 1/\sqrt{M}$（同温下）
$\displaystyle \frac{v_{\text{rms}}(H_2)}{v_{\text{rms}}(O_2)} = \sqrt{\frac{M_{O_2}}{M_{H_2}}} = \sqrt{\frac{32.0}{2.0}} = \sqrt{16} = 4$
氢分子的方均根速度是氧分子的 4 倍。
$\displaystyle \frac{v_{\text{rms}}(H_2)}{v_{\text{rms}}(O_2)} = \sqrt{\frac{M_{O_2}}{M_{H_2}}} = \sqrt{\frac{32.0}{2.0}} = \sqrt{16} = 4$
氢分子的方均根速度是氧分子的 4 倍。
$v_{\text{rms}} \propto 1/\sqrt{M}$ (at same T)
$\displaystyle \frac{v_{\text{rms}}(H_2)}{v_{\text{rms}}(O_2)} = \sqrt{\frac{M_{O_2}}{M_{H_2}}} = \sqrt{\frac{32.0}{2.0}} = \sqrt{16} = 4$
Hydrogen molecules have 4 times the rms speed of oxygen molecules.
$\displaystyle \frac{v_{\text{rms}}(H_2)}{v_{\text{rms}}(O_2)} = \sqrt{\frac{M_{O_2}}{M_{H_2}}} = \sqrt{\frac{32.0}{2.0}} = \sqrt{16} = 4$
Hydrogen molecules have 4 times the rms speed of oxygen molecules.
例 3.10 / Example 3.10
1 mol 理想气体在 27°C 时，所有分子的总平动动能是多少？
1 mol of ideal gas at 27°C. What is the total translational kinetic energy of all molecules?
解题过程 / Solution
$T = 27 + 273 = 300$ K
每个分子 $\bar{E}_k = \frac32 k_B T$
1 mol 有 $N_A = 6.02 \times 10^{23}$ 个分子
$E_{\text{总}} = N_A \cdot \frac32 k_B T = \frac32 RT$
$E_{\text{总}} = \frac32 \times 8.31 \times 300 = 3739.5$ J ≈ 3740 J
注意：也可以用 $E_{\text{总}} = \frac32 nRT$，$n=1$。
每个分子 $\bar{E}_k = \frac32 k_B T$
1 mol 有 $N_A = 6.02 \times 10^{23}$ 个分子
$E_{\text{总}} = N_A \cdot \frac32 k_B T = \frac32 RT$
$E_{\text{总}} = \frac32 \times 8.31 \times 300 = 3739.5$ J ≈ 3740 J
注意：也可以用 $E_{\text{总}} = \frac32 nRT$，$n=1$。
$T = 27 + 273 = 300$ K
Per molecule: $\bar{E}_k = \frac32 k_B T$
1 mol has $N_A = 6.02 \times 10^{23}$ molecules
$E_{\text{total}} = N_A \cdot \frac32 k_B T = \frac32 RT$
$E_{\text{total}} = \frac32 \times 8.31 \times 300 = 3739.5$ J ≈ 3740 J
Note: Alternatively $E_{\text{total}} = \frac32 nRT$, where $n=1$.
Per molecule: $\bar{E}_k = \frac32 k_B T$
1 mol has $N_A = 6.02 \times 10^{23}$ molecules
$E_{\text{total}} = N_A \cdot \frac32 k_B T = \frac32 RT$
$E_{\text{total}} = \frac32 \times 8.31 \times 300 = 3739.5$ J ≈ 3740 J
Note: Alternatively $E_{\text{total}} = \frac32 nRT$, where $n=1$.
🏆 挑战题 / Challenge Problems
多概念综合 / Multi-concept难度 / Hard
以下题目需要综合运用本章的多个知识点。建议先复习所有内容后再尝试。
These problems require integrating multiple concepts from this chapter. Review all sections before attempting.
挑战 1 / Challenge 1 热平衡中的相变 / Phase Change in Thermal Equilibrium
将 0.05 kg 的 100°C 水蒸气通入 0.5 kg 的 0°C 冰水混合物中（含有 0.2 kg 冰和 0.3 kg 水）。假设容器绝热且不吸热，求最终状态和温度。已知：$c_{\text{水}} = 4200$ J/kg·K，$L_f = 3.34 \times 10^5$ J/kg，$L_v = 2.26 \times 10^6$ J/kg。
Pass 0.05 kg of 100°C steam into 0.5 kg of ice-water mixture at 0°C (0.2 kg ice + 0.3 kg water). Insulated container, find final state and temperature. $c_{\text{water}} = 4200$ J/kg·K, $L_f = 3.34 \times 10^5$ J/kg, $L_v = 2.26 \times 10^6$ J/kg.
解题过程 / Solution
首先计算水蒸气液化放热：$Q_{\text{蒸汽}} = mL_v = 0.05 \times 2.26 \times 10^6 = 1.13 \times 10^5$ J
液化后 0.05 kg 水从 100°C 降到 0°C 放热：$Q_{\text{降温}} = mc\Delta T = 0.05 \times 4200 \times 100 = 2.1 \times 10^4$ J
水蒸气总放热：$Q_{\text{放}} = 1.13 \times 10^5 + 2.1 \times 10^4 = 1.34 \times 10^5$ J
熔化 0.2 kg 冰需要：$Q_{\text{熔}} = 0.2 \times 3.34 \times 10^5 = 6.68 \times 10^4$ J
$Q_{\text{放}} > Q_{\text{熔}}$，所以冰全部熔化，多余热量用于升温。
剩余热量：$Q_{\text{余}} = 1.34 \times 10^5 - 6.68 \times 10^4 = 6.72 \times 10^4$ J
总水量：$0.2 + 0.3 + 0.05 = 0.55$ kg
$Q_{\text{余}} = mc\Delta T$，$\Delta T = 6.72 \times 10^4 / (0.55 \times 4200) = 29.1$°C
最终温度：$T_f = 0 + 29.1 = 29.1$°C
结论：所有冰熔化成水，最终为 0.55 kg 水，温度 29.1°C。
液化后 0.05 kg 水从 100°C 降到 0°C 放热：$Q_{\text{降温}} = mc\Delta T = 0.05 \times 4200 \times 100 = 2.1 \times 10^4$ J
水蒸气总放热：$Q_{\text{放}} = 1.13 \times 10^5 + 2.1 \times 10^4 = 1.34 \times 10^5$ J
熔化 0.2 kg 冰需要：$Q_{\text{熔}} = 0.2 \times 3.34 \times 10^5 = 6.68 \times 10^4$ J
$Q_{\text{放}} > Q_{\text{熔}}$，所以冰全部熔化，多余热量用于升温。
剩余热量：$Q_{\text{余}} = 1.34 \times 10^5 - 6.68 \times 10^4 = 6.72 \times 10^4$ J
总水量：$0.2 + 0.3 + 0.05 = 0.55$ kg
$Q_{\text{余}} = mc\Delta T$，$\Delta T = 6.72 \times 10^4 / (0.55 \times 4200) = 29.1$°C
最终温度：$T_f = 0 + 29.1 = 29.1$°C
结论：所有冰熔化成水，最终为 0.55 kg 水，温度 29.1°C。
First, steam condenses: $Q_{\text{steam}} = mL_v = 0.05 \times 2.26 \times 10^6 = 1.13 \times 10^5$ J
Then condensed water cools from 100°C to 0°C: $Q_{\text{cool}} = mc\Delta T = 0.05 \times 4200 \times 100 = 2.1 \times 10^4$ J
Total heat released: $Q_{\text{released}} = 1.13 \times 10^5 + 2.1 \times 10^4 = 1.34 \times 10^5$ J
Melting 0.2 kg ice requires: $Q_{\text{melt}} = 0.2 \times 3.34 \times 10^5 = 6.68 \times 10^4$ J
$Q_{\text{released}} > Q_{\text{melt}}$, so all ice melts. Remaining heat warms the water.
Remaining heat: $Q_{\text{rem}} = 1.34 \times 10^5 - 6.68 \times 10^4 = 6.72 \times 10^4$ J
Total water: $0.2 + 0.3 + 0.05 = 0.55$ kg
$\Delta T = 6.72 \times 10^4 / (0.55 \times 4200) = 29.1$°C
Final temperature: $T_f = 0 + 29.1 = 29.1$°C
Conclusion: All ice melts, resulting in 0.55 kg of water at 29.1°C.
Then condensed water cools from 100°C to 0°C: $Q_{\text{cool}} = mc\Delta T = 0.05 \times 4200 \times 100 = 2.1 \times 10^4$ J
Total heat released: $Q_{\text{released}} = 1.13 \times 10^5 + 2.1 \times 10^4 = 1.34 \times 10^5$ J
Melting 0.2 kg ice requires: $Q_{\text{melt}} = 0.2 \times 3.34 \times 10^5 = 6.68 \times 10^4$ J
$Q_{\text{released}} > Q_{\text{melt}}$, so all ice melts. Remaining heat warms the water.
Remaining heat: $Q_{\text{rem}} = 1.34 \times 10^5 - 6.68 \times 10^4 = 6.72 \times 10^4$ J
Total water: $0.2 + 0.3 + 0.05 = 0.55$ kg
$\Delta T = 6.72 \times 10^4 / (0.55 \times 4200) = 29.1$°C
Final temperature: $T_f = 0 + 29.1 = 29.1$°C
Conclusion: All ice melts, resulting in 0.55 kg of water at 29.1°C.
挑战 2 / Challenge 2 混合气体的压强 / Pressure of Mixed Gases
一个 0.020 m³ 的容器中有 0.10 mol 的氢气（$H_2$）和 0.050 mol 的氧气（$O_2$），温度 27°C。求：(a) 容器总压强；(b) 两种气体的分压；(c) 两种气体的方均根速度之比。
A 0.020 m³ container has 0.10 mol H$_2$ and 0.050 mol O$_2$ at 27°C. Find: (a) total pressure; (b) partial pressures; (c) ratio of rms speeds.
解题过程 / Solution
(a) $n_{\text{总}} = 0.10 + 0.050 = 0.150$ mol，$T = 27 + 273 = 300$ K
$P_{\text{总}} = nRT/V = 0.150 \times 8.31 \times 300 / 0.020 = 1.87 \times 10^4$ Pa
(b) 道尔顿分压定律：分压与物质的量成正比
$P_{H_2} = (0.10/0.150) \times 1.87 \times 10^4 = 1.25 \times 10^4$ Pa
$P_{O_2} = (0.050/0.150) \times 1.87 \times 10^4 = 6.23 \times 10^3$ Pa
验证：$P_{H_2} + P_{O_2} = 1.87 \times 10^4$ Pa ✓
(c) $v_{\text{rms}} \propto 1/\sqrt{M}$（同温）
$M_{H_2} = 2.0$ g/mol，$M_{O_2} = 32.0$ g/mol
$\displaystyle \frac{v_{H_2}}{v_{O_2}} = \sqrt{\frac{32.0}{2.0}} = \sqrt{16} = 4$
$P_{\text{总}} = nRT/V = 0.150 \times 8.31 \times 300 / 0.020 = 1.87 \times 10^4$ Pa
(b) 道尔顿分压定律：分压与物质的量成正比
$P_{H_2} = (0.10/0.150) \times 1.87 \times 10^4 = 1.25 \times 10^4$ Pa
$P_{O_2} = (0.050/0.150) \times 1.87 \times 10^4 = 6.23 \times 10^3$ Pa
验证：$P_{H_2} + P_{O_2} = 1.87 \times 10^4$ Pa ✓
(c) $v_{\text{rms}} \propto 1/\sqrt{M}$（同温）
$M_{H_2} = 2.0$ g/mol，$M_{O_2} = 32.0$ g/mol
$\displaystyle \frac{v_{H_2}}{v_{O_2}} = \sqrt{\frac{32.0}{2.0}} = \sqrt{16} = 4$
(a) $n_{\text{total}} = 0.10 + 0.050 = 0.150$ mol, $T = 27 + 273 = 300$ K
$P_{\text{total}} = nRT/V = 0.150 \times 8.31 \times 300 / 0.020 = 1.87 \times 10^4$ Pa
(b) Dalton's law: partial pressures proportional to mole fraction
$P_{H_2} = (0.10/0.150) \times 1.87 \times 10^4 = 1.25 \times 10^4$ Pa
$P_{O_2} = (0.050/0.150) \times 1.87 \times 10^4 = 6.23 \times 10^3$ Pa
Check: $P_{H_2} + P_{O_2} = 1.87 \times 10^4$ Pa ✓
(c) $v_{\text{rms}} \propto 1/\sqrt{M}$ (same T)
$M_{H_2} = 2.0$ g/mol, $M_{O_2} = 32.0$ g/mol
$\displaystyle \frac{v_{H_2}}{v_{O_2}} = \sqrt{\frac{32.0}{2.0}} = \sqrt{16} = 4$
$P_{\text{total}} = nRT/V = 0.150 \times 8.31 \times 300 / 0.020 = 1.87 \times 10^4$ Pa
(b) Dalton's law: partial pressures proportional to mole fraction
$P_{H_2} = (0.10/0.150) \times 1.87 \times 10^4 = 1.25 \times 10^4$ Pa
$P_{O_2} = (0.050/0.150) \times 1.87 \times 10^4 = 6.23 \times 10^3$ Pa
Check: $P_{H_2} + P_{O_2} = 1.87 \times 10^4$ Pa ✓
(c) $v_{\text{rms}} \propto 1/\sqrt{M}$ (same T)
$M_{H_2} = 2.0$ g/mol, $M_{O_2} = 32.0$ g/mol
$\displaystyle \frac{v_{H_2}}{v_{O_2}} = \sqrt{\frac{32.0}{2.0}} = \sqrt{16} = 4$
挑战 3 / Challenge 3 气球上升与理想气体 / Balloon Ascent and Ideal Gas
一个气象气球在地面（温度 27°C，压强 $1.0 \times 10^5$ Pa）时体积为 5.0 m³。气球内气体物质的量不变。当气球上升到 10 km 高空时，外界压强为 $2.5 \times 10^4$ Pa，温度为 -50°C。假设气球内气体温度与外界相同，求此时气球的体积。如果气球材料能承受的最大体积为 20 m³，气球会破裂吗？
A weather balloon at ground level (27°C, $1.0 \times 10^5$ Pa) has volume 5.0 m³. Gas amount is constant. At 10 km altitude: $P = 2.5 \times 10^4$ Pa, $T = -50$°C. Assume gas temperature equals outside. Find volume. If the balloon bursts at 20 m³, will it survive?
解题过程 / Solution
$T_1 = 27 + 273 = 300$ K，$T_2 = -50 + 273 = 223$ K
$\displaystyle \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$
$V_2 = \frac{P_1 V_1 T_2}{T_1 P_2}$
$V_2 = \frac{1.0 \times 10^5 \times 5.0 \times 223}{300 \times 2.5 \times 10^4}$
$V_2 = \frac{1.115 \times 10^8}{7.5 \times 10^6} = 14.87$ m³ ≈ 14.9 m³
14.9 m³ < 20 m³，所以气球不会破裂。
分析：体积增大约 3 倍，主要是由于压强降低（$1/4$），但温度降低（$223/300$）部分抵消了膨胀效应。
$\displaystyle \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$
$V_2 = \frac{P_1 V_1 T_2}{T_1 P_2}$
$V_2 = \frac{1.0 \times 10^5 \times 5.0 \times 223}{300 \times 2.5 \times 10^4}$
$V_2 = \frac{1.115 \times 10^8}{7.5 \times 10^6} = 14.87$ m³ ≈ 14.9 m³
14.9 m³ < 20 m³，所以气球不会破裂。
分析：体积增大约 3 倍，主要是由于压强降低（$1/4$），但温度降低（$223/300$）部分抵消了膨胀效应。
$T_1 = 27 + 273 = 300$ K, $T_2 = -50 + 273 = 223$ K
$\displaystyle \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$
$V_2 = \frac{P_1 V_1 T_2}{T_1 P_2}$
$V_2 = \frac{1.0 \times 10^5 \times 5.0 \times 223}{300 \times 2.5 \times 10^4}$
$V_2 = \frac{1.115 \times 10^8}{7.5 \times 10^6} = 14.87$ m³ ≈ 14.9 m³
14.9 m³ < 20 m³, so the balloon does not burst.
Analysis: Volume increases about 3 times, mainly due to pressure drop (factor of $1/4$), but the temperature drop ($223/300$) partially offsets the expansion.
$\displaystyle \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$
$V_2 = \frac{P_1 V_1 T_2}{T_1 P_2}$
$V_2 = \frac{1.0 \times 10^5 \times 5.0 \times 223}{300 \times 2.5 \times 10^4}$
$V_2 = \frac{1.115 \times 10^8}{7.5 \times 10^6} = 14.87$ m³ ≈ 14.9 m³
14.9 m³ < 20 m³, so the balloon does not burst.
Analysis: Volume increases about 3 times, mainly due to pressure drop (factor of $1/4$), but the temperature drop ($223/300$) partially offsets the expansion.
挑战 4 / Challenge 4 能量守恒与气体动理论 / Energy Conservation and Kinetic Theory
一个绝热气缸中含有 0.50 mol 理想单原子气体，初始温度 300 K，体积 $1.0 \times 10^{-2}$ m³。活塞快速将体积压缩到 $5.0 \times 10^{-3}$ m³，过程中没有热交换（绝热压缩）。气体温度上升到 476 K。求：(a) 外界对气体做的功；(b) 最终压强；(c) 分子方均根速度的变化百分比。
An insulated cylinder has 0.50 mol monatomic ideal gas, initial T = 300 K, V = $1.0 \times 10^{-2}$ m³. Piston rapidly compresses to $5.0 \times 10^{-3}$ m³ (adiabatic). Gas temperature rises to 476 K. Find: (a) work done on gas; (b) final pressure; (c) percentage change in rms speed.
解题过程 / Solution
(a) 绝热过程 $Q = 0$，根据热力学第一定律 $\Delta U = Q + W = W$
单原子理想气体的内能变化：$\Delta U = \frac32 nR\Delta T$
$\Delta T = 476 - 300 = 176$ K
$W = \Delta U = \frac32 \times 0.50 \times 8.31 \times 176 = 1097$ J
正功表示外界对气体做功。
(b) 初始压强：$P_1 = nRT_1/V_1 = 0.50 \times 8.31 \times 300 / 0.010 = 1.247 \times 10^5$ Pa
最终压强：$P_2 = nRT_2/V_2 = 0.50 \times 8.31 \times 476 / 0.005 = 3.956 \times 10^5$ Pa
(c) $v_{\text{rms}} \propto \sqrt{T}$
$\displaystyle \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{476}{300}} = \sqrt{1.587} = 1.260$
增幅为 26.0%
单原子理想气体的内能变化：$\Delta U = \frac32 nR\Delta T$
$\Delta T = 476 - 300 = 176$ K
$W = \Delta U = \frac32 \times 0.50 \times 8.31 \times 176 = 1097$ J
正功表示外界对气体做功。
(b) 初始压强：$P_1 = nRT_1/V_1 = 0.50 \times 8.31 \times 300 / 0.010 = 1.247 \times 10^5$ Pa
最终压强：$P_2 = nRT_2/V_2 = 0.50 \times 8.31 \times 476 / 0.005 = 3.956 \times 10^5$ Pa
(c) $v_{\text{rms}} \propto \sqrt{T}$
$\displaystyle \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{476}{300}} = \sqrt{1.587} = 1.260$
增幅为 26.0%
(a) Adiabatic: $Q = 0$, first law $\Delta U = Q + W = W$
Monatomic gas internal energy: $\Delta U = \frac32 nR\Delta T$
$\Delta T = 476 - 300 = 176$ K
$W = \Delta U = \frac32 \times 0.50 \times 8.31 \times 176 = 1097$ J
Positive work means work done on the gas.
(b) Initial pressure: $P_1 = nRT_1/V_1 = 0.50 \times 8.31 \times 300 / 0.010 = 1.247 \times 10^5$ Pa
Final pressure: $P_2 = nRT_2/V_2 = 0.50 \times 8.31 \times 476 / 0.005 = 3.956 \times 10^5$ Pa
(c) $v_{\text{rms}} \propto \sqrt{T}$
$\displaystyle \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{476}{300}} = \sqrt{1.587} = 1.260$
Increase of 26.0%
Monatomic gas internal energy: $\Delta U = \frac32 nR\Delta T$
$\Delta T = 476 - 300 = 176$ K
$W = \Delta U = \frac32 \times 0.50 \times 8.31 \times 176 = 1097$ J
Positive work means work done on the gas.
(b) Initial pressure: $P_1 = nRT_1/V_1 = 0.50 \times 8.31 \times 300 / 0.010 = 1.247 \times 10^5$ Pa
Final pressure: $P_2 = nRT_2/V_2 = 0.50 \times 8.31 \times 476 / 0.005 = 3.956 \times 10^5$ Pa
(c) $v_{\text{rms}} \propto \sqrt{T}$
$\displaystyle \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{476}{300}} = \sqrt{1.587} = 1.260$
Increase of 26.0%
挑战 5 / Challenge 5 综合热力学分析 / Comprehensive Thermodynamic Analysis
一容器被隔板分成两部分：左侧体积 0.030 m³，含 1.0 mol 氦气（单原子），温度 400 K；右侧体积 0.020 m³，含 0.50 mol 氩气（单原子），温度 300 K。隔板绝热，抽出后气体混合。求：(a) 最终温度；(b) 最终压强；(c) 初始时哪种气体的方均根速度更大？大多少？
A container is divided: left side 0.030 m³, 1.0 mol He (monatomic), 400 K; right side 0.020 m³, 0.50 mol Ar (monatomic), 300 K. The partition is adiabatic. After removal, gases mix. Find: (a) final temperature; (b) final pressure; (c) which gas had higher initial rms speed and by how much?
解题过程 / Solution
(a) 能量守恒：混合前后总内能不变
单原子气体 $U = \frac32 nRT$
$U_{\text{前}} = \frac32 \times 1.0 \times 8.31 \times 400 + \frac32 \times 0.50 \times 8.31 \times 300$
$U_{\text{前}} = \frac32 \times 8.31 (400 + 150) = 1.5 \times 8.31 \times 550 = 6856$ J
$U_{\text{后}} = \frac32 (n_1 + n_2) R T_f = \frac32 \times 1.50 \times 8.31 \times T_f = 18.70 T_f$
$18.70 T_f = 6856$，$T_f = 366.7$ K ≈ 367 K
(b) 总体积 $V_{\text{总}} = 0.030 + 0.020 = 0.050$ m³
$P_f = n_{\text{总}} R T_f / V_{\text{总}} = 1.50 \times 8.31 \times 366.7 / 0.050$
$P_f = 4570 / 0.050 = 9.14 \times 10^4$ Pa
(c) $v_{\text{rms}} = \sqrt{3RT/M}$，同温下 $v_{\text{rms}} \propto 1/\sqrt{M}$
$M_{He} = 4.0$ g/mol，$M_{Ar} = 40.0$ g/mol
但这里温度不同，需分别计算：
$v_{\text{rms},He} = \sqrt{3 \times 8.31 \times 400 / 0.004} = \sqrt{2.493 \times 10^6} = 1579$ m/s
$v_{\text{rms},Ar} = \sqrt{3 \times 8.31 \times 300 / 0.040} = \sqrt{1.870 \times 10^5} = 432$ m/s
氦气方均根速度约为氩气的 $1579/432 = 3.66$ 倍。
这是由氦气温度更高且质量更轻共同导致的。
单原子气体 $U = \frac32 nRT$
$U_{\text{前}} = \frac32 \times 1.0 \times 8.31 \times 400 + \frac32 \times 0.50 \times 8.31 \times 300$
$U_{\text{前}} = \frac32 \times 8.31 (400 + 150) = 1.5 \times 8.31 \times 550 = 6856$ J
$U_{\text{后}} = \frac32 (n_1 + n_2) R T_f = \frac32 \times 1.50 \times 8.31 \times T_f = 18.70 T_f$
$18.70 T_f = 6856$，$T_f = 366.7$ K ≈ 367 K
(b) 总体积 $V_{\text{总}} = 0.030 + 0.020 = 0.050$ m³
$P_f = n_{\text{总}} R T_f / V_{\text{总}} = 1.50 \times 8.31 \times 366.7 / 0.050$
$P_f = 4570 / 0.050 = 9.14 \times 10^4$ Pa
(c) $v_{\text{rms}} = \sqrt{3RT/M}$，同温下 $v_{\text{rms}} \propto 1/\sqrt{M}$
$M_{He} = 4.0$ g/mol，$M_{Ar} = 40.0$ g/mol
但这里温度不同，需分别计算：
$v_{\text{rms},He} = \sqrt{3 \times 8.31 \times 400 / 0.004} = \sqrt{2.493 \times 10^6} = 1579$ m/s
$v_{\text{rms},Ar} = \sqrt{3 \times 8.31 \times 300 / 0.040} = \sqrt{1.870 \times 10^5} = 432$ m/s
氦气方均根速度约为氩气的 $1579/432 = 3.66$ 倍。
这是由氦气温度更高且质量更轻共同导致的。
(a) Energy conservation: total internal energy unchanged
Monatomic $U = \frac32 nRT$
$U_{\text{initial}} = \frac32 \times 1.0 \times 8.31 \times 400 + \frac32 \times 0.50 \times 8.31 \times 300$
$U_{\text{initial}} = 6856$ J
$U_{\text{final}} = \frac32 (n_1 + n_2) R T_f = 18.70 T_f$
$T_f = 366.7$ K ≈ 367 K
(b) Total volume $V_{\text{total}} = 0.030 + 0.020 = 0.050$ m³
$P_f = n_{\text{total}} R T_f / V_{\text{total}} = 1.50 \times 8.31 \times 366.7 / 0.050 = 9.14 \times 10^4$ Pa
(c) $v_{\text{rms}} = \sqrt{3RT/M}$
$M_{He} = 4.0$ g/mol, $M_{Ar} = 40.0$ g/mol
Different temperatures, so compute each:
$v_{\text{rms},He} = \sqrt{3 \times 8.31 \times 400 / 0.004} = 1579$ m/s
$v_{\text{rms},Ar} = \sqrt{3 \times 8.31 \times 300 / 0.040} = 432$ m/s
Helium rms speed is $1579/432 = 3.66$ times that of argon.
This results from helium being both hotter and lighter.
Monatomic $U = \frac32 nRT$
$U_{\text{initial}} = \frac32 \times 1.0 \times 8.31 \times 400 + \frac32 \times 0.50 \times 8.31 \times 300$
$U_{\text{initial}} = 6856$ J
$U_{\text{final}} = \frac32 (n_1 + n_2) R T_f = 18.70 T_f$
$T_f = 366.7$ K ≈ 367 K
(b) Total volume $V_{\text{total}} = 0.030 + 0.020 = 0.050$ m³
$P_f = n_{\text{total}} R T_f / V_{\text{total}} = 1.50 \times 8.31 \times 366.7 / 0.050 = 9.14 \times 10^4$ Pa
(c) $v_{\text{rms}} = \sqrt{3RT/M}$
$M_{He} = 4.0$ g/mol, $M_{Ar} = 40.0$ g/mol
Different temperatures, so compute each:
$v_{\text{rms},He} = \sqrt{3 \times 8.31 \times 400 / 0.004} = 1579$ m/s
$v_{\text{rms},Ar} = \sqrt{3 \times 8.31 \times 300 / 0.040} = 432$ m/s
Helium rms speed is $1579/432 = 3.66$ times that of argon.
This results from helium being both hotter and lighter.
附录：核心公式速查 / Formula Reference
| 知识点 / Topic | 公式 / Formula | 说明 / Notes |
|---|---|---|
| 温标转换 / Temp. Conversion | $T(\text{K}) = \theta(^{\circ}\text{C}) + 273.15$ | 温差 $\Delta T(\text{K}) = \Delta\theta(^{\circ}\text{C})$ |
| 比热容 / Specific Heat | $Q = mc\Delta T$ | $c$ 单位 J/(kg·K)，无相变时使用 |
| 潜热 / Latent Heat | $Q = mL$ | 相变时温度不变，$L_f$ 熔解、$L_v$ 汽化 |
| 热平衡 / Thermal Eq. | $Q_{\text{放}} = Q_{\text{吸}}$ | $m_1 c_1 (T_1 - T_f) = m_2 c_2 (T_f - T_2)$ |
| 理想气体 / Ideal Gas | $PV = nRT$ | $R = 8.31$ J/(mol·K)，$T$ 用 K |
| $PV = Nk_B T$ | $k_B = 1.38 \times 10^{-23}$ J/K | |
| 组合气体定律 | $\displaystyle \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$ | $n$ 不变时使用 |
| 气体压强公式 | $P = \frac13 \rho \overline{v^2}$ | $\rho$ 为密度，$\overline{v^2}$ 为均方速度 |
| 平均平动动能 | $\bar{E}_k = \frac32 k_B T$ | 每个分子，与气体种类无关 |
| 总平动动能 | $E_{\text{总}} = \frac32 nRT$ | $n$ mol 气体的总平动动能 |
| 方均根速度 / RMS | $v_{\text{rms}} = \sqrt{3k_B T/m} = \sqrt{3RT/M}$ | $M$ 为摩尔质量（kg/mol） |
| 热传导率 | $Q/t = kA(\Delta T/d)$ | $k$ 为导热系数，$d$ 为厚度 |
| 辐射功率 | $P = e\sigma A T^4$ | $\sigma = 5.67 \times 10^{-8}$ W/(m·K⁴) |
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