Unit B.3: Gas Laws单元 B.3：气体定律

Many gas-law manipulations are cleaner in terms of number density $n_V = N / V$ (molecules per cubic metre), not amount of substance $n$. Beware the notation clash: $n$ means moles, $n_V$ means molecules per volume. The molecular form $pV = N k_B T$ rearranges to $p = n_V k_B T$, which makes pressure manifestly a property of how crowded and how energetic the molecules are.

许多气体定律运算用数密度 $n_V = N / V$（每立方米的分子数）比用物质的量 $n$ 更简洁。注意记号冲突：$n$ 表示摩尔数，$n_V$ 表示单位体积的分子数。分子形式 $pV = N k_B T$ 可改写为 $p = n_V k_B T$，由此可见压强本质上取决于分子的密集程度与能量高低。

The Avogadro constant also links the two gas constants: $R = N_A k_B$, so $k_B = R / N_A = 8.31 / (6.02 \times 10^{23}) \approx 1.38 \times 10^{-23}\ \mathrm{J\,K^{-1}}$.

阿伏伽德罗常数还把两个气体常数联系起来：$R = N_A k_B$，故 $k_B = R / N_A = 8.31 / (6.02 \times 10^{23}) \approx 1.38 \times 10^{-23}\ \mathrm{J\,K^{-1}}$。

Plot $V$ against Celsius temperature at constant $p$ for any gas and you get a straight line. Extrapolating every such line backward, they all cross the temperature axis at the same point: $-273\ ^\circ\mathrm{C}$. Defining the Kelvin scale so that this intercept is $0\ \mathrm{K}$ makes the relation a pure proportionality $V \propto T$. The same construction on a $p$-$T$ graph (constant $V$) gives Gay-Lussac. This is the experimental origin of absolute zero.

在恒压下对任意气体作 $V$ 对摄氏温度的图，得到一条直线。把所有这样的直线向后外推，它们都交温度轴于同一点：$-273\ ^\circ\mathrm{C}$。把开尔文标度定义为该截距处为 $0\ \mathrm{K}$，便使关系成为纯比例 $V \propto T$。在 $p$-$T$ 图（恒容）上作同样构造得盖-吕萨克定律。这就是绝对零度的实验来源。

Crucially, $V \propto T$ only holds with $T$ in Kelvin. Doubling Celsius temperature from $20\ ^\circ\mathrm{C}$ to $40\ ^\circ\mathrm{C}$ does not double the volume; doubling absolute temperature from $293\ \mathrm{K}$ to $586\ \mathrm{K}$ does.

关键在于 $V \propto T$ 仅当 $T$ 取开尔文时成立。把摄氏温度从 $20\ ^\circ\mathrm{C}$ 翻倍到 $40\ ^\circ\mathrm{C}$ 不会使体积翻倍；把绝对温度从 $293\ \mathrm{K}$ 翻倍到 $586\ \mathrm{K}$ 才会。

For a fixed amount of gas, $nR$ is a constant, so $pV / T = nR = \text{const}$. Hold one variable fixed and the others reduce to an empirical law:

对固定气体量，$nR$ 为常数，故 $pV / T = nR = \text{常数}$。固定一个变量，其余即化为一条经验定律：

• Constant $T$: $pV = \text{const}$ (Boyle).
• $T$ 恒定：$pV = \text{常数}$（玻意耳）。
• Constant $p$: $V / T = \text{const}$ (Charles).
• $p$ 恒定：$V / T = \text{常数}$（查理）。
• Constant $V$: $p / T = \text{const}$ (Gay-Lussac).
• $V$ 恒定：$p / T = \text{常数}$（盖-吕萨克）。

The combined gas law $p_1 V_1 / T_1 = p_2 V_2 / T_2$ is just the statement that $pV/T$ keeps the same constant value before and after, which is why $n$ must not change between the two states.

合并气体定律 $p_1 V_1 / T_1 = p_2 V_2 / T_2$ 不过是说 $pV/T$ 在前后保持同一常数值，这也是两态间 $n$ 不能改变的原因。

Consider one molecule of mass $m$ moving with $x$-velocity $c_x$ in a cubic box of side $L$. Each elastic bounce off a wall reverses $c_x$, changing momentum by $2 m c_x$. The molecule returns to the same wall every $2L / c_x$ seconds, so the force it exerts averages $2 m c_x / (2L/c_x) = m c_x^{2} / L$.

设质量 $m$ 的一个分子在边长 $L$ 的立方盒内沿 $x$ 方向以速度 $c_x$ 运动。每次对壁面的弹性碰撞使 $c_x$ 反向，动量改变 $2 m c_x$。该分子每 $2L / c_x$ 秒回到同一壁面，故其平均作用力为 $2 m c_x / (2L/c_x) = m c_x^{2} / L$。

Summing over $N$ molecules and using $\overline{c_x^{2}} = \tfrac{1}{3}\overline{c^{2}}$ (isotropy) gives the total force, and dividing by the wall area $L^2$ with $V = L^3$ yields $pV = \tfrac{1}{3} N m \overline{c^{2}}$. The $\tfrac{1}{3}$ comes purely from the three equivalent directions of motion.

对 $N$ 个分子求和，并用各向同性 $\overline{c_x^{2}} = \tfrac{1}{3}\overline{c^{2}}$ 得总力，再除以壁面面积 $L^2$ 并令 $V = L^3$，即得 $pV = \tfrac{1}{3} N m \overline{c^{2}}$。其中 $\tfrac{1}{3}$ 纯粹来自三个等价的运动方向。

Equate the two expressions for $pV$. Kinetic theory gives $pV = \tfrac{1}{3} N m \overline{c^{2}}$; the ideal gas law gives $pV = N k_B T$. Setting them equal:

令 $pV$ 的两个表达式相等。动理论给出 $pV = \tfrac{1}{3} N m \overline{c^{2}}$；理想气体定律给出 $pV = N k_B T$。令二者相等：

Multiplying both sides by $\tfrac{3}{2}$ converts the left side into the mean translational KE $\tfrac{1}{2} m \overline{c^{2}}$:

两边乘以 $\tfrac{3}{2}$ 把左边化为平均平动动能 $\tfrac{1}{2} m \overline{c^{2}}$：

This is the central bridge of the unit: it identifies absolute temperature as a direct measure of mean molecular kinetic energy, independent of the gas's identity. For a monatomic gas there is no rotational or vibrational energy, so the internal energy is purely this translational sum, $U = \tfrac{3}{2} N k_B T$.

这是本单元的核心桥梁：它把绝对温度等同为平均分子动能的直接量度，与气体种类无关。对单原子气体没有转动或振动能量，故内能纯为这部分平动能量之和，$U = \tfrac{3}{2} N k_B T$。

Real-gas equations of state restore the two neglected effects. The molecules' own volume reduces the space available, effectively replacing $V$ with $(V - nb)$ where $b$ is a per-mole excluded volume. Intermolecular attraction reduces the measured pressure, effectively replacing $p$ with $(p + a\,n^2/V^2)$. Together these give the van der Waals equation $\left(p + a n^2/V^2\right)(V - nb) = nRT$.

真实气体的状态方程恢复这两个被忽略的效应。分子自身体积减少了可用空间，相当于把 $V$ 换为 $(V - nb)$，其中 $b$ 为每摩尔排斥体积。分子间吸引减小了测得压强，相当于把 $p$ 换为 $(p + a\,n^2/V^2)$。二者合起来给出范德瓦尔斯方程 $\left(p + a n^2/V^2\right)(V - nb) = nRT$。

The IB syllabus does not require this equation, but it does require knowing which assumption each correction repairs: the $b$ term fixes finite molecular volume (matters at high $p$), the $a$ term fixes intermolecular attraction (matters at low $T$).

IB 大纲不要求此方程，但要求知道每个修正项修复哪条假设：$b$ 项修复有限分子体积（高 $p$ 时重要），$a$ 项修复分子间吸引（低 $T$ 时重要）。