From states of matter to electron configurations, from the mole concept to ideal gases. This unit builds the foundational models that underpin all of chemistry.从物质三态到电子构型（electron configuration），从摩尔（mole）概念到理想气体（ideal gas）。本单元搭建起支撑整门化学的基础模型。
Introduction to the Particulate Nature of Matter物质粒子性导论
All matter is composed of particles. The three classifications of matter — elements, compounds, and mixtures — describe how those particles are organized and bonded. Understanding this hierarchy is the starting point for everything else in chemistry.一切物质都由粒子构成。物质的三种分类——元素（element）、化合物（compound）、混合物（mixture）——描述了这些粒子的组合方式与成键关系。理清这一层次结构是后续所有化学内容的起点。
Elements, Compounds, and MixturesElements cannot be chemically broken down into simpler substances. They consist of one type of atom.Compounds consist of atoms of different elements chemically bonded in a fixed ratio. Their properties differ from their constituent elements.Mixtures contain more than one element or compound in no fixed ratio, not chemically bonded — they can be separated by physical methods like filtration, distillation, evaporation, recrystallization, and chromatography.
Homogeneous vs. Heterogeneous Mixtures
A homogeneous mixture has uniform composition throughout (e.g., salt water, air). A heterogeneous mixture has visibly different components or phases (e.g., sand and water, oil and vinegar). The distinction matters when choosing separation techniques.
Kinetic Molecular Theory and States of Matter分子运动论与物质三态
The kinetic molecular theory models matter as particles in constant motion. The three states — solid, liquid, and gas — differ in the arrangement, spacing, and motion of their particles. Solids have fixed positions with vibrational motion. Liquids have particles close together but able to slide past each other. Gases have widely spaced particles moving rapidly in all directions.分子运动论（kinetic molecular theory）把物质看作不停运动的粒子集合。固、液、气三态在粒子排列、间距和运动方式上各不相同：固体粒子位置固定，仅作振动；液体粒子彼此靠近，但可相互滑动；气体粒子间距很大，向各方向高速运动。
Changes of State
Melting (solid to liquid), freezing (liquid to solid), vaporization (liquid to gas — includes evaporation and boiling), condensation (gas to liquid), sublimation (solid to gas), and deposition (gas to solid). Each involves energy transfer: breaking intermolecular forces requires energy (endothermic), forming them releases energy (exothermic).
State Symbols in Equations
Always include state symbols: (s) solid, (l) liquid, (g) gas, (aq) aqueous (dissolved in water). These are required in IB Chemistry equations and carry marks in exams.
Temperature in Kelvin (K) is a measure of the average kinetic energy ($E_k$) of particles. The kelvin has the same increment size as the Celsius degree. Converting between them:开尔文（K）温标度量的是粒子的平均动能（$E_k$）。开尔文与摄氏度的刻度大小相同。两者换算：
Celsius ↔ Kelvin Conversion摄氏度与开尔文的换算
$$T(K) = T(°C) + 273$$
At 0 K (absolute zero), particles have minimum kinetic energy. During a change of state, temperature remains constant even though heat is being added or removed — the energy goes into breaking or forming intermolecular forces.
During the melting of ice at 0 degrees C, which statement is correct?冰在 0 °C 熔化过程中，下列说法正确的是？
The kinetic energy of particles increases粒子的动能增加
The temperature rises steadily温度持续升高
Energy is used to overcome intermolecular forces能量用于克服分子间作用力
The particles stop moving粒子停止运动
Correct! During a phase change, added energy goes into overcoming intermolecular forces rather than increasing kinetic energy. The temperature stays constant until the phase change is complete.正确！物态变化期间，吸收的能量用于克服分子间作用力，而非增加动能。直到相变完成，温度始终保持不变。
During melting, the temperature remains constant. The added energy overcomes intermolecular forces rather than increasing particle speed. Answer: (C).熔化过程中温度保持不变。吸收的能量克服分子间作用力，而不是加快粒子运动。答案：(C)。
Going Deeper — Reading a Heating Curve
When a solid is heated at a steady rate, a graph of temperature against time shows sloping sections separated by two flat plateaus. On a sloping section a single state is warming and the average kinetic energy of the particles rises. The first plateau marks the melting point and the second marks the boiling point. During each plateau the temperature is constant because the absorbed energy is used to overcome intermolecular forces — potential energy rises while average kinetic energy stays fixed. The boiling plateau is longer than the melting plateau because vaporisation must separate the particles completely, which requires far more energy than merely loosening the lattice during melting.
Worked Example — Temperature Conversion and Kinetic Energy例题 — 温度换算与动能
Liquid nitrogen boils at $-196\;°$C. (a) Convert this temperature to kelvin. (b) A separate nitrogen sample is warmed from 77 K to 154 K. By what factor does the average kinetic energy of its particles change?
液氮在 $-196\;°$C 沸腾。(a) 将该温度换算为开尔文。(b) 另取一份氮气从 77 K 加热到 154 K，其粒子平均动能变为原来的几倍？
Part (a) — Celsius to kelvin
$$T = -196 + 273 = 77\;\text{K}$$
Absolute zero is the fixed reference point of the kelvin scale, so we simply add 273 to the Celsius value.
(a) — 摄氏转开尔文
$$T = -196 + 273 = 77\;\text{K}$$
绝对零度是开尔文标度的固定参考点，因此只需在摄氏数值上加 273。
Part (b) — Kinetic energy scales with absolute temperature
$$\bar{E}_k = \tfrac{3}{2}k_B T \implies \bar{E}_k \propto T$$
Average kinetic energy is directly proportional to the temperature in kelvin, so doubling $T$ from 77 K to 154 K doubles the average kinetic energy. This proportionality only holds in kelvin — taking the ratio in Celsius would be meaningless because the Celsius zero is arbitrary.
(b) — 动能与绝对温度成正比
$$\bar{E}_k = \tfrac{3}{2}k_B T \implies \bar{E}_k \propto T$$
平均动能与开尔文温度成正比，$T$ 由 77 K 翻倍到 154 K，平均动能也随之翻倍。该比例关系只在开尔文标度下成立——用摄氏度取比值毫无意义，因为摄氏零点是人为设定的。
Worked Example — Designing a Separation Scheme例题 — 设计分离方案
A solid mixture contains sand (insoluble in water), sodium chloride (soluble), and iodine (sublimes on gentle warming). Describe, in order, how to obtain a pure sample of each component.
Warm the mixture gently. Iodine sublimes directly from solid to vapour and re-deposits on a cold surface held above the dish, leaving sand and salt behind.
第 1 步 — 升华除去碘
轻微加热混合物，碘由固态直接升华为蒸气，在上方冷表面重新凝华收集，留下沙与盐。
Step 2 — Dissolve and filter
Add water and stir: the salt dissolves but the sand does not. Filter the mixture — sand is collected as the residue while the salt solution passes through as the filtrate.
第 2 步 — 溶解并过滤
加水搅拌：盐溶解、沙不溶。过滤混合物——沙作为残渣收集，盐溶液作为滤液通过。
Step 3 — Evaporate to recover the salt
Evaporate the filtrate to crystallise the sodium chloride. Every step exploits a different physical property and no chemical bonds are broken — confirming the sample is a mixture, not a compound.
Which of the following is best described as a homogeneous mixture?下列哪一项最适合描述为均相混合物？
Sand and water沙与水
Oil-and-vinegar salad dressing油醋沙拉汁
Granite rock花岗岩
Stainless steel不锈钢
Correct! Stainless steel is an alloy — a solid solution with uniform composition throughout, so it is homogeneous. The other three all show visibly distinct phases or grains.正确！不锈钢是合金——组成处处均匀的固溶体，属于均相混合物。其余三者都能看到明显不同的相或颗粒。
A homogeneous mixture has uniform composition with no visible boundaries between components. Only the alloy (stainless steel) qualifies. Answer: (D).均相混合物组成均匀，各组分间无可见界面。只有合金（不锈钢）符合。答案：(D)。
Structure 1.2
The Nuclear Atom核式原子模型
Atoms consist of a positively charged, dense nucleus containing protons and neutrons (collectively called nucleons), surrounded by negatively charged electrons. The nuclear symbol $_Z^A X$ encodes the identity and composition of any atom or ion.原子由带正电的致密原子核（含质子与中子，二者统称核子）和围绕其外、带负电的电子组成。核符号（nuclear symbol）$_Z^A X$ 编码了任意原子或离子的身份与组成。
Nuclear Symbol核符号
$$_Z^A X$$
$Z$ = atomic number (protons), $A$ = mass number (protons + neutrons). Neutrons = $A - Z$. For neutral atoms, electrons = protons = $Z$.
Isotopes are atoms of the same element (same number of protons) with different numbers of neutrons. They have identical chemical properties but different physical properties (mass, density, rate of diffusion). The relative atomic mass ($A_r$) of an element is a weighted average of its isotopic masses based on their natural abundances.同位素（isotope）是同一元素（质子数相同）但中子数不同的原子。它们化学性质相同，但物理性质（质量、密度、扩散速率）有差异。元素的相对原子质量（relative atomic mass，$A_r$）是按自然丰度对各同位素质量取的加权平均值。
Worked Example — Relative Atomic Mass from Isotopic Data例题 — 由同位素数据求相对原子质量
Chlorine has two isotopes: $^{35}$Cl (75.77%) and $^{37}$Cl (24.23%). Calculate the relative atomic mass.
Relative Isotopic Mass vs. Relative Atomic MassThese two terms are easy to conflate but mean different things. Relative isotopic mass is the mass of one specific isotope compared to $^{12}$C — e.g. $^{35}$Cl has a relative isotopic mass very close to 35 (a near-whole number, since it's dominated by whole numbers of protons and neutrons). Relative atomic mass ($A_r$) is the weighted average across all naturally occurring isotopes — chlorine's is 35.5, not close to any single isotope, precisely because it blends $^{35}$Cl and $^{37}$Cl in their natural proportions.
HL Only — Mass Spectrometry (1.2.3)
Mass spectra show peaks corresponding to each isotope. The $x$-axis shows mass-to-charge ratio ($m/z$), and the $y$-axis shows relative abundance. You can read off isotopic masses and their abundances directly from the spectrum and calculate $A_r$ from the data.
Worked Example — Relative Atomic Mass from a Mass Spectrum (HL)例题 — 由质谱图求相对原子质量（HL）
A mass spectrum of gallium shows two peaks: $m/z = 69$ with relative abundance 60.1%, and $m/z = 71$ with relative abundance 39.9%. Calculate the relative atomic mass of gallium.
The answer (69.80) lies between the two isotopic masses (69 and 71), closer to 69 — which makes sense since the lighter isotope is the more abundant one. Gallium's data-booklet $A_r$ is 69.72, matching closely.
When atoms absorb energy, electrons are excited to higher energy levels. When they fall back down, they emit photons of specific energies, producing a line emission spectrum. The hydrogen emission spectrum provides direct evidence that electrons exist in discrete (quantized) energy levels that converge at higher energies.当原子吸收能量时，电子被激发到更高的能级；当它们跃迁回较低能级时，发射出特定能量的光子，形成线状发射光谱（line emission spectrum）。氢原子发射光谱直接证明：电子处于离散（量子化）能级，且能级在高能端逐渐汇聚。
Continuous vs. Line Spectra
A continuous spectrum contains all wavelengths (like white light through a prism). A line spectrum contains only specific wavelengths, corresponding to specific energy transitions within atoms. Each element has a unique line spectrum — a chemical fingerprint.
The main energy level is given by the integer $n$ (1, 2, 3...) and can hold a maximum of $2n^2$ electrons. Within each main level, electrons occupy sublevels: s, p, d, and f, each with successively higher energy.主能级由整数 $n$（1、2、3……）标记，最多容纳 $2n^2$ 个电子。在每个主能级内，电子分布于亚层（subshell）：s、p、d、f，能量依次升高。
Sublevel Summary亚层总览
Sublevel
Orbitals
Max Electrons
Shape
s
1
2
Spherical
p
3
6
Dumbbell (3 orientations)
d
5
10
Complex (various)
f
7
14
Complex (various)
亚层
轨道数
最大电子数
形状
s
1
2
球形
p
3
6
哑铃形（3 种取向）
d
5
10
复杂（多种形状）
f
7
14
复杂（多种形状）
HL Only — Orbital Shapes and Degeneracy
An orbital is a region of space where there is a high probability of finding an electron — not a fixed orbit. All three $p$ orbitals in a sublevel ($p_x$, $p_y$, $p_z$) have identical dumbbell shapes but point along different axes; all five $d$ orbitals likewise have the same energy within a free atom. Orbitals of equal energy within the same sublevel are called degenerate — this is exactly why Hund's rule matters: electrons spread across degenerate orbitals singly before any pairing occurs, since pairing in the same orbital costs extra energy from electron–electron repulsion.
Three principles govern how electrons fill orbitals:三条原则规定了电子如何填充轨道（orbital）：
Filling RulesAufbau principle: Electrons fill the lowest energy orbitals first (1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p...).Hund's rule: Electrons occupy degenerate orbitals singly before pairing up, all with the same spin.Pauli exclusion principle: Each orbital holds a maximum of 2 electrons, which must have opposite spins.
Exceptions: Chromium and Copper
Cr (Z=24): expected [Ar] 3d$^4$ 4s$^2$, actual [Ar] 3d$^5$ 4s$^1$. Cu (Z=29): expected [Ar] 3d$^9$ 4s$^2$, actual [Ar] 3d$^{10}$ 4s$^1$. A half-filled or fully filled d-sublevel provides extra stability. You must know these two exceptions.
When transition metals form ions, electrons are removed from the 4s sublevel first (not 3d).
Fe$^{2+}$: [Ar] 3d$^6$
The 4s electrons are lost before the 3d electrons because in an ion, 3d is lower in energy than 4s.
Fe$^{2+}$（24 个电子）
过渡金属形成阳离子时，电子先从 4s 亚层失去，而不是 3d。
Fe$^{2+}$：[Ar] 3d$^6$
先失 4s 后失 3d 的原因是：在离子中 3d 的能量低于 4s。
HL Only — Ionization Energy from Spectra (1.3.6–1.3.7)
The convergence limit of spectral lines corresponds to ionization. The first ionization energy can be calculated from the frequency of the convergence limit: $IE = hf$ or $IE = hc/\lambda$. Successive IE data reveal the electron configuration — large jumps between successive IEs indicate electrons being removed from a different shell. IE trends and discontinuities across a period provide evidence for the existence of sublevels.
Worked Example — Ionization Energy from Convergence Frequency (HL)例题 — 由汇聚频率求电离能（HL）
The convergence limit of the Lyman series in hydrogen's emission spectrum occurs at a frequency of $3.29 \times 10^{15}\;\text{Hz}$. Calculate the first ionization energy of hydrogen in kJ mol$^{-1}$ ($h = 6.63\times10^{-34}\;\text{J s}$, $N_A = 6.022\times10^{23}\;\text{mol}^{-1}$).
Going Deeper — Reading a Successive Ionization Energy Graph
A graph of $\log(IE)$ against "electron removed" for a given element shows several roughly flat steps separated by sharp jumps. Each flat step corresponds to electrons being removed from within the same main energy level (or sublevel); a sharp jump means the next electron comes from a shell much closer to the nucleus, which is far harder to remove. Counting the number of electrons in each step before a jump directly reveals the electron arrangement into shells — e.g. a 2-8-1 pattern (steps of size 2, then 8, then 1 before running out of easy electrons) is exactly what you'd see for sodium, without ever citing its electron configuration by name.
Species with identical electron configurations ("iso" = same, "electronic" = electron arrangement) tend to share some physical properties — but not all, since nuclear charge still differs. Ionic radius is the clearest example: across this isoelectronic series, radius shrinks from O$^{2-}$ to Mg$^{2+}$ because the same 10 electrons are pulled in by an increasingly large nuclear charge (8+, 9+, 11+, 12+).
What is the electron configuration of Cu (Z=29)?Cu（Z=29）的电子构型是？
[Ar] 3d$^9$ 4s$^2$
[Ar] 3d$^8$ 4s$^2$ 4p$^1$
[Ar] 3d$^{10}$ 4s$^2$
[Ar] 3d$^{10}$ 4s$^1$
Correct! Copper is one of two required exceptions. A fully filled 3d$^{10}$ sublevel is more stable, so one 4s electron is promoted: [Ar] 3d$^{10}$ 4s$^1$.正确！铜是必考的两个例外之一。3d$^{10}$ 全充满更稳定，因此一个 4s 电子被"提"到 3d：[Ar] 3d$^{10}$ 4s$^1$。
Cu is an exception: [Ar] 3d$^{10}$ 4s$^1$, not the expected [Ar] 3d$^9$ 4s$^2$. Answer: (D).Cu 是例外：实际为 [Ar] 3d$^{10}$ 4s$^1$，不是 [Ar] 3d$^9$ 4s$^2$。答案：(D)。
Structure 1.4
Counting Particles by Mass: The Mole用质量数粒子：摩尔
The mole (mol) is the SI unit of amount of substance. One mole contains exactly the number of elementary entities given by the Avogadro constant ($N_A = 6.022 \times 10^{23}$ mol$^{-1}$). An elementary entity can be an atom, molecule, ion, electron, or any specified particle.摩尔（mole，mol）是 SI 单位制中物质的量的基本单位。1 mol 含有的基本粒子数由阿伏伽德罗常数（Avogadro's constant）给出：$N_A = 6.022 \times 10^{23}$ mol$^{-1}$。基本粒子可以是原子、分子、离子、电子或任何指明的粒子。
The Mole Triangle摩尔三角
$$n = \frac{m}{M} \qquad n = \frac{N}{N_A} \qquad n = C \times V$$
$n$ = amount (mol), $m$ = mass (g), $M$ = molar mass (g mol$^{-1}$), $N$ = number of particles, $N_A$ = Avogadro constant, $C$ = concentration (mol dm$^{-3}$), $V$ = volume (dm$^3$).
Masses of atoms are compared on a scale relative to $^{12}$C. The relative atomic mass ($A_r$) and relative formula mass ($M_r$) are dimensionless (no units). The molar mass ($M$) has units g mol$^{-1}$ and is numerically equal to $M_r$.原子质量以 $^{12}$C 为基准进行比较。相对原子质量（relative atomic mass，$A_r$）与相对式量（relative formula mass，$M_r$）是无量纲量。摩尔质量（molar mass，$M$）的单位是 g mol$^{-1}$，数值上等于 $M_r$。
Empirical and Molecular Formulas实验式与分子式
The empirical formula gives the simplest whole-number ratio of atoms. The molecular formula gives the actual number of atoms in a molecule. To find the molecular formula, divide the molar mass by the empirical formula mass to get a multiplier.实验式（empirical formula）给出原子数的最简整数比；分子式（molecular formula）给出分子中各原子的真实个数。用摩尔质量除以实验式式量，即可得到倍数，进而得到分子式。
Worked Example — Empirical Formula from Percentage Composition例题 — 由质量百分比求实验式
A compound contains 40.0% C, 6.7% H, and 53.3% O by mass. Its molar mass is 180 g mol$^{-1}$. Find the empirical and molecular formulas.
某化合物按质量计含 40.0% C、6.7% H、53.3% O，其摩尔质量为 180 g mol$^{-1}$。求其实验式与分子式。
A hydrated salt has the formula MgSO$_4\cdot x$H$_2$O. Heating 2.46 g of the hydrate to constant mass leaves 1.20 g of anhydrous MgSO$_4$ ($M = 120.4$ g mol$^{-1}$). Find $x$.
某水合盐的化学式为 MgSO$_4\cdot x$H$_2$O。将 2.46 g 该水合物加热至质量恒定，剩下 1.20 g 无水 MgSO$_4$（$M = 120.4$ g mol$^{-1}$）。求 $x$。
So the formula is MgSO$_4\cdot 7$H$_2$O (Epsom salt). This is the same empirical-formula logic as before, just with "moles of water" and "moles of salt" playing the role of two elements in a ratio.
Worked Example — Percentage Composition by Mass (the Reverse Direction)例题 — 由化学式求质量百分比（反方向计算）
Calculate the percentage by mass of nitrogen in ammonium nitrate, NH$_4$NO$_3$ ($M = 80.06$ g mol$^{-1}$).
求硝酸铵 NH$_4$NO$_3$（$M = 80.06$ g mol$^{-1}$）中氮的质量百分比。
Step 1 — Identify how many N atoms are in the formula
NH$_4$NO$_3$ contains two nitrogen atoms per formula unit — one in the ammonium ion, one in the nitrate ion. This is the step most often missed: counting only one N because the formula "looks like" it has one obvious nitrogen.
Percentage composition and empirical formula are inverse problems: one goes from formula to percentages, the other from percentages to formula, but both rely on exactly the same molar-mass bookkeeping.
Molar concentration is determined by the amount of solute and the volume of solution. Square brackets denote concentration: [NaOH] means the molar concentration of NaOH in mol dm$^{-3}$. You should also be able to convert between g dm$^{-3}$ and mol dm$^{-3}$.摩尔浓度由溶质的物质的量与溶液体积决定。方括号代表浓度：[NaOH] 即 NaOH 的摩尔浓度（mol dm$^{-3}$）。你还应能在 g dm$^{-3}$ 与 mol dm$^{-3}$ 之间互换。
Worked Example — Dilution Calculation例题 — 稀释计算
25.0 cm$^3$ of 2.00 mol dm$^{-3}$ HCl is diluted to a final volume of 250 cm$^3$. What is the new concentration?
Equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules. This means the mole ratio in a balanced equation directly gives the volume ratio for gaseous reactants and products.同温同压下，等体积的任何气体含有相同数目的分子（Avogadro's law）。因此，配平方程中的摩尔比就是气态反应物与产物的体积比。
Worked Example — Gas Volumes from a Balanced Equation例题 — 由配平方程求气体体积
Propane burns completely: C$_3$H$_8$(g) + 5O$_2$(g) → 3CO$_2$(g) + 4H$_2$O(g). At constant temperature and pressure, what volume of O$_2$ is needed to completely burn 3.00 dm$^3$ of propane, and what volume of CO$_2$ is produced?
No moles, mass, or molar volume are needed at all — because temperature and pressure are constant and equal for every gas in the reaction, the coefficients convert directly to volumes with no unit conversion.
An ideal gas is a theoretical model where particles have negligible volume and no intermolecular forces, and all collisions are elastic. Real gases approximate ideal behavior at high temperatures and low pressures (where particles are far apart and moving fast).理想气体（ideal gas）是一个理论模型：粒子自身体积可忽略、粒子间无作用力，且所有碰撞均为弹性碰撞。真实气体在高温低压（粒子彼此距离远、运动快）时近似服从理想气体行为。
When Real Gases Deviate
Real gases deviate from ideal behavior at low temperature (particles move slowly, intermolecular forces become significant) and high pressure (particles are close together, their volume is no longer negligible). Gases with stronger intermolecular forces (e.g., polar molecules, larger molecules) deviate more.
Going Deeper — Where the Ideal Gas Assumptions Break Down Mathematically
The van der Waals equation, $\left(P + \dfrac{an^2}{V^2}\right)(V - nb) = nRT$, patches the two ideal-gas assumptions directly. The term $\dfrac{an^2}{V^2}$ added to $P$ compensates for intermolecular attractions, which make the measured pressure slightly lower than an ideal gas would produce (attracted particles hit the walls a little less forcefully). The term $nb$ subtracted from $V$ compensates for the real, non-negligible volume the particles themselves occupy. Both correction terms shrink toward zero as $V$ becomes large (low pressure, particles far apart) and $T$ becomes high (fast-moving particles for which weak attractions barely matter) — exactly recovering $PV = nRT$ in the ideal limit.
Molar Volume and the Ideal Gas Equation摩尔体积与理想气体方程
At STP (273.15 K, 100 kPa), the molar volume of an ideal gas is 22.7 dm$^3$ mol$^{-1}$ (given in the data booklet). The ideal gas equation relates all four state variables:在 STP（STP，273.15 K、100 kPa）下，理想气体的摩尔体积（molar volume）为 22.7 dm$^3$ mol$^{-1}$（数据手册中给出）。理想气体方程把四个状态变量联系起来：
Ideal Gas Equation理想气体方程
$$PV = nRT$$
$P$ = pressure (Pa), $V$ = volume (m$^3$), $n$ = moles (mol), $R$ = 8.314 J K$^{-1}$ mol$^{-1}$, $T$ = temperature (K). Use SI units throughout.
Under which conditions does a real gas behave most like an ideal gas?真实气体在哪种条件下最接近理想气体行为？
High temperature and low pressure高温且低压
Low temperature and high pressure低温且高压
Low temperature and low pressure低温且低压
High temperature and high pressure高温且高压
Correct! At high temperature, particles move fast and intermolecular forces are insignificant. At low pressure, particles are far apart and their volume is negligible. Both conditions match the ideal gas assumptions.正确！高温下粒子运动快，分子间作用力可忽略；低压下粒子距离远，其自身体积可忽略——两者都恰好符合理想气体的假设。
Ideal gas behavior requires conditions where intermolecular forces and particle volume are negligible: high $T$, low $P$. Answer: (A).理想气体行为要求分子间作用力与粒子自身体积均可忽略，即高 $T$、低 $P$。答案：(A)。
Going Deeper — Graham's Law and Rates of Diffusion
Kinetic molecular theory predicts that at the same temperature, all gas particles have the same average kinetic energy ($\tfrac{1}{2}mv^2$) — but not the same speed. Lighter particles must move faster to have the same kinetic energy as heavier ones. This gives Graham's law: $\dfrac{\text{rate}_1}{\text{rate}_2} = \sqrt{\dfrac{M_2}{M_1}}$, where rate can mean either diffusion (spreading through another gas) or effusion (escaping through a tiny hole). A gas four times heavier diffuses at half the rate — not a quarter, because the relationship is a square root, not a direct ratio.
Note: because $P$ and $V$ only ever appear as a product/ratio here, kPa and dm$^3$ can be used directly without converting to Pa and m$^3$ — unlike the single-state $PV=nRT$ equation, which requires strict SI units.
This value is close to the molar mass of $\text{NO}_2$ or $\text{O}_3$ (both $\approx 46$–48 g mol$^{-1}$) — identifying the actual gas would need a second clue, such as its odor or color, but the calculation itself only ever needs mass, and the three PV=nRT variables.
这个数值接近 $\text{NO}_2$ 或 $\text{O}_3$ 的摩尔质量（两者都约为 46–48 g mol$^{-1}$）——要确定具体是哪种气体，还需要第二条线索（如气味或颜色），但计算本身只需要质量，以及 PV=nRT 中的其余三个变量。
Preparation备考
Exam Strategy考试策略
Paper 1 (Multiple Choice)Paper 1（Multiple Choice）
Know the subatomic particle table cold. For electron configurations, watch for ions (remove 4s before 3d for transition metals) and the Cr/Cu exceptions. Mole calculations are frequent — practise converting between mass, moles, particles, and concentration quickly.
亚原子粒子表必须背得滚瓜烂熟。电子构型题留意离子（过渡金属先失 4s 再失 3d），以及 Cr / Cu 两个例外。摩尔计算高频出现——练习快速在质量、物质的量、粒子数与浓度间互换。
Paper 2 (Extended Response)Paper 2（Extended Response）
Show all working in mole calculations: write the formula, substitute, and solve with units. For gas calculations, always convert to SI units before applying $PV = nRT$. Empirical formula questions: set up a clear table showing element, mass, moles, and ratio.
摩尔计算题写清全过程：先写公式，再代入数值，最后带单位求解。气体题在使用 $PV = nRT$ 之前先把全部数值换成 SI 单位。求实验式时列出清晰的"元素—质量—摩尔数—比"表。
Data BookletData Booklet
Key items to locate quickly: relative atomic masses (periodic table), Avogadro constant, molar gas volume at STP, gas constant $R$, the equations $n = m/M$, $n = CV$, $PV = nRT$, and the combined gas law.
Mistake 1 — Forgetting Unit Conversions for PV=nRT
$P$ must be in Pa (not kPa), $V$ must be in m$^3$ (not dm$^3$ or cm$^3$). 1 kPa = 1000 Pa. 1 dm$^3$ = 10$^{-3}$ m$^3$. Most errors in gas calculations come from mismatched units.
Mistake 2 — Removing 3d Electrons First When Forming Ions
When transition metals form cations, the 4s electrons are lost before 3d electrons. Fe loses its two 4s electrons first to become Fe$^{2+}$: [Ar] 3d$^6$, not [Ar] 4s$^2$ 3d$^4$.
Mistake 3 — Confusing $A_r$ and $M$
Relative atomic mass ($A_r$) and relative formula mass ($M_r$) have no units. Molar mass ($M$) has units of g mol$^{-1}$. Numerically they are the same, but examiners penalize missing or incorrect units.
错误 3 — 混淆 $A_r$ 与 $M$
相对原子质量（$A_r$）与相对式量（$M_r$）无单位；摩尔质量（$M$）单位为 g mol$^{-1}$。三者数值相同，但漏写或写错单位会被考官扣分。
Mistake 4 — Using Celsius in Gas Calculations
Temperature must always be in Kelvin for $PV = nRT$ and the combined gas law. Add 273 to convert from Celsius. Using Celsius gives wildly wrong answers (and zero marks).
Mistake 5 — Rounding Intermediate Mole Ratios Too Early
In empirical formula problems, dividing by the smallest mole value often gives numbers like 1.98 or 2.99 rather than exact integers. Students sometimes round these to whole numbers too early (before checking whether they're actually close to a simple fraction like 1.5 or 2.5), producing a formula with the wrong ratio. Always carry at least 2-3 decimal places through the division step, and if a ratio is close to $x.5$, multiply the whole set by 2 rather than rounding it away.
Equal volumes of all gases at the same $T$ and $P$ contain equal numbers of molecules.同 $T$ 同 $P$ 下，任何气体的等体积含有等数量的分子。
Electron config of Cr?Cr 的电子构型？
[Ar] 3d$^5$ 4s$^1$ (exception — not 3d$^4$ 4s$^2$). Half-filled d-sublevel is more stable.[Ar] 3d$^5$ 4s$^1$（例外，不是 3d$^4$ 4s$^2$）。d 亚层半充满更稳定。
When do real gases deviate most from ideal?真实气体在何条件下偏离理想气体最严重？
At low temperature and high pressure. Intermolecular forces and particle volume become significant.低温、高压。此时分子间作用力与粒子自身体积都不可忽略。
What is the molar volume at STP?STP 下的摩尔体积是？
22.7 dm$^3$ mol$^{-1}$ at 273.15 K and 100 kPa.在 273.15 K、100 kPa 下为 22.7 dm$^3$ mol$^{-1}$。
Empirical vs. molecular formula?实验式与分子式的区别？
Empirical = simplest whole-number ratio. Molecular = actual number of atoms. E.g., CH$_2$O vs. C$_6$H$_{12}$O$_6$.实验式 = 最简整数比；分子式 = 真实原子数。例：CH$_2$O 对应 C$_6$H$_{12}$O$_6$。
What happens to temperature during a phase change?物态变化过程中温度如何变化？
Temperature remains constant. Added energy overcomes intermolecular forces rather than increasing kinetic energy.温度保持不变。吸收的能量用于克服分子间作用力，而非增加动能。
Assessment测评
Unit Quiz单元测验
1. Which separation technique is most appropriate for separating a dissolved solid from a solution?1. 从溶液中分离已溶解的固体，最合适的分离方法是？
Filtration过滤（filtration）
Chromatography色谱（chromatography）
Evaporation蒸发（evaporation）
Distillation蒸馏（distillation）
Correct! Evaporation removes the solvent (water), leaving the dissolved solid behind. Filtration separates insoluble solids, not dissolved ones.正确！蒸发可去除溶剂（水），留下溶质固体。过滤只能分离不溶性固体，对已溶解的固体无效。
A dissolved solid passes through filter paper. Evaporation removes the water, leaving the solid. Answer: (C).已溶解的固体会随溶液穿过滤纸。蒸发掉水后即可得到固体。答案：(C)。
2. What is the electron configuration of V$^{3+}$ (Z=23)?2. V$^{3+}$（Z=23）的电子构型是？
[Ar] 3d$^2$ 4s$^2$ minus 3 from 4s and 3d[Ar] 3d$^2$ 4s$^2$ 然后从 4s 与 3d 各失去若干
[Ar] 3d$^2$
[Ar] 3d$^3$ 4s$^2$ minus 3 from 3d[Ar] 3d$^3$ 4s$^2$ 仅从 3d 失去 3 个
[Ar] 4s$^2$ 3d$^0$
Correct! V is [Ar] 3d$^3$ 4s$^2$. Removing 3 electrons: 2 from 4s first, then 1 from 3d, giving [Ar] 3d$^2$.正确！V 的电子构型为 [Ar] 3d$^3$ 4s$^2$。失去 3 个电子时，先失 2 个 4s，再失 1 个 3d，剩 [Ar] 3d$^2$。
V = [Ar] 3d$^3$ 4s$^2$. Remove 4s electrons first (2), then 1 from 3d: V$^{3+}$ = [Ar] 3d$^2$. Answer: (B).V = [Ar] 3d$^3$ 4s$^2$。先失 4s（2 个），再失 3d（1 个）：V$^{3+}$ = [Ar] 3d$^2$。答案：(B)。
3. What volume does 0.100 mol of an ideal gas occupy at STP (molar volume = 22.7 dm$^3$ mol$^{-1}$)?3. 在 STP（摩尔体积 = 22.7 dm$^3$ mol$^{-1}$）下，0.100 mol 理想气体占多大体积？