PART I · SHORT RESPONSE第一部分 · 短答题AP-style MCQ + ON/BC short answer · 23 marksAP 风格选择题 + 安/卑省考短答 · 共 23 分
Section A · Short ResponseA 部分 · 短答题
Mix of multiple-choice and short-answer items. For MCQs, circle the letter and show enough reasoning in the margin that a marker could verify. For short-answer items, state units in every answer and show all substitutions. Use $R = 8.314\ \text{L kPa mol}^{-1}\text{K}^{-1}$ and molar volume at STP $= 22.4\ \text{L mol}^{-1}$ unless otherwise stated. Temperature must always be converted to Kelvin.本节包含选择题与短答题。选择题请圈出字母答案，并在空白处写下足以让阅卷人核对的思路。短答题每道都要写出单位并展示所有代入过程。除非另有说明，取 $R = 8.314\ \text{L kPa mol}^{-1}\text{K}^{-1}$，标准状态下摩尔体积 $= 22.4\ \text{L mol}^{-1}$。温度必须换算为开尔文。
Q1EASY易🇺🇸 US美AP-style MCQAP 风格选择题§1 KMT and states of matter分子运动论与物质状态 · HS-PS1-7[3 marks][3 分]
Which statement best describes gas particles according to the Kinetic Molecular Theory (KMT)?根据分子运动论 (KMT)，以下哪项最能描述气体粒子的特征？
(A)Gas particles are stationary and held in fixed positions by strong forces.气体粒子静止不动，被强烈的力固定在固定位置。
(B)Gas particles are in constant random motion and the volume of the particles themselves is negligible compared to the container volume.气体粒子做持续无规则运动，粒子本身的体积与容器体积相比可忽略不计。
(C)Gas particles exert strong attractive forces on one another at all times.气体粒子之间始终存在强烈的吸引力。
(D)Gas particles have lower average kinetic energy than liquid particles at the same temperature.在相同温度下，气体粒子的平均动能低于液体粒子。
Q2EASY易🇨🇦 ON安AP-style MCQAP 风格选择题§2 Pressure units and temperature conversion压强单位与温度换算 · SCH3U F1[3 marks][3 分]
A gas sample has a pressure of $1.50\ \text{atm}$ and a temperature of $25.0\ ^\circ\text{C}$. What are the equivalent pressure in kilopascals and temperature in kelvin? ($1\ \text{atm} = 101.325\ \text{kPa}$)某气体样品压强为 $1.50\ \text{atm}$，温度为 $25.0\ ^\circ\text{C}$。换算为千帕和开尔文后分别是多少？（$1\ \text{atm} = 101.325\ \text{kPa}$）
A sealed syringe contains $4.00\ \text{L}$ of gas at a pressure of $150\ \text{kPa}$ and constant temperature. The plunger is pushed in until the pressure increases to $200\ \text{kPa}$.一个密封注射器在恒定温度下装有 $4.00\ \text{L}$ 气体，压强为 $150\ \text{kPa}$。推入活塞直到压强增大至 $200\ \text{kPa}$。
(a)State the name and mathematical form of the gas law that applies here.写出此处适用的气体定律的名称及其数学形式。[2]
(b)Calculate the new volume of the gas, with units.计算气体的新体积，并写出单位。[2]
(c)State one assumption required for this law to hold.写出该定律成立所需的一个假设条件。[1]
A balloon holds $2.00\ \text{L}$ of gas at $300\ \text{K}$ at constant pressure. It is heated until its temperature rises to $450\ \text{K}$. What is the new volume?一个气球在 $300\ \text{K}$、恒定压强下装有 $2.00\ \text{L}$ 气体。加热至温度升至 $450\ \text{K}$。新体积是多少？
(A) $1.33\ \text{L}$
(B) $2.00\ \text{L}$
(C) $3.00\ \text{L}$
(D) $4.00\ \text{L}$
Q5MEDIUM中🇨🇦 AB阿AB Diploma-style阿尔伯塔毕业考风格§4 Combined Gas Law综合气体定律 · Chemistry 20 Unit B GO1[8 marks][8 分]
A gas occupies $5.00\ \text{L}$ at $1.00\ \text{atm}$ ($101.325\ \text{kPa}$) and $25.0\ ^\circ\text{C}$. The gas is compressed to a new pressure of $2.50\ \text{atm}$ ($253.3\ \text{kPa}$) while its temperature is raised to $100.0\ ^\circ\text{C}$.某气体在 $1.00\ \text{atm}$（$101.325\ \text{kPa}$）、$25.0\ ^\circ\text{C}$ 条件下体积为 $5.00\ \text{L}$。将其压缩至新压强 $2.50\ \text{atm}$（$253.3\ \text{kPa}$），同时温度升至 $100.0\ ^\circ\text{C}$。
(a)Write the combined gas law and identify each variable.写出综合气体定律，并说明每个变量的含义。[2]
(b)Convert all temperatures to kelvin and list all known values with units.将所有温度换算为开尔文，并列出所有已知量及其单位。[2]
(c)Calculate the new volume $V_2$, showing all steps and units.计算新体积 $V_2$，展示所有步骤及单位。[3]
(d)Explain in one sentence whether the result is physically reasonable.用一句话解释该结果在物理上是否合理。[1]
PART II · EXTENDED RESPONSE第二部分 · 简答题AP-feeder FRQ + honors · 35 marksAP 衔接简答题 + 荣誉级 · 共 35 分
Section B · Extended ResponseB 部分 · 简答题
Show every step of reasoning. Identify the gas law or formula used before substituting values. State units in every final answer. Temperature must be in kelvin. Use $R = 8.314\ \text{L kPa mol}^{-1}\text{K}^{-1}$ and molar volume at STP $= 22.4\ \text{L mol}^{-1}$. Calculator permitted on Q6-Q9.每一步推理都要写出。在代入数值前先注明所用气体定律或公式。每个最终答案都要写单位。温度须用开尔文。取 $R = 8.314\ \text{L kPa mol}^{-1}\text{K}^{-1}$，标准状态下摩尔体积 $= 22.4\ \text{L mol}^{-1}$。Q6-Q9 可用计算器。
A rigid container holds $2.00\ \text{mol}$ of an ideal gas at $150\ \text{kPa}$ and $400\ \text{K}$.一个刚性容器装有 $2.00\ \text{mol}$ 理想气体，压强为 $150\ \text{kPa}$，温度为 $400\ \text{K}$。
(a)Write the ideal gas law and name each variable.写出理想气体定律，并说明每个变量的含义。[2]
(b)Calculate the volume of the gas. Show all steps with units.计算气体的体积。展示所有步骤及单位。[3]
(c)The temperature is then raised to $600\ \text{K}$ at constant volume. Calculate the new pressure.随后在体积不变的情况下，温度升至 $600\ \text{K}$。计算新压强。[2]
(d)State which individual gas law part (c) is equivalent to, and explain why.说明 (c) 等价于哪条单一气体定律，并解释原因。[1]
At STP ($0\ ^\circ\text{C}$, $101.325\ \text{kPa}$), the molar volume of an ideal gas is $22.4\ \text{L mol}^{-1}$.在标准状态（$0\ ^\circ\text{C}$，$101.325\ \text{kPa}$）下，理想气体的摩尔体积为 $22.4\ \text{L mol}^{-1}$。
(a)A sample of oxygen gas occupies $11.2\ \text{L}$ at STP. How many moles of $\text{O}_2$ are present?一份氧气样品在标准状态下体积为 $11.2\ \text{L}$。含有多少摩尔 $\text{O}_2$？[2]
(b)Calculate the mass of this $\text{O}_2$ sample. Molar mass of $\text{O}_2 = 32.00\ \text{g mol}^{-1}$.计算该 $\text{O}_2$ 样品的质量。$\text{O}_2$ 摩尔质量 $= 32.00\ \text{g mol}^{-1}$。[2]
(c)Verify the molar volume using the ideal gas law at STP. Show all work.利用理想气体定律在标准状态下验证摩尔体积。展示所有步骤。[3]
(d)State why real gases deviate from the molar-volume prediction at high pressures and low temperatures.说明为何真实气体在高压、低温下偏离摩尔体积预测值。[2]
Q8HARD难🇨🇦 BC卑BC Provincial-style卑诗省考风格§7 Gas Stoichiometry气体化学计量 · Chemistry 11[8 marks][8 分]
Hydrogen gas reacts with oxygen gas according to the equation: $2\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{H}_2\text{O(g)}$. At STP, $44.8\ \text{L}$ of $\text{H}_2$ is available to react.氢气与氧气按方程式 $2\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{H}_2\text{O(g)}$ 反应。在标准状态下，$44.8\ \text{L}$ 的 $\text{H}_2$ 可用于反应。
(a)Calculate the moles of $\text{H}_2$ available.计算可用 $\text{H}_2$ 的摩尔数。[2]
(b)Calculate the volume of $\text{O}_2$ required at STP to react completely with this $\text{H}_2$.计算在标准状态下与该 $\text{H}_2$ 完全反应所需 $\text{O}_2$ 的体积。[3]
(c)Calculate the mass of water vapour produced. Molar mass of $\text{H}_2\text{O} = 18.02\ \text{g mol}^{-1}$.计算生成水蒸气的质量。$\text{H}_2\text{O}$ 摩尔质量 $= 18.02\ \text{g mol}^{-1}$。[2]
(d)State the mole ratio of $\text{H}_2$ to $\text{O}_2$ and explain how Avogadro's Law links it to the volume ratio at constant temperature and pressure.写出 $\text{H}_2$ 与 $\text{O}_2$ 的摩尔比，并解释阿伏伽德罗定律如何在恒温恒压条件下将其与体积比联系起来。[1]
Q9HARD难Honors荣誉级🇺🇸 US美AP-feeder FRQAP 衔接简答题§7 Dalton's Law + Ideal Gas Law道尔顿分压定律 + 理想气体定律 · HS-PS1-7[10 marks][10 分]
A $10.0\ \text{L}$ rigid container at $298\ \text{K}$ holds a mixture of helium gas with partial pressure $80.0\ \text{kPa}$ and nitrogen gas with partial pressure $40.0\ \text{kPa}$.一个 $10.0\ \text{L}$ 刚性容器在 $298\ \text{K}$ 下装有氦气与氮气的混合气体，氦气分压为 $80.0\ \text{kPa}$，氮气分压为 $40.0\ \text{kPa}$。
(a)State Dalton's Law of Partial Pressures and calculate the total pressure of the mixture.写出道尔顿分压定律，并计算混合气体的总压强。[2]
(b)Use the ideal gas law to calculate the moles of He present. Show all steps.利用理想气体定律计算氦气的摩尔数。展示所有步骤。[3]
(c)Calculate the moles of $\text{N}_2$ present.计算氮气的摩尔数。[2]
(d)Calculate the mole fraction of He in the mixture and show that it equals the ratio of its partial pressure to the total pressure.计算混合气体中氦气的摩尔分数，并验证其等于氦气分压与总压强之比。[3]
PART III · MODELING / APPLIED第三部分 · 建模与应用AB Diploma + Universal · 23 marks阿省毕业考 + 通用题型 · 共 23 分
Section C · Modeling and ApplicationsC 部分 · 建模与应用
Define symbols (with units) at the start of each question. State the gas law or formula used before substituting. Conclude each question with a one-sentence answer in context. Calculator permitted throughout Part III. Temperature must be converted to kelvin.每题开始时定义符号（含单位）。代入数值前先写出所用气体定律或公式。每题以一句结合情境的完整句子作答。第三部分全程可用计算器。温度须换算为开尔文。
A car tyre contains air at $180\ \text{kPa}$ when measured at $20.0\ ^\circ\text{C}$ on a cold morning. After highway driving, the tyre temperature rises to $50.0\ ^\circ\text{C}$. Assume the tyre volume does not change significantly.一辆汽车轮胎在寒冷早晨 $20.0\ ^\circ\text{C}$ 时测得气压为 $180\ \text{kPa}$。行驶高速公路后，胎温升至 $50.0\ ^\circ\text{C}$。假设轮胎体积无明显变化。
(a)Identify the gas law that applies and write its formula.确定适用的气体定律并写出其公式。[2]
(b)Convert both temperatures to kelvin.将两个温度换算为开尔文。[1]
(c)Calculate the new tyre pressure after driving, with units.计算行驶后的新胎压，并写出单位。[3]
(d)Explain in terms of KMT why the pressure increases when the temperature increases at constant volume.从分子运动论角度解释，为何在体积不变时温度升高会导致压强增大。[2]
Q11MEDIUM中🇨🇦 AB阿AB Diploma-style阿尔伯塔毕业考风格§5 Ideal Gas Law (applied)理想气体定律（应用） · Chemistry 20 GO1[8 marks][8 分]
An industrial gas cylinder contains $5.00\ \text{mol}$ of compressed nitrogen gas at $20.0\ ^\circ\text{C}$ and has an internal volume of $50.0\ \text{L}$.一个工业气罐在 $20.0\ ^\circ\text{C}$ 下装有 $5.00\ \text{mol}$ 压缩氮气，内部体积为 $50.0\ \text{L}$。
(a)Calculate the pressure inside the cylinder using the ideal gas law.利用理想气体定律计算气罐内的压强。[3]
(b)The safety valve opens if pressure exceeds $300\ \text{kPa}$. At what temperature (in $^\circ\text{C}$) would the valve open?若压强超过 $300\ \text{kPa}$ 时安全阀开启。安全阀在什么温度（以 $^\circ\text{C}$ 表示）下会开启？[3]
(c)Calculate the number of nitrogen molecules in the cylinder. ($N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$)计算气罐中氮气分子的数目。（$N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$）[2]
Calcium carbonate reacts with excess hydrochloric acid: $\text{CaCO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{CaCl}_2\text{(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}$. A $25.0\ \text{g}$ sample of pure $\text{CaCO}_3$ (molar mass $= 100.09\ \text{g mol}^{-1}$) is used. The $\text{CO}_2$ produced is collected over water at $25.0\ ^\circ\text{C}$. The total gas pressure in the collection flask is $101.3\ \text{kPa}$ and the vapour pressure of water at $25.0\ ^\circ\text{C}$ is $3.17\ \text{kPa}$.碳酸钙与过量盐酸反应：$\text{CaCO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{CaCl}_2\text{(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}$。使用 $25.0\ \text{g}$ 纯 $\text{CaCO}_3$（摩尔质量 $= 100.09\ \text{g mol}^{-1}$）。生成的 $\text{CO}_2$ 在 $25.0\ ^\circ\text{C}$ 下排水集气，集气瓶中总气压为 $101.3\ \text{kPa}$，该温度下水的蒸气压为 $3.17\ \text{kPa}$。
(a)Calculate the moles of $\text{CO}_2$ produced.计算生成 $\text{CO}_2$ 的摩尔数。[2]
(b)Use Dalton's Law to find the partial pressure of $\text{CO}_2$ alone.利用分压定律求 $\text{CO}_2$ 的分压。[1]
(c)Calculate the volume of the collected $\text{CO}_2$ at $25.0\ ^\circ\text{C}$ and its partial pressure, using $PV = nRT$.利用 $PV = nRT$，计算在 $25.0\ ^\circ\text{C}$ 及其分压下收集到的 $\text{CO}_2$ 体积。[3]
(d)Explain why the collected volume differs from the volume the same $\text{CO}_2$ would occupy at STP.解释为何收集到的体积与相同 $\text{CO}_2$ 在标准状态下的体积不同。[1]
🇺🇸 US NGSS美国 NGSSHS-PS1-7
🇨🇦 Ontario安大略SCH3U Strand F · F1 · F2
🇨🇦 British Columbia不列颠哥伦比亚Chemistry 11: gases, gas laws, ideal gas law化学 11：气体、气体定律、理想气体定律
🇨🇦 Alberta阿尔伯塔Chemistry 20 Unit B · GO1
Full Syllabus Map lives in ../Study Guides/Unit_7_States_of_Matter_and_the_Gas_Laws.html. The individual gas laws (Boyle's, Charles's, Gay-Lussac's) and the combined gas law are NGSS-honors level; the ideal gas law and gas stoichiometry are core for AP Chemistry.完整大纲对照表见 ../Study Guides/Unit_7_States_of_Matter_and_the_Gas_Laws.html。各分气体定律（玻意耳、查理、盖-吕萨克）及综合气体定律为 NGSS 荣誉级；理想气体定律与气体化学计量为 AP 化学核心内容。