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States of Matter and the Gas Laws · Solutions物质状态与气体定律 · 详解

Companion to the Practice Set · Mark-by-mark walkthroughs · AP-Feeder / ON / BC / AB styles练习题配套答案 · 逐分讲解 · AP 衔接 / 安 / 卑 / 阿省考风格

EASY易 MEDIUM中 HARD难 🇺🇸 US美 🇨🇦 ON安 🇨🇦 BC卑 🇨🇦 AB阿 AP-style MCQAP 风格选择题 AP-feeder FRQAP 衔接简答题 ON Provincial-style安大略省考风格 BC Provincial-style卑诗省考风格 AB Diploma-style阿尔伯塔毕业考风格 Honors荣誉级
PART I  ·  SHORT RESPONSE · SOLUTIONS第一部分  ·  短答题 · 详解AP MCQ + ON/BC short answer · 23 marksAP 选择题 + 安/卑省考短答 · 共 23 分

Section A · Worked SolutionsA 部分 · 详细解答

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？

Answer:答案：  (B)

Identify the correct KMT postulate确定正确的 KMT 假设 A1·A1·A1

The KMT rests on five core postulates: (1) gas particles are in constant, rapid, random motion; (2) the volume of the particles themselves is negligible compared to the container volume; (3) collisions between particles and with container walls are perfectly elastic; (4) there are no intermolecular forces between gas particles (except during collisions); (5) the average kinetic energy of the particles is proportional to the absolute temperature and is the same for all ideal gases at the same temperature. Statement (B) correctly combines postulates 1 and 2.分子运动论有五条核心假设：(1) 气体粒子做持续、快速、无规则运动；(2) 粒子本身的体积与容器体积相比可忽略不计；(3) 粒子之间及与器壁的碰撞为完全弹性碰撞；(4) 粒子之间除碰撞瞬间外无分子间作用力；(5) 粒子的平均动能与绝对温度成正比，在同一温度下所有理想气体相同。选项 (B) 正确体现了假设 1 和 2。
Why the other options are wrong.其他选项错误原因。
(A) Describes a solid, not a gas; solid particles are held in fixed positions by strong intermolecular forces.描述的是固体，不是气体；固体粒子被强烈的分子间作用力固定在固定位置。
(C) Violates KMT postulate 4: ideal gas particles exert no significant attractive forces on one another.违反 KMT 假设 4：理想气体粒子之间无显著吸引力。
(D) False; at the same temperature all particles (gas, liquid, solid) have the same average kinetic energy. Temperature is the measure of average kinetic energy regardless of phase.错误；在相同温度下所有粒子（气体、液体、固体）的平均动能相同。温度是平均动能的量度，与相态无关。
KMT is the microscopic model that explains all macroscopic gas-law behaviour.分子运动论是从微观角度解释所有宏观气体定律行为的模型。 Every gas law follows from KMT. Boyle's law: compress the container and particles collide with the walls more often, raising pressure. Charles's law: raise the temperature and average kinetic energy rises, particles hit walls harder and more often, so at constant pressure the volume must expand. Gay-Lussac's law: at constant volume, higher temperature means harder wall collisions, so pressure rises. The negligible-volume and no-forces postulates are what make a gas "ideal"; real gases deviate when these break down at high pressure or low temperature.每条气体定律都源于分子运动论。玻意耳定律：压缩容器后粒子与器壁碰撞更频繁，压强升高。查理定律：升温后平均动能增加，粒子撞击器壁更有力、更频繁，因此在恒压下体积膨胀。盖-吕萨克定律：恒容时升温意味着撞击更有力，压强升高。体积可忽略和无作用力的假设使气体成为"理想"气体；在高压或低温下这些假设失效时，真实气体会发生偏离。
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 $25.0\ ^\circ\text{C}$. Equivalent values in kPa and K?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？

Answer:答案：  (A)  $152\ \text{kPa}$ and $298\ \text{K}$

Convert pressure from atm to kPa将压强从 atm 换算为 kPa M1·A1

$$ P \;=\; 1.50\ \text{atm} \times 101.325\ \frac{\text{kPa}}{\text{atm}} \;=\; 151.99\ \text{kPa} \;\approx\; 152\ \text{kPa}. $$

Convert temperature from Celsius to Kelvin将温度从摄氏度换算为开尔文 A1

$$ T \;=\; 25.0 + 273.15 \;=\; 298.15\ \text{K} \;\approx\; 298\ \text{K}. $$
Why the other options are wrong.其他选项错误原因。
(B) $248\ \text{K}$: subtracts 273 from 25 instead of adding, giving $25 - 273 = -248$, then drops the sign.用 $25 - 273$ 而非 $25 + 273$，得到 $-248$，再去掉负号。
(C) $67.6\ \text{kPa}$: divides 1 atm by 1.50 instead of multiplying 1.50 by 101.325.用 1 atm 除以 1.50，而非用 1.50 乘以 101.325。
(D) $101\ \text{kPa}$: reports the pressure of 1.00 atm, ignoring the factor of 1.50.报告的是 1.00 atm 对应的压强，忽略了系数 1.50。
Two conversions to memorize: $1\ \text{atm} = 101.325\ \text{kPa}$; $T(\text{K}) = T(^\circ\text{C}) + 273.15$.两个必记换算：$1\ \text{atm} = 101.325\ \text{kPa}$；$T(\text{K}) = T(^\circ\text{C}) + 273.15$。 All gas-law calculations require pressure in consistent units and temperature in kelvin. Never use Celsius in $PV = nRT$ or the combined gas law. The kelvin scale starts at absolute zero ($-273.15\ ^\circ\text{C}$), the temperature at which all molecular motion theoretically stops. A Celsius temperature can be zero or negative, but a kelvin temperature can never be negative; this is your consistency check. For pressure, the SI unit is the pascal (Pa), and $1\ \text{kPa} = 1000\ \text{Pa}$. The value $R = 8.314\ \text{L kPa mol}^{-1}\text{K}^{-1}$ requires pressure in kPa and volume in litres.所有气体定律计算要求压强单位一致，温度用开尔文。切勿在 $PV = nRT$ 或综合气体定律中使用摄氏度。开尔文标度从绝对零度（$-273.15\ ^\circ\text{C}$）开始，即理论上分子运动完全停止的温度。摄氏温度可以为零或负数，但开尔文温度永远不能为负，这是一致性校验。对于压强，SI 单位是帕斯卡（Pa），$1\ \text{kPa} = 1000\ \text{Pa}$。使用 $R = 8.314\ \text{L kPa mol}^{-1}\text{K}^{-1}$ 时，压强须用 kPa，体积须用升。
Q3MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §3 Boyle's Law玻意耳定律 · SCH3U F2 [5 marks][5 分]

Sealed syringe: $4.00\ \text{L}$ at $150\ \text{kPa}$, constant temperature; plunger pushed until pressure $= 200\ \text{kPa}$.密封注射器：恒温下 $4.00\ \text{L}$，$150\ \text{kPa}$；推入活塞直到压强 $= 200\ \text{kPa}$。

Answer:答案：  (a) Boyle's Law, $P_1 V_1 = P_2 V_2$玻意耳定律，$P_1 V_1 = P_2 V_2$  ·  (b) $V_2 = 3.00\ \text{L}$  ·  (c) constant temperature温度恒定

(a) Name and mathematical form of the law定律名称及数学形式 A1·A1

Boyle's Law states that at constant temperature, the pressure and volume of a fixed amount of gas are inversely proportional:玻意耳定律指出，在恒定温度下，固定量气体的压强与体积成反比： $$ P_1 V_1 \;=\; P_2 V_2 \qquad (T,\ n \text{ constant}). $$

(b) Calculate new volume计算新体积 M1·A1

Known: $P_1 = 150\ \text{kPa}$, $V_1 = 4.00\ \text{L}$, $P_2 = 200\ \text{kPa}$. Solving for $V_2$:已知：$P_1 = 150\ \text{kPa}$，$V_1 = 4.00\ \text{L}$，$P_2 = 200\ \text{kPa}$。求解 $V_2$： $$ V_2 \;=\; \frac{P_1 V_1}{P_2} \;=\; \frac{150 \times 4.00}{200} \;=\; \frac{600}{200} \;=\; 3.00\ \text{L}. $$

(c) Required assumption所需假设条件 A1

The temperature must remain constant (isothermal process). If temperature changed, all three variables ($P$, $V$, $T$) would change simultaneously and the combined gas law would be needed instead.温度必须保持恒定（等温过程）。若温度改变，则 $P$、$V$、$T$ 三个变量同时变化，此时需改用综合气体定律。
Boyle's Law: pressure and volume are inversely proportional at constant $T$ and $n$.玻意耳定律：在 $T$ 和 $n$ 恒定时，压强与体积成反比。 A useful sanity check: if pressure increases, volume must decrease, and vice versa. Here pressure increased from 150 to 200 kPa (a factor of $4/3$), so volume decreased from 4.00 to 3.00 L (a factor of $3/4$). The product $PV$ is constant: $150 \times 4.00 = 600 = 200 \times 3.00$. In a KMT picture, compressing the same number of molecules into a smaller space means more frequent wall collisions per unit time, hence higher pressure. Boyle's Law applies only if the gas behaves ideally; real gases can deviate significantly at very high pressures where intermolecular forces and particle volume become non-negligible.一个有用的合理性检验：压强增大时体积必须减小，反之亦然。此处压强从 150 增至 200 kPa（倍数为 $4/3$），因此体积从 4.00 减至 3.00 L（倍数为 $3/4$）。乘积 $PV$ 为常数：$150 \times 4.00 = 600 = 200 \times 3.00$。从分子运动论的角度看，将相同数量的分子压缩到更小的空间中意味着单位时间内器壁碰撞更频繁，从而压强升高。玻意耳定律仅适用于理想气体行为；在极高压下分子间作用力和粒子体积不可忽视时，真实气体会显著偏离。
Q4MEDIUM中 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §3 Charles's Law查理定律 · Chemistry 11 [4 marks][4 分]

Balloon: $2.00\ \text{L}$ at $300\ \text{K}$, constant pressure; heated to $450\ \text{K}$. New volume?气球：$300\ \text{K}$、恒压下 $2.00\ \text{L}$；加热至 $450\ \text{K}$。新体积？

Answer:答案：  (C)  $3.00\ \text{L}$

Apply Charles's Law套用查理定律 M1·A1·A1·A1

Charles's Law: at constant pressure and fixed amount, volume is directly proportional to absolute temperature:查理定律：在恒压且气体量固定时，体积与绝对温度成正比： $$ \frac{V_1}{T_1} \;=\; \frac{V_2}{T_2} \quad \Longrightarrow \quad V_2 \;=\; V_1 \cdot \frac{T_2}{T_1} \;=\; 2.00 \times \frac{450}{300} \;=\; 2.00 \times 1.50 \;=\; 3.00\ \text{L.} $$ Note: temperatures are already given in kelvin, so no conversion is needed here. Option (C).注意：温度已以开尔文给出，无需转换。选 (C)。
Why the other options are wrong.其他选项错误原因。
(A) $1.33\ \text{L}$: inverts the ratio, $V_2 = 2.00 \times (300/450) = 1.33$ L, treating heating as compression.颠倒了比例，$V_2 = 2.00 \times (300/450) = 1.33$ L，把加热误当作压缩。
(B) $2.00\ \text{L}$: assumes volume does not change, ignoring Charles's Law entirely.假设体积不变，完全忽略了查理定律。
(D) $4.00\ \text{L}$: doubles the volume incorrectly; perhaps from $(450 - 300)/300 \times 2 + 2 = 3$ was misread, or from treating the temperature ratio as 2 instead of 1.5.错误地将体积翻倍；可能是将温度比误算为 2 而非 1.5。
Charles's Law requires absolute temperature (K): volume is proportional to $T$, not to $^\circ\text{C}$.查理定律要求使用绝对温度（K）：体积与 $T$ 成正比，而非与 $^\circ\text{C}$ 成正比。 If the temperatures had been given in Celsius (say $27\ ^\circ\text{C}$ and $177\ ^\circ\text{C}$), students who forget to convert would compute $177/27 = 6.56$ instead of $450/300 = 1.50$, getting a wildly wrong answer. The direct-proportionality only holds on the kelvin scale. Physically: at constant pressure, giving particles more kinetic energy causes them to spread out further; the gas expands until the pressure balances again. A $50\%$ increase in $T$ (from 300 to 450 K) produces a $50\%$ increase in $V$ (from 2.00 to 3.00 L).如果温度以摄氏度给出（例如 $27\ ^\circ\text{C}$ 和 $177\ ^\circ\text{C}$），忘记转换的学生会计算 $177/27 = 6.56$ 而非 $450/300 = 1.50$，得到严重错误的答案。正比关系仅在开尔文标度上成立。从物理角度看：恒压下，给粒子更多动能使其扩散得更远；气体膨胀直到压强再次平衡。$T$ 增加 $50\%$（从 300 到 450 K）产生 $V$ 增加 $50\%$（从 2.00 到 3.00 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]

Gas: $5.00\ \text{L}$ at $101.325\ \text{kPa}$ and $25.0\ ^\circ\text{C}$; compressed to $253.3\ \text{kPa}$ while temperature rises to $100.0\ ^\circ\text{C}$.气体：$5.00\ \text{L}$，$101.325\ \text{kPa}$，$25.0\ ^\circ\text{C}$；压缩至 $253.3\ \text{kPa}$，同时温度升至 $100.0\ ^\circ\text{C}$。

Answer:答案：  (a) $\dfrac{P_1 V_1}{T_1} = \dfrac{P_2 V_2}{T_2}$  ·  (b) $T_1 = 298.15\ \text{K},\ T_2 = 373.15\ \text{K}$  ·  (c) $V_2 \approx 2.50\ \text{L}$  ·  (d) Reasonable: higher $P$ dominates the lower $V$, partially offset by higher $T$.合理：压强大幅升高主导体积减小，被温度升高部分抵消。

(a) Combined gas law and variable identification综合气体定律与变量说明 A1·A1

$$ \frac{P_1 V_1}{T_1} \;=\; \frac{P_2 V_2}{T_2} $$ where $P$ = absolute pressure (kPa), $V$ = volume (L), $T$ = absolute temperature (K). Subscripts 1 and 2 denote initial and final states. Amount $n$ is constant (sealed system).其中 $P$ = 绝对压强（kPa），$V$ = 体积（L），$T$ = 绝对温度（K）。下标 1 和 2 分别表示初态和末态。气体量 $n$ 恒定（密封系统）。

(b) Convert temperatures and list knowns换算温度并列出已知量 A1·A1

$$ T_1 \;=\; 25.0 + 273.15 \;=\; 298.15\ \text{K}, \qquad T_2 \;=\; 100.0 + 273.15 \;=\; 373.15\ \text{K}. $$ Known: $P_1 = 101.325\ \text{kPa}$, $V_1 = 5.00\ \text{L}$, $T_1 = 298.15\ \text{K}$, $P_2 = 253.3\ \text{kPa}$, $T_2 = 373.15\ \text{K}$. Unknown: $V_2$.已知：$P_1 = 101.325\ \text{kPa}$，$V_1 = 5.00\ \text{L}$，$T_1 = 298.15\ \text{K}$，$P_2 = 253.3\ \text{kPa}$，$T_2 = 373.15\ \text{K}$。未知：$V_2$。

(c) Calculate $V_2$计算 $V_2$ M1·A1·A1

Rearrange for $V_2$:整理求 $V_2$： $$ V_2 \;=\; \frac{P_1 V_1 T_2}{T_1 P_2} \;=\; \frac{101.325 \times 5.00 \times 373.15}{298.15 \times 253.3}. $$ Numerator: $101.325 \times 5.00 = 506.625$; $506.625 \times 373.15 = 189{,}016\ (\text{kPa}\cdot\text{L}\cdot\text{K})$.分子：$101.325 \times 5.00 = 506.625$；$506.625 \times 373.15 \approx 189{,}016$（kPa$\cdot$L$\cdot$K）。
Denominator: $298.15 \times 253.3 = 75{,}521\ (\text{kPa}\cdot\text{K})$.分母：$298.15 \times 253.3 \approx 75{,}521$（kPa$\cdot$K）。 $$ V_2 \;=\; \frac{189{,}016}{75{,}521} \;\approx\; 2.50\ \text{L.} $$

(d) Physical reasonableness物理合理性 A1

The result is physically reasonable: the pressure more than doubled (from $\approx 101$ to $253\ \text{kPa}$, a factor of $\approx 2.5$), which would halve the volume by Boyle's Law; the temperature increase from 298 K to 373 K (factor $\approx 1.25$) partially offsets this compression, so the final volume of $2.50\ \text{L}$ (half the original) is sensible.结果在物理上是合理的：压强增加超过两倍（从约 $101$ 到 $253\ \text{kPa}$，约为 $2.5$ 倍），按玻意耳定律会使体积减半；温度从 298 K 升至 373 K（约为 $1.25$ 倍）部分抵消了压缩效果，因此最终体积 $2.50\ \text{L}$（约为原来的一半）是合理的。
The combined gas law unifies Boyle's, Charles's, and Gay-Lussac's laws into one equation.综合气体定律将玻意耳定律、查理定律和盖-吕萨克定律统一为一个方程。 When one variable is held constant, the combined law reduces to the appropriate simple law: hold $T$ constant and get $P_1 V_1 = P_2 V_2$ (Boyle); hold $P$ constant and get $V_1/T_1 = V_2/T_2$ (Charles); hold $V$ constant and get $P_1/T_1 = P_2/T_2$ (Gay-Lussac). Always perform temperature conversion to kelvin before substituting, and verify that the final answer direction makes intuitive sense (higher $P$ drives $V$ down; higher $T$ drives $V$ up).当一个变量保持恒定时，综合气体定律简化为相应的简单定律：保持 $T$ 恒定得到 $P_1 V_1 = P_2 V_2$（玻意耳定律）；保持 $P$ 恒定得到 $V_1/T_1 = V_2/T_2$（查理定律）；保持 $V$ 恒定得到 $P_1/T_1 = P_2/T_2$（盖-吕萨克定律）。代入前务必将温度换算为开尔文，并验证最终答案方向是否符合直觉（压强升高使 $V$ 减小；温度升高使 $V$ 增大）。
PART II  ·  EXTENDED RESPONSE · SOLUTIONS第二部分  ·  简答题 · 详解AP-feeder FRQ + honors · 35 marksAP 衔接简答题 + 荣誉级 · 共 35 分

Section B · Worked SolutionsB 部分 · 详细解答

Q6MEDIUM中 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §5 Ideal Gas Law理想气体定律 · HS-PS1-7 [8 marks][8 分]

Rigid container: $2.00\ \text{mol}$ ideal gas at $150\ \text{kPa}$ and $400\ \text{K}$.刚性容器：$2.00\ \text{mol}$ 理想气体，$150\ \text{kPa}$，$400\ \text{K}$。

Answer:答案：  (a) $PV = nRT$  ·  (b) $V \approx 44.3\ \text{L}$  ·  (c) $P_2 = 225\ \text{kPa}$  ·  (d) Gay-Lussac's Law ($P \propto T$ at constant $V$ and $n$)盖-吕萨克定律（$V$ 和 $n$ 恒定时 $P \propto T$）

(a) Ideal gas law and variable identification理想气体定律与变量说明 A1·A1

$$ PV \;=\; nRT $$ where $P$ = absolute pressure (kPa), $V$ = volume (L), $n$ = amount of gas (mol), $R = 8.314\ \text{L kPa mol}^{-1}\text{K}^{-1}$ = universal gas constant, $T$ = absolute temperature (K).其中 $P$ = 绝对压强（kPa），$V$ = 体积（L），$n$ = 气体物质的量（mol），$R = 8.314\ \text{L kPa mol}^{-1}\text{K}^{-1}$ = 通用气体常数，$T$ = 绝对温度（K）。

(b) Calculate volume计算体积 M1·A1·A1

Solving for $V$:求解 $V$： $$ V \;=\; \frac{nRT}{P} \;=\; \frac{2.00 \times 8.314 \times 400}{150} \;=\; \frac{6651.2}{150} \;=\; 44.3\ \text{L.} $$

(c) New pressure at $T_2 = 600\ \text{K}$, constant $V$$T_2 = 600\ \text{K}$、体积不变时的新压强 M1·A1

At constant $V$ and $n$, $P \propto T$, so:在 $V$ 和 $n$ 恒定时，$P \propto T$，因此： $$ \frac{P_2}{P_1} \;=\; \frac{T_2}{T_1} \quad \Longrightarrow \quad P_2 \;=\; 150 \times \frac{600}{400} \;=\; 150 \times 1.50 \;=\; 225\ \text{kPa.} $$

(d) Equivalent individual gas law等价的单一气体定律 A1

Part (c) is equivalent to Gay-Lussac's Law: at constant volume and fixed amount of gas, pressure is directly proportional to absolute temperature ($P_1/T_1 = P_2/T_2$). This is the special case of the ideal gas law when $V$ and $n$ do not change.(c) 等价于盖-吕萨克定律：在体积和气体量固定时，压强与绝对温度成正比（$P_1/T_1 = P_2/T_2$）。这是在 $V$ 和 $n$ 不变时理想气体定律的特殊情况。
$PV = nRT$ contains all three simple gas laws as special cases.$PV = nRT$ 包含三条简单气体定律作为其特殊情况。 When $n$ and $T$ are constant, $PV = \text{const}$ (Boyle). When $n$ and $P$ are constant, $V/T = \text{const}$ (Charles). When $n$ and $V$ are constant, $P/T = \text{const}$ (Gay-Lussac). For the combined gas law, $n$ is constant but all three of $P$, $V$, $T$ can vary. For the ideal gas law in full, none of the four quantities $P$, $V$, $n$, $T$ need be constant. The ideal gas law is the master equation; the simpler laws are derived by constraining variables. A rigid container means $V$ is constant, so any temperature change drives a proportional pressure change.当 $n$ 和 $T$ 恒定时，$PV = \text{常数}$（玻意耳定律）。当 $n$ 和 $P$ 恒定时，$V/T = \text{常数}$（查理定律）。当 $n$ 和 $V$ 恒定时，$P/T = \text{常数}$（盖-吕萨克定律）。综合气体定律中 $n$ 恒定但 $P$、$V$、$T$ 均可变化。完整的理想气体定律中，$P$、$V$、$n$、$T$ 四个量无需任何一个恒定。理想气体定律是主方程；较简单的定律通过约束变量而推导得出。刚性容器意味着 $V$ 恒定，因此温度的任何变化都会引起成比例的压强变化。
Q7MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §6 Molar Volume and STP摩尔体积与标准状态 · SCH3U F2 [9 marks][9 分]

At STP ($0\ ^\circ\text{C}$, $101.325\ \text{kPa}$): molar volume $= 22.4\ \text{L mol}^{-1}$.标准状态（$0\ ^\circ\text{C}$，$101.325\ \text{kPa}$）：摩尔体积 $= 22.4\ \text{L mol}^{-1}$。

Answer:答案：  (a) $0.500\ \text{mol}$  ·  (b) $16.0\ \text{g}$  ·  (c) $22.4\ \text{L mol}^{-1}$ verified  ·  (d) real gas particle volume and intermolecular attractions become significant真实气体粒子体积和分子间引力变得显著

(a) Moles of $\text{O}_2$ at STP标准状态下 $\text{O}_2$ 的摩尔数 M1·A1

$$ n \;=\; \frac{V}{V_{\text{molar}}} \;=\; \frac{11.2\ \text{L}}{22.4\ \text{L mol}^{-1}} \;=\; 0.500\ \text{mol.} $$

(b) Mass of $\text{O}_2$ sample$\text{O}_2$ 样品质量 M1·A1

$$ m \;=\; n \times M \;=\; 0.500\ \text{mol} \times 32.00\ \text{g mol}^{-1} \;=\; 16.0\ \text{g.} $$

(c) Verify molar volume using $PV = nRT$ at STP用 $PV = nRT$ 在标准状态下验证摩尔体积 M1·A1·A1

STP conditions: $P = 101.325\ \text{kPa}$, $T = 273.15\ \text{K}$, $n = 1.00\ \text{mol}$.标准状态条件：$P = 101.325\ \text{kPa}$，$T = 273.15\ \text{K}$，$n = 1.00\ \text{mol}$。 $$ V_{\text{molar}} \;=\; \frac{nRT}{P} \;=\; \frac{1.00 \times 8.314 \times 273.15}{101.325} \;=\; \frac{2270.9}{101.325} \;=\; 22.4\ \text{L mol}^{-1}. \quad \checkmark $$

(d) Why real gases deviate at high $P$ and low $T$真实气体在高压低温下偏离的原因 A1·A1

Two KMT postulates break down: (1) at high pressure, molecules are forced close together and the volume of the molecules themselves is no longer negligible relative to the container volume, so the actual volume is greater than the ideal prediction; (2) at low temperature, molecules move slowly and intermolecular attractive forces (van der Waals attractions) become significant, pulling molecules together and reducing the measured volume below the ideal prediction.两条分子运动论假设失效：(1) 在高压下，分子被迫靠近，分子本身的体积相对于容器体积不再可忽略，实际体积大于理想预测值；(2) 在低温下，分子运动缓慢，分子间引力（范德华引力）变得显著，将分子吸引在一起，使测量体积低于理想预测值。
Molar volume at STP is a powerful shortcut: $22.4\ \text{L mol}^{-1}$ links volume directly to moles without using $R$ and $T$ explicitly.标准状态下的摩尔体积是有力的捷径：$22.4\ \text{L mol}^{-1}$ 无需显式使用 $R$ 和 $T$ 即可将体积直接与摩尔数联系起来。 At STP, any ideal gas occupies $22.4\ \text{L}$ per mole. This makes stoichiometry involving gases at STP extremely fast: moles $= V (\text{L}) / 22.4$, or $V = n \times 22.4$. It only applies at STP ($0\ ^\circ\text{C}$, $101.325\ \text{kPa}$); at any other condition use $PV = nRT$. Note that some texts use SATP ($25\ ^\circ\text{C}$, $100\ \text{kPa}$) where the molar volume is $24.8\ \text{L mol}^{-1}$. The verification in (c) shows exactly where $22.4\ \text{L mol}^{-1}$ comes from: it is the numerical result of $RT/P$ evaluated at the STP definition.在标准状态下，任何理想气体每摩尔占 $22.4\ \text{L}$。这使涉及标准状态气体的化学计量计算极为快捷：摩尔数 $= V(\text{L}) / 22.4$，或 $V = n \times 22.4$。该值仅适用于标准状态（$0\ ^\circ\text{C}$，$101.325\ \text{kPa}$）；其他任何条件下须使用 $PV = nRT$。注意部分教材使用 SATP（$25\ ^\circ\text{C}$，$100\ \text{kPa}$），此时摩尔体积为 $24.8\ \text{L mol}^{-1}$。(c) 中的验证精确地说明了 $22.4\ \text{L mol}^{-1}$ 的来源：它是在标准状态定义下对 $RT/P$ 求数值的结果。
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]

$2\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{H}_2\text{O(g)}$; $44.8\ \text{L}$ of $\text{H}_2$ at STP.$2\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{H}_2\text{O(g)}$；标准状态下 $44.8\ \text{L}$ 的 $\text{H}_2$。

Answer:答案：  (a) $2.00\ \text{mol}\ \text{H}_2$  ·  (b) $22.4\ \text{L}\ \text{O}_2$ at STP  ·  (c) $36.0\ \text{g}\ \text{H}_2\text{O}$  ·  (d) $2:1$ mole ratio $=$ $2:1$ volume ratio by Avogadro's Law

(a) Moles of $\text{H}_2$ available可用 $\text{H}_2$ 的摩尔数 M1·A1

$$ n(\text{H}_2) \;=\; \frac{44.8\ \text{L}}{22.4\ \text{L mol}^{-1}} \;=\; 2.00\ \text{mol.} $$

(b) Volume of $\text{O}_2$ required at STP在标准状态下所需 $\text{O}_2$ 的体积 M1·A1·A1

From the balanced equation, the mole ratio is $\text{H}_2 : \text{O}_2 = 2 : 1$.由配平方程，摩尔比为 $\text{H}_2 : \text{O}_2 = 2 : 1$。 $$ n(\text{O}_2) \;=\; 2.00\ \text{mol} \times \frac{1\ \text{mol O}_2}{2\ \text{mol H}_2} \;=\; 1.00\ \text{mol O}_2. $$ $$ V(\text{O}_2) \;=\; 1.00\ \text{mol} \times 22.4\ \text{L mol}^{-1} \;=\; 22.4\ \text{L.} $$

(c) Mass of water vapour produced生成水蒸气的质量 M1·A1

Mole ratio $\text{H}_2 : \text{H}_2\text{O} = 2 : 2 = 1 : 1$, so $n(\text{H}_2\text{O}) = 2.00\ \text{mol}$.摩尔比 $\text{H}_2 : \text{H}_2\text{O} = 2 : 2 = 1 : 1$，故 $n(\text{H}_2\text{O}) = 2.00\ \text{mol}$。 $$ m(\text{H}_2\text{O}) \;=\; 2.00\ \text{mol} \times 18.02\ \text{g mol}^{-1} \;=\; 36.0\ \text{g.} $$

(d) Mole ratio and Avogadro's Law摩尔比与阿伏伽德罗定律 A1

The mole ratio $\text{H}_2 : \text{O}_2 = 2 : 1$. By Avogadro's Law, at constant temperature and pressure equal volumes of gases contain equal numbers of moles, so the volume ratio at constant $T$ and $P$ equals the mole ratio: $V(\text{H}_2) : V(\text{O}_2) = 2 : 1$.摩尔比 $\text{H}_2 : \text{O}_2 = 2 : 1$。根据阿伏伽德罗定律，在恒温恒压下等体积气体含等摩尔数，因此恒 $T$ 恒 $P$ 下体积比等于摩尔比：$V(\text{H}_2) : V(\text{O}_2) = 2 : 1$。
Gas stoichiometry at STP: moles $\leftrightarrow$ volume via $22.4\ \text{L mol}^{-1}$; moles $\leftrightarrow$ mass via molar mass.标准状态下的气体化学计量：摩尔数通过 $22.4\ \text{L mol}^{-1}$ 与体积互换；摩尔数通过摩尔质量与质量互换。 The typical chain is: volume (at STP) $\to$ moles (divide by 22.4) $\to$ moles of product (mole ratio) $\to$ mass (multiply by molar mass). For gas-to-gas ratios at STP (or any constant $T$, $P$ condition), volume ratios equal mole ratios directly by Avogadro's Law, saving an intermediate step. The 2:1:2 ratio in this equation is also the volume ratio for $\text{H}_2:\text{O}_2:\text{H}_2\text{O}$ gas at constant $T$ and $P$. This is how Gay-Lussac's Law of Combining Volumes was originally observed experimentally.典型计算链：体积（标准状态）$\to$ 摩尔数（除以 22.4）$\to$ 产物摩尔数（摩尔比）$\to$ 质量（乘以摩尔质量）。对于标准状态（或任何恒 $T$、恒 $P$）下的气体对气体比，由阿伏伽德罗定律体积比直接等于摩尔比，省去中间步骤。该方程中 2:1:2 的比例也是恒 $T$ 恒 $P$ 下 $\text{H}_2:\text{O}_2:\text{H}_2\text{O}$ 气体的体积比。这正是盖-吕萨克气体化合体积定律最初通过实验观察到的方式。
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]

$10.0\ \text{L}$ rigid container at $298\ \text{K}$: He at $80.0\ \text{kPa}$ partial pressure, $\text{N}_2$ at $40.0\ \text{kPa}$ partial pressure.$10.0\ \text{L}$ 刚性容器，$298\ \text{K}$：氦气分压 $80.0\ \text{kPa}$，氮气分压 $40.0\ \text{kPa}$。

Answer:答案：  (a) $P_{\text{total}} = 120.0\ \text{kPa}$  ·  (b) $n(\text{He}) \approx 0.323\ \text{mol}$  ·  (c) $n(\text{N}_2) \approx 0.161\ \text{mol}$  ·  (d) $\chi_{\text{He}} = 0.667 = P_{\text{He}}/P_{\text{total}}$ ✓

(a) Dalton's Law and total pressure道尔顿分压定律与总压强 A1·A1

Dalton's Law of Partial Pressures: the total pressure of a mixture of non-reacting ideal gases equals the sum of the partial pressures of each component gas:道尔顿分压定律：不反应的理想气体混合物的总压强等于各组分气体分压之和： $$ P_{\text{total}} \;=\; P_{\text{He}} + P_{\text{N}_2} \;=\; 80.0 + 40.0 \;=\; 120.0\ \text{kPa.} $$

(b) Moles of He using $PV = nRT$用 $PV = nRT$ 计算氦气摩尔数 M1·A1·A1

Each gas behaves independently; use the partial pressure of He:每种气体独立行为；使用氦气的分压： $$ n(\text{He}) \;=\; \frac{P_{\text{He}} V}{RT} \;=\; \frac{80.0 \times 10.0}{8.314 \times 298} \;=\; \frac{800.0}{2477.6} \;=\; 0.3229\ \text{mol} \;\approx\; 0.323\ \text{mol.} $$

(c) Moles of $\text{N}_2$氮气摩尔数 M1·A1

$$ n(\text{N}_2) \;=\; \frac{P_{\text{N}_2} V}{RT} \;=\; \frac{40.0 \times 10.0}{8.314 \times 298} \;=\; \frac{400.0}{2477.6} \;=\; 0.1615\ \text{mol} \;\approx\; 0.161\ \text{mol.} $$

(d) Mole fraction of He and equivalence to pressure fraction氦气摩尔分数及其与压强分数的等价关系 M1·A1·A1

$$ n_{\text{total}} \;=\; 0.3229 + 0.1615 \;=\; 0.4844\ \text{mol.} $$ $$ \chi_{\text{He}} \;=\; \frac{n(\text{He})}{n_{\text{total}}} \;=\; \frac{0.3229}{0.4844} \;=\; 0.6667 \;\approx\; 0.667. $$ Pressure fraction:压强分数： $$ \frac{P_{\text{He}}}{P_{\text{total}}} \;=\; \frac{80.0}{120.0} \;=\; 0.6667. \quad \therefore\; \chi_{\text{He}} \;=\; \frac{P_{\text{He}}}{P_{\text{total}}}. \quad \checkmark $$
Mole fraction equals partial-pressure fraction for ideal gas mixtures: $\chi_i = P_i / P_{\text{total}}$.对于理想气体混合物，摩尔分数等于分压分数：$\chi_i = P_i / P_{\text{total}}$。 This equality follows directly from $PV = nRT$. For each component: $P_i V = n_i RT$, and for the total: $P_{\text{total}} V = n_{\text{total}} RT$. Dividing gives $P_i / P_{\text{total}} = n_i / n_{\text{total}} = \chi_i$. Dalton's Law is thus a consequence of ideal gas behaviour: each gas occupies the full volume and exerts its pressure independently of the others. In real gas mixtures, slight deviations occur because molecules of different species can interact with each other, but Dalton's Law is an excellent approximation at low pressures. Note that $n(\text{N}_2)$ is exactly half $n(\text{He})$ because $P_{\text{N}_2}$ is exactly half $P_{\text{He}}$, with $V$, $R$, $T$ identical for both.这一等式直接源于 $PV = nRT$。对每种组分：$P_i V = n_i RT$；对总体：$P_{\text{total}} V = n_{\text{total}} RT$。相除得 $P_i / P_{\text{total}} = n_i / n_{\text{total}} = \chi_i$。因此道尔顿定律是理想气体行为的推论：每种气体占据全部体积，独立于其他气体施加压强。在真实气体混合物中，由于不同种类分子之间可以相互作用，会有轻微偏差，但在低压下道尔顿定律是极好的近似。注意 $n(\text{N}_2)$ 恰好是 $n(\text{He})$ 的一半，因为 $P_{\text{N}_2}$ 恰好是 $P_{\text{He}}$ 的一半，而 $V$、$R$、$T$ 对两者相同。
PART III  ·  MODELING / APPLIED · SOLUTIONS第三部分  ·  建模与应用 · 详解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

Q10MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §3 Gay-Lussac's Law (applied)盖-吕萨克定律（应用） · Chemistry 20 GO1 [8 marks][8 分]

Car tyre: $180\ \text{kPa}$ at $20.0\ ^\circ\text{C}$; after highway driving temperature rises to $50.0\ ^\circ\text{C}$; tyre volume constant.汽车轮胎：$20.0\ ^\circ\text{C}$ 时 $180\ \text{kPa}$；高速行驶后温度升至 $50.0\ ^\circ\text{C}$；轮胎体积恒定。

Answer:答案：  (a) Gay-Lussac's Law, $P_1/T_1 = P_2/T_2$盖-吕萨克定律，$P_1/T_1 = P_2/T_2$  ·  (b) $293\ \text{K}$, $323\ \text{K}$  ·  (c) $P_2 \approx 198\ \text{kPa}$  ·  (d) higher $T$ means faster molecules colliding harder and more often with the tyre walls温度升高意味着分子运动更快，与轮胎内壁碰撞更有力、更频繁

(a) Gas law identification and formula确定气体定律及其公式 A1·A1

Gay-Lussac's Law: at constant volume and fixed amount of gas, pressure is directly proportional to absolute temperature:盖-吕萨克定律：在体积和气体量固定时，压强与绝对温度成正比： $$ \frac{P_1}{T_1} \;=\; \frac{P_2}{T_2} \qquad (V,\ n \text{ constant}). $$

(b) Temperature conversions to kelvin将温度换算为开尔文 A1

$$ T_1 \;=\; 20.0 + 273.15 \;=\; 293.15\ \text{K} \;\approx\; 293\ \text{K}, \qquad T_2 \;=\; 50.0 + 273.15 \;=\; 323.15\ \text{K} \;\approx\; 323\ \text{K}. $$

(c) New tyre pressure after driving行驶后的新胎压 M1·A1·A1

$$ P_2 \;=\; P_1 \cdot \frac{T_2}{T_1} \;=\; 180 \times \frac{323.15}{293.15} \;=\; 180 \times 1.1023 \;=\; 198.4\ \text{kPa} \;\approx\; 198\ \text{kPa.} $$

(d) KMT explanation for pressure increase at constant volume从分子运动论解释恒容时压强升高 A1·A1

At higher temperature, the average kinetic energy of the air molecules increases. The molecules move faster and collide with the inner tyre walls both more frequently and with greater force per collision. Since the tyre volume is essentially constant, this increase in collision rate and force per collision results in a higher measured pressure. No molecules escape; the same number of molecules simply transfer more momentum to the walls per unit time.温度升高时，空气分子的平均动能增加。分子运动更快，与轮胎内壁的碰撞既更频繁，每次碰撞的力也更大。由于轮胎体积基本不变，碰撞频率和每次碰撞力的增加导致测量压强升高。没有分子逸出；相同数量的分子仅仅是每单位时间向器壁传递更多动量。
A $30\ ^\circ\text{C}$ temperature rise produces only a $\sim\!10\%$ pressure increase because the Kelvin values are large relative to the change.温度升高 $30\ ^\circ\text{C}$ 只使压强升高约 $10\%$，因为开尔文值相对于变化量较大。 The ratio $T_2/T_1 = 323/293 = 1.102$ is modest because the reference temperature (293 K) is large relative to the change (30 K). Had the temperature been in Celsius, the ratio $50/20 = 2.5$ would give a wildly wrong answer. This is the reason all gas law calculations require absolute temperature: percentage changes in $T$ (K) equal percentage changes in $P$ (at constant $V$), and those percentage changes must be computed on the kelvin scale. Tyre manufacturers account for this in their recommended inflation pressures, often specifying "cold inflation" values.比值 $T_2/T_1 = 323/293 = 1.102$ 较小，因为参考温度（293 K）相对于变化量（30 K）较大。如果温度用摄氏度，比值 $50/20 = 2.5$ 会给出严重错误的答案。这正是所有气体定律计算都需要绝对温度的原因：$T$（K）的百分比变化等于恒 $V$ 时 $P$ 的百分比变化，这些百分比变化必须在开尔文标度上计算。轮胎制造商在推荐充气压强时考虑了这一效应，通常规定"冷充气"值。
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]

Industrial cylinder: $5.00\ \text{mol}$ $\text{N}_2$ at $20.0\ ^\circ\text{C}$, internal volume $50.0\ \text{L}$; safety valve opens above $300\ \text{kPa}$.工业气罐：$20.0\ ^\circ\text{C}$ 下 $5.00\ \text{mol}$ 氮气，内部体积 $50.0\ \text{L}$；安全阀在压强超过 $300\ \text{kPa}$ 时开启。

Answer:答案：  (a) $P \approx 244\ \text{kPa}$  ·  (b) $T \approx 361\ \text{K}\ (88\ ^\circ\text{C})$  ·  (c) $3.01 \times 10^{24}$ molecules个分子

(a) Pressure inside the cylinder气罐内的压强 M1·A1·A1

$T = 20.0 + 273.15 = 293.15\ \text{K}$. Using $PV = nRT$:$T = 20.0 + 273.15 = 293.15\ \text{K}$。使用 $PV = nRT$： $$ P \;=\; \frac{nRT}{V} \;=\; \frac{5.00 \times 8.314 \times 293.15}{50.0} \;=\; \frac{12{,}189}{50.0} \;=\; 243.8\ \text{kPa} \;\approx\; 244\ \text{kPa.} $$

(b) Temperature at which safety valve opens ($P = 300\ \text{kPa}$)安全阀开启时的温度（$P = 300\ \text{kPa}$） M1·A1·A1

Volume and moles are constant (rigid cylinder). Using Gay-Lussac's Law:体积和摩尔数恒定（刚性气罐）。使用盖-吕萨克定律： $$ \frac{P_1}{T_1} \;=\; \frac{P_2}{T_2} \quad \Longrightarrow \quad T_2 \;=\; T_1 \cdot \frac{P_2}{P_1} \;=\; 293.15 \times \frac{300}{243.8} \;=\; 293.15 \times 1.2305 \;=\; 360.9\ \text{K.} $$ $$ T_2(^\circ\text{C}) \;=\; 360.9 - 273.15 \;=\; 87.7\ ^\circ\text{C} \;\approx\; 88\ ^\circ\text{C.} $$

(c) Number of nitrogen molecules氮气分子数目 M1·A1

$$ N \;=\; n \times N_A \;=\; 5.00\ \text{mol} \times 6.022 \times 10^{23}\ \text{mol}^{-1} \;=\; 3.011 \times 10^{24}\ \text{molecules.} $$
$PV = nRT$ solves for any one of the four variables when the other three are known.$PV = nRT$ 可在已知其他三个变量时求解任意一个变量。 Part (a) solves for $P$; part (b) solves for $T$ using the rigid-container shortcut $P_2/P_1 = T_2/T_1$. The safety valve temperature ($\approx 88\ ^\circ\text{C}$) is only about $68\ ^\circ\text{C}$ above the storage temperature ($20\ ^\circ\text{C}$), illustrating that industrial gas storage must be kept well away from heat sources. For part (c), Avogadro's number $N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$ converts between the chemist's macroscopic unit (mol) and the actual count of particles. The result $3 \times 10^{24}$ underscores the enormous number of molecules in even a modest gas sample.(a) 求解 $P$；(b) 利用刚性容器捷径 $P_2/P_1 = T_2/T_1$ 求解 $T$。安全阀温度（约 $88\ ^\circ\text{C}$）仅比储存温度（$20\ ^\circ\text{C}$）高约 $68\ ^\circ\text{C}$，说明工业气体储存必须远离热源。对于 (c)，阿伏伽德罗数 $N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$ 在化学家的宏观单位（mol）和实际粒子数之间转换。结果 $3 \times 10^{24}$ 再次强调了即使是适量气体样品中分子数量之庞大。
Q12HARD难 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §7 Gas stoichiometry + Dalton's Law气体化学计量 + 分压定律 · HS-PS1-7 [7 marks][7 分]

$\text{CaCO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{CaCl}_2\text{(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}$; $25.0\ \text{g}$ $\text{CaCO}_3$ ($M = 100.09\ \text{g mol}^{-1}$); $\text{CO}_2$ collected over water at $25.0\ ^\circ\text{C}$; $P_{\text{total}} = 101.3\ \text{kPa}$; $P_{\text{H}_2\text{O}} = 3.17\ \text{kPa}$.$\text{CaCO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{CaCl}_2\text{(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}$；$25.0\ \text{g}$ $\text{CaCO}_3$（$M = 100.09\ \text{g mol}^{-1}$）；$\text{CO}_2$ 在 $25.0\ ^\circ\text{C}$ 下排水集气；$P_{\text{总}} = 101.3\ \text{kPa}$；$P_{\text{H}_2\text{O}} = 3.17\ \text{kPa}$。

Answer:答案：  (a) $0.250\ \text{mol}\ \text{CO}_2$  ·  (b) $P_{\text{CO}_2} = 98.1\ \text{kPa}$  ·  (c) $V \approx 6.31\ \text{L}$  ·  (d) higher $T$ and lower $P$ at collection vs STP both increase volume收集时温度较高、压强较低，两者均使体积大于标准状态体积

(a) Moles of $\text{CO}_2$ produced生成 $\text{CO}_2$ 的摩尔数 M1·A1

The mole ratio $\text{CaCO}_3 : \text{CO}_2 = 1 : 1$.摩尔比 $\text{CaCO}_3 : \text{CO}_2 = 1 : 1$。 $$ n(\text{CO}_2) \;=\; \frac{25.0\ \text{g}}{100.09\ \text{g mol}^{-1}} \;=\; 0.2498\ \text{mol} \;\approx\; 0.250\ \text{mol.} $$

(b) Partial pressure of $\text{CO}_2$ by Dalton's Law由分压定律求 $\text{CO}_2$ 的分压 A1

$$ P_{\text{CO}_2} \;=\; P_{\text{total}} - P_{\text{H}_2\text{O}} \;=\; 101.3 - 3.17 \;=\; 98.1\ \text{kPa.} $$

(c) Volume of collected $\text{CO}_2$ using $PV = nRT$用 $PV = nRT$ 计算收集到的 $\text{CO}_2$ 体积 M1·A1·A1

$T = 25.0 + 273.15 = 298.15\ \text{K}$; use $P = P_{\text{CO}_2} = 98.1\ \text{kPa}$ and $n = 0.2498\ \text{mol}$:$T = 25.0 + 273.15 = 298.15\ \text{K}$；使用 $P = P_{\text{CO}_2} = 98.1\ \text{kPa}$，$n = 0.2498\ \text{mol}$： $$ V \;=\; \frac{nRT}{P} \;=\; \frac{0.2498 \times 8.314 \times 298.15}{98.1} \;=\; \frac{619.4}{98.1} \;=\; 6.314\ \text{L} \;\approx\; 6.31\ \text{L.} $$

(d) Why collected volume differs from volume at STP收集体积与标准状态体积不同的原因 A1

At STP ($0\ ^\circ\text{C}$, $101.325\ \text{kPa}$), $0.250\ \text{mol}$ $\text{CO}_2$ would occupy $0.250 \times 22.4 = 5.60\ \text{L}$. The collected volume ($6.31\ \text{L}$) is larger for two compounding reasons: (1) the collection temperature ($25\ ^\circ\text{C} = 298\ \text{K}$) is higher than STP ($273\ \text{K}$), causing expansion; (2) the partial pressure of $\text{CO}_2$ at collection ($98.1\ \text{kPa}$) is lower than the STP pressure ($101.325\ \text{kPa}$), also causing additional expansion.在标准状态（$0\ ^\circ\text{C}$，$101.325\ \text{kPa}$）下，$0.250\ \text{mol}$ $\text{CO}_2$ 的体积为 $0.250 \times 22.4 = 5.60\ \text{L}$。收集体积（$6.31\ \text{L}$）更大，原因有两个叠加：(1) 收集温度（$25\ ^\circ\text{C} = 298\ \text{K}$）高于标准状态（$273\ \text{K}$），气体膨胀；(2) 收集时 $\text{CO}_2$ 的分压（$98.1\ \text{kPa}$）低于标准状态压强（$101.325\ \text{kPa}$），也导致额外膨胀。
Collecting gas over water introduces water vapour: always subtract $P_{\text{H}_2\text{O}}$ from $P_{\text{total}}$ before using $PV = nRT$.排水集气时引入水蒸气：在使用 $PV = nRT$ 前务必从总压中减去 $P_{\text{H}_2\text{O}}$。 The collection flask contains a mixture of $\text{CO}_2$ and water vapour. Dalton's Law separates them: $P_{\text{CO}_2} = P_{\text{total}} - P_{\text{H}_2\text{O}}$. The vapour pressure of water at a given temperature is a fixed physical property found in data tables. At $25\ ^\circ\text{C}$ it is $3.17\ \text{kPa}$; at $0\ ^\circ\text{C}$ it is only $0.61\ \text{kPa}$, which is why the effect is more noticeable at room temperature. Once $P_{\text{CO}_2}$ is known, use $PV = nRT$ with the actual $n$ (from stoichiometry) and the actual collection temperature to find the collected volume. This is a standard AP Chemistry lab calculation.集气瓶中含有 $\text{CO}_2$ 和水蒸气的混合物。分压定律将二者分离：$P_{\text{CO}_2} = P_{\text{总}} - P_{\text{H}_2\text{O}}$。给定温度下水的饱和蒸气压是固定的物理性质，可查数据表。$25\ ^\circ\text{C}$ 时为 $3.17\ \text{kPa}$；$0\ ^\circ\text{C}$ 时仅为 $0.61\ \text{kPa}$，这就是为何该效应在室温下更为明显。一旦知道 $P_{\text{CO}_2}$，使用实际的 $n$（来自化学计量）和实际收集温度代入 $PV = nRT$ 求收集体积。这是 AP 化学实验报告中的标准计算。