Companion to the IB-Style Practice SetIB 风格练习题的解析配套
Syllabus B3.1 to B3.6考纲 B3.1 至 B3.6PHYSICS HL
$88\ \mathrm{g}$ of $\mathrm{CO_2}$, $M = 44\ \mathrm{g\,mol^{-1}}$. (a) amount of substance; (b) number of molecules.$88\ \mathrm{g}$ 的 $\mathrm{CO_2}$，$M = 44\ \mathrm{g\,mol^{-1}}$。(a) 物质的量；(b) 分子数。
The amount of substance is mass divided by molar mass: $n = m/M$. (M1)
$$ n = \frac{88}{44} = 2.0\ \mathrm{mol}. $$(A1)
Multiply the amount of substance by the Avogadro constant: $N = n N_A$. (M1)
$$ N = (2.0)(6.02\times 10^{23}) = 1.2\times 10^{24}\ \text{molecules}. $$(A1)
物质的量等于质量除以摩尔质量：$n = m/M$。(M1)
$$ n = \frac{88}{44} = 2.0\ \mathrm{mol}. $$(A1)
把物质的量乘以阿伏伽德罗常数：$N = n N_A$。(M1)
$$ N = (2.0)(6.02\times 10^{23}) = 1.2\times 10^{24}\ \text{个分子}. $$(A1)
Fixed mass at $1.0\times 10^{5}\ \mathrm{Pa}$, $0.024\ \mathrm{m^{3}}$, compressed isothermally to $0.0060\ \mathrm{m^{3}}$. (a) which law; (b) new pressure; (c) internal energy.一定质量气体在 $1.0\times 10^{5}\ \mathrm{Pa}$、$0.024\ \mathrm{m^{3}}$，等温压缩到 $0.0060\ \mathrm{m^{3}}$。(a) 何定律；(b) 新压强；(c) 内能。
The change is isothermal (constant temperature) for a fixed mass of gas, so Boyle's law applies: $pV = \text{constant}$. (B1)
Apply $p_1 V_1 = p_2 V_2$, so $p_2 = p_1 V_1 / V_2$. (M1)
$$ p_2 = \frac{(1.0\times 10^{5})(0.024)}{0.0060} = 4.0\times 10^{5}\ \mathrm{Pa}. $$(A1)
For an ideal gas the internal energy depends only on temperature. The change is isothermal, so the internal energy is unchanged. (B1)
该变化对固定质量气体为等温（温度恒定），故适用玻意耳定律：$pV = \text{常数}$。(B1)
用 $p_1 V_1 = p_2 V_2$，故 $p_2 = p_1 V_1 / V_2$。(M1)
$$ p_2 = \frac{(1.0\times 10^{5})(0.024)}{0.0060} = 4.0\times 10^{5}\ \mathrm{Pa}. $$(A1)
理想气体的内能只依赖于温度。该变化为等温，故内能不变。(B1)
Rigid container of gas at $1.5\times 10^{5}\ \mathrm{Pa}$, $300\ \mathrm{K}$. (a) cooled to $250\ \mathrm{K}$, find new pressure; (b) separately, $0.50\ \mathrm{mol}$ ideal gas at $1.0\times 10^{5}\ \mathrm{Pa}$, $300\ \mathrm{K}$, find volume.刚性容器气体在 $1.5\times 10^{5}\ \mathrm{Pa}$、$300\ \mathrm{K}$。(a) 冷却到 $250\ \mathrm{K}$，求新压强；(b) 另取 $0.50\ \mathrm{mol}$ 理想气体于 $1.0\times 10^{5}\ \mathrm{Pa}$、$300\ \mathrm{K}$，求体积。
A rigid container fixes the volume, so Gay-Lussac's law applies: $p/T = \text{constant}$, i.e. $p_1/T_1 = p_2/T_2$. (M1)
Rearrange for $p_2 = p_1 (T_2/T_1)$: (M1)
$$ p_2 = (1.5\times 10^{5})\frac{250}{300} = 1.25\times 10^{5}\ \mathrm{Pa}. $$(A1)
Use $pV = nRT$ rearranged to $V = nRT/p$. The temperature is already $300\ \mathrm{K}$. (M1)
$$ V = \frac{(0.50)(8.31)(300)}{1.0\times 10^{5}}. $$(M1 for substitution)
$$ V = \frac{1246.5}{1.0\times 10^{5}} = 0.0125\ \mathrm{m^{3}} \approx 0.012\ \mathrm{m^{3}}. $$(A1)
刚性容器使体积固定，故适用盖-吕萨克定律：$p/T = \text{常数}$，即 $p_1/T_1 = p_2/T_2$。(M1)
解出 $p_2 = p_1 (T_2/T_1)$：(M1)
$$ p_2 = (1.5\times 10^{5})\frac{250}{300} = 1.25\times 10^{5}\ \mathrm{Pa}. $$(A1)
用 $pV = nRT$ 改写为 $V = nRT/p$。温度已是 $300\ \mathrm{K}$。(M1)
$$ V = \frac{(0.50)(8.31)(300)}{1.0\times 10^{5}}. $$(代入得 M1)
$$ V = \frac{1246.5}{1.0\times 10^{5}} = 0.0125\ \mathrm{m^{3}} \approx 0.012\ \mathrm{m^{3}}. $$(A1)
$pV = \tfrac{1}{3} N m \overline{c^{2}}$. (a) two ideal-gas assumptions; (b) why a gas exerts pressure; (c) compute $p$ from $n_V = 2.7\times 10^{25}\ \mathrm{m^{-3}}$, $m = 4.7\times 10^{-26}\ \mathrm{kg}$, $\overline{c^{2}} = 2.6\times 10^{5}\ \mathrm{m^{2}\,s^{-2}}$; (d) effect of doubling $n_V$.$pV = \tfrac{1}{3} N m \overline{c^{2}}$。(a) 两条理想气体假设；(b) 气体为何产生压强；(c) 由 $n_V = 2.7\times 10^{25}\ \mathrm{m^{-3}}$、$m = 4.7\times 10^{-26}\ \mathrm{kg}$、$\overline{c^{2}} = 2.6\times 10^{5}\ \mathrm{m^{2}\,s^{-2}}$ 算 $p$；(d) $n_V$ 翻倍的影响。
Any two of: molecules are point particles whose own volume is negligible compared with the container; there are no intermolecular forces except during collisions; collisions are perfectly elastic; molecular motion is random; the time of a collision is negligible compared with the time between collisions. (A1·A1)
Molecules move randomly and collide with the container walls. Each elastic collision reverses the molecule's momentum component, so the wall exerts an impulse on the molecule and, by Newton's third law, the molecule exerts an equal force on the wall. (M1)
The pressure is the total rate of momentum transfer per unit area summed over the very large number of such collisions. (A1)
Use $p = \tfrac{1}{3} n_V m \overline{c^{2}}$. (M1)
$$ p = \tfrac{1}{3}(2.7\times 10^{25})(4.7\times 10^{-26})(2.6\times 10^{5}). $$(M1 for substitution)
$$ p \approx 1.1\times 10^{5}\ \mathrm{Pa}. $$(A1)
Since $p \propto n_V$ at constant $\overline{c^{2}}$ (constant temperature), doubling the number density doubles the pressure. (B1)
以下任取两条：分子为质点，其自身体积相对容器可忽略；除碰撞瞬间外无分子间作用力；碰撞完全弹性；分子运动随机；碰撞持续时间相对碰撞间隔可忽略。(A1·A1)
分子随机运动并与器壁碰撞。每次弹性碰撞使分子的动量分量反向，故器壁对分子施加冲量，由牛顿第三定律，分子对器壁施加等大的力。(M1)
压强即对大量这类碰撞求和后单位面积上动量传递的总速率。(A1)
用 $p = \tfrac{1}{3} n_V m \overline{c^{2}}$。(M1)
$$ p = \tfrac{1}{3}(2.7\times 10^{25})(4.7\times 10^{-26})(2.6\times 10^{5}). $$(代入得 M1)
$$ p \approx 1.1\times 10^{5}\ \mathrm{Pa}. $$(A1)
在 $\overline{c^{2}}$ 不变（温度不变）时 $p \propto n_V$，故数密度翻倍则压强翻倍。(B1)
Monatomic ideal gas at $400\ \mathrm{K}$. (a) show $\bar{E}_k = \tfrac{3}{2} k_B T$; (b) mean KE per molecule; (c) internal energy of $3.0\ \mathrm{mol}$ and why it is purely kinetic.单原子理想气体于 $400\ \mathrm{K}$。(a) 证明 $\bar{E}_k = \tfrac{3}{2} k_B T$；(b) 每分子平均动能；(c) $3.0\ \mathrm{mol}$ 的内能及其为何纯为动能。
Kinetic theory gives $pV = \tfrac{1}{3} N m \overline{c^{2}}$; the ideal gas law gives $pV = N k_B T$. Equate the two expressions for $pV$: (M1)
$$ \tfrac{1}{3} N m \overline{c^{2}} = N k_B T \;\Rightarrow\; \tfrac{1}{3} m \overline{c^{2}} = k_B T. $$Multiply both sides by $\tfrac{3}{2}$ so the left side becomes the mean translational kinetic energy $\tfrac{1}{2} m \overline{c^{2}}$: (M1)
$$ \bar{E}_k = \tfrac{1}{2} m \overline{c^{2}} = \tfrac{3}{2} k_B T. $$(A1)
Substitute $T = 400\ \mathrm{K}$ into $\bar{E}_k = \tfrac{3}{2} k_B T$: (M1)
$$ \bar{E}_k = \tfrac{3}{2}(1.38\times 10^{-23})(400) \approx 8.3\times 10^{-21}\ \mathrm{J}. $$(A1)
For a monatomic ideal gas $U = \tfrac{3}{2} n R T$: (M1)
$$ U = \tfrac{3}{2}(3.0)(8.31)(400) \approx 1.5\times 10^{4}\ \mathrm{J}. $$(A1)
A monatomic gas has no rotational or vibrational modes and the ideal model has no intermolecular potential energy, so the internal energy is purely the sum of the molecules' translational kinetic energies. (B1)
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 $pV$ 的两式相等：(M1)
$$ \tfrac{1}{3} N m \overline{c^{2}} = N k_B T \;\Rightarrow\; \tfrac{1}{3} m \overline{c^{2}} = k_B T. $$两边乘以 $\tfrac{3}{2}$，使左边化为平均平动动能 $\tfrac{1}{2} m \overline{c^{2}}$：(M1)
$$ \bar{E}_k = \tfrac{1}{2} m \overline{c^{2}} = \tfrac{3}{2} k_B T. $$(A1)
把 $T = 400\ \mathrm{K}$ 代入 $\bar{E}_k = \tfrac{3}{2} k_B T$：(M1)
$$ \bar{E}_k = \tfrac{3}{2}(1.38\times 10^{-23})(400) \approx 8.3\times 10^{-21}\ \mathrm{J}. $$(A1)
单原子理想气体 $U = \tfrac{3}{2} n R T$：(M1)
$$ U = \tfrac{3}{2}(3.0)(8.31)(400) \approx 1.5\times 10^{4}\ \mathrm{J}. $$(A1)
单原子气体没有转动或振动自由度，且理想模型无分子间势能，故内能纯为各分子平动动能之和。(B1)
Constant-$T$ compression; $p$ vs $1/V$ data given. (a) show $p$ vs $1/V$ is linear through the origin and state the gradient; (b) gradient and the value of $pV$; (c) percentage uncertainty in the $2.40\times 10^{5}\ \mathrm{Pa}$ reading ($\pm 0.05\times 10^{5}$); (d) why points curve at high pressure.恒温压缩；给出 $p$ 对 $1/V$ 的数据。(a) 证明 $p$ 对 $1/V$ 为过原点直线并说明斜率；(b) 斜率与 $pV$ 值；(c) $2.40\times 10^{5}\ \mathrm{Pa}$ 读数（$\pm 0.05\times 10^{5}$）的百分比不确定度；(d) 高压下点为何弯曲。
Boyle's law at constant temperature and fixed mass: $pV = \text{constant}$. Rearrange to make $p$ the subject: (M1)
$$ p = (pV)\cdot\frac{1}{V}. $$This has the form $p = (\text{gradient})\times (1/V)$ with no intercept, so a plot of $p$ against $1/V$ is a straight line through the origin. (A1)
Comparing with $y = mx$, the gradient equals the constant $pV$. (A1)
Read the gradient from two well-separated points, e.g. $(25,\,0.60\times 10^{5})$ and $(100,\,2.40\times 10^{5})$: (M1)
$$ \text{gradient} = \frac{(2.40 - 0.60)\times 10^{5}}{100 - 25} = \frac{1.80\times 10^{5}}{75} = 2.4\times 10^{3}\ \mathrm{J}. $$(A1)
Since the gradient is the Boyle's-law constant, $pV = 2.4\times 10^{3}\ \mathrm{J}$. (A1)
Divide the absolute uncertainty by the reading: (M1)
$$ \frac{0.05\times 10^{5}}{2.40\times 10^{5}}\times 100\% \approx 2.1\% \approx 2\%. $$(A1)
At very high pressure the gas is no longer ideal: the finite volume of the molecules themselves becomes significant and intermolecular attractions are no longer negligible. (B1)
Boyle's law assumes ideal behaviour, so the real data deviate from the straight line as these assumptions fail. (R1)
恒温、固定质量下的玻意耳定律：$pV = \text{常数}$。改写为以 $p$ 为主体：(M1)
$$ p = (pV)\cdot\frac{1}{V}. $$此式形如 $p = (\text{斜率})\times (1/V)$，无截距，故 $p$ 对 $1/V$ 作图为过原点的直线。(A1)
与 $y = mx$ 比较，斜率等于常数 $pV$。(A1)
用相距较远的两点读斜率，如 $(25,\,0.60\times 10^{5})$ 与 $(100,\,2.40\times 10^{5})$：(M1)
$$ \text{斜率} = \frac{(2.40 - 0.60)\times 10^{5}}{100 - 25} = \frac{1.80\times 10^{5}}{75} = 2.4\times 10^{3}\ \mathrm{J}. $$(A1)
因斜率即玻意耳定律常数，故 $pV = 2.4\times 10^{3}\ \mathrm{J}$。(A1)
用绝对不确定度除以读数：(M1)
$$ \frac{0.05\times 10^{5}}{2.40\times 10^{5}}\times 100\% \approx 2.1\% \approx 2\%. $$(A1)
在极高压下气体不再理想：分子自身的有限体积变得显著，分子间吸引也不再可忽略。(B1)
玻意耳定律假设理想行为，故当这些假设失效时实验数据偏离直线。(R1)
Fixed amount of ideal gas; $pV$ (J) vs $T$ (K) data given. (a) show $pV$ vs $T$ is linear through the origin and state the gradient; (b) gradient from first and last points; (c) amount of substance $n$; (d) why through the origin and meaning of $pV$ at $T = 0$; (e) effect of recording $T$ in Celsius.固定量理想气体；给出 $pV$（J）对 $T$（K）的数据。(a) 证明 $pV$ 对 $T$ 为过原点直线并说明斜率；(b) 由首末点求斜率；(c) 物质的量 $n$；(d) 为何过原点及 $T = 0$ 处 $pV$ 的含义；(e) 用摄氏度记温的影响。
The ideal gas equation is $pV = nRT$. For a fixed amount of gas $n$ is constant, so $nR$ is a constant: (M1)
$$ pV = (nR)\,T. $$This has the form $pV = (\text{gradient})\times T$ with no intercept, so a plot of $pV$ against $T$ (in kelvin) is a straight line through the origin. (A1)
Comparing with $y = mx$, the gradient is $nR$. (A1)
Use the first and last data points $(200,\,830)$ and $(500,\,2080)$: (M1)
$$ \text{gradient} = \frac{2080 - 830}{500 - 200}. $$(M1 for substitution)
$$ \text{gradient} = \frac{1250}{300} \approx 4.2\ \mathrm{J\,K^{-1}}. $$(A1)
The gradient equals $nR$, so $n = \text{gradient}/R$: (M1)
$$ n = \frac{4.2}{8.31} \approx 0.50\ \mathrm{mol}. $$(A1)
Because $pV = nRT$ has no constant term, $pV$ must tend to zero as $T$ tends to $0\ \mathrm{K}$, which is why the line passes through the origin. (B1)
A value $pV = 0$ at $T = 0\ \mathrm{K}$ represents the molecules having zero average translational kinetic energy, so they exert no pressure: absolute zero. (R1)
Using Celsius shifts every temperature by $273$, so the line would no longer pass through the origin: it would cross the $pV$ axis at a positive intercept and the temperature axis near $-273\ ^\circ\mathrm{C}$. (A1)
The line stays straight with the same gradient $nR$, but the proportionality $pV \propto T$ holds only when $T$ is the absolute (kelvin) temperature. (R1)
理想气体方程为 $pV = nRT$。对固定气体量 $n$ 为常数，故 $nR$ 为常数：(M1)
$$ pV = (nR)\,T. $$此式形如 $pV = (\text{斜率})\times T$，无截距，故 $pV$ 对 $T$（开尔文）作图为过原点的直线。(A1)
与 $y = mx$ 比较，斜率为 $nR$。(A1)
用首末两点 $(200,\,830)$ 与 $(500,\,2080)$：(M1)
$$ \text{斜率} = \frac{2080 - 830}{500 - 200}. $$(代入得 M1)
$$ \text{斜率} = \frac{1250}{300} \approx 4.2\ \mathrm{J\,K^{-1}}. $$(A1)
斜率等于 $nR$，故 $n = \text{斜率}/R$：(M1)
$$ n = \frac{4.2}{8.31} \approx 0.50\ \mathrm{mol}. $$(A1)
因 $pV = nRT$ 无常数项，当 $T$ 趋于 $0\ \mathrm{K}$ 时 $pV$ 必趋于零，这正是直线过原点的原因。(B1)
$T = 0\ \mathrm{K}$ 处 $pV = 0$ 表示分子的平均平动动能为零，因而不产生压强：即绝对零度。(R1)
用摄氏度会把每个温度平移 $273$，故直线不再过原点：它将在 $pV$ 轴交于一个正截距，并在 $-273\ ^\circ\mathrm{C}$ 附近交温度轴。(A1)
直线仍为直线、斜率仍为 $nR$，但比例关系 $pV \propto T$ 仅当 $T$ 取绝对（开尔文）温度时成立。(R1)
Rigid cylinder $0.025\ \mathrm{m^{3}}$, monatomic ideal gas at $2.0\times 10^{5}\ \mathrm{Pa}$, $300\ \mathrm{K}$. (a) moles; (b) number of molecules; (c) new pressure heated to $500\ \mathrm{K}$ at constant $V$; (d) increase in internal energy $300 \to 500\ \mathrm{K}$; (e) change in mean KE per molecule.刚性气缸 $0.025\ \mathrm{m^{3}}$，单原子理想气体于 $2.0\times 10^{5}\ \mathrm{Pa}$、$300\ \mathrm{K}$。(a) 摩尔数；(b) 分子数；(c) 恒容加热到 $500\ \mathrm{K}$ 的新压强；(d) $300 \to 500\ \mathrm{K}$ 内能增量；(e) 每分子平均动能的变化。
Use $pV = nRT$, so $n = pV/(RT)$ with $T = 300\ \mathrm{K}$. (M1)
$$ n = \frac{(2.0\times 10^{5})(0.025)}{(8.31)(300)}. $$(M1 for substitution)
$$ n = \frac{5000}{2493} \approx 2.0\ \mathrm{mol}. $$(A1)
$N = n N_A$: (M1)
$$ N = (2.0)(6.02\times 10^{23}) \approx 1.2\times 10^{24}. $$(A1)
Constant volume, so $p \propto T$ (Gay-Lussac): $p_2 = p_1 (T_2/T_1)$. (M1)
$$ p_2 = (2.0\times 10^{5})\frac{500}{300} \approx 3.3\times 10^{5}\ \mathrm{Pa}. $$(A1)
For a monatomic ideal gas $U = \tfrac{3}{2} n R T$, so $\Delta U = \tfrac{3}{2} n R \,\Delta T$ with $\Delta T = 200\ \mathrm{K}$. (M1)
$$ \Delta U = \tfrac{3}{2}(2.0)(8.31)(200). $$(M1 for substitution)
$$ \Delta U \approx 5.0\times 10^{3}\ \mathrm{J}. $$(A1)
Since $\bar{E}_k = \tfrac{3}{2} k_B T$, the mean kinetic energy is proportional to absolute temperature, so it increases. (R1)
It changes by the factor $T_2/T_1 = 500/300 = 5/3 \approx 1.7$. (A1)
用 $pV = nRT$，故 $n = pV/(RT)$，其中 $T = 300\ \mathrm{K}$。(M1)
$$ n = \frac{(2.0\times 10^{5})(0.025)}{(8.31)(300)}. $$(代入得 M1)
$$ n = \frac{5000}{2493} \approx 2.0\ \mathrm{mol}. $$(A1)
$N = n N_A$：(M1)
$$ N = (2.0)(6.02\times 10^{23}) \approx 1.2\times 10^{24}. $$(A1)
体积恒定，故 $p \propto T$（盖-吕萨克）：$p_2 = p_1 (T_2/T_1)$。(M1)
$$ p_2 = (2.0\times 10^{5})\frac{500}{300} \approx 3.3\times 10^{5}\ \mathrm{Pa}. $$(A1)
单原子理想气体 $U = \tfrac{3}{2} n R T$，故 $\Delta U = \tfrac{3}{2} n R \,\Delta T$，其中 $\Delta T = 200\ \mathrm{K}$。(M1)
$$ \Delta U = \tfrac{3}{2}(2.0)(8.31)(200). $$(代入得 M1)
$$ \Delta U \approx 5.0\times 10^{3}\ \mathrm{J}. $$(A1)
因 $\bar{E}_k = \tfrac{3}{2} k_B T$，平均动能与绝对温度成正比，故增大。(R1)
它变为原来的 $T_2/T_1 = 500/300 = 5/3 \approx 1.7$ 倍。(A1)
Fixed mass: $1.0\times 10^{5}\ \mathrm{Pa}$, $0.020\ \mathrm{m^{3}}$, $300\ \mathrm{K}$ to $0.010\ \mathrm{m^{3}}$, $350\ \mathrm{K}$. (a) combined gas law and its condition; (b) final pressure; (c) four ideal-gas assumptions and the two that fail near condensation; (d) conditions for ideal behaviour and why.固定质量：$1.0\times 10^{5}\ \mathrm{Pa}$、$0.020\ \mathrm{m^{3}}$、$300\ \mathrm{K}$ 变到 $0.010\ \mathrm{m^{3}}$、$350\ \mathrm{K}$。(a) 合并气体定律及其条件；(b) 末压强；(c) 四条理想假设及凝结附近失效的两条；(d) 理想行为的条件及原因。
For a fixed amount of gas the quantity $pV/T$ is constant: (A1)
$$ \frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}. $$It applies only when the amount of gas $n$ does not change between the two states (the cylinder is sealed). (B1)
Rearrange for $p_2 = p_1 \dfrac{V_1}{V_2}\dfrac{T_2}{T_1}$, with all temperatures in kelvin. (M1)
$$ p_2 = (1.0\times 10^{5})\frac{0.020}{0.010}\cdot\frac{350}{300}. $$(M1 for substitution)
$$ p_2 = (1.0\times 10^{5})(2.0)(1.1\overline{6}) \approx 2.3\times 10^{5}\ \mathrm{Pa}. $$(A1)
An ideal gas assumes: molecules are point particles of negligible volume; there are no intermolecular forces except during collisions; collisions are perfectly elastic and motion is random; the collision time is negligible compared with the time between collisions. (A1·A1 for the four)
As the gas is strongly compressed and cooled towards condensation, the two that fail are the negligible-molecular-volume assumption (molecules are now crowded) and the no-intermolecular-forces assumption (slow molecules attract appreciably). (R1)
A real gas behaves most ideally at low pressure and high temperature. (A1)
Low pressure keeps the molecules far apart so their own volume is negligible, and high temperature keeps them fast so intermolecular attractions are negligible: both assumptions then hold. (R1)
对固定气体量，$pV/T$ 为常数：(A1)
$$ \frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}. $$它仅当两态间气体量 $n$ 不变时适用（气缸密封）。(B1)
解出 $p_2 = p_1 \dfrac{V_1}{V_2}\dfrac{T_2}{T_1}$，温度全用开尔文。(M1)
$$ p_2 = (1.0\times 10^{5})\frac{0.020}{0.010}\cdot\frac{350}{300}. $$(代入得 M1)
$$ p_2 = (1.0\times 10^{5})(2.0)(1.1\overline{6}) \approx 2.3\times 10^{5}\ \mathrm{Pa}. $$(A1)
理想气体假设：分子为体积可忽略的质点；除碰撞外无分子间作用力；碰撞完全弹性且运动随机；碰撞时间相对碰撞间隔可忽略。(四条得 A1·A1)
当气体被强烈压缩并冷却趋近凝结时，失效的两条是分子体积可忽略的假设（分子现已拥挤）与无分子间作用力的假设（慢分子间有明显吸引）。(R1)
真实气体在低压、高温下最接近理想。(A1)
低压使分子彼此远离故自身体积可忽略，高温使分子运动快故分子间吸引可忽略：两条假设便都成立。(R1)
Helium ($M = 4.0\ \mathrm{g\,mol^{-1}}$) at $300\ \mathrm{K}$, monatomic ideal. (a) mass of one atom; (b) mean translational KE; (c) rms speed from $\tfrac{1}{2}m\overline{c^{2}} = \tfrac{3}{2}k_BT$; (d) compare rms speed of argon ($M = 40$) at the same $T$.氦（$M = 4.0\ \mathrm{g\,mol^{-1}}$）于 $300\ \mathrm{K}$，单原子理想。(a) 单原子质量；(b) 平均平动动能；(c) 由 $\tfrac{1}{2}m\overline{c^{2}} = \tfrac{3}{2}k_BT$ 求方均根速率；(d) 比较同温下氩（$M = 40$）的方均根速率。
Convert the molar mass to $\mathrm{kg\,mol^{-1}}$ and divide by $N_A$: $m = M/N_A$. (M1)
$$ m = \frac{4.0\times 10^{-3}}{6.02\times 10^{23}} \approx 6.6\times 10^{-27}\ \mathrm{kg}. $$(A1)
Use $\bar{E}_k = \tfrac{3}{2} k_B T$: (M1)
$$ \bar{E}_k = \tfrac{3}{2}(1.38\times 10^{-23})(300) \approx 6.2\times 10^{-21}\ \mathrm{J}. $$(A1)
Set $\tfrac{1}{2} m \overline{c^{2}} = \bar{E}_k$, so $\overline{c^{2}} = 2\bar{E}_k/m$ and $c_{\mathrm{rms}} = \sqrt{\overline{c^{2}}}$. (M1)
$$ \overline{c^{2}} = \frac{2(6.21\times 10^{-21})}{6.64\times 10^{-27}} \approx 1.87\times 10^{6}\ \mathrm{m^{2}\,s^{-2}}, $$ $$ c_{\mathrm{rms}} = \sqrt{1.87\times 10^{6}} \approx 1.4\times 10^{3}\ \mathrm{m\,s^{-1}}. $$(A1)
At the same temperature both gases have the same mean translational kinetic energy $\tfrac{3}{2} k_B T$, so $\tfrac{1}{2} m \overline{c^{2}}$ is the same for both. (R1)
Therefore $c_{\mathrm{rms}} \propto 1/\sqrt{m}$. Argon is $10$ times heavier, so its rms speed is smaller by a factor of $\sqrt{10} \approx 3.2$. (A1)
把摩尔质量换成 $\mathrm{kg\,mol^{-1}}$ 再除以 $N_A$：$m = M/N_A$。(M1)
$$ m = \frac{4.0\times 10^{-3}}{6.02\times 10^{23}} \approx 6.6\times 10^{-27}\ \mathrm{kg}. $$(A1)
用 $\bar{E}_k = \tfrac{3}{2} k_B T$：(M1)
$$ \bar{E}_k = \tfrac{3}{2}(1.38\times 10^{-23})(300) \approx 6.2\times 10^{-21}\ \mathrm{J}. $$(A1)
令 $\tfrac{1}{2} m \overline{c^{2}} = \bar{E}_k$，故 $\overline{c^{2}} = 2\bar{E}_k/m$，$c_{\mathrm{rms}} = \sqrt{\overline{c^{2}}}$。(M1)
$$ \overline{c^{2}} = \frac{2(6.21\times 10^{-21})}{6.64\times 10^{-27}} \approx 1.87\times 10^{6}\ \mathrm{m^{2}\,s^{-2}}, $$ $$ c_{\mathrm{rms}} = \sqrt{1.87\times 10^{6}} \approx 1.4\times 10^{3}\ \mathrm{m\,s^{-1}}. $$(A1)
同温下两种气体平均平动动能相同，均为 $\tfrac{3}{2} k_B T$，故两者 $\tfrac{1}{2} m \overline{c^{2}}$ 相同。(R1)
因此 $c_{\mathrm{rms}} \propto 1/\sqrt{m}$。氩重 $10$ 倍，故其方均根速率小至 $1/\sqrt{10} \approx 1/3.2$。(A1)