Companion to the IB-Style Practice Set · HL onlyIB 风格练习题的解析配套 · 仅 HL
Syllabus B.4.1 to B.4.6考纲 B.4.1 至 B.4.6PHYSICS HL
A gas is given $360\ \mathrm{J}$ of heat while compressed, the surroundings doing $150\ \mathrm{J}$ of work on it. Using $Q = \Delta U + W$: (a) state $Q$ and $W$ with signs; (b) find $\Delta U$ and whether the gas warms.气体被压缩，外界对其做 $150\ \mathrm{J}$ 功，同时获得 $360\ \mathrm{J}$ 热量。用 $Q = \Delta U + W$：(a) 写出带符号的 $Q$、$W$；(b) 求 $\Delta U$ 并判断是否升温。
Heat is added to the gas, so $Q = +360\ \mathrm{J}$. (A1)
The gas is compressed, so the work done by the gas is negative: $W = -150\ \mathrm{J}$ (the $150\ \mathrm{J}$ is work done on the gas). (A1)
Rearrange the first law: $\Delta U = Q - W = 360 - (-150)$. (M1)
$$ \Delta U = 360 + 150 = +510\ \mathrm{J}. $$$\Delta U > 0$, so the internal energy rises and the gas warms. (A1)
气体吸热，故 $Q = +360\ \mathrm{J}$。(A1)
气体被压缩，故气体对外做的功为负：$W = -150\ \mathrm{J}$（这 $150\ \mathrm{J}$ 是外界对气体做的功）。(A1)
将第一定律变形：$\Delta U = Q - W = 360 - (-150)$。(M1)
$$ \Delta U = 360 + 150 = +510\ \mathrm{J}. $$$\Delta U > 0$，故内能升高、气体升温。(A1)
Gas at constant $1.8\times 10^{5}\ \mathrm{Pa}$ expands $2.0\times 10^{-3}\to 5.0\times 10^{-3}\ \mathrm{m^{3}}$; $900\ \mathrm{J}$ of heat supplied. (a) work done by the gas; (b) change in internal energy.气体在恒压 $1.8\times 10^{5}\ \mathrm{Pa}$ 下由 $2.0\times 10^{-3}\to 5.0\times 10^{-3}\ \mathrm{m^{3}}$ 膨胀；供给 $900\ \mathrm{J}$ 热量。(a) 气体做的功；(b) 内能变化。
Constant pressure, so use $W = p\,\Delta V$ with $\Delta V = (5.0 - 2.0)\times 10^{-3} = 3.0\times 10^{-3}\ \mathrm{m^{3}}$. (M1)
$$ W = (1.8\times 10^{5})(3.0\times 10^{-3}) = +540\ \mathrm{J}. $$Positive because the gas expands. (A1)
First law: $\Delta U = Q - W = 900 - 540$. (M1)
$$ \Delta U = +360\ \mathrm{J}. $$(A1)
恒压，故用 $W = p\,\Delta V$，$\Delta V = (5.0 - 2.0)\times 10^{-3} = 3.0\times 10^{-3}\ \mathrm{m^{3}}$。(M1)
$$ W = (1.8\times 10^{5})(3.0\times 10^{-3}) = +540\ \mathrm{J}. $$因气体膨胀而为正。(A1)
第一定律：$\Delta U = Q - W = 900 - 540$。(M1)
$$ \Delta U = +360\ \mathrm{J}. $$(A1)
Identify the process and reduce the first law, then find: (a) isothermal expansion absorbing $300\ \mathrm{J}$, work by gas; (b) rigid container gaining $300\ \mathrm{J}$, $\Delta U$; (c) insulated sudden compression, $250\ \mathrm{J}$ done on the gas, $\Delta U$ and warm/cool.判断过程并化简第一定律，然后求：(a) 等温膨胀吸 $300\ \mathrm{J}$，气体做功；(b) 刚性容器获 $300\ \mathrm{J}$，$\Delta U$；(c) 隔热突然压缩，外界做 $250\ \mathrm{J}$ 功，$\Delta U$ 及升降温。
Constant temperature for an ideal gas means $\Delta U = 0$, so the first law reduces to $Q = W$. (M1)
$$ W = Q = +300\ \mathrm{J}. $$(A1)
A rigid sealed container has $\Delta V = 0$, so $W = p\,\Delta V = 0$ and the first law reduces to $Q = \Delta U$. (M1)
$$ \Delta U = Q = +300\ \mathrm{J}. $$(A1)
Well insulated and sudden means $Q = 0$, so the first law reduces to $\Delta U = -W$. The surroundings do $250\ \mathrm{J}$ on the gas, so the work done by the gas is $W = -250\ \mathrm{J}$. (M1)
$$ \Delta U = -W = -(-250) = +250\ \mathrm{J}. $$$\Delta U > 0$, so the gas warms. (A1)
理想气体恒温意味着 $\Delta U = 0$，故第一定律化为 $Q = W$。(M1)
$$ W = Q = +300\ \mathrm{J}. $$(A1)
刚性密封容器 $\Delta V = 0$，故 $W = p\,\Delta V = 0$，第一定律化为 $Q = \Delta U$。(M1)
$$ \Delta U = Q = +300\ \mathrm{J}. $$(A1)
隔热良好且突然意味着 $Q = 0$，故第一定律化为 $\Delta U = -W$。外界对气体做 $250\ \mathrm{J}$ 功，故气体对外做的功为 $W = -250\ \mathrm{J}$。(M1)
$$ \Delta U = -W = -(-250) = +250\ \mathrm{J}. $$$\Delta U > 0$，故气体升温。(A1)
$6000\ \mathrm{J}$ flows hot ($600\ \mathrm{K}$) to cold ($300\ \mathrm{K}$), reservoirs unchanged. (a) entropy change of each; (b) total and whether spontaneous; (c) total if reversed and its implication.$6000\ \mathrm{J}$ 由热源（$600\ \mathrm{K}$）流向冷源（$300\ \mathrm{K}$），热源温度不变。(a) 各自熵变；(b) 总熵变及能否自发；(c) 若反向的总熵变及其含义。
The hot reservoir loses heat ($\Delta Q_{H} = -6000\ \mathrm{J}$), the cold gains it ($\Delta Q_{C} = +6000\ \mathrm{J}$); use $\Delta S = \Delta Q / T$. (M1)
$$ \Delta S_{H} = \frac{-6000}{600} = -10\ \mathrm{J\,K^{-1}}, \qquad \Delta S_{C} = \frac{+6000}{300} = +20\ \mathrm{J\,K^{-1}}. $$(A1 each)
$\Delta S_{\text{total}} = -10 + 20 = +10\ \mathrm{J\,K^{-1}}$. (A1)
The isolated two-reservoir system has $\Delta S_{\text{total}} > 0$, which satisfies the second law, so the flow can occur spontaneously. (R1)
Reversing every sign gives $\Delta S_{\text{total}} = +10 - 20 = -10\ \mathrm{J\,K^{-1}} < 0$, which the second law forbids: heat cannot flow spontaneously from cold to hot. (A1)
热源失热（$\Delta Q_{H} = -6000\ \mathrm{J}$），冷源得热（$\Delta Q_{C} = +6000\ \mathrm{J}$）；用 $\Delta S = \Delta Q / T$。(M1)
$$ \Delta S_{H} = \frac{-6000}{600} = -10\ \mathrm{J\,K^{-1}}, \qquad \Delta S_{C} = \frac{+6000}{300} = +20\ \mathrm{J\,K^{-1}}. $$(各 A1)
$\Delta S_{\text{total}} = -10 + 20 = +10\ \mathrm{J\,K^{-1}}$。(A1)
孤立的两热源系统 $\Delta S_{\text{total}} > 0$，符合第二定律，故该热流可以自发发生。(R1)
把每个符号反过来得 $\Delta S_{\text{total}} = +10 - 20 = -10\ \mathrm{J\,K^{-1}} < 0$，为第二定律所禁止：热量不能自发地由冷流向热。(A1)
Engine takes in $1500\ \mathrm{J}$, rejects $900\ \mathrm{J}$ per cycle. (a) work output and why $\Delta U = 0$ over a cycle; (b) efficiency, verified two ways.热机每循环吸 $1500\ \mathrm{J}$，排 $900\ \mathrm{J}$。(a) 输出功及为何循环 $\Delta U = 0$；(b) 效率，并两种方式验证。
Over a complete cycle the gas returns to its starting state, so $\Delta U = 0$ (internal energy is a state function). Energy conservation then gives $W = Q_{\mathrm{in}} - Q_{\mathrm{out}}$. (M1)
$$ W = 1500 - 900 = 600\ \mathrm{J}. $$(A1)
$\eta = \dfrac{W}{Q_{\mathrm{in}}} = \dfrac{600}{1500} = 0.40 = 40\%$. (M1)
Check with the rejected-heat form: $\eta = 1 - \dfrac{Q_{\mathrm{out}}}{Q_{\mathrm{in}}} = 1 - \dfrac{900}{1500} = 1 - 0.60 = 0.40$. Consistent. (A1)
一个完整循环后气体回到初态，故 $\Delta U = 0$（内能是态函数）。由能量守恒得 $W = Q_{\mathrm{in}} - Q_{\mathrm{out}}$。(M1)
$$ W = 1500 - 900 = 600\ \mathrm{J}. $$(A1)
$\eta = \dfrac{W}{Q_{\mathrm{in}}} = \dfrac{600}{1500} = 0.40 = 40\%$。(M1)
用排热式验证：$\eta = 1 - \dfrac{Q_{\mathrm{out}}}{Q_{\mathrm{in}}} = 1 - \dfrac{900}{1500} = 1 - 0.60 = 0.40$，一致。(A1)
Engine between $500\ \mathrm{K}$ and $300\ \mathrm{K}$. (a) maximum (Carnot) efficiency; (b) is a claimed $45\%$ possible; (c) one temperature change that raises the ceiling, with reason.热机在 $500\ \mathrm{K}$ 与 $300\ \mathrm{K}$ 之间。(a) 最大（卡诺）效率；(b) 声称的 $45\%$ 是否可能；(c) 一种提高上限的温度改动及理由。
Temperatures already in kelvin; use $\eta_{\text{Carnot}} = 1 - \dfrac{T_{\mathrm{cold}}}{T_{\mathrm{hot}}}$. (M1)
$$ \eta_{\text{Carnot}} = 1 - \frac{300}{500} = 1 - 0.60 = 0.40 = 40\%. $$(A1)
No. The Carnot value $40\%$ is the absolute ceiling for any engine between these reservoirs, and a real engine must fall below it. (A1)
A claimed $45\% > 40\%$ would violate the second law, so the claim is impossible. (R1)
Increase $T_{\mathrm{hot}}$ or decrease $T_{\mathrm{cold}}$. (A1)
Either change makes the ratio $T_{\mathrm{cold}}/T_{\mathrm{hot}}$ smaller, so $1 - T_{\mathrm{cold}}/T_{\mathrm{hot}}$ is larger: a bigger temperature gap raises the Carnot limit. (R1)
温度已是开尔文；用 $\eta_{\text{Carnot}} = 1 - \dfrac{T_{\mathrm{cold}}}{T_{\mathrm{hot}}}$。(M1)
$$ \eta_{\text{Carnot}} = 1 - \frac{300}{500} = 1 - 0.60 = 0.40 = 40\%. $$(A1)
不可能。卡诺值 $40\%$ 是这两热源间任何热机的绝对上限，真实热机必低于它。(A1)
声称的 $45\% > 40\%$ 会违反第二定律，故该说法不可能。(R1)
升高 $T_{\mathrm{hot}}$ 或降低 $T_{\mathrm{cold}}$。(A1)
任一改动都使比值 $T_{\mathrm{cold}}/T_{\mathrm{hot}}$ 变小，故 $1 - T_{\mathrm{cold}}/T_{\mathrm{hot}}$ 变大：温差越大，卡诺极限越高。(R1)
Monatomic gas: stage 1 (X→Y) $V\!: 1.0\to 4.0\times 10^{-3}\ \mathrm{m^{3}}$, $p$ falls linearly $4.0\to 1.0\times 10^{5}\ \mathrm{Pa}$; stage 2 (Y→Z) isobaric at $1.0\times 10^{5}\ \mathrm{Pa}$, $V\!: 4.0\to 6.0\times 10^{-3}\ \mathrm{m^{3}}$. (a) why $W = p\Delta V$ fails in stage 1; (b) stage-1 work (trapezium); (c) stage-2 work; (d) $\Delta U$ and heat in stage 2.单原子气体：第 1 段（X→Y）$V\!: 1.0\to 4.0\times 10^{-3}\ \mathrm{m^{3}}$，$p$ 线性 $4.0\to 1.0\times 10^{5}\ \mathrm{Pa}$；第 2 段（Y→Z）等压 $1.0\times 10^{5}\ \mathrm{Pa}$，$V\!: 4.0\to 6.0\times 10^{-3}\ \mathrm{m^{3}}$。(a) 第 1 段为何不能用 $W = p\Delta V$；(b) 第 1 段功（梯形）；(c) 第 2 段功；(d) 第 2 段 $\Delta U$ 与热量。
$W = p\,\Delta V$ assumes a single fixed pressure, but in stage 1 the pressure changes continuously from $4.0\times 10^{5}$ to $1.0\times 10^{5}\ \mathrm{Pa}$. (M1)
For a varying pressure the work done by the gas equals the area between the path and the $V$-axis on the $p$-$V$ diagram, $W = \int p\,dV$. (A1)
The straight-line path encloses a trapezium of parallel sides $p_{1} = 4.0\times 10^{5}$ and $p_{2} = 1.0\times 10^{5}\ \mathrm{Pa}$ and width $\Delta V = (4.0 - 1.0)\times 10^{-3} = 3.0\times 10^{-3}\ \mathrm{m^{3}}$. (M1)
$$ W_{1} = \tfrac{1}{2}(p_{1} + p_{2})\,\Delta V = \tfrac{1}{2}(4.0\times 10^{5} + 1.0\times 10^{5})(3.0\times 10^{-3}). $$(M1 for substitution)
$$ W_{1} = \tfrac{1}{2}(5.0\times 10^{5})(3.0\times 10^{-3}) = +750\ \mathrm{J}. $$Positive: the gas expands. (A1)
Constant pressure, so $W_{2} = p\,\Delta V = (1.0\times 10^{5})(6.0 - 4.0)\times 10^{-3}$. (M1)
$$ W_{2} = (1.0\times 10^{5})(2.0\times 10^{-3}) = +200\ \mathrm{J}. $$(A1)
For a monatomic ideal gas $\Delta U = \tfrac{3}{2}\,\Delta(pV)$. At constant $p = 1.0\times 10^{5}\ \mathrm{Pa}$: $\Delta(pV) = p\,\Delta V = 200\ \mathrm{J}$. (M1)
$$ \Delta U_{2} = \tfrac{3}{2}(200) = +300\ \mathrm{J}. $$(A1)
First law: $Q_{2} = \Delta U_{2} + W_{2} = 300 + 200 = +500\ \mathrm{J}$. (A1)
$W = p\,\Delta V$ 假设压强为单一定值，但第 1 段中压强从 $4.0\times 10^{5}$ 连续变到 $1.0\times 10^{5}\ \mathrm{Pa}$。(M1)
压强变化时，气体所做的功等于 $p$-$V$ 图上路径与 $V$ 轴之间的面积，即 $W = \int p\,dV$。(A1)
直线路径下方为梯形，两平行边为 $p_{1} = 4.0\times 10^{5}$ 与 $p_{2} = 1.0\times 10^{5}\ \mathrm{Pa}$，宽 $\Delta V = (4.0 - 1.0)\times 10^{-3} = 3.0\times 10^{-3}\ \mathrm{m^{3}}$。(M1)
$$ W_{1} = \tfrac{1}{2}(p_{1} + p_{2})\,\Delta V = \tfrac{1}{2}(4.0\times 10^{5} + 1.0\times 10^{5})(3.0\times 10^{-3}). $$(代入得 M1)
$$ W_{1} = \tfrac{1}{2}(5.0\times 10^{5})(3.0\times 10^{-3}) = +750\ \mathrm{J}. $$为正：气体膨胀。(A1)
恒压，故 $W_{2} = p\,\Delta V = (1.0\times 10^{5})(6.0 - 4.0)\times 10^{-3}$。(M1)
$$ W_{2} = (1.0\times 10^{5})(2.0\times 10^{-3}) = +200\ \mathrm{J}. $$(A1)
单原子理想气体 $\Delta U = \tfrac{3}{2}\,\Delta(pV)$。恒压 $p = 1.0\times 10^{5}\ \mathrm{Pa}$ 时：$\Delta(pV) = p\,\Delta V = 200\ \mathrm{J}$。(M1)
$$ \Delta U_{2} = \tfrac{3}{2}(200) = +300\ \mathrm{J}. $$(A1)
第一定律：$Q_{2} = \Delta U_{2} + W_{2} = 300 + 200 = +500\ \mathrm{J}$。(A1)
$4200\ \mathrm{J}$ transferred hot ($700\ \mathrm{K}$) to cold ($350\ \mathrm{K}$), constant $T$; table gives $\Delta Q$ per reservoir. (a) explain the signs; (b) entropy change of each and the total; (c) why hot→cold is spontaneous (second law); (d) percentage uncertainty in cold-reservoir $\Delta S$ given $T = 350 \pm 10\ \mathrm{K}$.$4200\ \mathrm{J}$ 由热源（$700\ \mathrm{K}$）传到冷源（$350\ \mathrm{K}$），温度恒定；表给各热源 $\Delta Q$。(a) 解释符号；(b) 各熵变与总熵变；(c) 为何热→冷自发（第二定律）；(d) 给定 $T = 350 \pm 10\ \mathrm{K}$，冷源 $\Delta S$ 的百分比不确定度。
The hot reservoir gives up heat, so its $\Delta Q$ is negative: $-4200\ \mathrm{J}$. (A1)
The cold reservoir receives that heat, so its $\Delta Q$ is positive: $+4200\ \mathrm{J}$. (A1)
Use $\Delta S = \Delta Q / T$ for each reservoir. (M1)
$$ \Delta S_{H} = \frac{-4200}{700} = -6.0\ \mathrm{J\,K^{-1}}. $$(A1)
$$ \Delta S_{C} = \frac{+4200}{350} = +12.0\ \mathrm{J\,K^{-1}}. $$(A1)
$$ \Delta S_{\text{total}} = -6.0 + 12.0 = +6.0\ \mathrm{J\,K^{-1}}. $$(A1)
The second law requires the total entropy of the isolated system not to decrease: $\Delta S_{\text{total}} \ge 0$. (M1)
For hot→cold the result is $+6.0\ \mathrm{J\,K^{-1}} > 0$, so the process is allowed. (A1)
Reversing it (cold→hot) would give $-6.0\ \mathrm{J\,K^{-1}} < 0$, which the second law forbids; hence heat flows spontaneously only from hot to cold. (R1)
$\Delta S_{C} = \Delta Q / T$, and here only $T$ carries an uncertainty, so the percentage uncertainty in $\Delta S_{C}$ equals the percentage uncertainty in $T$. (M1)
$$ \frac{\Delta T}{T}\times 100\% = \frac{10}{350}\times 100\%. $$(M1)
$$ = 2.86\% \approx 3\%. $$(A1)
热源放出热量，故其 $\Delta Q$ 为负：$-4200\ \mathrm{J}$。(A1)
冷源接收该热量，故其 $\Delta Q$ 为正：$+4200\ \mathrm{J}$。(A1)
对每个热源用 $\Delta S = \Delta Q / T$。(M1)
$$ \Delta S_{H} = \frac{-4200}{700} = -6.0\ \mathrm{J\,K^{-1}}. $$(A1)
$$ \Delta S_{C} = \frac{+4200}{350} = +12.0\ \mathrm{J\,K^{-1}}. $$(A1)
$$ \Delta S_{\text{total}} = -6.0 + 12.0 = +6.0\ \mathrm{J\,K^{-1}}. $$(A1)
第二定律要求孤立系统的总熵不减小：$\Delta S_{\text{total}} \ge 0$。(M1)
热→冷结果为 $+6.0\ \mathrm{J\,K^{-1}} > 0$，故过程被允许。(A1)
反向（冷→热）会得 $-6.0\ \mathrm{J\,K^{-1}} < 0$，为第二定律所禁止；故热量只自发地由热流向冷。(R1)
$\Delta S_{C} = \Delta Q / T$，此处只有 $T$ 带不确定度，故 $\Delta S_{C}$ 的百分比不确定度等于 $T$ 的百分比不确定度。(M1)
$$ \frac{\Delta T}{T}\times 100\% = \frac{10}{350}\times 100\%. $$(M1)
$$ = 2.86\% \approx 3\%. $$(A1)
Monatomic gas, clockwise rectangular cycle A(2.0e5, 2.0e-3) → B(2.0e5, 5.0e-3) → C(1.0e5, 5.0e-3) → D(1.0e5, 2.0e-3) → A (SI units). A→B and C→D isobaric; B→C and D→A isochoric. (a) work on each leg and net work; (b) net work $=$ enclosed area; (c) $\Delta U$ on each leg and cycle total; (d) heat-input legs, $Q_{\mathrm{in}}$, efficiency.单原子气体，顺时针矩形循环 A(2.0e5, 2.0e-3) → B(2.0e5, 5.0e-3) → C(1.0e5, 5.0e-3) → D(1.0e5, 2.0e-3) → A（SI 单位）。A→B 与 C→D 等压；B→C 与 D→A 等容。(a) 各段功与净功；(b) 净功 $=$ 所围面积；(c) 各段 $\Delta U$ 与循环总和；(d) 吸热段、$Q_{\mathrm{in}}$、效率。
Isobaric legs use $W = p\,\Delta V$; isochoric legs have $\Delta V = 0$ so $W = 0$. (M1)
$$ W_{AB} = (2.0\times 10^{5})(5.0 - 2.0)\times 10^{-3} = +600\ \mathrm{J}, \qquad W_{BC} = 0. $$ $$ W_{CD} = (1.0\times 10^{5})(2.0 - 5.0)\times 10^{-3} = -300\ \mathrm{J}, \qquad W_{DA} = 0. $$(A1 for the two isobaric values)
$$ W_{\text{net}} = 600 + 0 - 300 + 0 = +300\ \mathrm{J}. $$(A1)
The cycle is a rectangle of height $\Delta p = (2.0 - 1.0)\times 10^{5} = 1.0\times 10^{5}\ \mathrm{Pa}$ and width $\Delta V = (5.0 - 2.0)\times 10^{-3} = 3.0\times 10^{-3}\ \mathrm{m^{3}}$. (M1)
$$ \text{area} = (1.0\times 10^{5})(3.0\times 10^{-3}) = 300\ \mathrm{J} = W_{\text{net}}. $$Clockwise, so the net work is positive (engine). (A1)
Use $\Delta U = \tfrac{3}{2}\,\Delta(pV)$ with $pV$ in joules ($A\!:400$, $B\!:1000$, $C\!:500$, $D\!:200$). (M1)
$$ \Delta U_{AB} = \tfrac{3}{2}(1000 - 400) = +900\ \mathrm{J}, \quad \Delta U_{BC} = \tfrac{3}{2}(500 - 1000) = -750\ \mathrm{J}. $$ $$ \Delta U_{CD} = \tfrac{3}{2}(200 - 500) = -450\ \mathrm{J}, \quad \Delta U_{DA} = \tfrac{3}{2}(400 - 200) = +300\ \mathrm{J}. $$(A1 for the set)
Sum: $900 - 750 - 450 + 300 = 0$, confirming $U$ returns to its starting value over the cycle. (A1)
Apply $Q = \Delta U + W$ to each leg; heat enters where $Q > 0$. (M1)
$Q_{AB} = 900 + 600 = +1500\ \mathrm{J}$ (in); $Q_{DA} = 300 + 0 = +300\ \mathrm{J}$ (in); $Q_{BC} = -750\ \mathrm{J}$ and $Q_{CD} = -450 - 300 = -750\ \mathrm{J}$ (both out). (M1)
$$ Q_{\mathrm{in}} = Q_{AB} + Q_{DA} = 1500 + 300 = 1800\ \mathrm{J}. $$(A1)
$$ \eta = \frac{W_{\text{net}}}{Q_{\mathrm{in}}} = \frac{300}{1800} = 0.167 = 16.7\%. $$(A1)
等压段用 $W = p\,\Delta V$；等容段 $\Delta V = 0$ 故 $W = 0$。(M1)
$$ W_{AB} = (2.0\times 10^{5})(5.0 - 2.0)\times 10^{-3} = +600\ \mathrm{J}, \qquad W_{BC} = 0. $$ $$ W_{CD} = (1.0\times 10^{5})(2.0 - 5.0)\times 10^{-3} = -300\ \mathrm{J}, \qquad W_{DA} = 0. $$(两个等压值得 A1)
$$ W_{\text{net}} = 600 + 0 - 300 + 0 = +300\ \mathrm{J}. $$(A1)
循环为矩形，高 $\Delta p = (2.0 - 1.0)\times 10^{5} = 1.0\times 10^{5}\ \mathrm{Pa}$，宽 $\Delta V = (5.0 - 2.0)\times 10^{-3} = 3.0\times 10^{-3}\ \mathrm{m^{3}}$。(M1)
$$ \text{面积} = (1.0\times 10^{5})(3.0\times 10^{-3}) = 300\ \mathrm{J} = W_{\text{net}}. $$顺时针，故净功为正（热机）。(A1)
用 $\Delta U = \tfrac{3}{2}\,\Delta(pV)$，$pV$ 以焦耳计（$A\!:400$、$B\!:1000$、$C\!:500$、$D\!:200$）。(M1)
$$ \Delta U_{AB} = \tfrac{3}{2}(1000 - 400) = +900\ \mathrm{J}, \quad \Delta U_{BC} = \tfrac{3}{2}(500 - 1000) = -750\ \mathrm{J}. $$ $$ \Delta U_{CD} = \tfrac{3}{2}(200 - 500) = -450\ \mathrm{J}, \quad \Delta U_{DA} = \tfrac{3}{2}(400 - 200) = +300\ \mathrm{J}. $$(整组得 A1)
求和：$900 - 750 - 450 + 300 = 0$，确认 $U$ 经一循环回到初值。(A1)
对每段用 $Q = \Delta U + W$；$Q > 0$ 处热量进入。(M1)
$Q_{AB} = 900 + 600 = +1500\ \mathrm{J}$（吸）；$Q_{DA} = 300 + 0 = +300\ \mathrm{J}$（吸）；$Q_{BC} = -750\ \mathrm{J}$、$Q_{CD} = -450 - 300 = -750\ \mathrm{J}$（皆放）。(M1)
$$ Q_{\mathrm{in}} = Q_{AB} + Q_{DA} = 1500 + 300 = 1800\ \mathrm{J}. $$(A1)
$$ \eta = \frac{W_{\text{net}}}{Q_{\mathrm{in}}} = \frac{300}{1800} = 0.167 = 16.7\%. $$(A1)
Carnot engine between $600\ \mathrm{K}$ and $300\ \mathrm{K}$, absorbing $1200\ \mathrm{J}$ per cycle. (a) the four processes in order; (b) efficiency, work output, heat rejected; (c) entropy change of each reservoir and total; (d) why a real $30\%$ engine has a larger total entropy change.卡诺热机在 $600\ \mathrm{K}$ 与 $300\ \mathrm{K}$ 之间，每循环吸 $1200\ \mathrm{J}$。(a) 四个过程的顺序；(b) 效率、输出功、排热；(c) 各热源熵变与总熵变；(d) 为何真实 $30\%$ 热机的总熵变更大。
In order: (1) isothermal expansion at $T_{\mathrm{hot}}$ (absorbs $Q_{\mathrm{in}}$); (2) adiabatic expansion ($T_{\mathrm{hot}}\to T_{\mathrm{cold}}$, $Q = 0$); (A1)
(3) isothermal compression at $T_{\mathrm{cold}}$ (rejects $Q_{\mathrm{out}}$); (4) adiabatic compression ($T_{\mathrm{cold}}\to T_{\mathrm{hot}}$, $Q = 0$). (A1)
$\eta = 1 - \dfrac{T_{\mathrm{cold}}}{T_{\mathrm{hot}}} = 1 - \dfrac{300}{600} = 0.50 = 50\%$. (M1·A1)
$$ W = \eta\,Q_{\mathrm{in}} = 0.50(1200) = 600\ \mathrm{J}. $$(A1)
$$ Q_{\mathrm{out}} = Q_{\mathrm{in}} - W = 1200 - 600 = 600\ \mathrm{J}. $$(A1)
Heat is exchanged with each reservoir only during the isothermal steps, at constant $T$, so use $\Delta S = \Delta Q / T$. (M1)
$$ \Delta S_{H} = \frac{-1200}{600} = -2.0\ \mathrm{J\,K^{-1}}, \qquad \Delta S_{C} = \frac{+600}{300} = +2.0\ \mathrm{J\,K^{-1}}. $$(A1)
Total $\Delta S_{\text{total}} = -2.0 + 2.0 = 0$, as expected for a reversible (Carnot) cycle. (A1)
A real $30\%$ engine rejects more heat ($Q_{\mathrm{out}} = 1200 - 0.30(1200) = 840\ \mathrm{J}$), so $\Delta S_{\text{total}} = -1200/600 + 840/300 = +0.8\ \mathrm{J\,K^{-1}} > 0$. Irreversibilities (friction, finite-rate heat transfer) generate extra entropy, so its total entropy change exceeds the Carnot value of zero. (R1)
依次为：(1) 在 $T_{\mathrm{hot}}$ 等温膨胀（吸收 $Q_{\mathrm{in}}$）；(2) 绝热膨胀（$T_{\mathrm{hot}}\to T_{\mathrm{cold}}$，$Q = 0$）；(A1)
(3) 在 $T_{\mathrm{cold}}$ 等温压缩（排出 $Q_{\mathrm{out}}$）；(4) 绝热压缩（$T_{\mathrm{cold}}\to T_{\mathrm{hot}}$，$Q = 0$）。(A1)
$\eta = 1 - \dfrac{T_{\mathrm{cold}}}{T_{\mathrm{hot}}} = 1 - \dfrac{300}{600} = 0.50 = 50\%$。(M1·A1)
$$ W = \eta\,Q_{\mathrm{in}} = 0.50(1200) = 600\ \mathrm{J}. $$(A1)
$$ Q_{\mathrm{out}} = Q_{\mathrm{in}} - W = 1200 - 600 = 600\ \mathrm{J}. $$(A1)
只有等温两步在恒定 $T$ 下与各热源交换热量，故用 $\Delta S = \Delta Q / T$。(M1)
$$ \Delta S_{H} = \frac{-1200}{600} = -2.0\ \mathrm{J\,K^{-1}}, \qquad \Delta S_{C} = \frac{+600}{300} = +2.0\ \mathrm{J\,K^{-1}}. $$(A1)
总熵变 $\Delta S_{\text{total}} = -2.0 + 2.0 = 0$，正如可逆（卡诺）循环所应。(A1)
真实 $30\%$ 热机排出更多热量（$Q_{\mathrm{out}} = 1200 - 0.30(1200) = 840\ \mathrm{J}$），故 $\Delta S_{\text{total}} = -1200/600 + 840/300 = +0.8\ \mathrm{J\,K^{-1}} > 0$。不可逆性（摩擦、有限速率传热）产生额外熵，故其总熵变超过卡诺的零值。(R1)