📚 AP Physics 2 Free-Response Question Walkthrough: Thermodynamic Cycle | AP 物理2自由问答题真题解析：热力学循环
In AP Physics 2, free-response questions (FRQs) often present a multi-step thermodynamic cycle and ask you to analyse energy transfers, work, and efficiency. This walkthrough dissects a typical FRQ involving an ideal gas that undergoes an isothermal expansion, an isobaric compression, and an isochoric return to its initial state. You will see how to apply the first law of thermodynamics, the ideal gas law, and specific heat capacities for a monatomic gas. Every calculation is worked out step by step so that you can master the logic needed for the exam.
在 AP 物理2 中，自由问答题常常给出一组多步热力学循环，要求分析能量转移、功和效率。本文详细解析一道典型的自由问答题：理想气体先后经历等温膨胀、等压压缩和等容回到初态。你将看到如何运用热力学第一定律、理想气体状态方程以及单原子气体的比热容。每一步计算都完整展示，帮助你掌握考试所需的逻辑。
1. Problem Setup | 题目设置
mol of a monatomic ideal gas. The gas is taken through the cycle 1 → 2 → 3 → 1 as follows:
- 1 → 2: Isothermal expansion at temperature T₁ from volume V₁ to V₂ = 2V₁.
- 2 → 3: Isobaric compression at pressure P₂ back to the original volume V₃ = V₁.
- 3 → 1: Isochoric (constant-volume) heating that returns the gas to state 1.
The universal gas constant is R = 8.31 J·mol⁻¹·K⁻¹. Assume all processes are quasi-static.
一个无摩擦活塞气缸装有 n = 2.00 mol 单原子理想气体。气体按以下方式经历循环 1 → 2 → 3 → 1：
- 1 → 2：等温膨胀，温度为 T₁，体积从 V₁ 膨胀到 V₂ = 2V₁。
- 2 → 3：等压压缩，压强为 P₂，压缩回原体积 V₃ = V₁。
- 3 → 1：等容加热，使气体返回状态 1。
初态 1 的压强 P₁ = 4.00 × 10⁵ Pa，体积 V₁ = 2.00 × 10⁻² m³。普适气体常量 R = 8.31 J·mol⁻¹·K⁻¹。假设所有过程都是准静态的。
2. Drawing the P-V Diagram | 绘制 P-V 图
a: P = nRT₁/V, so it ends at V₂ = 2V₁ with P₂ = P₁/2. The isobaric compression 2 → 3 is a horizontal line leftward to V₃ = V₁, with P₃ = P₂. The isochoric process 3 → 1 is a vertical line upward, restoring the original pressure P₁. The enclosed area represents the net work done by the gas per cycle.
计算前一定要先画出循环示意图。在压强-体积图上，状态 1 位于 (V₁, P₁)。等温膨胀 1 → 2 沿双曲线进行：P = nRT₁/V，因此终点在 V₂ = 2V₁，P₂ = P₁/2。等压压缩 2 → 3 是一条水平向左的线段，到达 V₃ = V₁，P₃ = P₂。等容过程 3 → 1 是一条竖直向上的线段，恢复至初态压强 P₁。封闭曲线所围面积代表每循环气体对外做的净功。
3. Determining Temperatures at Each State | 确定各状态温度
Use the ideal gas law PV = nRT. For state 1:
) = (4.00×10⁵ Pa)(2.00×10⁻² m³) / (2.00 mol × 8.31 J·mol⁻¹·K⁻¹) = 8.00×10³ / 16.62 ≈ 481 K
≈ 2.00×10⁵ Pa, which matches P₁/2. For state 3, pressure is P₂ and volume is V₃ = V₁, so:
T₃ = P₂V₁ / (nR) = (2.00×10⁵ Pa)(2.00×10⁻² m³) / (2.00 × 8.31) = 4.00×10³ / 16.62 ≈ 241 K
T₁ = 481 K, T₂ = 481 K, T₃ = 241 K. These values will be used in all energy calculations.
利用理想气体状态方程 PV = nRT。对于状态 1：
) = (4.00×10⁵ Pa)(2.00×10⁻² m³) / (2.00 mol × 8.31 J·mol⁻¹·K⁻¹) = 8.00×10³ / 16.62 ≈ 481 K
等温过程中 T₂ = T₁ = 481 K。状态 2 的压强 P₂ = nRT₁/V₂ = (2.00×8.31×481)/(4.00×10⁻²) ≈ 2.00×10⁵ Pa，恰为 P₁/2。对于状态 3，压强为 P₂，体积为 V₃ = V₁，故：
T₃ = P₂V₁ / (nR) = (2.00×10⁵ Pa)(2.00×10⁻² m³) / (2.00 × 8.31) = 4.00×10³ / 16.62 ≈ 241 K
因此 T₁ = 481 K，T₂ = 481 K，T₃ = 241 K。这些数值将用于所有能量计算。
4. Isothermal Expansion 1 → 2 Analysis | 等温膨胀 1 → 2 分析
W the work done BY the gas) gives Q = W. The work done during an isothermal expansion is:
W₁₂ = nRT₁ ln(V₂ / V₁) = nRT₁ ln(2)
Substituting the numbers:
W₁₂ = 2.00 × 8.31 × 481 × 0.693 ≈ 5.55 × 10³ J
.55 × 10³ J. The heat absorbed from the hot reservoir is Q₁₂ = +5.55 × 10³ J, since Q = W.
对理想气体等温过程，ΔU = 0，因为内能只取决于温度。热力学第一定律 ΔU = Q - W（W 为气体对外做的功）得出 Q = W。等温膨胀过程的功为：
W₁₂ = nRT₁ ln(V₂ / V₁) = nRT₁ ln(2)
代入数值：
W₁₂ = 2.00 × 8.31 × 481 × 0.693 ≈ 5.55 × 10³ J
因此 W₁₂ = +5.55 × 10³ J。从高温热源吸收的热量 Q₁₂ = +5.55 × 10³ J。
5. Isobaric Compression 2 → 3 Analysis | 等压压缩 2 → 3 分析
aric process, the work done by the gas is W = PΔV. Here ΔV = V₃ - V₂ = V₁ - 2V₁ = -V₁, so:
W₂₃ = P₂ × (V₁ - 2V₁) = -P₂V₁ = -(2.00×10⁵ Pa)(2.00×10⁻² m³) = -4.00 × 10³ J
l energy uses the molar heat capacity at constant volume for a monatomic gas, Cv = (3/2)R:
ΔU₂₃ = nCv (T₃ - T₂) = n × (3/2)R × (241 K - 481 K) = 3.00 × 8.31 × (-240) ≈ -5.98 × 10³ J
Now apply the first law to find Q₂₃:
Q₂₃ = ΔU₂₃ + W₂₃ = (-5.98 × 10³ J) + (-4.00 × 10³ J) ≈ -9.98 × 10³ J
ore, the gas releases about 9.98 × 10³ J of heat to the cold reservoir during compression.
等压过程中，气体对外做功为 W = PΔV。此处 ΔV = V₃ - V₂ = V₁ - 2V₁ = -V₁，于是：
W₂₃ = P₂ × (V₁ - 2V₁) = -P₂V₁ = -(2.00×10⁵ Pa)(2.00×10⁻² m³) = -4.00 × 10³ J
负号表示外界对气体做功。内能变化用单原子气体的等容摩尔热容 Cv = (3/2)R：
ΔU₂₃ = nCv (T₃ - T₂) = n × (3/2)R × (241 K - 481 K) = 3.00 × 8.31 × (-240) ≈ -5.98 × 10³ J
用第一定律求 Q₂₃：
Q₂₃ = ΔU₂₃ + W₂₃ = (-5.98 × 10³ J) + (-4.00 × 10³ J) ≈ -9.98 × 10³ J
因此，压缩过程中气体向低温热源释放约 9.98 × 10³ J 的热量。
6. Isochoric Process 3 → 1 Analysis | 等容过程 3 → 1 分析
At constant volume, no work is done: W₃₁ = 0. The change in internal energy is:
nCv (T₁ - T₃) = 2.00 × (3/2) × 8.31 × (481 K - 241 K) = 3.00 × 8.31 × 240 ≈ +5.99 × 10³ J
orbs this amount of energy as heat while its pressure rises from P₂ to P₁ at fixed volume.
等容过程中，不做功：W₃₁ = 0。内能变化为：
nCv (T₁ - T₃) = 2.00 × (3/2) × 8.31 × (481 K - 241 K) = 3.00 × 8.31 × 240 ≈ +5.99 × 10³ J
因 W = 0，第一定律给出 Q₃₁ = ΔU₃₁ = +5.99 × 10³ J。在体积不变的情况下，气体吸收这些热量，压强从 P₂ 升至 P₁。
7. Energy Conservation Check | 能量守恒验证
Over a complete cycle, the net change in internal energy must be zero. Summing the ΔU values:
ΔUcycle = 0 + (-5.98 × 10³ J) + (5.99 × 10³ J) ≈ 0
The tiny difference is due to rounding. The net work done by the gas per cycle is Wnet = 5.55 × 10³ J + (-4.00 × 10³ J) + 0 = 1.55 × 10³ J. The net heat absorbed per cycle is Qnet = 5.55 × 10³ J + (-9.98 × 10³ J) + 5.99 × 10³ J ≈ 1.56 × 10³ J. Within rounding errors, Wnet = Qnet, confirming the first law.
在一个完整循环中，内能净变化必须为零。把 ΔU 加起来：
ΔUcycle = 0 + (-5.98 × 10³ J) + (5.99 × 10³ J) ≈ 0
微小的差异源于四舍五入。每循环气体做出的净功为 Wnet = W₁₂ + W₂₃ + W₃₁ = 5.55 × 10³ J + (-4.00 × 10³ J) + 0 = 1.55 × 10³ J。每循环净吸热 Qnet = Q₁₂ + Q₂₃ + Q₃₁ = 5.55 × 10³ J + (-9.98 × 10³ J) + 5.99 × 10³ J ≈ 1.56 × 10³ J。在舍入误差范围内 Wnet = Qnet，验证了第一定律。
8. Thermal Efficiency of the Cycle | 循环热效率
at. Steps 1 → 2 (Q₁₂ > 0) and 3 → 1 (Q₃₁ > 0) both take in energy, while 2 → 3 releases heat. Therefore, Qin = Q₁₂ + Q₃₁ = 5.55 × 10³ J + 5.99 × 10³ J = 1.154 × 10⁴ J. The efficiency is:
e = Wnet / Qin = (1.55 × 10³ J) / (1.154 × 10⁴ J) ≈ 0.134 = 13.4%
This value is lower than the efficiency of a Carnot engine operating between the same extreme temperatures (Th = 481 K, Tc = 241 K), which would be 1 - Tc/Th difference arises because the isochoric heat addition is not a reversible isothermal step.
热效率 e 定义为净功输出与来自高温热源的总吸热量之比。我们需要找出哪些过程吸热。1 → 2（Q₁₂ > 0）和 3 → 1（Q₃₁ > 0）都吸收能量，而 2 → 3 放热。因此 Qin = Q₁₂ + Q₃₁ = 5.55 × 10³ J + 5.99 × 10³ J = 1.154 × 10⁴ J。效率为：
