Thermodynamics and statistical mechanics · 热力学与统计力学
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| entropy/ˈentrəpi/ | 熵 | shāng |
| partition function/pɑːˈtɪʃn ˈfʌŋkʃn/ | 配分函数 | pèi fēn hán shù |
A decision before an answer
- Two engines can obey energy conservation, yet only one obeys the entropy limit on efficiency.
- Your goal: Apply the first and second laws with a sign convention.
答案前的判断
- 两台发动机均可遵循能量守恒，但仅有一台符合效率的熵限。
- 你的目标：结合符号约定应用热力学第一定律和第二定律。
Read the relationship
- State the first-law sign convention: ΔU=Q−W when W is work done by the system.
- Relate ensembles, Boltzmann factors and entropy.
阅读关系
- 阐述第一定律符号约定：当 W 表示系统对外做功时，ΔU=Q−W。
- 关联系综、玻尔兹曼因子和熵。
A Carnot engine between 400 K and 300 K has efficiency:
1−300/400=0.25.
Use the defining rule
- For a reversible engine between reservoirs, efficiency is bounded by 1−Tc/Th using absolute temperatures.
- Analyse ideal gases, heat engines and quantum statistics.
运用定义规则
- 对于在热库之间运行的可逆发动机，其效率受 1−Tc/Th 限制，其中使用绝对温度。
- 分析理想气体、热机和量子统计。
A higher-energy state’s Boltzmann weight at fixed T is:
The negative exponential decreases with energy.
Check the conditions
- Boltzmann weights are exp(−E/kT). Probabilities require division by the partition function.
- Analyse ideal gases, heat engines and quantum statistics.
Reservoirs at 600 K and 300 K give a maximum reversible efficiency 1−300/600=0.5. An engine taking 1000 J can produce at most 500 J of work under these conditions. Conservation alone would not impose that bound.
检查条件
- 玻尔兹曼权重为 exp(−E/kT)。概率需除以配分函数进行归一化。
- 分析理想气体、热机和量子统计。
温度为 600 K 和 300 K 的热库所能达到的最大可逆效率为 1−300/600=0.5。在此条件下，吸收 1000 J 热量的发动机最多可做 500 J 的功。仅凭能量守恒无法得出该上限。
A system receives 100 J heat and does 40 J work. ΔU=____ J.
ΔU=Q−W=100−40.
Apply the task format
- Distinguish Maxwell–Boltzmann, Bose–Einstein and Fermi–Dirac assumptions. Fermions cannot share an identical single-particle state.
- Analyse ideal gases, heat engines and quantum statistics.
Celsius temperature ratios cannot be used in the Carnot efficiency formula.
应用题目格式
- 区分麦克斯韦-玻尔兹曼、玻色-爱因斯坦和费米-狄拉克统计假设。费米子不能占据相同的单粒子态。
- 分析理想气体、热机和量子统计。
摄氏温标下的温度比值不能用于卡诺效率公式。
Which answer fits this case? · 哪个答案符合此案例？
Apply the first and second laws with a sign convention · 结合符号法则应用热力学第一定律和第二定律
An engine may convert all heat from one reservoir into work in a complete cycle with no other effect.
That violates the second law.
Keep the distinctions
- entropy 熵 — A state quantity connected to multiplicity and reversible heat transfer.
- partition function 配分函数 — The sum of statistical weights used to normalise probabilities.
- Apply the first and second laws with a sign convention.
- Relate ensembles, Boltzmann factors and entropy.
- Analyse ideal gases, heat engines and quantum statistics.
保持区别
- 熵 — 一种与微观状态数（多重性）及可逆热传递相关的状态量。
- 配分函数 — 用于对概率进行归一化的统计权重之和。
- 应用热力学第一定律和第二定律，并注意符号约定。
- 关联系综、玻尔兹曼因子和熵。
- 分析理想气体、热机和量子统计。
Match each term with its precise meaning in this lesson.
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
The order follows the stated process; check each stage before the next.