The first law of thermodynamics · 热力学第一定律
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| first law of thermodynamics/fɜːst lɔː ɒv ˌθɜːməʊdaɪˈnæmɪks/ | 热力学第一定律 | rè lì xué dì yí dìng lǜ |
| sign convention/saɪn kənˈvenʃn/ | 符号约定 | fú hào yuē dìng |
| conservation of energy/ˌkɒnsəˈveɪʃn ɒv ˈenədʒi/ | 能量守恒 | néng liàng shǒu héng |
| isothermal/ˌaɪsəˈθɜːml/ | 等温 | děng wēn |
| adiabatic/ˌædiəˈbætɪk/ | 绝热 | jué rè |
Why a bicycle pump gets hot
- Pump up a tyre quickly and the barrel becomes too hot to hold comfortably. Nothing burned, nothing was heated, and no energy was supplied except by your arm.
- The arm did work on the air, compressing it. The compression was fast enough that no heat had time to escape, so all that work had nowhere to go but into the air's internal energy.
- Energy in equals energy stored. That single sentence, written carefully with signs, is the first law of thermodynamics.
- This lesson is $W = p\Delta V$, the sign convention 符号约定, the first law 热力学第一定律, and the four standard processes.
打气筒为什么会发烫
- 快速给轮胎打气,筒身会烫得握不太住。没有东西在燃烧,没有东西在加热,除了你的胳膊没有供给任何能量。
- 胳膊对空气做了功,压缩了它。压缩快到没有热量来得及逃走,所以那些功除了变成空气的内能无处可去。
- 进来的能量等于储存的能量。这一句话,连同符号仔细写清楚,就是热力学第一定律。
- 这一课讲 $W = p\Delta V$、符号约定(sign convention)、热力学第一定律(first law of thermodynamics),以及四个标准过程。
Work done when a gas changes volume
- When a gas changes volume against an outside pressure, mechanical work is done. At constant pressure $p$ with volume change $\Delta V$:
- On a $p$-$V$ graph this is the area under the line, which is why a constant-pressure change gives a simple rectangle.
- A gas that expands does work on its surroundings. A gas that is compressed has work done on it.
Force times distance, rewritten as pressure times swept volume
气体体积变化时做的功
- 当气体克服外界压强改变体积时,就有机械功。在恒定压强 $p$ 下体积变化 $\Delta V$ 时:
- 在 $p$-$V$ 图上这是线下的面积,这就是恒压变化给出一个简单矩形的原因。
- 膨胀的气体对周围做功。被压缩的气体则被做功。
力乘距离,改写成压强乘扫过的体积
The work done by a gas expanding at constant pressure is: · 气体在恒压下膨胀所做的功是:
Force = $pA$, distance = $\Delta x$, so work $= pA\Delta x = p\,\Delta V$. · 力 = $pA$,距离 = $\Delta x$,所以功 $= pA\Delta x = p\,\Delta V$。
The sign convention
- This syllabus writes the first law with $W$ as the work done on the gas.
- Compressed: $\Delta V$ is negative, the work done on the gas is positive, and the gas gains energy.
- Expanding: $\Delta V$ is positive, the work done on the gas is negative, and the gas loses energy to the surroundings.
- Read the question carefully. "Work done on the gas" and "work done by the gas" are equal in size and opposite in sign, and swapping them is the commonest error in this topic.
One quantity, two names, opposite signs
符号约定
- 本大纲把第一定律写成以 $W$ 表示对气体做的功。
- 被压缩:$\Delta V$ 为负,对气体做的功为正,气体获得能量。
- 膨胀:$\Delta V$ 为正,对气体做的功为负,气体把能量交给周围。
- 仔细读题。"对气体做的功"和"气体做的功"大小相等、符号相反,把它们弄反是本单元最常见的错误。
同一个量,两种叫法,相反的符号
When a gas expands, it does work on its surroundings. · 当气体膨胀时,它对周围做功。
Yes — it pushes the piston/atmosphere outward. So the work done on the gas is negative. · 是的——它把活塞/大气向外推。所以对气体所做的功是负的。
A gas expands. In this syllabus's convention, what is the sign of W, the work done ON the gas? · 气体膨胀。按本大纲的约定,对气体做的功 W 的符号是什么?
Work done BY the gas and work done ON it are equal in size and opposite in sign. Confusing the two is the commonest error in this topic. · 气体做的功与对气体做的功大小相等、符号相反。混淆两者是本单元最常见的错误。
The first law
- $\Delta U$ is the increase in internal energy, $q$ is the energy transferred to the system by heating, and $W$ is the work done on the system.
- The two-mark statement must define every symbol with its direction: "to the system" and "on the system" are where the marks are.
- It is conservation of energy 能量守恒 applied to a gas: energy arrives by heating or by work, and what arrives is stored.
第一定律
- $\Delta U$ 是内能的增加量,$q$ 是通过加热传递给系统的能量,$W$ 是对系统做的功。
- 两分的陈述必须给每个符号都定义方向:"给系统"和"对系统做"正是得分所在。
- 它是应用于气体的能量守恒(conservation of energy):能量通过加热或做功到来,而到来的能量被储存起来。
Work done on a gas · 对气体做的功
Push the piston in and you do work on the gas (W = pΔV); the first law says that work plus the heat added equals the rise in internal energy. · 把活塞推入,你就对气体做了功(W = pΔV);第一定律指出,这个功加上加入的热量等于内能的增加。
The first law of thermodynamics: $\Delta U = q +$ ____. · 热力学第一定律:$\Delta U = q +$ ____。
$\Delta U = q + W$, where $W$ is the work done on the gas and $q$ is the heat added. · $\Delta U = q + W$,其中 $W$ 是对气体 所做 的功,$q$ 是加入的热。
A two-mark statement of the first law must define its symbols with direction. Which are correct? Select all · 所有 that apply. · 第一定律的两分陈述必须给符号定义方向。哪些是正确的?选出所有适用的。
"To the system" and "on the system" are where the marks are. Defining W as the work done by the system flips the sign of the whole equation. · "给系统"和"对系统做"正是得分所在。把 W 定义成系统做的功会让整个方程的符号反过来。
Worked example: heat in, work out
- A gas absorbs $500\ \text{J}$ of heat while it expands and does $200\ \text{J}$ of work on its surroundings. Find the change in its internal energy.
- The heating is into the gas, so $q = +500\ \text{J}$.
- The gas does work, so the work done on it is $W = -200\ \text{J}$. This is the step the sign convention exists for.
- $\Delta U = q + W = 500 + (-200) = +300\ \text{J}$: the internal energy rises by 300 J.
例题:热进来,功出去
- 气体在膨胀的同时吸收 $500\ \text{J}$ 热量,并对周围做了 $200\ \text{J}$ 的功。求它内能的变化。
- 加热是进入气体的,所以 $q = +500\ \text{J}$。
- 气体做了功,所以对它做的功是 $W = -200\ \text{J}$。符号约定的存在正是为了这一步。
- $\Delta U = q + W = 500 + (-200) = +300\ \text{J}$:内能上升 300 J。
$100\ \text{J}$ of heat is added to a gas while $30\ \text{J}$ of work is done on it. What is the rise in internal energy? · 给一个气体加入 $100\ \text{J}$ 的热,同时对它 做 $30\ \text{J}$ 的功。内能增加多少?
$\Delta U = q + W = 100 + 30 = 130\ \text{J}$. · $\Delta U = q + W = 100 + 30 = 130\ \text{J}$。
Worked example: the same rise, two ways
- An ideal gas is heated at a constant pressure of $2.0 \times 10^5\ \text{Pa}$; its internal energy rises by $7600\ \text{J}$ while its volume increases by $1.2 \times 10^{-2}\ \text{m}^3$. Find the thermal energy supplied. Then find it again for the same rise in $U$ at constant volume.
- Work done by the gas is $p\Delta V = 2.0 \times 10^5 \times 1.2 \times 10^{-2} = 2400\ \text{J}$, so the work done on it is $W = -2400\ \text{J}$.
- First law: $q = \Delta U - W = 7600 - (-2400) = 1.0 \times 10^4\ \text{J}$.
- At constant volume no work is done, so $q = \Delta U = 7600\ \text{J}$.
- The extra 2400 J in the first case is exactly the energy the gas spent pushing its surroundings back. Same $\Delta U$, different $q$, because the work differed.
例题:同样的增量,两条路
- 某理想气体在 $2.0 \times 10^5\ \text{Pa}$ 的恒压下被加热;它的内能上升 $7600\ \text{J}$,体积增加 $1.2 \times 10^{-2}\ \text{m}^3$。求供给的热能。然后求在恒容下产生同样 $U$ 增量时的热能。
- 气体做的功是 $p\Delta V = 2.0 \times 10^5 \times 1.2 \times 10^{-2} = 2400\ \text{J}$,所以对它做的功是 $W = -2400\ \text{J}$。
- 第一定律:$q = \Delta U - W = 7600 - (-2400) = 1.0 \times 10^4\ \text{J}$。
- 在恒容下不做功,所以 $q = \Delta U = 7600\ \text{J}$。
- 第一种情况多出来的 2400 J,正是气体推开周围所花掉的能量。同样的 $\Delta U$,不同的 $q$,因为功不同。
An ideal gas at constant pressure 2.0 x 10^5 Pa gains 7600 J of internal energy while expanding by 1.2 x 10^-2 m^3. How many joules of thermal energy were supplied? · 某理想气体在 2.0 x 10^5 Pa 恒压下内能增加 7600 J,同时膨胀 1.2 x 10^-2 m^3。供给了多少焦耳的热能?
Work done by the gas is 2400 J, so W on the gas is -2400 J and q = 7600 - (-2400) = 10 000 J. At constant volume the same rise would need only 7600 J. · 气体做的功是 2400 J,所以对气体做的功 W 是 -2400 J,q = 7600 - (-2400) = 10 000 J。恒容下同样的增量只需 7600 J。
The four standard processes
| Process | Constant | $\Delta U$ | $W$ on gas | $q$ |
|---|---|---|---|---|
| isothermal 等温 | $T$ | $0$ | $W$ | $-W$ |
| constant volume | $V$ | $\tfrac32 nR\Delta T$ | $0$ | $\Delta U$ |
| constant pressure | $p$ | $\tfrac32 nR\Delta T$ | $-p\Delta V$ | $\Delta U - W$ |
| adiabatic 绝热 | no heat | $W$ | $W$ | $0$ |
- Each row is just $\Delta U = q + W$ with one quantity zero. Isothermal: $\Delta T = 0$, so $\Delta U = 0$ and $q = -W$, meaning heat in comes straight back out as work. Adiabatic: $q = 0$, so $\Delta U = W$, which is the bicycle pump.
- For an ideal gas $\Delta U = \tfrac32 nR\Delta T$ always, because $U$ depends only on temperature.
Four paths, one law
四个标准过程
| 过程 | 不变量 | $\Delta U$ | 对气体做的 $W$ | $q$ |
|---|---|---|---|---|
| 等温(isothermal) | $T$ | $0$ | $W$ | $-W$ |
| 恒容 | $V$ | $\tfrac32 nR\Delta T$ | $0$ | $\Delta U$ |
| 恒压 | $p$ | $\tfrac32 nR\Delta T$ | $-p\Delta V$ | $\Delta U - W$ |
| 绝热(adiabatic) | 无热交换 | $W$ | $W$ | $0$ |
- 每一行都不过是 $\Delta U = q + W$ 中有一项为零。等温:$\Delta T = 0$,所以 $\Delta U = 0$ 且 $q = -W$,即进来的热又原样以功的形式出去。绝热:$q = 0$,所以 $\Delta U = W$,这就是打气筒。
- 对理想气体,$\Delta U = \tfrac32 nR\Delta T$ 恒成立,因为 $U$ 只取决于温度。
四条路径,一条定律
Match each process to what it makes zero. · 把每个过程与它使之为零的量配对。
Constant $T$ → no change in internal energy; constant $V$ → no work; adiabatic → no heat flow. · 恒定 $T$ → 内能不变;恒定 $V$ → 不做功;绝热 → 没有热流动。
In an isothermal change of an ideal gas, the internal energy does not change. · 在理想气体的等温变化中,内能不改变。
For an ideal gas $U \propto T$, so constant $T$ means $\Delta U = 0$ (and then $q = -W$). · 对于理想气体 $U \propto T$,所以恒定 $T$ 意味着 $\Delta U = 0$(因而 $q = -W$)。
Worked example: explain with the first law
- A three-mark "use the first law to explain" answer has three steps: state $q$ with its sign, state $W$ with its sign, then combine for $\Delta U$.
- Why a bicycle pump gets hot: the air is compressed, so work is done on it, $W > 0$; the compression is fast, so no thermal energy has time to leave, $q \approx 0$; therefore $\Delta U = W > 0$, the temperature rises and the pump warms.
- Water evaporating on a hot day: thermal energy is transferred to the water, $q > 0$; the vapour occupies a far larger volume, so the system does work on the atmosphere, $W < 0$; $q$ exceeds the work done, so $\Delta U$ still increases, as potential energy, the molecules having been separated.
例题:用第一定律解释
- 三分的"用第一定律解释"的答案有三步:说出带符号的 $q$、说出带符号的 $W$,再合起来得到 $\Delta U$。
- 打气筒为什么发烫:空气被压缩,所以对它做了功,$W > 0$;压缩很快,热能来不及离开,$q \approx 0$;因此 $\Delta U = W > 0$,温度上升,筒身变热。
- 炎热天里蒸发的水:热能传递给水,$q > 0$;蒸汽占的体积大得多,所以系统对大气做功,$W < 0$;$q$ 超过所做的功,所以 $\Delta U$ 仍然增加,增加的是势能——分子被分开了。
Put the first-law explanation of why a bicycle pump gets hot in order. · 把用第一定律解释打气筒为什么发烫的步骤按顺序排列。
State q with its sign, state W with its sign, combine, then say what the change means. That is the three-mark shape for every first-law explanation. · 说出带符号的 q、说出带符号的 W、合起来,再说明这个变化意味着什么。每个第一定律解释的三分结构都是这样。
Marks that slip away
- $W$ in this syllabus is the work done on the gas. Expansion makes it negative.
- The two-mark statement of the law must define each symbol with direction: heating to the system, work on the system.
- At constant volume $W = 0$, so all the heat becomes internal energy. At constant pressure some of it leaves as work.
- $p\Delta V$ matters for gases. For a heated solid the work done against the atmosphere is a millionth of the heating, so $\Delta U \approx q$.
容易丢掉的分
- 本大纲中的 $W$ 是对气体做的功。膨胀使它为负。
- 该定律的两分陈述必须给每个符号定义方向:加热是给系统,功是对系统做的。
- 恒容时 $W = 0$,所以全部热量都变成内能。恒压时其中一部分以功的形式离开。
- $p\Delta V$ 对气体才要紧。对被加热的固体,克服大气所做的功只有加热量的百万分之一,所以 $\Delta U \approx q$。
You've got it
- $W = p\Delta V$ at constant pressure, the area under a $p$-$V$ line; expansion does work on the surroundings
- the sign convention: $W$ is the work done on the gas, positive when compressed and negative when expanding
- the first law $\Delta U = q + W$ is conservation of energy, with $q$ the heating to the system and $W$ the work on it, and every symbol defined with its direction
- isothermal $\Delta U = 0$ so $q = -W$; constant volume $W = 0$ so $q = \Delta U$; adiabatic $q = 0$ so $\Delta U = W$
你掌握了
- 恒压下 $W = p\Delta V$,即 $p$-$V$ 线下的面积;膨胀是对周围做功
- 符号约定:$W$ 是对气体做的功,压缩为正、膨胀为负
- 第一定律 $\Delta U = q + W$ 是能量守恒,其中 $q$ 是给系统的加热,$W$ 是对它做的功,每个符号都要带方向定义
- 等温 $\Delta U = 0$ 故 $q = -W$;恒容 $W = 0$ 故 $q = \Delta U$;绝热 $q = 0$ 故 $\Delta U = W$