热力学

Thermodynamics 热力学 studies heat and energy in large collections of particles. The kinetic theory 分子运动论 explains gas behavior from the random motion of its molecules:

• In the ideal-gas model, absolute Temperature 温度 is proportional to the molecules' average translational kinetic energy: $K_{\text{avg}}=\tfrac{3}{2}k_B T$ (with $T$ in kelvin). Hotter means faster-moving molecules.
• Pressure 压强 comes from molecules colliding with the container walls – more frequent or harder collisions give more pressure.

Worked example. Find the average kinetic energy of a gas molecule at $300\ \text{K}$, using $k_B=1.38\times10^{-23}\ \text{J/K}$:

It depends only on temperature, not on the type of gas – at the same $T$, light and heavy molecules share the same average kinetic energy (so the light ones move faster).

An ideal gas 理想气体 obeys

linking pressure $P$, volume $V$, amount ($n$ moles or $N$ molecules), and absolute temperature $T$. Use it to predict how a gas responds when you change one quantity and hold others fixed (e.g. heating at constant volume raises pressure).

Worked example. A sealed rigid container of gas is at $1.0\times10^{5}\ \text{Pa}$ and $300\ \text{K}$. It is heated to $450\ \text{K}$. Because the volume and amount are fixed, $P/T$ is constant:

Always convert temperatures to kelvin before using a gas law – using Celsius here would give nonsense.

Heat 热量 $Q$ is energy transferred because of a temperature difference; it flows from hot to cold. Two objects in contact reach thermal equilibrium 热平衡 when they share the same temperature, and net heat flow stops. The three transfer methods are conduction 传导, convection 对流, and radiation 辐射.

The first law 热力学第一定律 is conservation of energy for a gas:

where $\Delta U$ is the change in internal energy 内能 (depending only on temperature for an ideal gas), $Q$ is heat added to the gas, and $W$ is work done on the gas. On a pressure–volume diagram, the work done by the gas is the area under the process curve. Watch signs: compressing a gas does positive work on it.

Worked example. A gas absorbs $500\ \text{J}$ of heat and, as it expands, does $200\ \text{J}$ of work on its surroundings. Find the change in its internal energy. Work done on the gas is $W=-200\ \text{J}$ (it does work, so it loses that energy):

The internal energy rises by $300\ \text{J}$, so the gas ends up hotter. Getting the sign of $W$ right is the whole game in first-law problems.

Worked example (a cycle on a $PV$ diagram). A sample of monatomic ideal gas goes round the cycle A → B → C → D → A shown below.

• A → B (constant pressure, expanding): work done by the gas $= P\Delta V = (2.0\times10^5)(2.0\times10^{-3}) = 400\ \text{J}$, so $W = -400\ \text{J}$.
• B → C and D → A (constant volume): no area under the line, so $W = 0$.
• C → D (constant pressure, compressed): $W = +(1.0\times10^5)(2.0\times10^{-3}) = +200\ \text{J}$ done on the gas.
• Whole cycle: $W_{\text{net}} = -400 + 200 = -200\ \text{J}$, so the gas does a net $200\ \text{J}$ of work, the enclosed area. It returns to A, so $\Delta U = 0$ and $Q_{\text{net}} = \Delta U - W = +200\ \text{J}$: a clockwise cycle takes in heat and gives out work, like an engine.
• Temperatures without numbers for $n$: $T \propto PV$. In joules, $PV$ is $200$ at A, $600$ at B, $300$ at C and $100$ at D, so B is the hottest state and D the coldest.
• Heat in A → B: for a monatomic gas $U = \tfrac32 nRT = \tfrac32 PV$, so $\Delta U = \tfrac32(600 - 200) = 600\ \text{J}$, and $Q = \Delta U - W = 600 - (-400) = 1000\ \text{J}$.

• Specific heat 比热容 $c$ is the heat needed to raise one kilogram by one degree: $Q=mc\,\Delta T$. A high specific heat (like water's) means a substance resists temperature change.
• Thermal conductivity 热导率 measures how fast heat conducts through a material; the rate of conduction rises with area and temperature difference and falls with thickness.

Worked example. How much heat raises the temperature of $2.0\ \text{kg}$ of water from $20\,{}^{\circ}\text{C}$ to $80\,{}^{\circ}\text{C}$? Water's specific heat is $c=4200\ \text{J/(kg}\,{}^{\circ}\text{C)}$:

Water's large specific heat is why it is used as a coolant and why coastal climates are mild.

Entropy 熵 measures the disorder, or the number of ways to arrange, a system. The second law 热力学第二定律: the total entropy of an isolated system never decreases – energy naturally spreads out. This sets the direction of processes: heat flows hot→cold on its own, never the reverse, and no cyclic heat engine can convert heat taken from a single reservoir entirely into work without another effect.

Answer each before reading the answers below. Take $R=8.31\ \text{J/(mol K)}$.

1. A gas at $2.0\times10^{5}\ \text{Pa}$ fills $3.0\times10^{-3}\ \text{m}^3$ at $300\ \text{K}$. How many moles of gas are there?
2. The gas in question 1 is heated at constant pressure to $450\ \text{K}$. Find its new volume and the work done by the gas.
3. A gas is compressed: $150\ \text{J}$ of work is done on it while it gives out $60\ \text{J}$ of heat. Find the change in its internal energy. Does its temperature rise or fall?
4. $0.50\ \text{kg}$ of water at $80\,{}^{\circ}\text{C}$ is mixed with $1.0\ \text{kg}$ of water at $20\,{}^{\circ}\text{C}$ in an insulated flask. Find the final temperature.
5. A student says: "Hydrogen and oxygen in the same room are at the same temperature, so their molecules move at the same average speed." Correct the student.

1. $n=\dfrac{PV}{RT}=\dfrac{(2.0\times10^{5})(3.0\times10^{-3})}{(8.31)(300)}=\dfrac{600}{2493}=0.24\ \text{mol}$.
2. At constant $P$, $V\propto T$: $V_2=3.0\times10^{-3}\times\dfrac{450}{300}=4.5\times10^{-3}\ \text{m}^3$. Work by the gas $=P\Delta V=(2.0\times10^{5})(1.5\times10^{-3})=300\ \text{J}$, so $W=-300\ \text{J}$ in $\Delta U=Q+W$.
3. $Q=-60\ \text{J}$ and $W=+150\ \text{J}$, so $\Delta U=-60+150=+90\ \text{J}$. The internal energy rises, so the temperature rises.
4. Heat lost by the hot water = heat gained by the cold water: $0.50\,c\,(80-T)=1.0\,c\,(T-20)$. Then $40-0.50T=T-20$, so $1.5T=60$ and $T=40\,{}^{\circ}\text{C}$. The $c$ cancels; the final temperature sits nearer the larger mass's start.
5. The same temperature means the same average kinetic energy, $\tfrac32k_BT$, not the same speed. A hydrogen molecule has $\tfrac1{16}$ the mass of an oxygen molecule, so $v^2$ must be 16 times larger: hydrogen molecules move about 4 times faster on average.

• Always convert temperatures to kelvin ($T_{\text{K}}=T_{\text{C}}+273$) before using any gas law.
• In the first law $\Delta U = Q + W$, get the sign of $W$ right: work done on the gas is positive; when the gas expands and does work on its surroundings, $W$ is negative.
• Temperature measures the average kinetic energy of particles, so at the same $T$ lighter molecules move faster.
• On a PV diagram the work done by the gas is the area under the process curve.
• Use $Q=mc\,\Delta T$ for temperature change; the second law fixes the direction — heat flows hot→cold and total entropy never decreases.

Common mistakes

• Using degrees Celsius in $PV = nRT$ or $K_{\text{avg}} = \tfrac32 k_B T$. Temperatures must be in kelvin.
• Getting the sign of $W$ wrong in $\Delta U = Q + W$. In this course $W$ is work done on the gas: positive when the gas is compressed, negative when it expands.
• Saying a gas expanding at constant temperature has no heat flow. $\Delta U = 0$, so $Q = -W$: the gas absorbs exactly as much heat as the work it does.
• Reading work from a $P$–$V$ graph as $P \times V$ at one point. The work is the area under the process curve; along a closed cycle it is the enclosed area.
• Confusing temperature with heat. Temperature measures the average kinetic energy of the molecules; heat is energy transferred because of a temperature difference.