Ideal gases

Amount of substance 物质的量 is an SI base quantity. Its unit is the mole 摩尔 (mol) — one of the seven SI base units, with the kilogram, metre, second, ampere and kelvin 开尔文 from Topic 1.

One mole contains the Avogadro number of particles

One mole of any substance has a number of particles equal to the Avogadro constant 阿伏伽德罗常量:

A "particle" means whatever you are counting — atoms 原子 for a monatomic 单原子 element like helium, molecules 分子 for $\text{O}_{2}$ or $\text{H}_{2}\text{O}$. Always say what you are counting.

For $n$ moles, the number of particles is $N = n N_{\text{A}}$.

The molar mass 摩尔质量 $M_{\text{m}}$ is the mass of one mole ($\text{kg mol}^{-1}$ or $\text{g mol}^{-1}$). Mass of $n$ moles is $M = n M_{\text{m}}$. Mass of one particle is $m_{0} = M_{\text{m}} / N_{\text{A}}$.

Compressed gas cylinders store a fixed mass of gas at high pressure.

An ideal gas 理想气体 obeys $pV \propto T$ exactly, where $T$ is the thermodynamic temperature 热力学温度.

The equation of state 状态方程 can be written two equal ways:

Here:

• $p$ — pressure 压强 (Pa).
• $V$ — volume 体积 (m³).
• $T$ — thermodynamic temperature in kelvin (never °C).
• $n$ — number of moles; $N$ — number of molecules.
• $R$ — molar gas constant 摩尔气体常量, $R = 8.31\ \text{J mol}^{-1}\ \text{K}^{-1}$.
• $k$ — Boltzmann constant 玻尔兹曼常量, $k = 1.38 \times 10^{-23}\ \text{J K}^{-1}$.

Since $N = n N_{\text{A}}$, we get $k = R/N_{\text{A}}$: $k$ is the gas constant per molecule, as $R$ is per mole.

Using the equation of state

List the variables you have, find the unknown, and choose the form that matches your "amount" (moles → $nRT$; molecules → $NkT$). Always use SI units: Pa, m³, K.

Worked example. A cylinder of volume $0.020\ \text{m}^{3}$ holds gas at $27\ ^{\circ}\text{C}$ and a pressure of $2.0 \times 10^{5}\ \text{Pa}$. How many moles of gas are there? ($R = 8.31\ \text{J mol}^{-1}\ \text{K}^{-1}$.)

Convert to kelvin: $T = 27 + 273 = 300\ \text{K}$. Then from $pV = nRT$,

If a fixed amount of gas changes from state 1 to state 2:

Worked example. A fixed mass of gas at $300\ \text{K}$ occupies $0.50\ \text{m}^{3}$. It is heated to $450\ \text{K}$ at constant pressure. Find the new volume.

At constant pressure $V/T$ is constant, so

Special cases:

• constant temperature (Boyle's law 玻意耳定律): $p_{1} V_{1} = p_{2} V_{2}$.
• constant pressure (Charles's law 查理定律): $V / T = \text{constant}$.
• constant volume (pressure law 气体压强定律): $p / T = \text{constant}$.

A common mistake is using °C instead of K — $pV \propto T$ only holds with $T$ in kelvin.

Boyle's law: at constant temperature $pV$ is constant, so a $p$–$V$ graph is a hyperbola

At constant pressure the volume of a gas rises linearly with temperature, reaching zero at absolute zero (Charles's law)

A scuba diver breathes compressed gas; its pressure, volume and temperature are all linked.

The kinetic theory 分子动理论 explains a gas's large-scale behaviour from the random motion 无规则运动 of its molecules.

Assumptions

For an ideal gas:

1. a large number of identical molecules in continuous random motion.
2. the molecules' own volume is too small to matter compared with the container.
3. the time of each collision is too short to matter compared with the time between collisions.
4. intermolecular 分子间 forces are ignored except during collisions (molecules go in straight lines between them).
5. collisions (with the walls and with each other) are elastic — an elastic collision 弹性碰撞 loses no kinetic energy, so the gas does not cool down by itself.
6. Newton's laws apply.

The key assumptions of the kinetic theory of an ideal gas

These assumptions become poor at very high pressure (molecular volume matters) or very low temperature (intermolecular forces matter).

Pressure of a gas — outline of the derivation

Take a cubic box of side $L$ with $N$ molecules, each of mass $m$. Look at one molecule moving along the $x$-axis with velocity $u_{1}$.

The pressure derivation considers one molecule's velocity component $u_x$ normal to a face of a cube of gas

• one collision with the right wall: velocity reverses to $-u_{1}$, change in momentum 动量 $\Delta p_{x} = -2 m u_{1}$. By Newton's third law the wall gets an impulse 冲量 of $+2 m u_{1}$.
• time between hits on that wall: travel $2L$ there and back, so $\Delta t = 2L/u_{1}$.
• average force from this molecule: $F_{1} = \Delta p / \Delta t = m u_{1}^{2} / L$.
• add over all molecules: $F = (Nm/L)\langle u_{x}^{2} \rangle$, where $\langle u_{x}^{2} \rangle$ is the mean square 均方 of the $x$-velocity.
• pressure: $p = F/L^{2} = N m \langle u_{x}^{2} \rangle / V$.

In 3-D, by symmetry $\langle u_{x}^{2} \rangle = \tfrac{1}{3} \langle c^{2} \rangle$, where $\langle c^{2} \rangle$ is the mean-square speed 均方速率. So

Root-mean-square speed

The square root of $\langle c^{2} \rangle$ is the root-mean-square 均方根 (r.m.s.) speed:

It is a useful single measure of how fast the molecules move, slightly larger than the mean speed (squaring weights fast molecules more).

Root-mean-square speed: square each speed, take the mean, then the square root

The spread of molecular speeds: a higher temperature broadens the curve and shifts it to faster speeds

Compare the two expressions for $pV$:

Set them equal, cancel $N$, and multiply by $\tfrac{3}{2}$:

The right side is the average translational kinetic energy 平动动能 $\langle E_{\text{k}} \rangle$ of one molecule. So

This is a key result: the average translational kinetic energy of an ideal-gas molecule depends only on the thermodynamic temperature, not on the type of gas or its pressure.

Average molecular KE is proportional to thermodynamic temperature: $\langle E_k \rangle = \tfrac32 kT$

Worked example. Find the root-mean-square speed of oxygen molecules at $300\ \text{K}$. (Mass of one $\text{O}_{2}$ molecule $= 5.3 \times 10^{-26}\ \text{kg}$, $k = 1.38 \times 10^{-23}\ \text{J K}^{-1}$.)

From $\tfrac{1}{2}m\langle c^{2}\rangle = \tfrac{3}{2}kT$, the mean-square speed is $\langle c^{2}\rangle = 3kT/m$:

Consequences

• doubling the absolute temperature doubles the average KE of each molecule, so $\langle c^{2} \rangle$ doubles and $c_{\text{r.m.s.}}$ grows by $\sqrt{2}$.
• for two gases at the same temperature, the lighter gas has a larger $\langle c^{2} \rangle$. Hydrogen molecules move faster on average than oxygen molecules in the same room.
• total translational KE of $N$ molecules: $\tfrac{3}{2} N k T = \tfrac{3}{2} n R T$.

Internal energy of an ideal gas

For an ideal gas the molecules are point particles with no intermolecular potential energy and (in this simple model) no rotation or vibration. So the internal energy 内能 is just the total kinetic energy 动能:

So the internal energy of an ideal gas is proportional to the thermodynamic temperature — doubling $T$ doubles $U$.

Changing pressure at fixed temperature

Doubling $p$ at fixed $T$ (by squeezing the gas to half its volume) does not change $\langle E_{\text{k}} \rangle$ — that depends only on $T$. There are more wall collisions per second, but each molecule has the same average kinetic energy.

• Use $pV = nRT$ ($n$ in mol) or $pV = NkT$ ($N$ molecules), with temperature in kelvin.
• Learn the kinetic-theory assumptions (random motion, negligible molecular volume, elastic collisions, no intermolecular forces).
• Mean translational KE $= \frac{3}{2}kT$ — it depends only on temperature.