Kinetic Theory of Gases

Open a bottle of perfume in one corner of a room and you smell it in the far corner a little later. Gas molecules are always moving, and they spread out to fill whatever space they get. Kinetic theory turns that picture into numbers. It works because it makes a few simple assumptions about the molecules:

1. A gas is made of a very large number of identical molecules. Each one behaves like a tiny hard sphere.
2. The molecules themselves take up negligible volume compared with the container.
3. They move randomly in all directions and obey Newton’s laws.
4. There are no forces between molecules except during collisions, so they move in straight lines between collisions. Gravity on them is neglected.
5. Collisions with each other and with the walls are perfectly elastic, so kinetic energy is conserved.
6. A collision lasts a negligible time compared with the time between collisions.

A gas that obeys these assumptions exactly is an ideal gas. Real gases come close at low pressure and high temperature, where molecules are far apart and fast.

Hold two of the four quantities (n, P, V, T) fixed and the equation of state gives one of the named laws.

• Boyle’s law (n, T fixed): \(PV=\) constant. The P–V graph is a rectangular hyperbola. A hotter isotherm lies farther from the origin. A graph of PV against P is a horizontal line at height nRT.
• Charles’s law (n, P fixed): \(V\propto T\). The V–T graph is a straight line through the origin, with slope nR/P. A steeper line means a lower pressure.
• Gay-Lussac’s (pressure) law (n, V fixed): \(P\propto T\). The P–T graph is a straight line through the origin, with slope nR/V. A steeper line means a smaller volume.
• Avogadro’s law (P, T fixed): \(V\propto n\). Equal volumes of all ideal gases at the same P and T hold the same number of molecules. One mole occupies about 22.4 L at 0 °C and 1 atm.

If T is plotted in °C instead of K, the V–T and P–T lines no longer pass through the origin. Extended backwards, they all meet the temperature axis at −273.15 °C.

To compare two states on any graph, draw lines from the origin. On a P–T graph, the state on the steeper line from O has the smaller volume, because \(V=nR\,(T/P)\).

Write \(n=m/M\) to bring in mass and density: \(P=\dfrac{\rho RT}{M}\). At fixed P and T, gas density is proportional to molar mass. Write \(N=PV/(k_BT)\) to count molecules: the number density is \(P/(k_BT)\), the same for every ideal gas at the same P and T.

Real gases depart from PV = nRT at high pressure and low temperature. Molecules then occupy a noticeable part of the volume and attract each other. The van der Waals equation \(\left(P+\dfrac{an^2}{V^2}\right)(V-nb)=nRT\) corrects for both effects: b for molecular volume and a for attraction. For NEET you only need the idea: a real gas behaves most ideally at low pressure and high temperature.

Take a cube of side L containing N molecules, each of mass m. Look at one molecule with velocity components \((v_x, v_y, v_z)\) and follow its motion along x.

1. It hits the right wall (wall A, area L²) and bounces back elastically. Its x-momentum changes from \(+mv_x\) to \(-mv_x\). The change is \(-2mv_x\), so the wall receives an impulse \(2mv_x\). The y and z components do not change.
2. Before it hits wall A again it must cross to the far wall and come back, a distance 2L. Time between hits on wall A: \(\Delta t=2L/v_x\).
3. Average force on wall A from this molecule: \(\dfrac{2mv_x}{2L/v_x}=\dfrac{mv_x^2}{L}\).
4. Add all N molecules: \(F=\dfrac{m}{L}\sum v_x^2=\dfrac{mN}{L}\langle v_x^2\rangle\).
5. Pressure: \(P=\dfrac{F}{L^2}=\dfrac{mN}{V}\langle v_x^2\rangle\).
6. Motion is random, so no direction is special: \(\langle v_x^2\rangle=\langle v_y^2\rangle=\langle v_z^2\rangle\). Since \(v^2=v_x^2+v_y^2+v_z^2\), each equals \(\langle v^2\rangle/3\).
7. Therefore \(P=\dfrac13\dfrac{mN}{V}\langle v^2\rangle=\dfrac13\rho v_{rms}^2\), with \(\rho=mN/V\).

Collisions between molecules do not spoil the result. Momentum is conserved in each collision, so on average the momentum delivered to the wall is unchanged. The shape of the container does not matter either; the cube just keeps the algebra simple.

Multiply \(P=\frac13\frac{mN}{V}v_{rms}^2\) by V: \(PV=\frac13Nmv_{rms}^2=\frac23\left(\frac12Nmv_{rms}^2\right)=\frac23E\), where E is the total translational kinetic energy. So pressure equals two-thirds of the translational kinetic energy per unit volume.

Now compare with the ideal gas law \(PV=Nk_BT\). Then \(\frac23E=Nk_BT\), so the average translational kinetic energy of one molecule is \(\frac12mv_{rms}^2=\frac32k_BT\).

This is the kinetic meaning of temperature: absolute temperature measures the average translational kinetic energy of the molecules. It does not depend on the mass of the molecule, the pressure or the kind of gas. A hydrogen molecule and an oxygen molecule in the same room have the same average translational kinetic energy.

The “rms” in \(v_{rms}\) means root of the mean of the squares. For speeds 1, 2, 3, 4 and 5 km/s, the mean is 3 km/s but \(v_{rms}=\sqrt{(1+4+9+16+25)/5}=\sqrt{11}\approx3.32\) km/s. Squaring gives more weight to the fast molecules, so \(v_{rms}\) is always at least the mean speed.

From the pressure derivation, every ideal-gas molecule has average translational kinetic energy \(\frac32k_BT\), whatever its mass. At 300 K this is \(1.5\times1.38\times10^{-23}\times300\approx6.2\times10^{-21}\) J. One mole contains \(N_A\) molecules, so its translational kinetic energy is \(\frac32RT\).

Since \(\frac12mv_{rms}^2=\frac32k_BT\), we get \(v_{rms}=\sqrt{3k_BT/m}=\sqrt{3RT/M}\). Using \(P=\rho RT/M\) this is also \(v_{rms}=\sqrt{3P/\rho}\).

• At a fixed temperature, \(v_{rms}\propto1/\sqrt{M}\): lighter gases move faster.
• For a given gas, \(v_{rms}\propto\sqrt{T}\). Changing pressure at constant temperature does not change \(v_{rms}\), because density changes in the same ratio as pressure.
• Classically, all molecular motion would stop at T = 0 K. That is the kinetic meaning of absolute zero.

The molecules in a gas do not all move at the same speed. Three summaries of the distribution are used:

• Most probable speed \(v_{mp}=\sqrt{2RT/M}\): the speed at the peak of the distribution.
• Average (mean) speed \(\bar v=\sqrt{8RT/(\pi M)}\): the ordinary average of the speeds.
• Rms speed \(v_{rms}=\sqrt{3RT/M}\): the square root of the average of v².

All three have the form \(\sqrt{\text{number}\times RT/M}\), so their ratio is fixed for every gas at every temperature:

\[v_{rms}:\bar v:v_{mp}=\sqrt3:\sqrt{8/\pi}:\sqrt2\approx1.73:1.60:1.41\]

So \(v_{rms}>\bar v>v_{mp}\). The mean vector velocity is zero, because molecules move equally in all directions. Never use it in place of a speed.

The Maxwell distribution gives the fraction of molecules per unit speed interval. Its shape:

• It starts at zero (very few molecules are almost at rest), rises to a peak at \(v_{mp}\), then falls with a long tail of fast molecules.
• The area under the curve is the total number (or fraction) of molecules, so it stays the same when T changes.
• At higher T the peak moves to a higher speed and becomes lower, and the curve spreads out. The area stays the same.
• At the same T, a heavier gas has a narrower, taller curve peaked at a lower speed. Raising M has the same effect as lowering T, because only T/M appears.

Why Earth has almost no hydrogen. Escape speed from Earth is 11.2 km/s. At 300 K, \(v_{rms}\) of H₂ is only about 1.9 km/s. But the long tail always contains some molecules faster than the escape speed, and for light molecules this fraction is much larger. Over millions of years H₂ and He leak away, while N₂ and O₂ stay. The Moon’s escape speed is only about 2.4 km/s, so it has lost almost all its atmosphere.

Sound versus molecules. The speed of sound in a gas is \(v_s=\sqrt{\gamma RT/M}\). Dividing, \(v_s/v_{rms}=\sqrt{\gamma/3}\). Since γ ≤ 5/3, sound is always slower than the rms speed. For air (γ = 1.4) the ratio is about 0.68.

A molecule can store energy in several independent ways. Each way that adds a squared term to the energy is called a degree of freedom. Examples: \(\frac12mv_x^2\) for motion along x, \(\frac12I\omega^2\) for rotation about one axis, and \(\frac12kx^2\) for the stretch of a vibrating bond.

The law of equipartition says that in thermal equilibrium at temperature T, each such squared term has the same average energy \(\frac12k_BT\) per molecule.

• Translation: three terms (x, y, z), so \(\frac32k_BT\). This matches the kinetic theory result for every gas.
• If a molecule has f degrees of freedom, its average energy is \(\frac f2k_BT\). One mole has \(\frac f2RT\), and n moles have \(U=\frac f2nRT\).
• Ideal-gas molecules do not attract each other, so there is no intermolecular potential energy. U depends only on T, not on P or V.

A point atom moves in three directions, so N free atoms have 3N coordinates. Every rigid link between atoms removes one. A useful rule is \(f=3N-k\), where k is the number of independent constraints.

A diatomic molecule has only two rotational degrees of freedom. Its moment of inertia about the bond axis is almost zero, so rotation about that axis stores no energy. The same is true for any linear molecule. A non-linear molecule can rotate about all three axes.

One vibrational mode counts twice: once for the kinetic energy of the vibrating atoms and once for the potential energy of the stretched bond.

Equipartition is a classical result. Quantum mechanics allows rotational and vibrational energy only in steps. When \(k_BT\) is much smaller than the step size, that mode cannot be excited and contributes nothing: it is “frozen”. Vibrational steps are large, so diatomic vibrations are frozen at room temperature, and f = 5 is the right value there. The next section, on specific heats, shows how this makes heat capacities depend on temperature.

Heat capacity is \(C=\dfrac{Q}{n\Delta T}\), and Q depends on the process. Two processes are standard:

• Constant volume: no work is done, so \(Q=\Delta U=nC_V\Delta T\). For an ideal gas, \(C_V=\dfrac f2R\).
• Constant pressure: the gas expands and does work, so \(C_P>C_V\).

Other processes give other values. In an isothermal process heat flows in but ΔT = 0, so C is infinite. In an adiabatic process Q = 0 while T changes, so C = 0. A process can even have a negative heat capacity. Per-mole and per-kilogram values are related by \(C=Mc\).

1. Heat n moles of ideal gas at constant volume through ΔT: \(Q_V=nC_V\Delta T=\Delta U\).
2. Heat the same gas through the same ΔT at constant pressure: \(Q_P=nC_P\Delta T=\Delta U+P\Delta V\).
3. ΔU depends only on ΔT for an ideal gas, so it is the same in both cases: \(\Delta U=nC_V\Delta T\).
4. At constant P, \(P\Delta V=nR\Delta T\) from PV = nRT.
5. So \(nC_P\Delta T=nC_V\Delta T+nR\Delta T\), giving \(C_P-C_V=R\).

The same steps show how heat supplied at constant pressure is shared: the fraction going into internal energy is \(\dfrac{\Delta U}{Q}=\dfrac{C_V}{C_P}=\dfrac1\gamma\), and the fraction turned into work is \(\dfrac{W}{Q}=1-\dfrac1\gamma=\dfrac{2}{f+2}\). For a diatomic gas these are 5/7 and 2/7.

Since \(\gamma=1+2/f\), more degrees of freedom mean a smaller γ. γ always lies between 1 and 5/3 for an ideal gas. Useful inverses: \(C_V=\dfrac{R}{\gamma-1}\) and \(C_P=\dfrac{\gamma R}{\gamma-1}\).

Mixtures in one line. C_V values add with mole weights, so \(\dfrac{n_1+n_2}{\gamma_{mix}-1}=\dfrac{n_1}{\gamma_1-1}+\dfrac{n_2}{\gamma_2-1}\). The mixtures section derives this; never average γ directly.

Heat capacities are not fixed numbers. For hydrogen gas, C_V is about 3R/2 at very low temperature, because rotations are frozen. It rises to 5R/2 near room temperature when rotations switch on, and towards 7R/2 at a few thousand kelvin when vibration switches on. Classical equipartition cannot explain these steps; quantum theory does.

Solids (Dulong–Petit law). Each atom in a solid vibrates about a fixed point in three directions. Each direction has a kinetic and a potential energy term, so there are 6 squared terms and an energy of \(3k_BT\) per atom. One mole has \(U=3RT\), so \(C=3R\approx25\) J/mol K. For a solid, C_P ≈ C_V because its expansion work is tiny. This value holds at ordinary and high temperatures; at low temperatures the heat capacity falls towards zero.

In an ideal gas the molecules do not exert forces on each other between collisions. So each gas in a mixture hits the walls exactly as it would if it were alone in the container. Its pressure is \(P_i=\frac13\rho_iv_{rms,i}^2=n_iRT/V\), and the total pressure is the sum of these partial pressures.

All the gases share one temperature, so they share the same average translational kinetic energy per molecule, \(\frac32k_BT\). Their rms speeds still differ: the lighter gas moves faster.

The partial pressure fraction equals the mole fraction. If masses are given, divide each by its molar mass first.

A mixture behaves like a single ideal gas with an average molar mass, total mass divided by total moles:

\[M_{mix}=\frac{n_1M_1+n_2M_2}{n_1+n_2}\]

Example: dry air is roughly 79% N₂ and 21% O₂ by moles, so \(M_{mix}\approx0.79\times28+0.21\times32\approx28.8\) g/mol. The mixture density is then \(\rho=PM_{mix}/(RT)\).

Two containers are joined by a thin tube and no gas escapes. Moles are conserved: \(\dfrac{PV_{total}}{RT}=\dfrac{P_1V_1}{RT_1}+\dfrac{P_2V_2}{RT_2}\).

• If both start at the same temperature and it stays fixed, the final pressure is \(P=\dfrac{P_1V_1+P_2V_2}{V_1+V_2}\).
• If the system is insulated and does no work, the total internal energy is conserved: \(\frac{f_1}{2}n_1RT_1+\frac{f_2}{2}n_2RT_2=\left(\frac{f_1}{2}n_1+\frac{f_2}{2}n_2\right)RT\). So \(T=\dfrac{f_1n_1T_1+f_2n_2T_2}{f_1n_1+f_2n_2}\). For the same kind of gas this becomes \(T=\dfrac{n_1T_1+n_2T_2}{n_1+n_2}\).

Internal energies add, so the mixture has \(U=\frac{f_1}{2}n_1RT+\frac{f_2}{2}n_2RT\). Writing this as \(\frac{f_{mix}}{2}(n_1+n_2)RT\) gives

\[f_{mix}=\frac{n_1f_1+n_2f_2}{n_1+n_2},\qquad C_{V,mix}=\frac{n_1C_{V1}+n_2C_{V2}}{n_1+n_2},\qquad C_{P,mix}=C_{V,mix}+R\]

Since \(C_V=R/(\gamma-1)\), the same weighting gives \(\dfrac{n_1+n_2}{\gamma_{mix}-1}=\dfrac{n_1}{\gamma_1-1}+\dfrac{n_2}{\gamma_2-1}\). Never average the γ values directly. The specific heats section has a worked example.

Perfume molecules move at hundreds of metres per second, yet the smell takes many seconds to cross a room. The reason is collisions: each molecule zig-zags, changing direction billions of times every second.

1. Treat molecules as hard spheres of diameter d. Two molecules collide if their centres come within a distance d of each other.
2. Let one molecule move with average speed \(\bar v\) and imagine the others at rest. In time t it sweeps out a cylinder of radius d and length \(\bar vt\). Its volume is \(\pi d^2\bar vt\).
3. Every molecule whose centre lies inside this cylinder is hit. With \(n_N\) molecules per unit volume, the number of collisions is \(n_N\pi d^2\bar vt\).
4. Mean free path = distance travelled ÷ number of collisions = \(\dfrac{\bar vt}{n_N\pi d^2\bar vt}=\dfrac{1}{\pi d^2n_N}\).
5. The other molecules are moving too. The average relative speed is \(\sqrt2\,\bar v\), so there are \(\sqrt2\) times more collisions. This gives \(\lambda=\dfrac{1}{\sqrt2\pi d^2n_N}\).

The quantity \(\pi d^2\) is the collision cross-section. It uses the diameter, not the radius. If a question gives radius r, use \(\pi d^2=4\pi r^2\).

Using \(n_N=P/(k_BT)\), \(\lambda=\dfrac{k_BT}{\sqrt2\pi d^2P}\). The collision frequency (collisions per second for one molecule) is \(\nu=\bar v/\lambda=\sqrt2\pi d^2n_N\bar v\), and the mean time between collisions is \(\tau=1/\nu\).

For air at room conditions, λ is about 10⁻⁷ m. This is roughly 300 molecular diameters and about 30 times the average spacing between molecules. A molecule collides about 5×10⁹ times per second.