Kinetic Theory of Gases
Ideal gas equation, pressure of a gas, kinetic interpretation of temperature & equipartition
Ideal Gas and the Kinetic Model
Gas Laws and the Ideal Gas EquationTopic 1
An ideal gas is a simplified model in which molecules are treated as tiny, perfectly elastic point particles with no forces between them except during collisions. Real gases behave almost ideally at low pressure and high temperature, which is exactly the regime most NEET problems live in. The behaviour of such a gas is summarised by three experimental laws that combine into a single equation of state.
Boyle's law says that at constant temperature pressure and volume are inversely related, $PV = \text{constant}$. Charles's law says that at constant pressure volume is directly proportional to absolute temperature, $V \propto T$. Gay-Lussac's law says that at constant volume pressure is proportional to absolute temperature, $P \propto T$. Each requires temperature in kelvin, the most common slip in NEET calculations.
Combining the three gives the ideal gas equation $PV = nRT$, where $n$ is the number of moles and $R = 8.31\ \text{J mol}^{-1}\text{K}^{-1}$ is the universal gas constant. Writing it per molecule gives $PV = NkT$, where $N$ is the number of molecules and $k = R/N_A = 1.38\times10^{-23}\ \text{J K}^{-1}$ is Boltzmann's constant. These two forms are interchangeable and choosing the right one quickly is a useful exam habit.
The equation also captures Avogadro's law — equal volumes of gases at the same temperature and pressure contain equal numbers of molecules — and Dalton's law of partial pressures, which states that the total pressure of a non-reacting mixture is the sum of the pressures each gas would exert alone. Both appear regularly in NEET as one-line conceptual questions.
| Law | Relation (constant) |
|---|---|
| Boyle's law | $PV = $ const ($T$ fixed) |
| Charles's law | $V \propto T$ ($P$ fixed) |
| Gay-Lussac's law | $P \propto T$ ($V$ fixed) |
| Ideal gas equation | $PV = nRT = NkT$ |
A gas at $300\ \text{K}$ has volume $2\ \text{L}$ at constant pressure. What is its volume at $600\ \text{K}$?
Show solution
Charles's law: $V \propto T$, so $V_2 = V_1\dfrac{T_2}{T_1} = 2\times\dfrac{600}{300} = 4\ \text{L}$.
At constant temperature, a gas at pressure $P$ and volume $V$ is compressed to $V/4$. Find the new pressure.
Show solution
Boyle's law: $P_1 V_1 = P_2 V_2$, so $P_2 = P\dfrac{V}{V/4} = 4P$. Quartering the volume quadruples the pressure.
The ideal gas equation is:
Boyle's law holds at constant:
Boltzmann's constant $k$ equals:
Equal volumes of gases at the same $T$ and $P$ contain equal numbers of molecules. This is:
The total pressure of a non-reacting gas mixture equals the sum of partial pressures. This is:
NEET tip: Always use absolute (kelvin) temperature in the gas laws. Pick $PV = nRT$ for moles and $PV = NkT$ when the number of molecules is given.
Kinetic Theory of Pressure and rms SpeedTopic 2
The triumph of the kinetic model is that it derives the gas laws from mechanics. Picture molecules in random motion, bouncing elastically off the container walls. Each collision transfers momentum to the wall, and the average rate of momentum transfer is the force, hence the pressure. Working this out gives the central result $P = \dfrac{1}{3}\dfrac{mN}{V}\overline{v^{2}} = \dfrac{1}{3}\rho\,\overline{v^{2}}$, where $\overline{v^{2}}$ is the mean square speed and $\rho$ the density.
The square root of the mean square speed is the root-mean-square (rms) speed, the single most-used quantity here. Combining the pressure result with $PV = nRT$ gives $v_{rms} = \sqrt{\dfrac{3RT}{M}} = \sqrt{\dfrac{3kT}{m}}$, where $M$ is the molar mass and $m$ the mass of one molecule. Two consequences are tested constantly: rms speed rises with the square root of temperature, and at a given temperature lighter molecules move faster — which is why hydrogen and helium escape Earth's atmosphere while heavier gases stay.
It helps to distinguish three speeds that NEET likes to compare. The most probable speed $v_p = \sqrt{2RT/M}$ is the speed at the peak of the distribution; the average speed $v_{avg} = \sqrt{8RT/\pi M}$; and the rms speed $v_{rms} = \sqrt{3RT/M}$. Their fixed ordering $v_p < v_{avg} < v_{rms}$ is a frequently asked one-mark fact, and all three scale the same way with $\sqrt{T/M}$.
A subtle but examinable point: pressure depends on the mean square speed, not on the square of the mean speed. Because faster molecules contribute disproportionately, the rms speed always slightly exceeds the simple average speed. The kinetic picture also explains diffusion, effusion (Graham's law, where rate $\propto 1/\sqrt{M}$), and why a gas exerts the same pressure in all directions.
| Quantity | Expression |
|---|---|
| Pressure (kinetic) | $P = \dfrac{1}{3}\rho\,\overline{v^{2}}$ |
| rms speed | $v_{rms} = \sqrt{3RT/M}$ |
| Most probable speed | $v_p = \sqrt{2RT/M}$ |
| Average speed | $v_{avg} = \sqrt{8RT/\pi M}$ |
By what factor does the rms speed of a gas change when its absolute temperature is increased four-fold?
Show solution
$v_{rms} \propto \sqrt{T}$, so a four-fold rise in $T$ multiplies the rms speed by $\sqrt{4} = 2$.
At the same temperature, compare the rms speeds of oxygen ($M = 32$) and hydrogen ($M = 2$).
Show solution
$v_{rms} \propto 1/\sqrt{M}$, so $\dfrac{v_{H_2}}{v_{O_2}} = \sqrt{\dfrac{32}{2}} = \sqrt{16} = 4$. Hydrogen moves four times faster.
The pressure of an ideal gas is given by:
The rms speed of gas molecules is proportional to:
At the same temperature, a lighter gas has an rms speed that is:
The correct ordering of molecular speeds is:
Graham's law of effusion states the rate is proportional to:
NEET tip: Memorise $v_{rms} = \sqrt{3RT/M}$ and the ordering $v_p < v_{avg} < v_{rms}$. All three speeds scale as $\sqrt{T/M}$.
Energy, Degrees of Freedom and Mean Free Path
Kinetic Interpretation of Temperature and EquipartitionTopic 3
Kinetic theory gives temperature a beautifully simple meaning. Comparing $P = \tfrac{1}{3}\rho\,\overline{v^{2}}$ with $PV = NkT$ shows that the average translational kinetic energy of a single molecule is $\overline{E} = \dfrac{3}{2}kT$. So absolute temperature is a direct measure of the average kinetic energy of molecules — at the same temperature, molecules of every gas have the same average translational kinetic energy, regardless of their mass. This is one of the most powerful one-line results in the chapter.
A consequence is that translational kinetic energy, and hence molecular motion, would cease at absolute zero ($0\ \text{K}$). Heavier molecules simply move more slowly to carry the same energy, which is the deeper reason behind the $1/\sqrt{M}$ dependence of rms speed.
Molecules can store energy in more than just straight-line motion. A degree of freedom is an independent way a molecule can hold energy. A monatomic gas has $3$ translational degrees of freedom; a diatomic gas adds $2$ rotational ones (giving $5$ at ordinary temperatures), and more at high temperature when vibration switches on. The law of equipartition of energy states that each degree of freedom carries, on average, an energy $\tfrac{1}{2}kT$ per molecule.
Equipartition fixes the specific heats. The internal energy per mole is $U = \dfrac{f}{2}RT$ for $f$ degrees of freedom, giving $C_V = \dfrac{f}{2}R$, $C_P = C_V + R$, and the ratio $\gamma = \dfrac{C_P}{C_V} = 1 + \dfrac{2}{f}$. Hence $\gamma = 5/3$ for a monatomic gas and $\gamma = 7/5$ for a diatomic gas — values that appear directly in NEET adiabatic and specific-heat questions.
| Gas type | $f$, $C_V$, $\gamma$ |
|---|---|
| Monatomic | $f = 3,\ C_V = \tfrac{3}{2}R,\ \gamma = \tfrac{5}{3}$ |
| Diatomic | $f = 5,\ C_V = \tfrac{5}{2}R,\ \gamma = \tfrac{7}{5}$ |
| Energy per d.o.f. | $\tfrac{1}{2}kT$ per molecule |
Find the average translational kinetic energy of a molecule of any ideal gas at $300\ \text{K}$. ($k = 1.38\times10^{-23}\ \text{J K}^{-1}$)
Show solution
$\overline{E} = \dfrac{3}{2}kT = \dfrac{3}{2}\times(1.38\times10^{-23})\times300 \approx 6.2\times10^{-21}\ \text{J}$ — the same for every gas at this temperature.
What is the ratio $\gamma = C_P/C_V$ for a diatomic gas (translational + rotational)?
Show solution
With $f = 5$, $\gamma = 1 + \dfrac{2}{f} = 1 + \dfrac{2}{5} = \dfrac{7}{5} = 1.4$.
The average translational KE of a molecule is:
At the same temperature, the average KE of $\text{H}_2$ and $\text{O}_2$ molecules is:
Each degree of freedom contributes energy:
A monatomic ideal gas has $\gamma$ equal to:
For a diatomic gas at ordinary temperature, the number of degrees of freedom is:
NEET tip: Average KE $= \tfrac{3}{2}kT$ is the same for all gases at a given $T$. Use $\gamma = 1 + 2/f$: monatomic $5/3$, diatomic $7/5$.
Mean Free Path and Real GasesTopic 4
Although molecules travel fast — hundreds of metres per second at room temperature — a gas mixes slowly, because each molecule constantly collides with others and zig-zags rather than moving straight. The average distance a molecule travels between successive collisions is the mean free path $\lambda = \dfrac{1}{\sqrt{2}\,\pi d^{2} n}$, where $d$ is the molecular diameter and $n$ the number of molecules per unit volume. The key NEET reading is the proportionality, not the constant in front.
The formula tells a clear physical story. The mean free path is shorter when the gas is denser (larger $n$) or the molecules are bigger (larger $d$), and longer at low density. Since $n = P/kT$, the mean free path increases with temperature at fixed pressure and decreases with pressure at fixed temperature — so in a good vacuum a molecule can travel a long way before hitting anything, which matters for vacuum technology and for how sound and heat move in thin gases.
This collision picture also explains transport phenomena: diffusion (transport of molecules), viscosity (transport of momentum) and thermal conduction in gases (transport of energy) all depend on the mean free path. A larger mean free path generally means faster transport, linking this topic back to the heat-transfer ideas of the previous chapter.
Finally, the ideal gas is only a model. Real gases deviate from $PV = nRT$ at high pressure and low temperature, where molecules are close enough for their finite size and mutual attraction to matter. The van der Waals equation corrects for both effects, and gases liquefy below a characteristic critical temperature — phenomena NEET treats qualitatively but which are worth recognising as the limits of the ideal model.
| Factor | Effect on mean free path $\lambda$ |
|---|---|
| Number density $n$ | $\lambda \propto 1/n$ (denser to shorter) |
| Molecular diameter $d$ | $\lambda \propto 1/d^{2}$ |
| Temperature (fixed $P$) | $\lambda \propto T$ |
| Pressure (fixed $T$) | $\lambda \propto 1/P$ |
How does the mean free path of a gas change if its pressure is doubled at constant temperature?
Show solution
$\lambda \propto 1/n \propto 1/P$ at constant $T$. Doubling the pressure halves the mean free path.
Under what conditions does a real gas behave most like an ideal gas?
Show solution
At low pressure and high temperature, where molecules are far apart and intermolecular forces and finite size are negligible.
The mean free path of a gas molecule is:
The mean free path varies with number density $n$ as:
At constant temperature, increasing pressure makes the mean free path:
Real gases behave most ideally at:
The van der Waals equation corrects the ideal gas law for:
NEET tip: Read $\lambda$ through its proportionalities: $\lambda \propto 1/n \propto T/P$. Ideal behaviour is best at low pressure and high temperature.
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