Kinetic Theory of an Ideal Gases

Q: What are the main assumptions of the kinetic theory of an ideal gas?

The gas is a huge number of identical point molecules in random motion. Their own volume is negligible, there are no forces between them except during very short elastic collisions, and they obey Newton's laws. From these assumptions pressure and temperature can be derived.

Q: How does kinetic theory explain gas pressure?

Each molecule that hits a wall reverses its momentum and gives the wall a small impulse. Billions of hits every second add up to a steady average force per unit area. The result is that pressure equals one third of density times the mean square speed.

Q: What is the kinetic interpretation of temperature?

Absolute temperature is proportional to the average translational kinetic energy of the molecules. Each molecule carries on average three halves k T of translational kinetic energy, whatever the gas, so a hotter gas simply has faster molecules on average.

Q: What is the difference between rms speed, average speed and most probable speed?

The most probable speed is at the peak of the Maxwell curve, the average speed is the ordinary mean, and the rms speed is the square root of the mean of the squared speeds. They are in the ratio 1 to 1.128 to 1.225.

Q: What does the Maxwell distribution curve show?

It shows how molecular speeds are shared out at a given temperature. The area between two speeds is the fraction of molecules in that range and the total area is one. On heating, the peak moves to higher speed and becomes lower and broader.

Q: What is mean free path and what does it depend on?

Mean free path is the average distance a molecule travels between successive collisions. It is inversely proportional to the number density and to the square of the molecular diameter, so at constant temperature it doubles when the pressure is halved.

Q: Which kinetic theory questions come in NEET?

NEET usually asks the ratio of rms speeds of two gases, the change of rms speed with temperature, the average kinetic energy per molecule, or pressure from density and rms speed. Remember to use kelvin and that every speed varies as the square root of T over M.

Q: How is kinetic theory tested in JEE Main?

JEE Main asks numericals on rms, average and most probable speeds, the relation between pressure and kinetic energy, Maxwell distribution graphs at different temperatures or masses, and mean free path. It is often combined with equipartition and specific heats in one question.

PhysicsKinetic TheoryFor JEE aspirants

The kinetic theory of an ideal gas explains pressure and temperature from the motion of molecules. Treating a gas as a huge number of tiny molecules in random motion that collide elastically, it proves $P V = \frac{1}{3} N m \overline{v^{2}}$ , shows that the average kinetic energy of a molecule is $\frac{3}{2} k T$ , and gives the rms, average and most probable speeds, the Maxwell speed distribution and the mean free path. The kinetic theory of an ideal gas is a regular source of direct questions in JEE Main and NEET.

On this page1Postulates2Pressure of a gas3Meaning of temperature4Gas laws from KTG5Molecular speeds6Maxwell distribution7Mean free path

Key Formulas - Quick Reference

★ Must learnPressure of an ideal gas: $P V = \frac{1}{3} N m \overline{v^{2}} = \frac{1}{3} N m v_{r m s}^{2}$ , or $P = \frac{1}{3} ρ v_{r m s}^{2}$
Pressure and energy: $P = \frac{2}{3} \frac{E}{V}$ , where $E$ is the total translational kinetic energy
★ Must learnAverage translational KE of a molecule: $\overline{K E} = \frac{1}{2} m \overline{v^{2}} = \frac{3}{2} k T$ ; per mole $\frac{3}{2} R T$ ; for $n$ moles $E = \frac{3}{2} n R T$
Ideal gas equation: $P V = n R T = N k T$ , with $k = \frac{R}{N _{A}} = 1.38 \times 1 0^{- 23} J K^{- 1}$
★ Must learnSpeeds: $v_{r m s} = \frac{3 R T}{M}$ , $v_{a v} = \frac{8 R T}{π M}$ , $v_{m p} = \frac{2 R T}{M}$ ( $M$ in $kg mol^{- 1}$ )
★ Must learn $v_{m p} : v_{a v} : v_{r m s} = 2 : 8/ π : 3 = 1 : 1.128 : 1.225$
Maxwell distribution: $f (v) = 4 π (\frac{M}{2 π R T})^{3/2} v^{2} e^{- M v^{2} /2 R T}$ , total area $= 1$
★ Must learnMean free path: $λ = \frac{1}{2 π d ^{2} n} = \frac{k T}{2 π d ^{2} P}$

1. Molecular Nature of Matter and the Postulates

Every gas is made of molecules. At ordinary pressures they are far apart compared with their size, so they move freely in straight lines between collisions. Boyle's, Charles' and Avogadro's laws are what we see; kinetic theory explains them from the motion of the molecules, using only Newton's laws and averages.

For $n$ moles containing $N = n N_{A}$ molecules, the ideal gas equation can be written with the gas constant per mole ( $R$ ) or per molecule, the Boltzmann constant $k = \frac{R}{N _{A}}$ :

P V = n R T = N k T, k = \frac{8.314}{6.022 \times 1 0 ^{23}} = 1.38 \times 1 0^{- 23} J K^{- 1}

Figure 1: The kinetic model. A gas is a huge number of tiny, identical molecules moving randomly in all directions. At STP the average spacing

(\frac{V}{N})^{1/3} \approx 3.3 nm

is about ten times the molecular size, so the molecules themselves fill less than

0.1%

of the volume.

★ Must learn

Postulates of the kinetic theory of an ideal gas

A gas consists of a very large number of identical molecules (same mass and size).
The volume of the molecules is negligible compared with the volume of the container (point molecules).
The molecules move randomly in all directions with all possible speeds, colliding with one another and with the walls.
The collisions are perfectly elastic: total kinetic energy and momentum are conserved.
There are no intermolecular forces except during collisions, which last a negligible time. Between collisions a molecule moves in a straight line with constant velocity.
Newton's laws apply to the motion of each molecule; gravity has negligible effect.
The molecules are uniformly distributed, and every direction of motion is equally likely.

A gas that obeys all these assumptions exactly is an ideal gas. Real gases come close at low pressure and high temperature, when molecules are far apart and the attractive forces and molecular volume matter little.

Key idea

Kinetic theory replaces the gas by point molecules in random, elastic motion with no forces between them. Everything else on this page follows from Newton's laws and averaging.

2. Pressure Exerted by an Ideal Gas

Pressure is the average force per unit area that the molecules exert on the walls by bouncing off them. Take a cube of side $l$ containing $N$ molecules, each of mass $m$ (Figure 2).

Figure 2: Set-up for the pressure derivation. A molecule with velocity component

v_{x}

hits face A elastically: its

x

-momentum reverses, so each hit gives the wall an impulse

2 m v_{x}

. The

y

and

z

components are unchanged.

A molecule with velocity $(v_{x}, v_{y}, v_{z})$ hits face A (area $l^{2}$ ) elastically. Its $x$ -momentum changes from $+ m v_{x}$ to $- m v_{x}$ , so the wall receives an impulse $2 m v_{x}$ .
It travels to the opposite face and back, a distance $2 l$ , before hitting A again. Time between hits: $Δ t = \frac{2 l}{v _{x}}$ .
Average force by this molecule on A: $\frac{2 m v _{x}}{2 l / v _{x}} = \frac{m v _{x}^{2}}{l}$ .
Adding all $N$ molecules: $F = \frac{m}{l} (v_{x 1}^{2} + v_{x 2}^{2} + \dots + v_{x N}^{2}) = \frac{m N}{l} \overline{v_{x}^{2}}$ .
Motion is random, so no direction is special: $\overline{v_{x}^{2}} = \overline{v_{y}^{2}} = \overline{v_{z}^{2}}$ . Since $v^{2} = v_{x}^{2} + v_{y}^{2} + v_{z}^{2}$ , we get $\overline{v_{x}^{2}} = \frac{1}{3} \overline{v^{2}}$ .
Pressure $P = \frac{F}{l ^{2}} = \frac{1}{3} \frac{m N}{l ^{3}} \overline{v^{2}}$ , and $l^{3} = V$ :

★ Must learn

P V = \frac{1}{3} N m \overline{v^{2}} or P = \frac{1}{3} ρ v_{r m s}^{2}

Here $ρ = \frac{N m}{V}$ is the density of the gas and $v_{r m s} = \overline{v^{2}} = \frac{v _{1}^{2} + v _{2}^{2} + \dots + v _{N}^{2}}{N}$ is the root mean square speed. Collisions between molecules do not change the result: they only swap momenta, and the average stays the same in steady state.

Figure 3: One molecule hits face A once every

Δ t = \frac{2 l}{v _{x}}

(top). The force on the wall is a train of spikes of impulse

2 m v_{x}

(bottom); averaged over time it is

\frac{2 m v _{x}}{2 l / v _{x}} = \frac{m v _{x}^{2}}{l}

. With

\sim 1 0^{23}

molecules the spikes merge into a steady pressure.

Pressure in terms of energy. The total translational kinetic energy is $E = \frac{1}{2} N m \overline{v^{2}}$ , so

P = \frac{2}{3} \frac{E}{V}

Pressure equals two-thirds of the translational kinetic energy per unit volume. The same idea gives Dalton's law: in a mixture each gas contributes its own $\frac{2}{3} \frac{E _{i}}{V}$ , so $P = P_{1} + P_{2} + \dots$

Exam Trick

Units of $P = \frac{1}{3} ρ v_{r m s}^{2}$ : $kg m^{- 3} \times m^{2} s^{- 2} = N m^{- 2}$ . So if a question gives density and pressure, the rms speed is $v_{r m s} = \frac{3 P}{ρ}$ directly, with no need for $T$ or $M$ . For air at STP ( $ρ = 1.29 kg m^{- 3}$ ) this is about $485 m s^{- 1}$ .

Quick Recall: tap to check

Why does the factor

\frac{1}{3}

appear in

P V = \frac{1}{3} N m \overline{v^{2}}

Random motion shares

\overline{v^{2}}

equally among

x

y

and

z

\overline{v_{x}^{2}} = \frac{1}{3} \overline{v^{2}}

Does the shape of the container change the result?

No. The cube only makes the algebra easy; the result holds for any shape.

v_{r m s}

doubles at constant volume, what happens to

P

P \propto v_{r m s}^{2}

, so

P

becomes four times.

3. Kinetic Interpretation of Temperature

Compare the kinetic result with the experimental gas equation:

P V = \frac{2}{3} N (\frac{1}{2} m \overline{v^{2}}) and P V = N k T \Rightarrow \frac{1}{2} m \overline{v^{2}} = \frac{3}{2} k T

★ Must learn

The average translational kinetic energy of a gas molecule depends only on the absolute temperature:

\overline{K E} = \frac{1}{2} m \overline{v^{2}} = \frac{3}{2} k T

Absolute temperature is a measure of the average translational kinetic energy of the molecules. It does not depend on the nature of the gas, its pressure or its volume.

Figure 4:

\overline{K E} = \frac{3}{2} k T

is a straight line through the origin with slope

\frac{3}{2} k

. It is the same for every gas: at

300 K

a helium atom and a

C O_{2}

molecule both carry

6.21 \times 1 0^{- 21} J

of translational KE on average. Absolute temperature is a measure of this energy.

Quantity	Formula	At 300 K
Per molecule	$\frac{3}{2} k T$	$6.21 \times 1 0^{- 21} J$ ( $0.039 eV$ )
Per mole	$\frac{3}{2} R T$	$3.74 kJ mol^{- 1}$
For $n$ moles	$E = \frac{3}{2} n R T = \frac{3}{2} P V$	depends on $n$
Per unit mass	$\frac{3 R T}{2 M}$	depends on $M$

Per molecule:

\frac{3}{2} k T

Same for every gas at the same $T$ . A helium atom and an oxygen molecule at $300 K$ have the same average translational KE.

Per unit mass:

\frac{3 R T}{2 M}

Depends on the gas. At the same $T$ , $1 g$ of hydrogen has 16 times the translational KE of $1 g$ of oxygen, because it has 16 times as many molecules.

At absolute zero this classical picture gives zero kinetic energy. Real gases liquefy long before that, and quantum mechanics leaves a small zero-point energy, but the ideal-gas result is exact enough for all exam problems.

4. Gas Laws from Kinetic Theory

Write the kinetic equation as $P V = \frac{2}{3} N \overline{K E}$ with $\overline{K E} \propto T$ . Each gas law follows by holding some quantities fixed:

Law	Held fixed	Kinetic reasoning	Result
Boyle's law	$N$ , $T$	$\overline{K E}$ fixed, so $P V$ fixed	$P V = constant$
Charles' law	$N$ , $P$	$V = \frac{2 N K E}{3 P} \propto T$	$V \propto T$
Gay-Lussac's law	$N$ , $V$	$P \propto \overline{K E} \propto T$	$P \propto T$
Avogadro's law	$P$ , $V$ , $T$	$N = \frac{3 P V}{2 K E}$ is the same for all gases	equal volumes hold equal numbers of molecules
Graham's law	$P$ , $T$	$ρ_{1} v_{1}^{2} = ρ_{2} v_{2}^{2}$ from $P = \frac{1}{3} ρ v_{r m s}^{2}$	$\frac{r _{1}}{r _{2}} = \frac{ρ _{2}}{ρ _{1}} = \frac{M _{2}}{M _{1}}$
Dalton's law	$V$ , $T$	each gas adds $\frac{2}{3} \frac{E _{i}}{V}$	$P = P_{1} + P_{2} + \dots$

Key idea

One equation,

P V = \frac{2}{3} N \overline{K E}

with

\overline{K E} = \frac{3}{2} k T

, contains Boyle, Charles, Gay-Lussac, Avogadro, Graham and Dalton.

5. Molecular Speeds: rms, Average and Most Probable

Molecules do not all move at the same speed, so three different "typical" speeds are used. Each depends on $T / M$ ; only the numerical factor differs.

5.1 Root mean square speed

From $\frac{1}{2} m \overline{v^{2}} = \frac{3}{2} k T$ :

v_{r m s} = \overline{v^{2}} = \frac{3 k T}{m} = \frac{3 R T}{M} = \frac{3 P}{ρ}

$m$ is the mass of one molecule; $M = m N_{A}$ is the molar mass in $kg mol^{- 1}$ . It depends on the nature of the gas through $M$ .

5.2 Average (mean) speed

The arithmetic mean of the speeds, $v_{a v} = \frac{v _{1} + v _{2} + \dots + v _{N}}{N}$ . Using the Maxwell distribution (Section 6):

v_{a v} = \frac{8 k T}{π m} = \frac{8 R T}{π M}

5.3 Most probable speed

The speed possessed by the largest fraction of molecules at a given temperature (the peak of the distribution):

v_{m p} = \frac{2 k T}{m} = \frac{2 R T}{M}

Figure 5: The three speeds for

O_{2}

300 K

are fixed multiples of

R T / M = 279 m s^{- 1}

v_{m p} : v_{a v} : v_{r m s} = 2 : 8/ π : 3 = 1 : 1.128 : 1.225

. The order never changes.

★ Must learn

v_{m p} : v_{a v} : v_{r m s} = 2 : \frac{8}{π} : 3 = 1 : 1.128 : 1.225

Always $v_{m p} < v_{a v} < v_{r m s}$ , for every gas at every temperature.

Change	Effect on each speed
$T$ becomes $4 T$	doubles
$M$ becomes $4 M$ (same $T$ )	halves
$P$ changes at constant $T$	no change ( $P / ρ$ stays constant)
Same $\frac{T}{M}$ for two gases	same speeds

Exam Trick

Every speed scales as $T / M$ . For ratio questions never compute the speed: $\frac{v _{1}}{v _{2}} = \frac{T _{1} M _{2}}{T _{2} M _{1}}$ . Hydrogen at $300 K$ has the same rms speed as oxygen at $300 \times 16 = 4800 K$ . Remember $R T / M$ for air at room temperature is about $290 m s^{- 1}$ ; multiply by $1.41$ , $1.60$ , $1.73$ .

Figure 6:

v_{r m s} = \frac{3 R T}{M}

grows as

T

and falls as

\frac{1}{M}

. Hydrogen at

300 K

already moves at

1.93 km s^{- 1}

(its rms speed reaches the Moon's escape speed at about

450 K

), and a fraction of the Maxwell tail is far faster. That is why the Moon (escape speed

2.4 km s^{- 1}

) has no atmosphere and Earth (

11.2 km s^{- 1}

) has lost almost all its free hydrogen and helium.

v_{r m s}

Square, average, then square root. Linked to energy: $\frac{1}{2} m v_{r m s}^{2} = \frac{3}{2} k T$ . Used for pressure and kinetic energy.

v_{a v}

Plain average of speeds. About $8%$ smaller than $v_{r m s}$ . Used for collision rates, effusion and mean free path time.

6. Maxwell Distribution of Molecular Speeds

Collisions constantly change individual speeds, but at a fixed temperature the fraction of molecules in each speed range stays constant. Maxwell found this distribution using probability theory. If $f (v) d v$ is the fraction of molecules with speeds between $v$ and $v + d v$ :

f (v) = 4 π (\frac{M}{2 π R T})^{3/2} v^{2} e^{- M v^{2} /2 R T}

Figure 7: Maxwell distribution for

N_{2}

300 K

(computed from

f (v) = 4 π (\frac{M}{2 π R T})^{3/2} v^{2} e^{- M v^{2} /2 R T}

). Because of the long high-speed tail,

v_{m p} = 422 < v_{a v} = 476 < v_{r m s} = 517 m s^{- 1}

. The height of the curve times a small speed interval

d v

is the fraction of molecules in that interval.

It is a statistical result: it describes the whole collection of molecules, not one molecule.
Speeds range from $0$ to $\infty$ , but very few molecules are very slow or very fast.
The area under the curve between $v_{1}$ and $v_{2}$ is the fraction of molecules with speeds in that range; the total area is 1 at every temperature.
The peak is at the most probable speed $v_{m p}$ . The curve is not symmetric: the long tail pulls $v_{a v}$ and $v_{r m s}$ to the right of the peak.

Figure 8: Heating from

300 K

900 K

shifts the peak to higher speed (

v_{m p} \propto T

, here

\times 3

), lowers it and flattens the curve; the area stays 1. The fraction of molecules slower than

u_{1} = 400 m s^{- 1}

(shaded) drops from

38%

10%

Effect of temperature. On heating, the peak moves to a higher speed and becomes lower, and the curve spreads out (the area must stay 1). The fraction of molecules slower than any given speed decreases, and the fraction faster than it increases. This is why reaction rates and evaporation rise sharply with temperature: they depend on the fast tail.

Figure 9: At the same temperature,

v_{m p} = \frac{2 R T}{M}

: the heaviest gas (

C O_{2}

) peaks at the lowest speed with the tallest, narrowest curve; helium peaks near

1100 m s^{- 1}

. The order of molar masses is

M_{C O_{2}} > M_{N_{2}} > M_{He}

Effect of molar mass. At the same temperature a heavier gas has a narrower, taller curve that peaks at a lower speed; a light gas like helium has a low, broad curve.

JEE Advanced

The speeds follow from averages over $f (v)$ (use $\int_{0}^{\infty} v^{2 n + 1} e^{- a v^{2}} d v = \frac{n !}{2 a ^{n + 1}}$ and its even-power partner, with $a = \frac{M}{2 R T}$ ):

$\frac{df}{d v} = 0$ gives $2 v - 2 a v^{3} = 0$ , so $v_{m p} = 1/ a = \frac{2 R T}{M}$ .
$v_{a v} = \int_{0}^{\infty} v f (v) d v = \frac{8 R T}{π M}$ and $\overline{v^{2}} = \int_{0}^{\infty} v^{2} f (v) d v = \frac{3 R T}{M}$ .
Each velocity component has a Gaussian distribution with mean zero: $g (v_{x}) = \frac{a}{π} e^{- a v_{x}^{2}}$ , so $\overline{v_{x}} = 0$ but $\overline{v_{x}^{2}} = \frac{R T}{M}$ .
Effusion through a small hole: molecules escape per unit area per second at $\frac{1}{4} n v_{a v}$ , which is the origin of Graham's law.

Quick Recall: tap to check

What does the total area under a Maxwell curve represent?

The fraction of molecules with any speed from zero to infinity, which is 1 at all temperatures.

On heating, does the peak of the Maxwell curve rise or fall?

It falls and moves to higher speed, because the curve spreads out while keeping area 1.

Which is largest:

v_{m p}

v_{a v}

v_{r m s}

v_{r m s}

. The order is always

v_{m p} < v_{a v} < v_{r m s}

7. Mean Free Path

Although molecules move at hundreds of metres per second, a smell spreads across a room slowly. The reason is collisions: a molecule's path is a zigzag of short straight pieces. The average distance travelled between two successive collisions is the mean free path $λ$ .

Figure 10: Mean free path. A molecule collides with any molecule whose centre comes within

d

of its own centre, so in time

t

it sweeps a cylinder of volume

π d^{2} ⟨ v ⟩ t

and makes

nπ d^{2} ⟨ v ⟩ t

collisions. Allowing for the motion of the others (factor

2

λ = \frac{1}{2 π d ^{2} n}

Model molecules as spheres of diameter $d$ . Two molecules collide if their centres come within $d$ of each other.
In time $t$ a molecule sweeps a cylinder of cross-section $π d^{2}$ and length $⟨ v ⟩ t$ . With $n$ molecules per unit volume, the number of collisions is $nπ d^{2} ⟨ v ⟩ t$ .
Mean free path = distance / number of collisions $= \frac{1}{nπ d ^{2}}$ . The other molecules also move; using the relative speed ( $2$ times larger on average) gives:

★ Must learn

λ = \frac{1}{2 π d ^{2} n} = \frac{k T}{2 π d ^{2} P}

using $n = \frac{P}{k T}$ . At fixed $T$ , $λ \propto \frac{1}{P}$ ; at fixed $P$ , $λ \propto T$ ; at fixed density, $λ$ does not change with $T$ . Collision frequency $= \frac{v _{a v}}{λ}$ .

For nitrogen at STP ( $d \approx 3 \times 1 0^{- 10} m$ ), $λ \approx 9 \times 1 0^{- 8} m$ , about 300 molecular diameters, and each molecule collides about five billion times per second.

Key idea

Fast molecules, slow spreading: a molecule collides billions of times a second, and

λ = \frac{1}{2 π d ^{2} n}

is only about

0.1 μ m

in air at STP.

8. Choosing the Formula and Revision Map

Use the flowchart to pick the formula, then the mind map to revise the whole concept.

Figure 11: Picking the right kinetic-theory formula. Most slips come from using

M

g mol^{- 1}

(use

kg mol^{- 1}

) or

T

^{\circ} C

(use kelvin).

Exam Trick

Energy from pressure in one step: the translational KE of any ideal gas is $E = \frac{3}{2} P V$ . A $2 L$ cylinder at $10 atm$ holds $\frac{3}{2} (10 \times 1.013 \times 1 0^{5}) (2 \times 1 0^{- 3}) \approx 3.0 kJ$ of translational KE, whatever the gas and temperature.

Figure 12: Mind map of this concept. Cover a branch, recall its three points and formulas, then check.

9. Solved Examples

Solved Example 1

In a container of capacity

1 L

there are

1 0^{23}

molecules, each of mass

1 0^{- 22} g

. If the root mean square speed is

1 0^{5} cm s^{- 1}

, calculate the pressure of the gas.

Solution:

Given: $V = 1 0^{- 3} m^{3}$ , $N = 1 0^{23}$ , $m = 1 0^{- 25} kg$ , $v_{r m s} = 1 0^{3} m s^{- 1}$ , so $\overline{v^{2}} = 1 0^{6} m^{2} s^{- 2}$ .

$P = \frac{N m v ^{2}}{3 V} = \frac{1 0 ^{23} \times 1 0 ^{- 25} \times 1 0 ^{6}}{3 \times 1 0 ^{- 3}} = \frac{1}{3} \times 1 0^{7}$ .

Answer: $P \approx 3.33 \times 1 0^{6} Pa$ (about 33 atm).

Solved Example 2

For the gas of Example 1: (a) find the pressure in CGS units, (b) the total translational kinetic energy in calories, and (c) the temperature of the gas.

Solution:

(a) $P = \frac{N m v ^{2}}{3 V} = \frac{1 0 ^{23} \times 1 0 ^{- 22} g \times ( 1 0 ^{5} cm s ^{- 1} ) ^{2}}{3 \times 1000 cm ^{3}} = 3.33 \times 1 0^{7} dyn cm^{- 2}$ (same as $3.33 \times 1 0^{6} Pa$ ).

(b) $E = \frac{1}{2} N m \overline{v^{2}} = \frac{1}{2} (1 0^{23}) (1 0^{- 25}) (1 0^{6}) = 5000 J = \frac{5000}{4.184} cal \approx 1195 cal$ .

(c) $\frac{1}{2} m \overline{v^{2}} = \frac{3}{2} k T \Rightarrow T = \frac{m v ^{2}}{3 k} = \frac{1 0 ^{- 25} \times 1 0 ^{6}}{3 \times 1.38 \times 1 0 ^{- 23}}$ .

Answer: (a) $3.33 \times 1 0^{7} dyn cm^{- 2}$ ; (b) $5000 J \approx 1195 cal$ ; (c) $T \approx 2.41 \times 1 0^{3} K$ .

Solved Example 3

For two gases of molar masses

M_{A}

and

M_{B}

at temperatures

T_{A}

and

T_{B}

T_{A} M_{B} = T_{B} M_{A}

. Which property has the same value for both gases?
(A) Density
(B) Pressure
(C) KE per mole
(D) rms speed

Solution:

Answer: (D). $T_{A} M_{B} = T_{B} M_{A}$ means $\frac{T _{A}}{M _{A}} = \frac{T _{B}}{M _{B}}$ , and $v_{r m s} = \frac{3 R T}{M}$ depends only on $\frac{T}{M}$ . KE per mole ( $\frac{3}{2} R T$ ) needs equal temperatures; density and pressure need more data.

Solved Example 4

Hydrogen is thought to have escaped from the early Earth because of the high temperature. Taking the escape speed as

1.1 \times 1 0^{6} cm s^{- 1}

and assuming the average speed of

H_{2}

equalled it, estimate the temperature of the Earth at that time.

Solution:

Given: $v_{a v} = 1.1 \times 1 0^{4} m s^{- 1}$ , $M = 2 \times 1 0^{- 3} kg mol^{- 1}$ .

$v_{a v} = \frac{8 R T}{π M} \Rightarrow T = \frac{π M v _{a v}^{2}}{8 R} = \frac{3.1416 \times 2 \times 1 0 ^{- 3} \times ( 1.1 \times 1 0 ^{4} ) ^{2}}{8 \times 8.314}$ .

Answer: $T \approx 1.14 \times 1 0^{4} K$ ( $\approx 1115 7^{\circ} C$ ).

This is a rough estimate. In reality a planet loses a gas slowly, over geological time (billions of years), once its rms speed exceeds only about one-sixth of the escape speed, because collisions keep refilling the fast tail of the Maxwell distribution.

Solved Example 5

Find the rms, average and most probable speeds of oxygen molecules at

2 7^{\circ} C

. (

R = 8.314 J mol^{- 1} K^{- 1}

)

Solution:

$T = 300 K$ , $M = 0.032 kg mol^{- 1}$ , so $\frac{R T}{M} = \frac{8.314 \times 300}{0.032} = 7.79 \times 1 0^{4} m^{2} s^{- 2}$ .

$v_{r m s} = 3 \times 7.79 \times 1 0^{4}$ , $v_{a v} = \frac{8}{π} \times 7.79 \times 1 0^{4}$ , $v_{m p} = 2 \times 7.79 \times 1 0^{4}$ .

Answer: $v_{r m s} \approx 484 m s^{- 1}$ , $v_{a v} \approx 446 m s^{- 1}$ , $v_{m p} \approx 395 m s^{- 1}$ .

Solved Example 6

At what temperature will the rms speed of oxygen molecules equal the rms speed of hydrogen molecules at

2 7^{\circ} C

Solution:

Equal rms speeds need equal $\frac{T}{M}$ : $\frac{T _{O}}{32} = \frac{300}{2}$ .

Answer: $T_{O} = 4800 K$ ( $452 7^{\circ} C$ ).

Solved Example 7

Find the total translational kinetic energy of the molecules in

1 m^{3}

of an ideal gas at a pressure of

1 atm

(

1.013 \times 1 0^{5} Pa

Solution:

$P = \frac{2}{3} \frac{E}{V} \Rightarrow E = \frac{3}{2} P V = \frac{3}{2} (1.013 \times 1 0^{5}) (1)$ .

Answer: $E \approx 1.52 \times 1 0^{5} J$ , whatever the gas or its temperature.

Solved Example 8

Estimate the mean free path and the collision frequency of a nitrogen molecule at STP (

T = 273 K

P = 1.013 \times 1 0^{5} Pa

). Take the molecular diameter

d = 3 \times 1 0^{- 10} m

Solution:

Number density $n = \frac{P}{k T} = \frac{1.013 \times 1 0 ^{5}}{1.38 \times 1 0 ^{- 23} \times 273} = 2.69 \times 1 0^{25} m^{- 3}$ .

λ = \frac{1}{2 π d ^{2} n} = \frac{1}{1.414 \times 3.1416 \times ( 9 \times 1 0 ^{- 20} ) \times ( 2.69 \times 1 0 ^{25} )} = 9.3 \times 1 0^{- 8} m

$v_{a v} = \frac{8 R T}{π M} = \frac{8 \times 8.314 \times 273}{3.1416 \times 0.028} = 454 m s^{- 1}$ ; frequency $= \frac{v _{a v}}{λ}$ .

Answer: $λ \approx 9.3 \times 1 0^{- 8} m$ (about 310 diameters); about $4.9 \times 1 0^{9}$ collisions per second.

Solved Example 9

The temperature of a gas is raised from

2 7^{\circ} C

92 7^{\circ} C

. Its rms speed becomes
(A)

\frac{927}{27}

times
(B) 2 times
(C) 4 times
(D) half

Solution:

Answer: (B). Use kelvin: $\frac{v _{2}}{v _{1}} = \frac{1200}{300} = 4 = 2$ . Option (A) is the trap of using Celsius.

Solved Example 10

The density of nitrogen at

0^{\circ} C

and

1.013 \times 1 0^{5} Pa

1.25 kg m^{- 3}

. Find the rms speed of its molecules using only these data.

Solution:

$P = \frac{1}{3} ρ v_{r m s}^{2} \Rightarrow v_{r m s} = \frac{3 P}{ρ} = \frac{3 \times 1.013 \times 1 0 ^{5}}{1.25}$ .

Answer: $v_{r m s} \approx 493 m s^{- 1}$ (check: $\frac{3 R T}{M}$ with $T = 273 K$ gives the same).

Solved Example 11

Hydrogen and oxygen are kept at the same temperature. Which quantity is the same for a molecule of each?
(A) rms speed
(B) momentum
(C) average translational kinetic energy
(D) most probable speed

Solution:

Answer: (C). $\overline{K E} = \frac{3}{2} k T$ depends only on temperature. All the speeds depend on $M$ (hydrogen is 4 times faster), and so does the momentum.

Practice Questions

Find the rms speed of helium atoms at $300 K$ .Answer: $\approx 1368 m s^{- 1}$
Find the ratio of rms speeds of $H_{2}$ and $O_{2}$ at the same temperature.Answer: $4 : 1$
At what temperature is the rms speed of nitrogen molecules $1 km s^{- 1}$ ?Answer: $\approx 1123 K$
Find the total translational kinetic energy of $2 mol$ of an ideal gas at $300 K$ .Answer: $\approx 7.48 kJ$
Five molecules have speeds $1$ , $2$ , $3$ , $4$ and $5 km s^{- 1}$ . Find their average and rms speeds.Answer: $3 km s^{- 1}$ ; $11 \approx 3.32 km s^{- 1}$
The pressure of a gas is halved at constant temperature. What happens to its mean free path and its rms speed?Answer: mean free path doubles; rms speed unchanged
The rms speed of a gas at a certain temperature is $v$ . If the gas is replaced by one with four times the molar mass and the temperature is doubled, what is the new rms speed?Answer: $\frac{v}{2}$

Common Mistakes to Avoid

Watch out

Using $T$ in $^{\circ} C$ in $3 R T / M$ or $\frac{3}{2} k T$ . Kinetic theory always needs kelvin.
Using $M$ in $g mol^{- 1}$ with $R = 8.314$ . Convert to $kg mol^{- 1}$ ( $O_{2}$ : $0.032$ ), or the speed comes out $1000$ times too small.
Writing $v_{a v}$ without the $π$ : it is $\frac{8 R T}{π M}$ , not $\frac{8 R T}{M}$ .
Thinking $\overline{v}^{2} = \overline{v^{2}}$ . The square of the average speed is smaller than the average of the squares.
Believing all molecules move at $v_{r m s}$ . It is only one kind of average; speeds are spread over the whole Maxwell curve.
Saying the Maxwell peak rises on heating. It falls and shifts right, because the total area stays 1.
Thinking pressure changes $v_{r m s}$ at constant temperature. $P$ and $ρ$ change together; $\frac{P}{ρ}$ and the speeds stay the same.
Assuming the average translational KE of a molecule depends on the gas. At the same $T$ it is $\frac{3}{2} k T$ for every gas; only the energy per gram differs.

Frequently Asked Questions

What are the main assumptions of the kinetic theory of an ideal gas?

The gas is a huge number of identical point molecules in random motion. Their own volume is negligible, there are no forces between them except during very short elastic collisions, and they obey Newton's laws. From these assumptions pressure and temperature can be derived.

How does kinetic theory explain gas pressure?

Each molecule that hits a wall reverses its momentum and gives the wall a small impulse. Billions of hits every second add up to a steady average force per unit area. The result is that pressure equals one third of density times the mean square speed.

What is the kinetic interpretation of temperature?

Absolute temperature is proportional to the average translational kinetic energy of the molecules. Each molecule carries on average three halves k T of translational kinetic energy, whatever the gas, so a hotter gas simply has faster molecules on average.

What is the difference between rms speed, average speed and most probable speed?

The most probable speed is at the peak of the Maxwell curve, the average speed is the ordinary mean, and the rms speed is the square root of the mean of the squared speeds. They are in the ratio 1 to 1.128 to 1.225.

What does the Maxwell distribution curve show?

It shows how molecular speeds are shared out at a given temperature. The area between two speeds is the fraction of molecules in that range and the total area is one. On heating, the peak moves to higher speed and becomes lower and broader.

What is mean free path and what does it depend on?

Mean free path is the average distance a molecule travels between successive collisions. It is inversely proportional to the number density and to the square of the molecular diameter, so at constant temperature it doubles when the pressure is halved.

Which kinetic theory questions come in NEET?

NEET usually asks the ratio of rms speeds of two gases, the change of rms speed with temperature, the average kinetic energy per molecule, or pressure from density and rms speed. Remember to use kelvin and that every speed varies as the square root of T over M.

How is kinetic theory tested in JEE Main?

JEE Main asks numericals on rms, average and most probable speeds, the relation between pressure and kinetic energy, Maxwell distribution graphs at different temperatures or masses, and mean free path. It is often combined with equipartition and specific heats in one question.

Previous year questions on Kinetic Theory of an Ideal Gases

28 questions from past papers, each with a step-by-step solution.

Show all 28 questions

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