NotesShala Class 11 · Physics

Chapter12

Kinetic Theory

Molecular Nature of Matter · Gas Laws · Kinetic Theory of an Ideal Gas · Equipartition of Energy and Mean Free Path

Reading key Definition Note Remember Derivation Strategy Formulas sit on graph paper · coloured words are the key terms

Part 1

Molecular Nature of Matter

The kinetic theory of gases explains how a gas behaves. It pictures a gas as a huge number of tiny atoms or molecules that move very fast. The forces between atoms act only over short distances. In a gas the molecules are far apart, so these forces can be neglected. That is why gas molecules move so quickly and freely.

1.1Molecular Nature of Matter

Richard Feynman was a great physicist of the 20th century. He rated one idea among the most important discoveries of science. The idea is that matter is made up of atoms.

The modern atomic theory comes from John Dalton (1766–1844), an English chemist. He used it to explain two laws that elements obey when they join to form compounds:

iLaw of Definite Proportions

A given compound always contains its elements in a fixed ratio by mass.

iiLaw of Multiple Proportions

Two elements may join to form more than one compound. Take a fixed mass of the first element. The masses of the second element that join with it are then in a ratio of small whole numbers.

Concept

Dalton's Atomic Theory

About 200 years ago, Dalton said that the smallest parts of an element are atoms. All atoms of one element are identical. They differ from the atoms of every other element. A few atoms of each element join together to make a molecule of a compound.

Atoms join to form molecules. So this theory is also called the molecular theory of matter.

Gay Lussac's Law (19th century)

Gases may react and form another gas. Then the volumes of the gases are in a ratio of small whole numbers.

Avogadro's Law

Given by Amedeo Avogadro (1776–1856). Equal volumes of all gases at the same temperature and pressure contain the same number of molecules.

Note

Avogadro's law, used together with Dalton's theory, explains Gay Lussac's law. Avogadro's idea made it easy to understand how different gases combine.

1.2Atomic Hypothesis and States of Matter

Concept

Atomic Hypothesis

The atomic hypothesis says that everything is made of atoms. Atoms are tiny particles that are always moving (perpetual motion). They attract each other when they are a little distance apart. They repel each other when they are pressed into one another.

Today we can actually see molecules. The electron microscope and the scanning tunnelling microscope make this possible.

Size of an atom (order of magnitude)

10^{-10}\ \text{m}

This length is one angstrom

1\ \text{Å} = 10^{-10}\ \text{m}

States of Matter

Solids

Have a definite shape, size and volume.

Spacing between atoms is the smallest (about 2 Å).

Attraction between atoms is the strongest.

Liquids

Have a definite volume but no definite shape or size.

Spacing is more than in solids and less than in gases.

Attraction is weaker than in solids. So a liquid can flow.

Gases

Have no fixed shape, size or volume.

Spacing between molecules is the largest (about 20 Å).

Attraction is the weakest of the three states.

1.3Motion of Molecules in a Gas

Definition

The mean free path is the average distance a molecule travels without a collision, that is, between two successive collisions. Topic 4.10 covers it in detail.

Mean free path of molecules in a gas (order)Thousands of Å

So gas molecules are quite free to move. They can travel long distances before they hit another molecule. This is why a gas spreads out (disperses) when it is not kept in a closed container.

Atoms Are Not Elementary

Modern theories show that atoms are not indivisible. They are made of still smaller particles:

Atom→Nucleus + Electrons→Protons + Neutrons→Quarks

Note

In future, quarks may also turn out to be made of some still smaller elementary particles.

1.4Behaviour of Gases

Gases are the easiest state of matter to understand. Their molecules are far from each other. So the forces between them are negligible, except when two molecules collide.

Take a gas at low pressure and high temperature. The temperature must be far above the point where the gas liquefies (or solidifies). Such a gas obeys this relation:

pV = KT \qquad \ldots(i)

Here $p$ = pressure, $V$ = volume and $T$ = absolute temperature (in kelvin). $K$ is constant for a given sample. Its value depends on the size of the sample, that is, on the number of molecules $N$ in it:

K = Nk \qquad \ldots(ii)

Experiments show that $k$ has the same value for all gases. This universal constant is written as $k_B$ . So Eq. (ii) becomes:

K = Nk_B \qquad \ldots(iii)

k_B

= Boltzmann constant (SI value)

1.38 \times 10^{-23}\ \text{J K}^{-1}

Putting Eq. (iii) in Eq. (i) gives:

Gas Equation with Boltzmann Constant

pV = Nk_BT \qquad\text{or}\qquad \dfrac{pV}{NT} = k_B = \text{constant}

where, $N$ = number of molecules in the sample, $k_B$ = Boltzmann constant

So for different gases, we can write:

\dfrac{p_1V_1}{N_1T_1} = \dfrac{p_2V_2}{N_2T_2} = \dfrac{p_3V_3}{N_3T_3} = \cdots = k_B = \text{constant}

Law

Avogadro's Hypothesis

Equal volumes of all gases at the same temperature and pressure contain an equal number of molecules.

It follows from the relation above. If $p$ , $V$ and $T$ are the same for two gases, then $N$ must also be the same.

22.4 L of any gas at STP (273 K and 1 atm) has the same number of molecules

N_A

Avogadro's number

N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}

1.5Perfect Gas Equation

The gas equation can also be written with the number of moles. This form is called the perfect gas equation (or ideal gas equation).

Perfect Gas Equation

pV = \mu RT

where, $\mu$ = number of moles, $R = N_Ak_B$ = universal gas constant, $T$ = absolute temperature (in kelvin)

The number of moles of a gas is:

\mu = \dfrac{M}{M_0} = \dfrac{N}{N_A}

Here $M$ = mass of the gas, $M_0$ = molar mass of the gas and $N$ = number of molecules.

Universal gas constant

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

Perfect Gas Equation in Terms of Density

Put $\mu = \dfrac{M}{M_0}$ in $pV = \mu RT$ and divide by $V$ . The mass per unit volume $\dfrac{M}{V}$ is the density $\rho$ . So:

Gas Equation with Density

p = \dfrac{\rho RT}{M_0}

where, $\rho$ = mass density of the gas, $M_0$ = molar mass (mass of one mole of the gas)

Note

Take 22.4 L of any gas at STP. Its mass in grams is equal to its molecular weight. This amount of gas is one mole.
One mole of any substance has $6.022 \times 10^{23}$ atoms or molecules.
So 22.4 L of any gas at STP holds 1 mole of atoms or molecules.

Part 2

Gas Laws

An ideal gas obeys a few simple laws. This part covers these laws, how real gases differ from them, and the work done when a gas is compressed or expanded.

2.1Real Gas and Ideal Gas

Definition

An ideal gas is a theoretical model of a gas. No real gas is perfectly ideal. A real gas behaves most like an ideal gas at low pressure and high temperature.

The reason is simple. At low pressure and high temperature, the molecules are far apart. So the attraction between them becomes negligible.

For an ideal gas, $pV = \mu RT$ . Here $\mu$ is the number of moles and $R = N_Ak_B$ is the universal constant. It is called the gas constant for one gram mole. So:

Ideal Gas

\therefore \quad \dfrac{pV}{\mu T} = R = \text{constant}

Plot $\dfrac{pV}{\mu T}$ against $p$ . For an ideal gas the graph is a horizontal straight line (solid line).

Now plot the same graph for a real gas at three temperatures $T_1 > T_2 > T_3$ . We get the dashed curves.

Every real-gas curve comes close to the ideal line at low pressure. The curve for the highest temperature $T_1$ stays closest to it. So a real gas behaves like an ideal gas at low pressure and high temperature.

Solid line: ideal gas. Dashed curves: a real gas at

T_1 > T_2 > T_3

2.2Boyle's Law

Law

Boyle's Law

For a given mass of a gas kept at constant temperature, the volume of the gas is inversely proportional to its pressure.

V \propto \dfrac{1}{p}

The ideal gas equation is $pV = \mu RT$ . If the temperature $T$ does not change, the right side stays constant. So $pV$ = constant. This is Boyle's law:

Boyle's Law

p_1V_1 = p_2V_2 = p_3V_3 = \cdots = \text{constant}

The figure compares two sets of $p$ – $V$ curves at three temperatures $T_1 > T_2 > T_3$ :

Solid lines: curves found by experiment for a real gas.
Dashed lines: curves predicted by Boyle's law.

At low pressure (large volume) the two sets agree. At high pressure the real gas moves away from Boyle's law.

Experimental

p

–

V

curves (solid) and Boyle's law curves (dashed)

2.3Charles' Law

Law

Charles' Law

For a given mass of an ideal gas kept at constant pressure, the volume $V$ of the gas is directly proportional to its absolute temperature $T$ .

V \propto T

Rearrange the ideal gas equation:

\dfrac{V}{T} = \dfrac{\mu R}{p}

If the pressure does not change, the right side is constant. So $\dfrac{V}{T}$ = constant. This is Charles' law:

Charles' Law

\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2} = \dfrac{V_3}{T_3} = \cdots = \text{constant}

The figure compares two sets of $T$ – $V$ curves at three pressures $p_1 > p_2 > p_3$ :

Solid lines: curves found by experiment for a real gas.
Dashed lines: straight lines predicted by Charles' law.

At high temperature the two sets agree. At low temperature the real gas moves away from Charles' law.

Experimental

T

–

V

curves (solid) and Charles' law lines (dashed)

2.4Dalton's Law of Partial Pressures

Law

Dalton's Law of Partial Pressures

In a mixture of non-interacting ideal gases, the total pressure equals the sum of the partial pressures of all the gases present.

p = p_1 + p_2 + p_3 + \cdots

Setup — Given Information

A vessel holds a mixture of non-interacting ideal gases.
Gas 1 has $\mu_1$ moles, gas 2 has $\mu_2$ moles, and so on.
Volume of the mixture $= V$ , temperature $= T$ , pressure $= p$ .

Derivation

Write the equation of state for the whole mixture:
$pV = (\mu_1 + \mu_2 + \cdots)RT$
Divide both sides by $V$ . The pressure splits into one part for each gas:
$p = \dfrac{\mu_1RT}{V} + \dfrac{\mu_2RT}{V} + \cdots$
Each term is the partial pressure of one gas. So:
$p = p_1 + p_2 + p_3 + \cdots$

Note

Here $p_1 = \dfrac{\mu_1RT}{V}$ is the partial pressure of gas 1. It is the pressure that gas 1 would exert if it alone filled the same volume at the same temperature.

2.5Graham's Law of Diffusion

Law

Graham's Law of Diffusion

The rate of diffusion $r$ of a gas varies inversely as the square root of its density $\rho$ .

r \propto \dfrac{1}{\sqrt{\rho}} \qquad \ldots(i)

Setup — Given Information

Ideal gas equation: $pV = \mu RT = \dfrac{M}{M_0}RT$
$M$ = mass of the gas, $M_0$ = molecular (molar) mass of the gas.

Derivation

Rearrange the ideal gas equation to get $\dfrac{M}{V}$ :
$\dfrac{M}{V} = \dfrac{pM_0}{RT} \qquad \ldots(ii)$
Mass per unit volume is the density, $\rho = \dfrac{M}{V}$ . So Eq. (ii) gives:
$\rho = \dfrac{pM_0}{RT} \qquad \ldots(iii)$
At the same $p$ and $T$ , density is proportional to molecular mass. Put this in Eq. (i):
$r \propto \dfrac{1}{\sqrt{M_0}}$

Graham's Law for Two Gases

\dfrac{r_1}{r_2} = \sqrt{\dfrac{\rho_2}{\rho_1}} = \sqrt{\dfrac{M_{02}}{M_{01}}}

where, $r_1, r_2$ = rates of diffusion, $\rho_1, \rho_2$ = densities and $M_{01}, M_{02}$ = molecular masses of gas 1 and gas 2

2.6Work Done on Compressing a Gas

When an ideal gas expands or is compressed, it follows relation (i) below. Here $n$ is a constant for the process, called the polytropic index. The gas also obeys the combined gas law (ii):

pV^n = \text{constant} \quad \ldots(i) \qquad\qquad \dfrac{pV}{T} = \text{constant} \quad \ldots(ii)

Relations between

T

V

and

p

Derivation

Divide Eq. (i) by Eq. (ii):
$pV^n \div \dfrac{pV}{T} = pV^n \cdot \dfrac{T}{pV} = \text{constant}$
Simplify. This is the temperature–volume relation:
$TV^{\,n-1} = \text{constant} \qquad \ldots(iii)$
From Eq. (ii), $V = \text{constant} \times \dfrac{T}{p}$ . Put this value in Eq. (i):
$p\left(\dfrac{T}{p}\right)^{n} = \dfrac{pT^n}{p^n} = \text{constant}$
Take the $n$ ^th root. This is the temperature–pressure relation:
$\dfrac{T}{p^{\frac{n-1}{n}}} = \text{constant} \qquad \ldots(iv)$

Work Done in Compression

Let the gas be compressed from state 1 to state 2. The work done on the gas is:

W = -\displaystyle\int_1^2 p\,dV \qquad \ldots(v)

Derivation

From Eq. (i), write $p$ in terms of $V$ . Here $C = pV^n$ is the constant:
$p = \dfrac{C}{V^n}$
Put this value in Eq. (v):
$W = -C\displaystyle\int_1^2 V^{-n}\,dV$
Integrate:
$W = -C\left[\dfrac{V^{-n+1}}{-n+1}\right]_1^2$
Put back $C = pV^n$ and simplify:
$W = -\left[pV^n\,\dfrac{V^{1-n}}{1-n}\right]_1^2 = -\left[\dfrac{pV}{1-n}\right]_1^2$
So the work done in compression is:
$W = -\,\dfrac{p_2V_2 - p_1V_1}{1-n} \qquad \ldots(vi)$
For an ideal gas, $p_1V_1 = \mu RT_1$ and $p_2V_2 = \mu RT_2$ . Put these in Eq. (vi):
$W = -\,\dfrac{\mu R\,(T_2 - T_1)}{1-n}$

Work Done on a Gas in Compression

W = -\,\dfrac{p_2V_2 - p_1V_1}{1-n} = -\,\dfrac{\mu R\,(T_2 - T_1)}{1-n}

Note

Work Done in Expansion

For expansion, start from the work done by the gas, $W = \displaystyle\int_1^2 p\,dV$ . The same steps give:

W = \dfrac{p_1V_1 - p_2V_2}{n-1} \qquad\text{and}\qquad W = \dfrac{\mu R\,(T_1 - T_2)}{n-1}

Part 3

Kinetic Theory of an Ideal Gas

The kinetic theory of gases explains the gas laws from the motion of molecules. It links what we measure, such as pressure and temperature, with how fast the molecules move.

3.1Assumptions of Kinetic Theory

The kinetic theory of gases was built mainly by two scientists:

James Clerk Maxwell

(1831–1879), a Scottish physicist.

Ludwig Boltzmann

(1844–1906), a physicist from Austria.

Kinetic theory rests on the molecular picture of matter. It connects two kinds of properties of a gas:

Macroscopic: pressure, temperature⟷Microscopic: molecular speed, kinetic energy

Postulates (Assumptions) of Kinetic Theory of Gases

Large number of identical molecules. Any amount of gas contains a huge number of molecules, about $10^{23}$ (the order of Avogadro's number). All the molecules are identical in every way.
Random motion. The molecules move randomly and non-stop in all directions with different speeds. Between collisions, each molecule moves freely in a straight line, as Newton's first law says.
Negligible size. A molecule is much smaller than the average distance between molecules. At ordinary pressure and temperature, a molecule is about 2 Å in size. The average distance between molecules is about 20 Å.
No intermolecular forces. Molecules exert no force on each other, except during a collision.
Elastic collisions. Collisions between molecules, and between molecules and the walls, are perfectly elastic. The total momentum and the total kinetic energy are conserved. Only the velocities of the molecules change.
Instantaneous collisions. A collision lasts a negligible time compared with the time between two successive collisions.
Uniform distribution. The density of the gas and the spread of its molecules are uniform throughout the container.

3.2Pressure of an Ideal Gas

Setup — Given Information

An ideal gas has $N$ molecules in a cube of edge $l$ . So its volume is $V = l^3$ .
One molecule has mass $m$ and velocity $(v_x, v_y, v_z)$ .

(a) A molecule in a cube of edge

l

moves towards face

A

. (b) In an elastic hit on wall

A

, only

v_x

reverses

Derivation

The molecule hits wall $A$ (parallel to the $yz$ -plane) elastically. Its $x$ -component of velocity reverses. Its $y$ - and $z$ -components do not change. So the velocity changes from $(v_x, v_y, v_z)$ to $(-v_x, v_y, v_z)$ in the collision.
Change in momentum of the molecule along the $x$ -direction:
$\Delta p_x = (p_x)_{\text{final}} - (p_x)_{\text{initial}} = (-mv_x) - (mv_x) = -2mv_x$
Impulse = change in momentum. Let $F$ be the average force on the molecule by the wall:
$F\,\Delta t = \Delta p_x = -2mv_x \qquad \ldots(i)$
To hit the same wall again, the molecule must go to the opposite wall and come back. This is a distance $2l$ along $x$ . So the time between two hits on the same wall is:
$\Delta t = \dfrac{2l}{v_x}$
So the average force on the molecule by the wall, for each collision, is:
$F = \dfrac{-2mv_x}{\Delta t} = \dfrac{-2mv_x}{(2l/v_x)} = \dfrac{-mv_x^2}{l}$
By Newton's third law, the molecule pushes the wall with a force $\dfrac{mv_x^2}{l}$ . For all $N$ molecules, the total force on the wall is:
$F_{\text{wall}} = \dfrac{m}{l}\left(v_{x_1}^2 + v_{x_2}^2 + \cdots + v_{x_N}^2\right) = \dfrac{Nm}{l}\,\overline{v_x^2} \qquad \ldots(ii)$
Here $\overline{v_x^2} = \dfrac{v_{x_1}^2 + v_{x_2}^2 + \cdots + v_{x_N}^2}{N}$ is the average value of $v_x^2$ .
Velocity has three components, so $\overline{v^2} = \overline{v_x^2} + \overline{v_y^2} + \overline{v_z^2}$ . The motion is completely random, so the three averages are equal:
$\overline{v_x^2} = \overline{v_y^2} = \overline{v_z^2} \quad\Rightarrow\quad \overline{v^2} = 3\,\overline{v_x^2} \quad\text{or}\quad \overline{v_x^2} = \dfrac{1}{3}\,\overline{v^2} \qquad \ldots(iii)$
From Eqs. (ii) and (iii), the normal force on the wall is:
$F_{\text{wall}} = \dfrac{N}{3}\left(\dfrac{m}{l}\,\overline{v^2}\right) \qquad \ldots(iv)$
Pressure = force ÷ area. The wall's area is $l^2$ and the volume is $V = l^3$ :
$p = \dfrac{F_{\text{wall}}}{l^2} = \dfrac{1}{l^2}\left(\dfrac{N}{3}\,\dfrac{m\overline{v^2}}{l}\right) = \dfrac{1}{l^3}\left(\dfrac{Nm\overline{v^2}}{3}\right) = \dfrac{1}{V}\left(\dfrac{Nm\overline{v^2}}{3}\right)$
Write $n = \dfrac{N}{V}$ , the number of molecules per unit volume:
$p = \dfrac{1}{3}\left(\dfrac{N}{V}\right)m\overline{v^2} = \dfrac{1}{3}\,nm\overline{v^2} \qquad \ldots(v)$

Kinetic Gas Equation (Pressure of an Ideal Gas)

p = \dfrac{1}{3}\,nm\overline{v^2}

where, $p$ = pressure on the wall, $n$ = number density (number of molecules per unit volume), $m$ = mass of one molecule, $\overline{v^2}$ = mean square speed

Conclusions from the Derivation

The shape of the vessel does not matter. The area $A$ and the time $\Delta t$ are absent from the final answer. For a vessel of any shape, take a tiny flat area of its wall and repeat the same steps.
By Pascal's law, a gas in equilibrium has the same pressure in every part of the container.
We ignored collisions between molecules. Such collisions do not change the result if they are not too frequent and if each lasts a negligible time compared with the time between collisions.

3.3Relation between Pressure and Density

Setup — Given Information

A vessel holds an ideal gas.
Total mass of all the molecules $= M$ , volume $= V$ , density $= \rho$ .

Derivation

Multiply the number density $n$ by the mass of one molecule $m$ . The product is the total mass of the molecules in a unit volume:
$n \times m = \dfrac{M}{V}$
Mass per unit volume is the density of the gas $\left[\because\ \rho = \dfrac{\text{Mass}}{\text{Volume}}\right]$ . So:
$n \times m = \rho \qquad \ldots(vi)$
Put Eq. (vi) in the kinetic gas equation (v):
$p = \dfrac{1}{3}\,\rho\,\overline{v^2}$

Pressure in Terms of Density

p = \dfrac{1}{3}\,\rho\,\overline{v^2}

where, $\rho$ = density of the gas, $\overline{v^2}$ = mean square speed of its molecules

3.4Kinetic Interpretation of Temperature

Concept

The average kinetic energy of a gas molecule depends only on the absolute temperature of the gas. This idea is called the kinetic interpretation of temperature.

Setup — Given Information

A sample of an ideal gas has $N$ molecules.
Volume $= V$ , pressure $= p$ , temperature $= T$ .

Derivation

Start with the pressure of an ideal gas:
$p = \dfrac{1}{3}\,nm\overline{v^2} \qquad \ldots(i)$
Multiply both sides by $V$ . Here $nV = N$ , the number of molecules in the sample:
$pV = \dfrac{1}{3}(nV)\,m\overline{v^2} = \dfrac{2}{3}\,N\left(\dfrac{1}{2}m\overline{v^2}\right) \qquad \ldots(ii)$
$\tfrac{1}{2}m\overline{v^2}$ is the average kinetic energy of one molecule. So the total energy (internal energy) of the gas is:
$E = N\left(\dfrac{1}{2}m\overline{v^2}\right) \qquad \ldots(iii)$
From Eqs. (ii) and (iii):
$pV = \dfrac{2}{3}E \qquad \ldots(iv)$
For an ideal gas, $pV = \mu RT = \mu k_BN_AT = k_B(\mu N_A)T$ . Here $\mu N_A = N$ , the total number of molecules. So:
$pV = k_BNT \qquad \ldots(v)$
Compare Eqs. (iv) and (v):
$\dfrac{2}{3}E = k_BNT \quad\Rightarrow\quad \dfrac{E}{N} = \dfrac{3}{2}k_BT \qquad \ldots(vi)$
$\dfrac{E}{N}$ is the average kinetic energy of one molecule. So:
$\dfrac{1}{2}m\overline{v^2} = \dfrac{3}{2}\,k_BT \qquad \ldots(vii)$

Average Kinetic Energy of a Molecule

\dfrac{1}{2}m\overline{v^2} = \dfrac{3}{2}\,k_BT \qquad\Rightarrow\qquad \dfrac{1}{2}m\overline{v^2} \propto T

Important

A Basic Result

A molecule's average kinetic energy is directly proportional to the absolute temperature $T$ . It does not depend on the pressure, the volume or the kind of ideal gas. This result links the temperature of a gas with the average kinetic energy of its molecules.

Definition

Absolute Zero

Put $T = 0$ in Eq. (vii). Then $\tfrac{1}{2}m\overline{v^2} = 0$ . Since $m \neq 0$ , we get $\overline{v^2} = 0$ .

Absolute zero is the temperature at which the mean square speed of gas molecules becomes zero.

3.5Root Mean Square Speed

Definition

The root mean square (rms) speed of gas molecules is the square root of the mean square speed $\overline{v^2}$ . That is, $v_{\text{rms}} = \sqrt{\overline{v^2}}$ .

Derivation

From the kinetic interpretation of temperature:
$\dfrac{1}{2}m\overline{v^2} = \dfrac{3}{2}k_BT$
So the mean square speed is:
$\overline{v^2} = \dfrac{3k_BT}{m}$
Take the square root of both sides:
$\sqrt{\overline{v^2}} = \sqrt{\dfrac{3k_BT}{m}}$

Root Mean Square Speed

v_{\text{rms}} = \sqrt{\dfrac{3k_BT}{m}}

where, $m$ = mass of one molecule, $k_B$ = Boltzmann constant, $T$ = absolute temperature

3.6Maxwell's Speed Distribution

In a gas, the molecules do not all have the same speed. Their motion is random, in all directions. Still, the way the speeds are shared among the molecules (the distribution of speeds) stays fixed. Maxwell was the first to study this. He derived an equation for the speed distribution of molecules.

Let $dN$ be the number of molecules with speeds between $v$ and $v + dv$ . Then:

Maxwell's Speed Distribution

dN = 4\pi N\left(\dfrac{m}{2\pi k_BT}\right)^{3/2} v^2\, e^{-\frac{mv^2}{2k_BT}}\, dv

where, $v$ = speed of a molecule, $m$ = mass of one molecule, $T$ = temperature (in kelvin), $N$ = total number of molecules, $k_B$ = Boltzmann constant

iArea Under the Curve

Strip area = fraction of molecules in

dv

iiEffect of Temperature

Hotter gas: flatter curve, peak at a higher speed

In figure (i), the fraction of molecules with speeds between $v$ and $v + dv$ is the area of the thin strip:

\dfrac{dN}{N} = f(v)\,dv = \text{area of the strip}

The total area under the curve is 1 (unity), because it covers all the molecules.

Figure (ii) shows $f(v)$ for a gas at three temperatures. Each curve has a peak. The peak shows the speed that the largest fraction of molecules have. This speed $v_{\text{max}}$ is called the most probable speed.

Note

As the temperature rises, the curve becomes flatter and its peak shifts towards higher speed.

Most Probable Speed

The most probable speed is the speed that the largest number of molecules of a gas have at a given temperature.

v_{\text{mp}} = \sqrt{\dfrac{2k_BT}{m}}

Mean (Average) Speed

The mean speed is the arithmetic mean of the speeds of all the molecules of a gas at a given temperature.

v_{\text{av}} = \sqrt{\dfrac{8k_BT}{\pi m}}

Part 4

Equipartition of Energy and Mean Free Path

Energy is shared equally among the ways a molecule can move. This gives the specific heats of gases, solids and water.

4.1Degrees of Freedom

Definition

The number of degrees of freedom of a system is the total number of independent coordinates (or quantities) needed to describe its position and configuration completely. It is written as $f$ .

Degrees of Freedom

f = 3N - K

where, $N$ = number of particles in the system, $K$ = number of independent relations (constraints) between the particles

Monoatomic Gas

e.g. Helium, neon, argon. Only translational motion is possible.

Here $N = 1$ and $K = 0$ . So $f = 3 \times 1 - 0$ .

f = 3

Diatomic Gas

e.g. Oxygen, hydrogen, nitrogen. Each molecule has 3 translational and 2 rotational degrees of freedom.

f = 5

Triatomic Gas

e.g. SO₂, H₂S. Each molecule has three atoms.

f = 7

(linear molecule)

f = 6

(non-linear molecule)

4.2Law of Equipartition of Energy

Law

Law of Equipartition of Energy

For a system in thermal equilibrium, the total energy is shared equally among all its degrees of freedom. The average energy of a molecule for each degree of freedom is $\dfrac{1}{2}k_BT$ .

Here $k_B$ = Boltzmann constant $= 1.38 \times 10^{-23}\ \text{J K}^{-1}$ and $T$ = absolute temperature (kelvin scale).

Translational Energy

Take a monoatomic gas in thermal equilibrium at temperature $T$ . The average translational energy of one molecule is:

\langle E_t \rangle = \left\langle \tfrac{1}{2}mv_x^2 \right\rangle + \left\langle \tfrac{1}{2}mv_y^2 \right\rangle + \left\langle \tfrac{1}{2}mv_z^2 \right\rangle = \dfrac{3}{2}k_BT

No direction is special. So each of the three terms gets an equal share:

\left\langle \tfrac{1}{2}mv_x^2 \right\rangle = \left\langle \tfrac{1}{2}mv_y^2 \right\rangle = \left\langle \tfrac{1}{2}mv_z^2 \right\rangle = \dfrac{1}{2}k_BT

KeyEach molecule has an energy of $\tfrac{1}{2}k_BT$ per degree of freedom. This is the law of equipartition of energy.

Rotational Energy of a Diatomic Molecule

A diatomic molecule can also rotate. So its total energy is the sum of the translational energy $E_t$ and the rotational energy $E_r$ :

E_t + E_r = \left(\tfrac{1}{2}mv_x^2 + \tfrac{1}{2}mv_y^2 + \tfrac{1}{2}mv_z^2\right) + \left(\tfrac{1}{2}I_1\omega_1^2 + \tfrac{1}{2}I_2\omega_2^2\right)

The molecule can rotate about two independent axes, called axis 1 and axis 2. Both axes pass through its centre and are perpendicular to the line joining the two atoms.

Here $\omega_1$ and $\omega_2$ are the angular speeds about axes 1 and 2. $I_1$ and $I_2$ are the moments of inertia about these axes.

Rotation of a diatomic molecule about two independent axes

Vibrational Energy

A diatomic or polyatomic molecule is not always a rigid rotator. Its atoms may also vibrate. For example, the CO molecule vibrates even at moderate temperatures. This adds a vibrational energy $E_v$ to the total energy:

E_v = \dfrac{1}{2}m\left(\dfrac{dy}{dt}\right)^2 + \dfrac{1}{2}ky^2

Here $k$ = force constant of the oscillator and $y$ = vibrational coordinate.

Total Energy of a Molecule

E = E_t + E_r + E_v

translational energy + rotational energy + vibrational energy

4.3Specific Heat Capacity of Monoatomic Gases

The law of equipartition of energy lets us find the specific heat capacity of gases with different degrees of freedom.

A monoatomic molecule has 3 translational degrees of freedom. Each one has an average energy of $\tfrac{1}{2}k_BT$ :

3 degrees of freedom×

\tfrac{1}{2}k_BT

each=

\tfrac{3}{2}k_BT

per molecule

Derivation

One mole has $N_A$ molecules. Using $R = N_Ak_B$ , the internal energy of one mole is:
$U = \dfrac{3}{2}k_BT \times N_A = \dfrac{3}{2}RT \qquad \ldots(i)$
Let $dU$ be the heat needed to raise the temperature of 1 mole of the gas by $dT$ at constant volume. Then the molar specific heat at constant volume is:
$C_V = \dfrac{dU}{dT} \qquad \ldots(ii)$
Put $U = \tfrac{3}{2}RT$ in Eq. (ii):
$C_V = \dfrac{d}{dT}\left(\dfrac{3}{2}RT\right) = \dfrac{3}{2}R$
Use Mayer's relation, $C_p - C_V = R$ , for the molar specific heat at constant pressure:
$C_p = \dfrac{3}{2}R + R = \dfrac{5}{2}R$
The ratio of specific heats is:
$\gamma = \dfrac{C_p}{C_V} = \dfrac{(5/2)R}{(3/2)R} = \dfrac{5}{3} = 1.67$

C_V

C_V = \dfrac{3}{2}R

C_p

C_p = \dfrac{5}{2}R

\gamma

\gamma = \dfrac{5}{3} = 1.67

4.4Specific Heat Capacity of Diatomic Gases

Ignore the vibrational mode. At room temperature, a diatomic molecule then has:

3 translational+2 rotational=5 degrees of freedom (

f = 5

)

Derivation

By the law of equipartition, the internal energy of one mole is $\left[\because\ R = k_BN_A\right]$ :
$U = 5 \times \left(\dfrac{1}{2}k_BT\right) \times N_A = \dfrac{5}{2}RT$
The molar specific heat at constant volume is:
$C_V = \dfrac{dU}{dT} = \dfrac{d}{dT}\left(\dfrac{5}{2}RT\right) = \dfrac{5}{2}R$
Use Mayer's relation, $C_p = C_V + R$ :
$C_p = \dfrac{5}{2}R + R = \dfrac{7}{2}R$
The ratio of specific heats is:
$\gamma = \dfrac{C_p}{C_V} = \dfrac{(7/2)R}{(5/2)R} = \dfrac{7}{5} = 1.40$

C_V

C_V = \dfrac{5}{2}R

C_p

C_p = \dfrac{7}{2}R

\gamma

\gamma = \dfrac{7}{5} = 1.40

4.5Specific Heat Capacity of Triatomic Gases

A linear triatomic molecule has seven degrees of freedom ( $f = 7$ ).

Derivation

By the law of equipartition, the internal energy of one mole is $\left[\because\ R = k_BN_A\right]$ :
$U = 7 \times \left(\dfrac{1}{2}k_BT\right) \times N_A = \dfrac{7}{2}RT$
The molar specific heat at constant volume is:
$C_V = \dfrac{dU}{dT} = \dfrac{d}{dT}\left(\dfrac{7}{2}RT\right) = \dfrac{7}{2}R$
Use Mayer's relation, $C_p = C_V + R$ :
$C_p = \dfrac{7}{2}R + R = \dfrac{9}{2}R$
The ratio of specific heats is:
$\gamma = \dfrac{C_p}{C_V} = \dfrac{(9/2)R}{(7/2)R} = \dfrac{9}{7} \approx 1.29$

C_V

C_V = \dfrac{7}{2}R

C_p

C_p = \dfrac{9}{2}R

\gamma

\gamma = \dfrac{9}{7} \approx 1.29

4.6Specific Heat Capacity of Polyatomic Gases

In general, a polyatomic molecule has:

3 translational+3 rotational+

f

vibrational modes

Here $f$ is the number of vibrational modes. Each mode has kinetic and potential energy, so it gets $2 \times \tfrac{1}{2}k_BT = k_BT$ .

Derivation

By the law of equipartition, the internal energy of one mole is:
$U = \left[\dfrac{3}{2}k_BT + \dfrac{3}{2}k_BT + f\,k_BT\right] \times N_A \quad\Rightarrow\quad U = (3+f)\,RT$
The molar specific heat at constant volume is:
$C_V = \dfrac{dU}{dT} = \dfrac{d\,[(3+f)RT]}{dT} = (3+f)\,R$
Use Mayer's relation, $C_p = C_V + R = (3+f)R + R$ :
$C_p = (4+f)\,R$
The ratio of specific heats is:
$\gamma = \dfrac{C_p}{C_V} = \dfrac{4+f}{3+f}$

Taking

f = 6

\gamma = \dfrac{4+6}{3+6} = \dfrac{10}{9}

\gamma = 1.11

4.7Determination of γ from Degrees of Freedom

Derivation

Let a gas molecule have $f$ degrees of freedom. The internal energy of one mole of the gas is:
$U = f \times \left(\dfrac{1}{2}k_BT\right) \times N_A = \dfrac{f}{2}RT$
The molar specific heat at constant volume is:
$C_V = \dfrac{dU}{dT} = \dfrac{d}{dT}\left(\dfrac{f}{2}RT\right) = \dfrac{f}{2}R$
Use Mayer's relation, $C_p = C_V + R = \dfrac{f}{2}R + R$ :
$C_p = \left(\dfrac{f}{2} + 1\right)R$
So the ratio of specific heats is:
$\gamma = \dfrac{C_p}{C_V} = \dfrac{\left(\dfrac{f}{2}+1\right)R}{\dfrac{f}{2}R} \quad\Rightarrow\quad \gamma = 1 + \dfrac{2}{f}$

Relation between γ and f

\gamma = 1 + \dfrac{2}{f}

where, $f$ = degrees of freedom of one molecule

Summary for Gases

Gas	$f$	$C_V$	$C_p$	$\gamma = C_p/C_V$
Monoatomic	3	$\tfrac{3}{2}R$	$\tfrac{5}{2}R$	$\tfrac{5}{3} = 1.67$
Diatomic	5	$\tfrac{5}{2}R$	$\tfrac{7}{2}R$	$\tfrac{7}{5} = 1.40$
Linear triatomic	7	$\tfrac{7}{2}R$	$\tfrac{9}{2}R$	$\tfrac{9}{7} \approx 1.29$

4.8Specific Heat Capacity of Solids

Setup — Given Information

In a solid, each atom vibrates about its mean position.
One mole of a solid has $N_A$ atoms.
Each atom oscillates in 3 dimensions.

Derivation

An oscillation in one dimension has kinetic and potential energy. By the law of equipartition, the average energy per atom for one dimension is:
$2 \times \left(\dfrac{1}{2}k_BT\right) = k_BT$
So in three dimensions, the average energy per atom is:
$3k_BT$
The total energy of one mole of the solid is:
$U = (3k_BT) \times N_A = 3RT \qquad [\because\ k_BN_A = R]$
By the first law of thermodynamics, $\Delta Q = \Delta U + \Delta W = \Delta U + p\,\Delta V$ . For a solid, $\Delta V$ is negligible. So:
$\Delta Q \approx \Delta U$
So the molar specific heat of a solid is:
$C = \dfrac{\Delta Q}{\Delta T} = \dfrac{\Delta U}{\Delta T} = 3R$

Molar Specific Heat of a Solid

C = 3R = 3 \times 8.314 \approx 24.94\ \text{J mol}^{-1}\text{K}^{-1}

using $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$

Specific Heat Capacities of Some Solids

Substance	Specific heat (J kg $^{-1}$ K $^{-1}$ )	Molar specific heat (J mol $^{-1}$ K $^{-1}$ )
Aluminium	900.0	24.4
Carbon	506.5	6.1
Copper	386.4	24.5
Lead	127.7	26.5
Silver	236.1	25.5
Tungsten	134.4	24.9

Values at room temperature and atmospheric pressure.

The table shows that the predicted value $3R$ generally agrees with experiment at ordinary temperatures. Carbon is an exception.

Note

Each solid has a characteristic temperature called its Debye temperature. Above this temperature, the molar specific heat $C_V$ of the solid becomes nearly constant (close to $3R$ ).

4.9Specific Heat Capacity of Water

Setup — Given Information

Treat a water molecule like a solid of 3 atoms (2 hydrogen + 1 oxygen). Each atom can vibrate in 3 dimensions about its mean position.

Derivation

By the law of equipartition, the average energy per atom for one dimension is:
$2 \times \left(\dfrac{1}{2}k_BT\right) = k_BT$
In three dimensions, the average energy per atom is:
$3k_BT$
A water molecule has 3 atoms. So the total energy of one molecule is:
$3 \times (3k_BT) = 9k_BT$
The total energy of 1 mole of water is:
$U = 3 \times 3k_BT \times N_A = 9RT$
So the molar specific heat of water is:
$C = \dfrac{\Delta Q}{\Delta T} = \dfrac{\Delta U}{\Delta T} = 9R$

Molar Specific Heat of Water

C = 9R = 9 \times 8.314 \approx 74.83\ \text{J mol}^{-1}\text{K}^{-1}

using $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$

4.10Mean Free Path

A gas has a huge number of molecules that collide very often. So a molecule cannot travel straight for long. The molecules are always in continuous random motion. Their collisions with each other are perfectly elastic.

Definition

The mean free path ( $\lambda$ ) of a gas molecule is the average distance it travels between two successive collisions.

Each straight piece of the path between two collisions is one free path. The mean free path is the average of these free paths.

Free paths

\lambda_1, \lambda_2, \lambda_3, \lambda_4

of one molecule between collisions

Expression for Mean Free Path

Setup — Given Information

Let $d$ be the diameter of each molecule.
A molecule collides with every molecule whose centre comes within a distance $d$ of its own centre.
$\langle v \rangle$ = average speed of the molecule, $n$ = number of molecules per unit volume.

Cylinder of radius

d

swept in time

\Delta t

. Red molecules lie inside it and collide

Derivation

In a small time $\Delta t$ , the molecule sweeps a cylinder of radius $d$ and length $\langle v\rangle\Delta t$ . It collides with every molecule whose centre lies in this volume:
$\text{Volume swept} = \pi d^2 \langle v \rangle \Delta t$
There are $n$ molecules per unit volume. So the number of collisions in time $\Delta t$ is:
$\pi d^2 \langle v \rangle \Delta t \times n$
So the number of collisions per second is:
$\dfrac{\pi d^2 \langle v \rangle \Delta t \times n}{\Delta t} = n\pi d^2 \langle v \rangle$
The average time between two successive collisions is:
$\tau = \dfrac{1}{n\pi d^2 \langle v \rangle}$
The mean free path is the average distance covered between two successive collisions:
$\lambda = \langle v \rangle\,\tau = \dfrac{\langle v \rangle}{n\pi d^2 \langle v \rangle} = \dfrac{1}{n\pi d^2}$

Mean Free Path

\lambda = \dfrac{1}{n\pi d^2}

where, $d$ = diameter of each molecule, $n$ = number of molecules per unit volume

4.11Assumptions and Accurate Value of Mean Free Path

The derivation in topic 4.10 rests on these assumptions:

The gas molecules are hard spheres, each of diameter $d$ .
Collisions between gas molecules are perfectly elastic.
All the molecules are at rest, except the one being studied.
The chosen molecule collides with every molecule whose centre is at a distance $d$ from its own centre.

In the derivation we took the other molecules to be at rest. In reality, all the molecules are moving. Including this gives a more accurate value:

Mean Free Path (Accurate Value)

\lambda = \dfrac{1}{\sqrt{2}\,n\pi d^2}

where, $n$ = number of molecules per unit volume, $d$ = diameter of a molecule, $\lambda$ = mean free path

From $pV = Nk_BT$ , we get $p = nk_BT$ , that is, $n = \dfrac{p}{k_BT}$ . Put this in the formula above:

Mean Free Path in Terms of

T

and

p

\lambda = \dfrac{k_BT}{\sqrt{2}\,\pi d^2 p}

where, $T$ = absolute temperature (in kelvin), $p$ = pressure of the gas

At Constant Pressure

$\lambda$ is directly proportional to the absolute temperature.

\lambda \propto T

At Constant Temperature

$\lambda$ is inversely proportional to the pressure.

\lambda \propto \dfrac{1}{p}

Note

Kinetic theory easily links bulk properties that we can measure, like viscosity, heat conductivity and the rate of diffusion, with microscopic quantities such as the size of a molecule.

4.12Brownian Motion

In 1827, the Scottish botanist Robert Brown was looking at the pollen grains of a flower under a microscope. He saw the grains moving all the time in a zig-zag, random way.

Definition

Brownian motion is the continuous zig-zag motion of microscopic particles suspended in water, air or any other fluid. Kinetic theory explains this motion.

A tiny particle (size about $10^{-5}$ m) in a fluid is hit from all sides by the fluid molecules, all the time. The impulses and torques from these hits do not add up exactly to zero. So at every moment there is a net impulse or torque in some direction. This keeps the particle moving randomly.

Brownian motion depends on

iSize

Size of the suspended particle.

iiDensity

Density of the fluid.

iiiTemperature

Temperature of the medium.

ivViscosity

Viscosity of the medium.

Note

In 1987, the Egyptian scientist Ahmed Zewail observed the detailed interactions of molecules. He lit them with very short flashes of laser light, each lasting about 10 femtoseconds ( $1\ \text{femtosecond} = 10^{-15}\ \text{s}$ ).

End of Chapter 12 · notesshala.in