## Kinetic Theory

### 12.1 INTRODUCTION

Boyle discovered the law named after him in 1661. Boyle, Newton and several others tried to explain the behaviour of gases by considering that gases are made up of tiny atomic particles. The actual atomic theory got established more than 150 years later. Kinetic theory explains the behaviour of gases based on the idea that the gas consists of rapidly moving atoms or molecules. This is possible as the inter-atomic forces, which are short range forces that are important for solids and liquids, can be neglected for gases. The kinetic theory was developed in the nineteenth century by Maxwell, Boltzmann and others. It has been remarkably successful. It gives a molecular interpretation of pressure and temperature of a gas, and is consistent with gas laws and Avogadro’s hypothesis. It correctly explains specific heat capacities of many gases. It also relates measurable properties of gases such as viscosity, conduction and diffusion with molecular parameters, yielding estimates of molecular sizes and masses. This chapter gives an introduction to kinetic theory.

### 12.2 MOLECULAR NATURE OF MATTER

Richard Feynman, one of the great physicists of 20th century considers the discovery that “Matter is made up of atoms” to be a very significant one. Humanity may suffer annihilation (due to nuclear catastrophe) or extinction (due to environmental disasters) if we do not act wisely. If that happens, and all of scientific knowledge were to be destroyed then Feynman would like the ‘Atomic Hypothesis’ to be communicated to the next generation of creatures in the universe. Atomic Hypothesis: All things are made of atoms little particles that move around in perpetual motion, attracting each other when they are a little distance apart, but repelling upon being squeezed into one another.

Speculation that matter may not be continuous, existed in many places and cultures. Kanada in India and Democritus in Greece had suggested that matter may consist of indivisible constituents. The scientific ‘Atomic Theory’ is usually credited to John Dalton. He proposed the atomic theory to explain the laws of definite and multiple proportions obeyed by elements when they combine into compounds. The first law says that any given compound has, a fixed proportion by mass of its constituents. The second law says that when two elements form more than one compound, for a fixed mass of one element, the masses of the other elements are in ratio of small integers.

To explain the laws Dalton suggested, about 200 years ago, that the smallest constituents of an element are atoms. Atoms of one element are identical but differ from those of other elements. A small number of atoms of each element combine to form a molecule of the compound. Gay Lussac’s law, also given in early $19^{\text {th }}$ century, states: When gases combine chemically to yield another gas, their volumes are in the ratios of small integers. Avogadro’s law (or hypothesis) says: Equal volumes of all gases at equal temperature and pressure have the same number of molecules. Avogadro’s law, when combined with Dalton’s theory explains Gay Lussac’s law. Since the elements are often in the form of molecules, Dalton’s atomic theory can also be referred to as the molecular theory of matter. The theory is now well accepted by scientists. However even at the end of the nineteenth century there were famous scientists who did not believe in atomic theory!

From many observations, in recent times we now know that molecules (made up of one or more atoms) constitute matter. Electron microscopes and scanning tunnelling microscopes enable us to even see them. The size of an atom is about an angstrom $\left(10^{-10} \mathrm{~m}\right)$. In solids, which are tightly packed, atoms are spaced about a few angstroms $(2 \mathring{A})$ apart. In liquids the separation between atoms is also about the same. In liquids the atoms are not as rigidly fixed as in solids, and can move around. This enables a liquid to flow. In gases the interatomic distances are in tens of angstroms. The average distance a molecule can travel without colliding is called the mean free path. The mean free path, in gases, is of the order of thousands of angstroms. The atoms are much freer in gases and can travel long distances without colliding. If they are not enclosed, gases disperse away. In solids and liquids the closeness makes the interatomic force important. The force has a long range attraction and a short range repulsion. The atoms attract when they are at a few angstroms but repel when they come closer. The static appearance of a gas is misleading. The gas is full of activity and the equilibrium is a dynamic one. In dynamic equilibrium, molecules collide and change their speeds during the collision. Only the average properties are constant.

Atomic theory is not the end of our quest, but the beginning. We now know that atoms are not indivisible or elementary. They consist of a nucleus and electrons. The nucleus itself is made up of protons and neutrons. The protons and neutrons are again made up of quarks. Even quarks may not be the end of the story. There may be string like elementary entities. Nature always has surprises for us, but the search for truth is often enjoyable and the discoveries beautiful. In this chapter, we shall limit ourselves to understanding the behaviour of gases (and a little bit of solids), as a collection of moving molecules in incessant motion.

**Atomic Hypothesis in Ancient India and Greece**

Though John Dalton is credited with the introduction of atomic viewpoint in modern science, scholars in ancient India and Greece conjectured long before the existence of atoms and molecules. In the Vaiseshika school of thought in India founded by Kanada (Sixth century B.C.) the atomic picture was developed in considerable detail. Atoms were thought to be eternal, indivisible, infinitesimal and ultimate parts of matter. It was argued that if matter could be subdivided without an end, there would be no difference between a mustard seed and the Meru mountain. The four kinds of atoms (Paramanu - Sanskrit word for the smallest particle) postulated were Bhoomi (Earth), Ap (water), Tejas (fire) and Vayu (air) that have characteristic mass and other attributes, were propounded. Akasa (space) was thought to have no atomic structure and was continuous and inert. Atoms combine to form different molecules (e.g. two atoms combine to form a diatomic molecule dvyanuka, three atoms form a tryanuka or a triatomic molecule), their properties depending upon the nature and ratio of the constituent atoms. The size of the atoms was also estimated, by conjecture or by methods that are not known to us. The estimates vary. In Lalitavistara, a famous biography of the Buddha written mainly in the second century B.C., the estimate is close to the modern estimate of atomic size, of the order of $10^{-10} \mathrm{~m}$.

In ancient Greece, Democritus (Fourth century B.C.) is best known for his atomic hypothesis. The word ‘atom’ means ‘indivisible’ in Greek. According to him, atoms differ from each other physically, in shape, size and other properties and this resulted in the different properties of the substances formed by their combination. The atoms of water were smooth and round and unable to ‘hook’ on to each other, which is why liquid / water flows easily. The atoms of earth were rough and jagged, so they held together to form hard substances. The atoms of fire were thorny which is why it caused painful burns. These fascinating ideas, despite their ingenuity, could not evolve much further, perhaps because they were intuitive conjectures and speculations not tested and modified by quantitative experiments - the hallmark of modern science.

### 12.3 BEHAVIOUR OF GASES

Properties of gases are easier to understand than those of solids and liquids. This is mainly because in a gas, molecules are far from each other and their mutual interactions are negligible except when two molecules collide. Gases at low pressures and high temperatures much above that at which they liquefy (or solidify) approximately satisfy a simple relation between their pressure, temperature and volume given by (see Chapter 10)

$$ \begin{equation*} P V=K T \tag{12.1} \end{equation*} $$

for a given sample of the gas. Here $T$ is the temperature in kelvin or (absolute) scale. $K$ is a constant for the given sample but varies with the volume of the gas. If we now bring in the idea of atoms or molecules, then $K$ is proportional to the number of molecules, (say) $N$ in the sample. We can write $K=N k$. Observation tells us that this $k$ is same for all gases. It is called Boltzmann constant and is denoted by $k_{\mathrm{B}}$.

$$ \text{ As} \frac{P_{1} V_{1}}{N_{1} T_{1}}=\frac{P_{2} V_{2}}{N_{2} T_{2}}= \text{constant} =k_{\mathrm{B}} \tag{12.2}$$

if $P, V$ and $T$ are same, then $N$ is also same for all gases. This is Avogadro’s hypothesis, that the number of molecules per unit volume is the same for all gases at a fixed temperature and pressure. The number in 22.4 litres of any gas is $6.02 \times 10^{23}$. This is known as Avogadro number and is denoted by $N_{\mathrm{A}}$. The mass of 22.4 litres of any gas is equal to its molecular weight in grams at S.T.P (standard temperature $273 \mathrm{~K}$ and pressure $1 \mathrm{~atm}$). This amount of substance is called a mole (see Chapter 1 for a more precise definition). Avogadro had guessed the equality of numbers in equal volumes of gas at a fixed temperature and pressure from chemical reactions. Kinetic theory justifies this hypothesis.

The perfect gas equation can be written as

$$ \begin{equation*} P V=\mu R T \tag{12.3} \end{equation*} $$

where $\mu$ is the number of moles and $R=N_{\mathrm{A}}$ $k_{\mathrm{B}}$ is a universal constant. The temperature $T$ is absolute temperature. Choosing kelvin scale for absolute temperature, $R=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$. Here

$$ \begin{equation*} \mu=\frac{M}{M_{0}}=\frac{N}{N_{A}} \tag{12.4} \end{equation*} $$

where $M$ is the mass of the gas containing $N$ molecules, $M_{0}$ is the molar mass and $N_{\mathrm{A}}$ the Avogadro’s number. Using Eqs. (12.4) and (12.3) can also be written as

$$P V=k_{\mathrm{B}} N T \quad \text { or } \quad P=k_{\mathrm{B}} n T$$

Fig.12.1 Real gases approach ideal gas behaviour at low pressures and high temperatures.

where $n$ is the number density, i.e. number of molecules per unit volume. $k_{\mathrm{B}}$ is the Boltzmann constant introduced above. Its value in SI units is $1.38 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1}$.

Another useful form of Eq. (12.3) is

$$ \begin{equation*} P=\frac{\rho R T}{M_{0}} \tag{12.5} \end{equation*} $$

where $\rho$ is the mass density of the gas.

A gas that satisfies Eq. (12.3) exactly at all pressures and temperatures is defined to be an ideal gas. An ideal gas is a simple theoretical model of a gas. No real gas is truly ideal. Fig. 12.1 shows departures from ideal gas behaviour for a real gas at three different temperatures. Notice that all curves approach the ideal gas behaviour for low pressures and high temperatures.

At low pressures or high temperatures the molecules are far apart and molecular interactions are negligible. Without interactions the gas behaves like an ideal one.

If we fix $\mu$ and $T$ in Eq. (12.3), we get

$$ \begin{equation*} P V=\text { constant } \tag{12.6} \end{equation*} $$

i.e., keeping temperature constant, pressure of a given mass of gas varies inversely with volume. This is the famous Boyle’s law. Fig. 12.2 shows comparison between experimental $P-V$ curves and the theoretical curves predicted by Boyle’s law. Once again you see that the agreement is good at high temperatures and low pressures. Next, if you fix $P$, Eq. (12.1) shows that $V \propto T$ i.e., for a fixed pressure, the volume of a gas is proportional to its absolute temperature $T$ (Charles’ law). See Fig. 12.3.

Fig.12.2 Experimental P-V curves (solid lines) for steam at three temperatures compared with Boyle’s law (dotted lines). P is in units of 22 atm and V in units of 0.09 litres

Finally, consider a mixture of non-interacting ideal gases: $\mu_{1}$ moles of gas $1, \mu_{2}$ moles of gas 2 , etc. in a vessel of volume $V$ at temperature $T$ and pressure $P$. It is then found that the equation of state of the mixture is :

$$ \begin{align*} & P V=\left(\mu_{1}+\mu_{2}+\ldots\right) R T \tag{12.7}\\ & \text { i.e. } P=\mu_{1} \frac{R T}{V}+\mu_{2} \frac{R T}{V}+\ldots \tag{12.8}\\ & =P_{1}+P_{2}+\ldots \tag{12.9} \end{align*} $$

Clearly $P_{1}=\mu_{1} R T / V$ is the pressure that gas 1 would exert at the same conditions of volume and temperature if no other gases were present. This is called the partial pressure of the gas. Thus, the total pressure of a mixture of ideal gases is the sum of partial pressures. This is Dalton’s law of partial pressures.

Fig. 12.3 Experimental T-V curves (solid lines) for CO2 at three pressures compared with Charles’ law (dotted lines). T is in units of 300 K and V in units of 0.13 litres

We next consider some examples which give us information about the volume occupied by the molecules and the volume of a single molecule.

**Example 12.1** The density of water is 1000 $\mathrm{kg} \mathrm{m}^{-3}$. The density of water vapour at $100{ }^{\circ} \mathrm{C}$ and $1 \mathrm{~atm}$ pressure is $0.6 \mathrm{~kg} \mathrm{~m}^{-3}$. The volume of a molecule multiplied by the total number gives, what is called, molecular volume. Estimate the ratio (or fraction) of the molecular volume to the total volume occupied by the water vapour under the above conditions of temperature and pressure.

**Answer** For a given mass of water molecules, the density is less if volume is large. So the volume of the vapour is $1000 / 0.6=1 /\left(6 \times 10^{-4}\right)$ times larger. If densities of bulk water and water molecules are same, then the fraction of molecular volume to the total volume in liquid state is 1. As volume in vapour state has increased, the fractional volume is less by the same amount, i.e. $6 \times 10^{-4}$.

**Example 12.2** Estimate the volume of a water molecule using the data in Example 12.1 .

**Answer** In the liquid (or solid) phase, the molecules of water are quite closely packed. The density of water molecule may therefore, be regarded as roughly equal to the density of bulk water $=1000 \mathrm{~kg} \mathrm{~m}^{-3}$. To estimate the volume of a water molecule, we need to know the mass of a single water molecule. We know that 1 mole of water has a mass approximately equal to

$(2+16) \mathrm{g}=18 \mathrm{~g}=0.018 \mathrm{~kg}$.

Since 1 mole contains about $6 \times 10^{23}$ molecules (Avogadro’s number), the mass of a molecule of water is $(0.018) /\left(6 \times 10^{23}\right) \mathrm{kg}=$ $3 \times 10^{-26} \mathrm{~kg}$. Therefore, a rough estimate of the volume of a water molecule is as follows :

Volume of a water molecule

$$ \begin{aligned} & =\left(3 \times 10^{-26} \mathrm{~kg}\right) /\left(1000 \mathrm{~kg} \mathrm{~m}^{-3}\right) \\ & =3 \times 10^{-29} \mathrm{~m}^{3} \\ & =(4 / 3) \pi \text { (Radius) }^{3} \end{aligned} $$

Hence, Radius $\approx 2 \times 10^{-10} \mathrm{~m}=2 \mathring{A}$

**Example 12.3** What is the average distance between atoms (interatomic distance) in water? Use the data given in Examples 12.1 and 12.2 .

**Answer** A given mass of water in vapour state has $1.67 \times 10^{3}$ times the volume of the same mass of water in liquid state (Ex. 12.1). This is also the increase in the amount of volume available for each molecule of water. When volume increase by $10^{3}$ times the radius increases by $V^{1 / 3}$ or 10 times, i.e., $10 \times 2 \mathring{A}=20 \mathring{A}$. So the average distance is $2 \times 20=40 \mathring{A}$.

**Example 12.4** A vessel contains two nonreactive gases : neon (monatomic) and oxygen (diatomic). The ratio of their partial pressures is 3:2. Estimate the ratio of (i) number of molecules and (ii) mass density of neon and oxygen in the vessel. Atomic mass of $\mathrm{Ne}=20.2 \mathrm{u}$, molecular mass of $\mathrm{O}_{2}$ $=32.0 \mathrm{u}$

**Answer** Partial pressure of a gas in a mixture is the pressure it would have for the same volume and temperature if it alone occupied the vessel. (The total pressure of a mixture of non-reactive gases is the sum of partial pressures due to its constituent gases.) Each gas (assumed ideal) obeys the gas law. Since $V$ and $T$ are common to the two gases, we have $P_{1} V=\mu_{1} R T$ and $P_{2} V=$ $\mu_{2} R T$, i.e. $\left(P_{1} / P_{2}\right)=\left(\mu_{1} / \mu_{2}\right)$. Here 1 and 2 refer to neon and oxygen respectively. Since

$$ \left(P _{1} / P _{2}\right)=(3 / 2) \text { (Given), }\left(\mu _{1} / \mu _{2}\right)=3 / 2 $$

(i) By definition $\mu_{1}=\left(N_{1} / N_{\mathrm{A}}\right)$ and $\mu_{2}=\left(N_{2} / N_{\mathrm{A}}\right)$ where $N_{1}$ and $N_{2}$ are the number of molecules of 1 and 2 , and $N_{\mathrm{A}}$ is the Avogadro’s number. Therefore,

$\left(N_{1} / N_{2}\right)=\left(\mu_{1} / \mu_{2}\right)=3 / 2$.

(ii) We can also write $\mu_{1}=\left(m_{1} / M_{1}\right)$ and $\mu_{2}=$ $\left(m_{2} / M_{2}\right)$ where $m_{1}$ and $m_{2}$ are the masses of 1 and 2; and $M_{1}$ and $M_{2}$ are their molecular masses. (Both $m_{1}$ and $M_{1}$; as well as $m_{2}$ and $M_{2}$ should be expressed in the same units). If $\rho_{1}$ and $\rho_{2}$ are the mass densities of 1 and 2 respectively, we have

$$ \begin{aligned} & \frac{\rho_{1}}{\rho_{2}}=\frac{m_{1} / V}{m_{2} / V}=\frac{m_{1}}{m_{2}}=\frac{\mu_{1}}{\mu_{2}} \times \frac{M_{1}}{M_{2}} \\ & =\frac{3}{2} \times \frac{20.2}{32.0}=0.947 \end{aligned} $$

### 12.4 KINETIC THEORY OF AN IDEAL GAS

Kinetic theory of gases is based on the molecular picture of matter. A given amount of gas is a collection of a large number of molecules (typically of the order of Avogadro’s number) that are in incessant random motion. At ordinary pressure and temperature, the average distance between molecules is a factor of 10 or more than the typical size of a molecule ( $2 \mathring{A}$ ). Thus, interaction between molecules is negligible and we can assume that they move freely in straight lines according to Newton’s first law. However, occasionally, they come close to each other, experience intermolecular forces and their velocities change. These interactions are called collisions. The molecules collide incessantly against each other or with the walls and change their velocities. The collisions are considered to be elastic. We can derive an expression for the pressure of a gas based on the kinetic theory.

We begin with the idea that molecules of a gas are in incessant random motion, colliding against one another and with the walls of the container. All collisions between molecules among themselves or between molecules and the walls are elastic. This implies that total kinetic energy is conserved. The total momentum is conserved as usual.

#### 12.4.1 Pressure of an Ideal Gas

Consider a gas enclosed in a cube of side l. Take the axes to be parallel to the sides of the cube, as shown in Fig. 12.4. A molecule with velocity $\left(V_{x}, V_{y}, V_{z}\right)$ hits the planar wall parallel to $y z^{-}$ plane of area $A\left(=I^{2}\right)$. Since the collision is elastic, the molecule rebounds with the same velocity; its $y$ and $z$ components of velocity do not change in the collision but the $x$-component reverses sign. That is, the velocity after collision is $\left(-V_{x}, V_{y}, V_{z}\right)$. The change in momentum of the molecule is: $-m v_{x}-\left(m v_{x}\right)=-2 m v_{x}$. By the principle of conservation of momentum, the momentum imparted to the wall in the collision $=2 m v_{x}$.

Fig. 12.4 Elastic collision of a gas molecule with the all of the container

To calculate the force (and pressure) on the wall, we need to calculate momentum imparted to the wall per unit time. In a small time interval $\Delta t$, a molecule with $x$-component of velocity $v_{x}$ will hit the wall if it is within the distance $v_{x} \Delta t$ from the wall. That is, all molecules within the volume $A v_{x} \Delta t$ only can hit the wall in time $\Delta t$. But, on the average, half of these are moving towards the wall and the other half away from the wall. Thus, the number of molecules with velocity $\left(v_{x}, V_{y}, V_{Z}\right)$ hitting the wall in time $\Delta t$ is $1 / 2 A v_{X} \Delta t n$, where $n$ is the number of molecules per unit volume. The total momentum transferred to the wall by these molecules in time $\Delta t$ is :

$ \begin{equation*} Q=\left(2 m v _{x}\right)\left(-n A v _{x} \Delta t\right) \tag{12.10} \end{equation*} $

The force on the wall is the rate of momentum transfer $Q / \Delta t$ and pressure is force per unit area :

$$ \begin{equation*} P=Q /(A \Delta t)=n m v_{x}^{2} \tag{12.11} \end{equation*} $$

Actually, all molecules in a gas do not have the same velocity; there is a distribution in velocities. The above equation, therefore, stands for pressure due to the group of molecules with speed $V_{X}$ in the $x$-direction and $\mathrm{n}$ stands for the number density of that group of molecules. The total pressure is obtained by summing over the contribution due to all groups:

$$ \begin{equation*} P=n m \overline{v_{x}^{2}} \tag{12.12} \end{equation*} $$

where $\overline{v_{x}^{2}}$ is the average of $v_{x}^{2}$. Now the gas is isotropic, i.e. there is no preferred direction of velocity of the molecules in the vessel. Therefore, by symmetry,

$$ \begin{align*} & \overline{v_{x}^{2}}=\overline{v_{y}^{2}}=\overline{v_{z}^{2}} \\ & =(1 / 3)\left[\overline{v_{x}^{2}}+\overline{v_{y}^{2}}+\overline{v_{z}^{2}}\right]=(1 / 3) \overline{v^{2}} \tag{12.13} \end{align*} $$

where $v$ is the speed and $\overline{v^{2}}$ denotes the mean of the squared speed. Thus

$$ \begin{equation*} P=(1 / 3) n m \overline{v^{2}} \tag{12.14} \end{equation*} $$

Some remarks on this derivation. First, though we choose the container to be a cube, the shape of the vessel really is immaterial. For a vessel of arbitrary shape, we can always choose a small infinitesimal (planar) area and carry through the steps above. Notice that both A and $\Delta t$ do not appear in the final result. By Pascal’s law, given in Ch. 9, pressure in one portion of the gas in equilibrium is the same as anywhere else. Second, we have ignored any collisions in the derivation. Though this assumption is difficult to justify rigorously, we can qualitatively see that it will not lead to erroneous results. The number of molecules hitting the wall in time $\Delta t$ was found to be $1 / 2 n A v_{x} \Delta t$. Now the collisions are random and the gas is in a steady state. Thus, if a molecule with velocity $\left(v_{x}, v_{y}, v_{z}\right)$ acquires a different velocity due to collision with some molecule, there will always be some other molecule with a different initial velocity which after a collision acquires the velocity $\left(V_{X}, V_{y}, V_{z}\right)$. If this were not so, the distribution of velocities would not remain steady. In any case we are finding $\overline{v_{x}^{2}}$. Thus, on the whole, molecular collisions (if they are not too frequent and the time spent in a collision is negligible compared to time between collisions) will not affect the calculation above.

#### 12.4.2 Kinetic Interpretation of Temperature

Equation (13.14) can be written as

$$ \begin{align*} & P V=(1 / 3) n V m \overline{v^{2}} \tag{12.15a}\\ & P V=(2 / 3) N X^{1 / 2} m \overline{v^{2}} \tag{12.15b} \end{align*} $$

where $N(=n V)$ is the number of molecules in the sample.

The quantity in the bracket is the average translational kinetic energy of the molecules in the gas. Since the internal energy $E$ of an ideal gas is purely kinetic*,

$$ \begin{equation*} E=N(1 / 2) m \overline{v^{2}} \tag{12.16} \end{equation*} $$

Equation (12.15) then gives :

$$P V=(2 / 3) E \tag{12.17}$$

We are now ready for a kinetic interpretation of temperature. Combining Eq. (12.17) with the ideal gas Eq. (12.3), we get

$$ \begin{equation*} E=(3 / 2) \quad k_{B} N T \tag{12.18} \end{equation*} $$

$$ \text{or} \quad E / N=1 / 2 m \overline{v^{2}}=(3 / 2) k_{B} T\tag{12.19}$$

i.e., the average kinetic energy of a molecule is proportional to the absolute temperature of the gas; it is independent of pressure, volume or the nature of the ideal gas. This is a fundamental result relating temperature, a macroscopic measurable parameter of a gas (a thermodynamic variable as it is called) to a molecular quantity, namely the average kinetic energy of a molecule. The two domains are connected by the Boltzmann constant. We note in passing that Eq. (12.18) tells us that internal energy of an ideal gas depends only on temperature, not on pressure or volume. With this interpretation of temperature, kinetic theory of an ideal gas is completely consistent with the ideal gas equation and the various gas laws based on it.

For a mixture of non-reactive ideal gases, the total pressure gets contribution from each gas in the mixture. Equation (12.14) becomes

$$ \begin{equation*} P=(1 / 3)\left[n_{1} m_{1} \overline{v_{1}^{2}}+n_{2} m_{2} \overline{v_{2}^{2}}+\ldots\right] \tag{12.20} \end{equation*} $$

In equilibrium, the average kinetic energy of the molecules of different gases will be equal. That is,

$1 / 2 m_{1} \overline{v_{1}^{2}}=1 / 2 m_{2} \overline{v_{2}^{2}}=(3 / 2) k_{B} T$

so that

$$ \begin{equation*} P=\left(n_{1}+n_{2}+\ldots\right) k_{B} T \tag{12.21} \end{equation*} $$

which is Dalton’s law of partial pressures.

From Eq. (12.19), we can get an idea of the typical speed of molecules in a gas. At a temperature $T=300 \mathrm{~K}$, the mean square speed of a molecule in nitrogen gas is :

$ \begin{gathered} m=\frac{M_{N_{2}}}{N_{A}}=\frac{28}{6.02 \times 10^{26}}=4.65 \times 10^{-26} \mathrm{~kg} .\end{gathered} $

$ \begin{gathered} \overline{v^{2}}=3 k_{B} T / \mathrm{m}=(516)^{2} \mathrm{~m}^{2} \mathrm{~s}^{-2} \end{gathered} $

The square root of $\overline{v^{2}}$ is known as root mean square (rms) speed and is denoted by $V_{\text {rms }}$,

(We can also write $\overline{v^{2}}$ as $ < v^{2}>$.)

$V_{\text {rms }}=516 \mathrm{~m} \mathrm{~s}^{-1}$

The speed is of the order of the speed of sound in air. It follows from Eq. (12.19) that at the same temperature, lighter molecules have greater rms speed.

**Example 12.5** A flask contains argon and chlorine in the ratio of $2: 1$ by mass. The temperature of the mixture is $27 \mathrm{C}$. Obtain the ratio of (i) average kinetic energy per molecule, and (ii) root mean square speed $v_{\text {rms }}$ of the molecules of the two gases. Atomic mass of argon $=39.9 \mathrm{u}$; Molecular mass of chlorine $=70.9 \mathrm{u}$.

**Answer** The important point to remember is that the average kinetic energy (per molecule) of any (ideal) gas (be it monatomic like argon, diatomic like chlorine or polyatomic) is always equal to $(3 / 2) k_{B} T$. It depends only on temperature, and is independent of the nature of the gas.

(i) Since argon and chlorine both have the same temperature in the flask, the ratio of average kinetic energy (per molecule) of the two gases is $1: 1$.

(ii) Now $1 / 2 m v_{\text {rms }}^{2}=$ average kinetic energy per molecule $=(3 / 2)) k_{\mathrm{B}} T$ where $m$ is the mass of a molecule of the gas. Therefore,

$$\dfrac{\mathbf{v} _{r m s _{\mathrm{Ar}}}^2}{\mathbf{v} _{r m s _{\mathrm{Cl}}}^2}\quad \dfrac{m _{\mathrm{Cl}}}{m _{\mathrm{Ar}}} \quad \dfrac{M _{\mathrm{Cl}}}{M _{\mathrm{Ar}}}=\dfrac{70.9}{39.9}=1.77$$

where $M$ denotes the molecular mass of the gas. (For argon, a molecule is just an atom of argon.) Taking square root of both sides,

$$\frac{v_{r m s}{Ar}} {v_{rms }{Cl}}=1.33$$

You should note that the composition of the mixture by mass is quite irrelevant to the above calculation. Any other proportion by mass of argon and chlorine would give the same answers to (i) and (ii), provided the temperature remains unaltered.

**Example 12.6** Uranium has two isotopes of masses 235 and 238 units. If both are present in Uranium hexafluoride gas which would have the larger average speed? If atomic mass of fluorine is 19 units, estimate the percentage difference in speeds at any temperature.

**Answer** At a fixed temperature the average energy $=1 / 2 m\left\langle v^{2}\right\rangle$ is constant. So smaller the mass of the molecule, faster will be the speed. The ratio of speeds is inversely proportional to the square root of the ratio of the masses. The masses are 349 and 352 units. So

$$V_{349} / V_{352}=(352 / 349)^{1 / 2}=1.0044$$.

Hence difference $\frac{\Delta V}{V}=0.44 %$.

[${ }^{235} \mathrm{U}$ is the isotope needed for nuclear fission. To separate it from the more abundant isotope ${ }^{238} \mathrm{U}$, the mixture is surrounded by a porous cylinder. The porous cylinder must be thick and narrow, so that the molecule wanders through individually, colliding with the walls of the long pore. The faster molecule will leak out more than the slower one and so there is more of the lighter molecule (enrichment) outside the porous cylinder (Fig. 12.5). The method is not very efficient and has to be repeated several times for sufficient enrichment.].

When gases diffuse, their rate of diffusion is inversely proportional to square root of the masses (see Exercise 12.12). Can you guess the explanation from the above answer?

Fig. 12.5 Molecules going through a porous wall.

**Example 12.7** (a) When a molecule (or an elastic ball) hits a ( massive) wall, it rebounds with the same speed. When a ball hits a massive bat held firmly, the same thing happens. However, when the bat is moving towards the ball, the ball rebounds with a different speed. Does the ball move faster or slower? (Ch. 5 will refresh your memory on elastic collisions.)

(b) When gas in a cylinder is compressed by pushing in a piston, its temperature rises. Guess at an explanation of this in terms of kinetic theory using (a) above.

(c) What happens when a compressed gas pushes a piston out and expands. What would you observe?

(d) Sachin Tendulkar used a heavy cricket bat while playing. Did it help him in anyway?

**Answer** (a) Let the speed of the ball be $u$ relative to the wicket behind the bat. If the bat is moving towards the ball with a speed $V$ relative to the wicket, then the relative speed of the ball to bat is $V+u$ towards the bat. When the ball rebounds (after hitting the massive bat) its speed, relative to bat, is $V+u$ moving away from the bat. So relative to the wicket the speed of the rebounding ball is $V+(V+u)=2 V+u$, moving away from the wicket. So the ball speeds up after the collision with the bat. The rebound speed will be less than $u$ if the bat is not massive. For a molecule this would imply an increase in temperature.

You should be able to answer (b) (c) and (d) based on the answer to (a).

(Hint: Note the correspondence, piston $\rightarrow$ bat, cylinder $\rightarrow$ wicket, molecule $\rightarrow$ ball.)

### 12.5 LAW OF EQUIPARTITION OF ENERGY

The kinetic energy of a single molecule is

$$ \begin{equation*} \varepsilon_{t}=\frac{1}{2} m v_{x}^{2}+\frac{1}{2} m v_{y}^{2}+\frac{1}{2} m v_{z}^{2} \tag{12.22} \end{equation*} $$

For a gas in thermal equilibrium at temperature $T$ the average value of energy denoted by $<\varepsilon_{t}>$ is

$$\left\langle\varepsilon_{t}\right\rangle=\left\langle\frac{1}{2} m v_{x}^{2}\right\rangle+\left\langle\frac{1}{2} m v_{y}^{2}\right\rangle+\left\langle\frac{1}{2} m v_{z}^{2}\right\rangle=\frac{3}{2} k_{B} T \tag{12.23}$$

Since there is no preferred direction, Eq. (12.23) implies

$\left\langle\frac{1}{2} m v_{x}^{2}\right\rangle=\frac{1}{2} k_{B} T,\left\langle\frac{1}{2} m v_{y}^{2}\right\rangle=\frac{1}{2} k_{B} T$,

$$\left\langle\frac{1}{2} m v_{z}^{2}\right\rangle=\frac{1}{2} k_{B} T \tag{12.24}$$

A molecule free to move in space needs three coordinates to specify its location. If it is constrained to move in a plane it needs two; and if constrained to move along a line, it needs just one coordinate to locate it. This can also be expressed in another way. We say that it has one degree of freedom for motion in a line, two for motion in a plane and three for motion in space. Motion of a body as a whole from one point to another is called translation. Thus, a molecule free to move in space has three translational degrees of freedom. Each translational degree of freedom contributes a term that contains square of some variable of motion, e.g., $1 / 2 m v_{x}^{2}$ and similar terms in $v_{y}$ and $v_{z}$. In, Eq. (12.24) we see that in thermal equilibrium, the average of each such term is $1 / 2 k_{B} T$.

Molecules of a monatomic gas like argon have only translational degrees of freedom. But what about a diatomic gas such as $\mathrm{O}_2$ or $\mathrm{N}_2$ ? A molecule of $\mathrm{O}_2$ has three translational degrees of freedom. But in addition it can also rotate about its centre of mass. Figure 12.6 shows the two independent axes of rotation 1 and 2 , normal to the axis joining the two oxygen atoms about which the molecule can rotate*. The molecule thus has two rotational degrees of freedom, each of which contributes a term to the total energy consisting of translational energy $\varepsilon_t$ and rotational energy $\varepsilon_r$.

$$ \begin{equation*} \varepsilon_{t}+\varepsilon_{r}=\frac{1}{2} m v_{x}^{2}+\frac{1}{2} m v_{y}^{2}+\frac{1}{2} m v_{z}^{2}+\frac{1}{2} I_{1} \omega_{1}^{2}+\frac{1}{2} I_{2} \omega_{2}^{2} \tag{12.25} \end{equation*} $$

Fig. 12.6 The two independent axes of rotation of a diatomic molecule

where $\omega_{1}$ and $\omega_{2}$ are the angular speeds about the axes 1 and 2 and $I_{1}, I_{2}$ are the corresponding moments of inertia. Note that each rotational degree of freedom contributes a term to the energy that contains square of a rotational variable of motion.

We have assumed above that the $\mathrm{O}_2$ molecule is a ‘rigid rotator’, i.e., the molecule does not vibrate. This assumption, though found to be true (at moderate temperatures) for $\mathrm{O}_2$, is not always valid. Molecules, like $\mathrm{CO}$, even at moderate temperatures have a mode of vibration, i.e., its atoms oscillate along the interatomic axis like a one-dimensional oscillator, and contribute a vibrational energy term $\varepsilon_V$ to the total energy:

$$ \begin{align*} & \varepsilon_{v}=\frac{1}{2} m \frac{\mathrm{d} y^{2}}{\mathrm{~d} t}+\frac{1}{2} k y^{2} \\ & \varepsilon=\varepsilon_{t}+\varepsilon_{r}+\varepsilon_{v} \tag{12.26} \end{align*} $$

where $\mathrm{k}$ is the force constant of the oscillator and y the vibrational co-ordinate.

Once again the vibrational energy terms in Eq. (12.26) contain squared terms of vibrational variables of motion $y$ and $\mathrm{d} y / \mathrm{d} t$.[^0]

At this point, notice an important feature in Eq.(12.26). While each translational and rotational degree of freedom has contributed only one ‘squared term’ in Eq.(12.26), one vibrational mode contributes two ‘squared terms’ : kinetic and potential energies.

Each quadratic term occurring in the expression for energy is a mode of absorption of energy by the molecule. We have seen that in thermal equilibrium at absolute temperature $\mathrm{T}$, for each translational mode of motion, the average energy is $1 / 2 k_{B} T$. The most elegant principle of classical statistical mechanics (first proved by Maxwell) states that this is so for each mode of energy: translational, rotational and vibrational. That is, in equilibrium, the total energy is equally distributed in all possible energy modes, with each mode having an average energy equal to $1 / 2 k_{B} T$. This is known as the law of equipartition of energy. Accordingly, each translational and rotational degree of freedom of a molecule contributes $1 / 2 k_{B} T$ to the energy, while each vibrational frequency contributes $2 \times 1 / 2 k_{B} T=k_{B} T$, since a vibrational mode has both kinetic and potential energy modes.

The proof of the law of equipartition of energy is beyond the scope of this book. Here, we shall apply the law to predict the specific heats of gases theoretically. Later, we shall also discuss briefly, the application to specific heat of solids.

### 12.6 SPECIFIC HEAT CAPACITY

#### 12.6.1 Monatomic Gases

The molecule of a monatomic gas has only three translational degrees of freedom. Thus, the average energy of a molecule at temperature $T$ is $(3 / 2) k_{\mathrm{B}} T$. The total internal energy of a mole of such a gas is

$$ \begin{equation*} U=\frac{3}{2} k_{B} T \times N_{A}=\frac{3}{2} R T \tag{12.27} \end{equation*} $$

The molar specific heat at constant volume, $C_{v}$, is

$ \begin{equation*} C_{V}(\text { monatomic gas })=\frac{\mathrm{d} U}{\mathrm{~d} T}=\frac{3}{2} R T \tag{12.28} \end{equation*} $

For an ideal gas,

$$ \begin{equation*} C_{p}-C_{V}=R \tag{12.29} \end{equation*} $$

where $C_{p}$ is the molar specific heat at constant pressure. Thus,

$$ \begin{equation*} C_{p}=\frac{5}{2} R \tag{12.30} \end{equation*} $$

The ratio of specific heats

$$\gamma=\frac{C_{\mathrm{p}}}{C_{\mathrm{v}}}=\frac{5}{3}\tag{12.31}$$

#### 12.6.2 Diatomic Gases

As explained earlier, a diatomic molecule treated as a rigid rotator, like a dumbbell, has 5 degrees of freedom: 3 translational and 2 rotational. Using the law of equipartition of energy, the total internal energy of a mole of such a gas is

$$ \begin{equation*} U=\frac{5}{2} k_{B} T \times N_{A}=\frac{5}{2} R T \tag{12.32} \end{equation*} $$

The molar specific heats are then given by

$ \begin{align*} & C_{V}(\text { rigid diatomic })=\frac{5}{2} R, C_{p}=\frac{7}{2} R \tag{12.33} \end{align*} $

$ \begin{align*} & \gamma(\text { rigid diatomic })=\frac{7}{5} \tag{12.34} \end{align*} $

If the diatomic molecule is not rigid but has in addition a vibrational mode

$$ \begin{align*} & U=\left(\frac{5}{2} k_{B} T+k_{B} T\right) N_{A}=\frac{7}{2} R T \\ & C_{v}=\frac{7}{2} R, C_{p}=\frac{9}{2} R, \quad \gamma=\frac{9}{7} R \tag{12.35} \end{align*} $$

#### 12.6.3 Polyatomic Gases

In general a polyatomic molecule has 3 translational, 3 rotational degrees of freedom and a certain number ( $f$ ) of vibrational modes. According to the law of equipartition of energy, it is easily seen that one mole of such a gas has

$ \begin{align*} & U=\left(\frac{3}{2} k_{B} T+\frac{3}{2} k_{B} T+f k_{B} T\right) N_{A} & \text { i.e., } C_{V}=(3+f) R, C_{p}=(4+f) R \end{align*} $

$ \begin{equation*} \gamma=\frac{(4+f)}{(3+f)} \tag{12.36} \end{equation*} $

Note that $C_{p}-C_{v}=R$ is true for any ideal gas, whether mono, di or polyatomic.

Table 12.1 summarises the theoretical predictions for specific heats of gases ignoring any vibrational modes of motion. The values are in good agreement with experimental values of specific heats of several gases given in Table 12.2. Of course, there are discrepancies between predicted and actual values of specific heats of several other gases (not shown in the table), such as $\mathrm{Cl}_2, \mathrm{C}_2 \mathrm{H}_6$ and many other polyatomic gases. Usually, the experimental values for specific heats of these gases are greater than the predicted values as given in Table 12.1 suggesting that the agreement can be improved by including vibrational modes of motion in the calculation. The law of equipartition of energy is, thus, well verified experimentally at ordinary temperatures.

Table 12.1 Predicted values of specific heat capacities of gases (ignoring vibrational modes)

Nature of Gas |
$\mathbf{C}_{\mathbf{v}}$ $\left(\mathbf{J ~ m o l}^{-1} \mathbf{K}^{-1}\right)$ |
$\mathbf{C}_{\mathrm{p}}$ $\left(\mathrm{J} \mathrm{~mol}^{-1} \mathbf{K}^{-1}\right)$ |
$\mathbf{C}_p-\mathbf{C}_v$ $\left(\mathrm{J} \mathrm{~mol}^{-1} \mathbf{K}^{-1}\right)$ |
$\gamma$ |
---|---|---|---|---|

Monatomic | 12.5 | 20.8 | 8.31 | 1.67 |

Diatomic | 20.8 | 29.1 | 8.31 | 1.40 |

Triatomic | 24.93 | 33.24 | 8.31 | 1.33 |

$\mathbf{C}_p-\mathbf{C}_v$

$\left(\mathrm{J} \mathrm{mol}^{-1} \mathbf{K}^{-1}\right)$

Table12.2 Measured values of specific heat capacities of some gases

Nature of gas |
Gas | $\mathbf{C}_{\mathbf{v}}$ $\left(\mathrm{J} \mathrm{~mol}^{-1} \mathbf{K}^{-1}\right)$ |
$\mathbf{C}_{\mathbf{p}}$ $\left(\mathrm{J} \mathrm{~mol}^{-1} \mathbf{K}^{-1}\right)$ |
$\mathbf{C}_p-\mathbf{C}_v$ $\left(\mathrm{J} \mathrm{~mol}^{-1} \mathbf{K}^{-1}\right)$ |
$\gamma$ |
---|---|---|---|---|---|

Monatomic | $\mathrm{He}$ | 12.5 | 20.8 | 8.30 | 1.66 |

Monatomic | $\mathrm{Ne}$ | 12.7 | 20.8 | 8.12 | 1.64 |

Monatomic | $\mathrm{Ar}$ | 12.5 | 20.8 | 8.30 | 1.67 |

Diatomic | $\mathrm{H}_{2}$ | 20.4 | 28.8 | 8.45 | 1.41 |

Diatomic | $\mathrm{O}_{2}$ | 21.0 | 29.3 | 8.32 | 1.40 |

Diatomic | $\mathrm{N}_{2}$ | 20.8 | 29.1 | 8.32 | 1.40 |

Triatomic | $\mathrm{H}_{2} \mathrm{O}$ | 27.0 | 35.4 | 8.35 | 1.31 |

Polyatomic | $\mathbf{C H}_{4}$ | $\mathbf{2 7 . 1}$ | $\mathbf{3 5 . 4}$ | $\mathbf{8 . 3 6}$ | $\mathbf{1 . 3 1}$ |

**Example 12.8** A cylinder of fixed capacity 44.8 litres contains helium gas at standard temperature and pressure. What is the amount of heat needed to raise the temperature of the gas in the cylinder by $15.0^{\circ} \mathrm{C}$ ? $\left(R=8.31 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)$.

**Answer** Using the gas law $P V=\mu R T$, you can easily show that $1 \mathrm{~mol}$ of any (ideal) gas at standard temperature $(273 \mathrm{~K})$ and pressure ( $1 \mathrm{~atm}=1.01 \times 10^{5} \mathrm{~Pa}$ ) occupies a volume of 22.4 litres. This universal volume is called molar volume. Thus the cylinder in this example contains $2 \mathrm{~mol}$ of helium. Further, since helium is monatomic, its predicted (and observed) molar specific heat at constant volume, $C_{V}=(3 / 2) R$, and molar specific heat at constant pressure, $C_{p}=(3 / 2) R+R=(5 / 2) R$. Since the volume of the cylinder is fixed, the heat required is determined by $C_{v}$. Therefore,

Heat required $=$ no. of moles $\times$ molar specific heat rise in temperature

$$ \begin{aligned} & =2 \times 1.5 R \times 15.0=45 R \\ & =45 \times 8.31=374 \mathrm{~J} . \end{aligned} $$

#### 12.6.4 Specific Heat Capacity of Solids

We can use the law of equipartition of energy to determine specific heats of solids. Consider a solid of $N$ atoms, each vibrating about its mean position. An oscillation in one dimension has average energy of $2 \times 1 / 2 k_{B} T=k_{B} T$. In three dimensions, the average energy is $3 k_{B} T$. For a mole of solid, $N=N_{A}$, and the total energy is

$$ U=3 k_{B} T \times N_{A}=3 R T $$

Now at constant pressure $\Delta Q=\Delta U+P \Delta V$ $=\Delta U$, since for a solid $\Delta V$ is negligible. Hence,

$$ \begin{equation*} C=\frac{\Delta Q}{\Delta T}=\frac{\Delta U}{\Delta T}=3 R \tag{12.37} \end{equation*} $$

Table 12.3 Specific Heat Capacity of some solids at room temperature and atmospheric pressure

Substance | Specific heat $\left(\mathbf{J} \mathbf{k g}^{-1} \mathbf{k}^{-1}\right)$ |
Molar specific heat $\left(\mathbf{J ~} \mathbf{~ m}^{-1} \mathbf{k}^{-1}\right)$ |
---|---|---|

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 |

As Table 12.3 shows the prediction generally agrees with experimental values at ordinary temperature (Carbon is an exception).

### 12.7 MEAN FREE PATH

Molecules in a gas have rather large speeds of the order of the speed of sound. Yet a gas leaking from a cylinder in a kitchen takes considerable time to diffuse to the other corners of the room. The top of a cloud of smoke holds together for hours. This happens because molecules in a gas have a finite though small size, so they are bound to undergo collisions. As a result, they cannot move straight unhindered; their paths keep getting incessantly deflected.

Fig. 12.7 The volume swept by a molecule in time ∆t in which any molecule will collide with it.

Suppose the molecules of a gas are spheres of diameter $d$. Focus on a single molecule with the average speed $< v>$. It will suffer collision with any molecule that comes within a distance $d$ between the centres. In time $\Delta t$, it sweeps a volume $\pi d^2 < v > \Delta t$ wherein any other molecule will collide with it (see Fig. 12.7). If $n$ is the number of molecules per unit volume, the molecule suffers $n \pi d^2 < v > \Delta t$ collisions in time $\Delta t$. Thus the rate of collisions is $n \pi d^2 < v >$ or the time between two successive collisions is on the average,

$$ \begin{equation*} \tau=1 /\left(n \pi < v > d^2\right) \tag{12.38} \end{equation*} $$

The average distance between two successive collisions, called the mean free path 1 , is :

$$ \begin{equation*} 1= < v > =1 /\left(n \pi d^2\right) \tag{12.39} \end{equation*} $$

In this derivation, we imagined the other molecules to be at rest. But actually all molecules are moving and the collision rate is determined by the average relative velocity of the molecules. Thus we need to replace $\langle v\rangle$ by $\langle V\rangle$ in Eq. (12.38). A more exact treatment gives

$$ \begin{equation*} l=1 /\left(\sqrt{2} n \pi d^{2}\right) \tag{12.40} \end{equation*} $$

Let us estimate $l$ and $\tau$ for air molecules with average speeds $\langle v\rangle=(485 \mathrm{~m} / \mathrm{s})$. At STP

$$ \begin{align*} & n=\frac{\left(0.02 \times 10^{23}\right)}{\left(22.4 \times 10^{-3}\right)} \\ & =2.7 \times 10^{25} \mathrm{~m}^{-3} . \\ & \text { Taking, } d=2 \times 10^{-10} \mathrm{~m}, \\ & \tau=6.1 \times 10^{-10} \mathrm{~s} \\ & \text { and } l=2.9 \times 10^{-7} \mathrm{~m} \approx 1500 d \tag{12.41} \end{align*} $$

As expected, the mean free path given by Eq. (12.40) depends inversely on the number density and the size of the molecules. In a highly evacuated tube $n$ is rather small and the mean free path can be as large as the length of the tube.

**Example 12.9** Estimate the mean free path for a water molecule in water vapour at $373 \mathrm{~K}$. Use information from Exercises 12.1 and Eq. (12.41) above.

**Answer** The $d$ for water vapour is same as that of air. The number density is inversely proportional to absolute temperature.

So $n=2.7 \times 10^{25} \times \frac{273}{373}=2 \times 10^{25} \mathrm{~m}^{-3}$

Hence, mean free path $l=4 \times 10^{-7} \mathrm{~m}$

Note that the mean free path is 100 times the interatomic distance $\sim 40 \mathring{A}=4 \times 10^{-9} \mathrm{~m}$ calculated earlier. It is this large value of mean free path that leads to the typical gaseous behaviour. Gases can not be confined without a container.

Using, the kinetic theory of gases, the bulk measurable properties like viscosity, heat conductivity and diffusion can be related to the microscopic parameters like molecular size. It is through such relations that the molecular sizes were first estimated.

### Summary

**1.** The ideal gas equation connecting pressure $(P)$, volume $(V)$ and absolute temperature $(T)$ is

$$ P V=\mu R T \quad=k_{B} N T $$

where $\mu$ is the number of moles and $N$ is the number of molecules. $R$ and $k_{B}$ are universal constants.

$R=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}, \quad k_{B}=\frac{R}{N_{A}}=1.38 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1}$

Real gases satisfy the ideal gas equation only approximately, more so at low pressures and high temperatures.

**2.** Kinetic theory of an ideal gas gives the relation

$$ P=\frac{1}{3} n m \overline{v^{2}} $$

where $n$ is number density of molecules, $m$ the mass of the molecule and $\overline{v^{2}}$ is the mean of squared speed. Combined with the ideal gas equation it yields a kinetic interpretation of temperature.

$$ \frac{1}{2} m \overline{v^{2}}=\frac{3}{2} k_{B} T, \quad v_{r m s}=\left(\overline{v^{2}}\right)^{1 / 2}=\sqrt{\frac{3 k_{B} T}{m}} $$

This tells us that the temperature of a gas is a measure of the average kinetic energy of a molecule, independent of the nature of the gas or molecule. In a mixture of gases at a fixed temperature the heavier molecule has the lower average speed.

**3.** The translational kinetic energy

$$ E=\frac{3}{2} k_{B} N T $$

This leads to a relation

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

**4.** The law of equipartition of energy states that if a system is in equilibrium at absolute temperature $T$, the total energy is distributed equally in different energy modes of absorption, the energy in each mode being equal to $1 / 2 k_{B} T$. Each translational and rotational degree of freedom corresponds to one energy mode of absorption and has energy $1 / 2 k_{B} T$. Each vibrational frequency has two modes of energy (kinetic and potential) with corresponding energy equal to
$2 \times 1 / 2 k_{B} T=k_{B} T$.

**5.** Using the law of equipartition of energy, the molar specific heats of gases can be determined and the values are in agreement with the experimental values of specific heats of several gases. The agreement can be improved by including vibrational modes of motion.

**6.** The mean free path $l$ is the average distance covered by a molecule between two successive collisions :

$$ l=\frac{1}{\sqrt{2} n \pi d^{2}} $$

where $n$ is the number density and $d$ the diameter of the molecule.

### POINTS TO PONDER

**1.** Pressure of a fluid is not only exerted on the wall. Pressure exists everywhere in a fluid. Any layer of gas inside the volume of a container is in equilibrium because the pressure is the same on both sides of the layer.

**2.** We should not have an exaggerated idea of the intermolecular distance in a gas. At ordinary pressures and temperatures, this is only 10 times or so the interatomic distance in solids and liquids. What is different is the mean free path which in a gas is 100 times the interatomic distance and 1000 times the size of the molecule.

**3.** The law of equipartition of energy is stated thus: the energy for each degree of freedom in thermal equilibrium is $1 / 2 k_{B} T$. Each quadratic term in the total energy expression of a molecule is to be counted as a degree of freedom. Thus, each vibrational mode gives 2 (not 1) degrees of freedom (kinetic and potential energy modes), corresponding to the energy $2 \times 1 / 2 k_{B} T=k_{B} T$.

**4.** Molecules of air in a room do not all fall and settle on the ground (due to gravity) because of their high speeds and incessant collisions. In equilibrium, there is a very slight increase in density at lower heights (like in the atmosphere). The effect is small since the potential energy ( $m g h)$ for ordinary heights is much less than the average kinetic energy $1 / 2 m v^{2}$ of the molecules.

**5.** $< v^{2} >$ is not always equal to $(< v > )^{2}$. The average of a squared quantity is not necessarily the square of the average. Can you find examples for this statement.

### Exercises

**12.1** Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be $3 \mathring{A}$.

## Show Answer

**Answer**

Diameter of an oxygen molecule, $d=3 \mathring{A}$

Radius, $r=\frac{d}{2}=\frac{3}{2}=1.5 \mathring{A}=1.5 \times 10^{-8} cm$

Actual volume occupied by 1 mole of oxygen gas at $STP=22400 cm^{3}$

Molecular volume of oxygen gas,

$ V=\frac{4}{3} \pi r^{3} \cdot N $

Where, $N$ is Avogadro’s number $=6.023 \times 10^{23}$ molecules $/ mole$

$\therefore V=\frac{4}{3} \times 3.14 \times(1.5 \times 10^{-8})^{3} \times 6.023 \times 10^{23}=8.51 cm^{3}$

Ratio of the molecular volume to the actual volume of oxygen $=\frac{8.51}{22400}$

$=3.8 \times 10^{-4}$

**12.2** Molar volume is the volume occupied by $1 \mathrm{~mol}$ of any (ideal) gas at standard temperature and pressure (STP : 1 atmospheric pressure, $0^{\circ} \mathrm{C}$ ). Show that it is 22.4 litres.

## Show Answer

**Answer**

The ideal gas equation relating pressure $(P)$, volume $(V)$, and absolute temperature $(T)$ is given as: $P V=n R T$

Where,

$R$ is the universal gas constant $=8.314 J mol^{-1} K^{-1}$

$n=$ Number of moles $=1$

$T=$ Standard temperature $=273 K$

$P=$ Standard pressure $=1 atm=1.013 \times 10^{5} Nm^{-2}$

$\therefore V=\frac{n R T}{P}$

$=\frac{1 \times 8.314 \times 273}{1.013 \times 10^{5}}$

$=0.0224 m^{3}$

$=22.4$ litres

Hence, the molar volume of a gas at STP is 22.4 litres.

**12.3** Figure 12.8 shows plot of $P V / T$ versus $P$ for $1.00 \times 10^{-3} \mathrm{~kg}$ of oxygen gas at two different temperatures.

Fig 12.8

(a) What does the dotted plot signify?

(b) Which is true: $T_{1}>T_{2}$ or $T_{1}<T_{2}$ ?

(c) What is the value of $P V / T$ where the curves meet on the $y$-axis?

(d) If we obtained similar plots for $1.00 \times 10^{-3} \mathrm{~kg}$ of hydrogen, would we get the same value of $P V / T$ at the point where the curves meet on the $y$-axis? If not, what mass of hydrogen yields the same value of $P V / T$ (for low pressure high temperature region of the plot) ? (Molecular mass of $\mathrm{H}_2=2.02 \mathrm{u}$, of $\mathrm{O}_2=32.0 \mathrm{u}$, $R=8.31 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$.)

## Show Answer

**Answer**

**(a)** The dotted plot in the graph signifies the ideal behaviour of the gas, i.e., the ratio $\frac{P V}{T}$ is equal. $\mu R$ ( $\mu$ is the number of moles and $R$ is the universal gas constant) is a constant quality. It is not dependent on the pressure of the gas.

**(b)** The dotted plot in the given graph represents an ideal gas. The curve of the gas at temperature $T_1$ is closer to the dotted plot than the curve of the gas at temperature $T_2$. A real gas approaches the behaviour of an ideal gas when its temperature increases.

Therefore, $T_1>T_2$ is true for the given plot.

**(c)** The value of the ratio $P V / T$, where the two curves meet, is $\mu R$. This is because the ideal gas equation is given as:

$P V=\mu R T$

$\frac{P V}{T}=\mu R$

Where,

$P$ is the pressure

$T$ is the temperature

$V$ is the volume

$\mu$ is the number of moles

$R$ is the universal constant

Molecular mass of oxygen $=32.0 g$

Mass of oxygen $=1 \times 10^{-3} kg=1 g$

$R=8.314 J mole^{-1} K^{-1}$

$\therefore \frac{P V}{T}=\frac{1}{32} \times 8.314$ $=0.26 J K^{-1}$

Therefore, the value of the ratio $P V / T$, where the curves meet on the $y$-axis, is

$0.26 J K^{-1}$.

**(d)** If we obtain similar plots for $1.00 \times 10^{-3} kg$ of hydrogen, then we will not get the same value of $P V / T$ at the point where the curves meet the $y$-axis. This is because the molecular mass of hydrogen $(2.02 u)$ is different from that of oxygen $(32.0 u)$.

We have:

$\frac{P V}{T}=0.26 J K^{-1}$

$R=8.314 J mole^{-1} K^{-1}$

Molecular mass $(M)$ of $H_2=2.02 u$

$\frac{P V}{T}=\mu R$ at constant temperature

Where, $\mu=\frac{m}{M}$

$m=$ Mass of $H_2$

$\therefore \quad m=\frac{P V}{T} \times \frac{M}{R}$

$=\frac{0.26 \times 2.02}{8.31}$

$=6.3 \times 10^{-2} g=6.3 \times 10^{-5} kg$

Hence, $6.3 \times 10^{-5} kg$ of $H_2$ will yield the same value of $P V / T$.

**12.4** An oxygen cylinder of volume 30 litre has an initial gauge pressure of $15 \mathrm{~atm}$ and a temperature of $27^{\circ} \mathrm{C}$. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to $11 \mathrm{~atm}$ and its temperature drops to $17^{\circ} \mathrm{C}$. Estimate the mass of oxygen taken out of the cylinder $\left(R=8.31 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right.$, molecular mass of $\left.\mathrm{O}_{2}=32 \mathrm{u}\right)$.

## Show Answer

**Answer**

Volume of oxygen, $V_1=30$ litres $=30 \times 10^{-3} m^{3}$

Gauge pressure, $P_1=15 atm=15 \times 1.013 \times 10^{5} Pa$

Temperature, $T_1=27^{\circ} C=300 K$

Universal gas constant, $R=8.314 J mole^{-1} K^{-1}$

Let the initial number of moles of oxygen gas in the cylinder be $n_1$.

The gas equation is given as:

$P_1 V_1=n_1 R T_1$

$\therefore n_1=\frac{P_1 V_1}{R T_1}$

$=\frac{15.195 \times 10^{5} \times 30 \times 10^{-3}}{(8.314) \times 300}=18.276$

But, $n_1=\frac{m_1}{M}$

Where,

$m_1=$ Initial mass of oxygen

$M=$ Molecular mass of oxygen $=32 g$

$\therefore m_1=n_1 M=18.276 \times 32=584.84 g$

After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces.

Volume, $V_2=30$ litres $=30 \times 10^{-3} m^{3}$

Gauge pressure, $P_2=11 atm=11 \times 1.013 \times 10^{5} Pa$

Temperature, $T_2=17^{\circ} C=290 K$

Let $n_2$ be the number of moles of oxygen left in the cylinder.

The gas equation is given as:

$P_2 V_2=n_2 R T_2$

$\therefore n_2=\frac{P_2 V_2}{R T_2}$

$=\frac{11.143 \times 10^{5} \times 30 \times 10^{-3}}{8.314 \times 290}=13.86$

But, $n_2=\frac{m_2}{M}$

Where,

$m_2$ is the mass of oxygen remaining in the cylinder

$\therefore m_2=n_2 M=13.86 \times 32=453.1 g$

The mass of oxygen taken out of the cylinder is given by the relation:

Initial mass of oxygen in the cylinder - Final mass of oxygen in the cylinder

$=m_1-m_2$

$=584.84 g-453.1 g$

$=131.74 g$

$=0.131 kg$

Therefore, $0.131 kg$ of oxygen is taken out of the cylinder.

**12.5** An air bubble of volume $1.0 \mathrm{~cm}^{3}$ rises from the bottom of a lake $40 \mathrm{~m}$ deep at a temperature of $12{ }^{\circ} \mathrm{C}$. To what volume does it grow when it reaches the surface, which is at a temperature of $35^{\circ} \mathrm{C}$ ?

## Show Answer

**Answer**

Volume of the air bubble, $V_1=1.0 cm^{3}=1.0 \times 10^{-6} m^{3}$

Bubble rises to height, $d=40 m$

Temperature at a depth of $40 m, T_1=12^{\circ} C=285 K$

Temperature at the surface of the lake, $T_2=35^{\circ} C=308 K$

The pressure on the surface of the lake:

$P_2=1 atm=1 \times 1.013 \times 10^{5} Pa$

The pressure at the depth of $40 m$ :

$P_1=1 atm+d \rho g$

Where,

$\rho$ is the density of water $=10^{3} kg / m^{3}$

$g$ is the acceleration due to gravity $=9.8 m / s^{2}$

$\therefore P_1=1.013 \times 10^{5}+40 \times 10^{3} \times 9.8=493300 Pa$

We have: $\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}$

Where, $V_2$ is the volume of the air bubble when it reaches the surface

$V_2=\frac{P_1 V_1 T_2}{T_1 P_2}$

$=\frac{(493300)(1.0 \times 10^{-6}) 308}{285 \times 1.013 \times 10^{5}}$

$=5.263 \times 10^{-6} m^{3}$ or $5.263 cm^{3}$

Therefore, when the air bubble reaches the surface, its volume becomes $5.263 cm^{3}$.

**12.6** Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity $25.0 \mathrm{~m}^{3}$ at a temperature of $27^{\circ} \mathrm{C}$ and $1 \mathrm{~atm}$ pressure.

## Show Answer

**Answer**

Volume of the room, $V=25.0 m^{3}$

Temperature of the room, $T=27^{\circ} C=300 K$

Pressure in the room, $P=1 atm=1 \times 1.013 \times 10^{5} Pa$

The ideal gas equation relating pressure $(P)$, Volume $(V)$, and absolute temperature $(T)$ can be written as:

$P V=k_B N T$

Where,

$K_B$ is Boltzmann constant $=1.38 \times 10^{-23} m^{2} kg s^{-2} K^{-1}$

$N$ is the number of air molecules in the room

$ \begin{aligned} & \quad N=\frac{P V}{k_B T} \\ & =\frac{1.013 \times 10^{5} \times 25}{1.38 \times 10^{-23} \times 300}=6.11 \times 10^{26} \text{ molecules } \end{aligned} $

Therefore, the total number of air molecules in the given room is $6.11 \times 10^{26}$.

**12.7** Estimate the average thermal energy of a helium atom at (i) room temperature $\left(27^{\circ} \mathrm{C}\right)$, (ii) the temperature on the surface of the Sun ($6000 \mathrm{~K}$), (iii) the temperature of 10 million kelvin (the typical core temperature in the case of a star).

## Show Answer

**Answer**

At room temperature, $T=27^{\circ} C=300 K$

Average thermal energy $=\frac{3}{2} k T$

Where $k$ is Boltzmann constant $=1.38 \times 10^{-23} m^{2} kg s^{-2} K^{-1}$

$\therefore \frac{3}{2} k T=\frac{3}{2} \times 1.38 \times 10^{-38} \times 300$

$=6.21 \times 10^{-21} J$

Hence, the average thermal energy of a helium atom at room temperature $(27^{\circ} C)$ is $6.21 \times$ $10^{-21} J$.

On the surface of the sun, $T=6000 K$

Average thermal energy $=\frac{3}{2} k T$

$=\frac{3}{2} \times 1.38 \times 10^{-38} \times 6000$

$=1.241 \times 10^{-19} J$

Hence, the average thermal energy of a helium atom on the surface of the sun is $1.241 \times$ $10^{-19} J$.

At temperature, $T=10^{7} K$

Average thermal energy $=\frac{3}{2} k T$

$=\frac{3}{2} \times 1.38 \times 10^{-23} \times 10^{7}$

$=2.07 \times 10^{-16} J$

Hence, the average thermal energy of a helium atom at the core of a star is $2.07 \times 10^{-16} J$.

**12.8** Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules? Is the root mean square speed of molecules the same in the three cases? If not, in which case is $V_{\mathrm{rms}}$ the largest?

## Show Answer

**Answer**

Yes. All contain the same number of the respective molecules.

No. The root mean square speed of neon is the largest.

Since the three vessels have the same capacity, they have the same volume.

Hence, each gas has the same pressure, volume, and temperature.

According to Avogadro’s law, the three vessels will contain an equal number of the respective molecules. This number is equal to Avogadro’s number, $N=6.023 \times 10^{23}$.

The root mean square speed ( $v_{rms}$ ) of a gas of mass $m$, and temperature $T$, is given by the relation:

$ v_{rms}=\sqrt{\frac{3 k T}{m}} $

Where, $k$ is Boltzmann constant

For the given gases, $k$ and $T$ are constants.

Hence $v_{\text{rms }}$ depends only on the mass of the atoms, i.e.,

$ v_{rms} \propto \sqrt{\frac{1}{m}} $

Therefore, the root mean square speed of the molecules in the three cases is not the same. Among neon, chlorine, and uranium hexafluoride, the mass of neon is the smallest. Hence, neon has the largest root mean square speed among the given gases.

**12.9** At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at $-20^{\circ} \mathrm{C}$ ? (atomic mass of $\mathrm{Ar}$ $=39.9 \mathrm{u}$, of $\mathrm{He}=4.0 \mathrm{u}$ ).

## Show Answer

**Answer**

Temperature of the helium atom, $T_{He}=-20^{\circ} C=253 K$

Atomic mass of argon, $M_{Ar}=39.9 u$

Atomic mass of helium, $M_{He}=4.0 u$

Let, $(v_{rms})_{Ar}$ be the rms speed of argon.

Let $(v_{rms})_{He}$ be the rms speed of helium.

The rms speed of argon is given by:

$(v_{rms})_{Ar} $

$=\sqrt{\frac{3 R T_{Ar}}{M_{Ar}}}\ldots(i)$

Where,

$R$ is the universal gas constant

$T_{Ar}$ is temperature of argon gas

The rms speed of helium is given by:

$(v_{rms})_{He}$

$=\sqrt{\frac{3 R T_{He}}{M_{He}}} \ldots($ ii $)$

It is given that:

$(v_{\text{rms }})_{Ar}$

$=(v_{rms})_{He}$

$ \begin{aligned} & \sqrt{\frac{3 R T_{Ar}}{M_{Ar}}}=\sqrt{\frac{3 R T_{He}}{M_{He}}} \\ \\ & \frac{T_{Ar}}{M_{Ar}}=\frac{T_{He}}{M_{He}} \\ \\ & T_{Ar}=\frac{T_{He}}{M_{He}} \times M_{Ar} \\ \\ & =\frac{253}{4} \times 39.9 \\ \\ & =2523.675=2.52 \times 10^{3} K \end{aligned} $

Therefore, the temperature of the argon atom is $2.52 \times 10^{3} K$.

**12.10** Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at $2.0 \mathrm{~atm}$ and temperature $17^{\circ} \mathrm{C}$. Take the radius of a nitrogen molecule to be roughly $1.0 \mathring{A}$. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of $\mathrm{N}_{2}=28.0 \mathrm{u}$ ).

## Show Answer

**Answer**

Mean free path $=1.11 \times 10^{-7} m$

Collision frequency $=4.58 \times 10^{9} s^{-1}$

Successive collision time $\approx 500 \times($ Collision time $)$

Pressure inside the cylinder containing nitrogen, $P=2.0 atm=2.026 \times 10^{5} Pa$

Temperature inside the cylinder, $T=17^{\circ} C=290 K$

Radius of a nitrogen molecule, $r=1.0 \mathring{A}=1 \times 10^{10} m$

Diameter, $d=2 \times 1 \times 10^{10}=2 \times 10^{10} m$

Molecular mass of nitrogen, $M=28.0 g=28 \times 10^{-3} kg$

The root mean square speed of nitrogen is given by the relation: $v_{\text{rms }}=\sqrt{\frac{3 R T}{M}}$

Where,

$R$ is the universal gas constant $=8.314 J mole^{-1} K^{-1}$

$\therefore v_{\text{rms }}=\sqrt{\frac{3 \times 8.314 \times 290}{28 \times 10^{-3}}}=508.26 m / s$

The mean free path $(l)$ is given by the relation:

$l=\frac{k T}{\sqrt{2} \times d^{2} \times P}$

Where,

$k$ is the Boltzmann constant $=1.38 \times 10^{-23} kg m^{2} s^{-2} K^{-1}$

$\therefore l=\frac{1.38 \times 10^{-23} \times 290}{\sqrt{2} \times 3.14 \times(2 \times 10^{-10})^{2} \times 2.026 \times 10^{5}}$

$=1.11 \times 10^{-7} m$

Collision frequency $=\frac{v_{\text{rms }}}{l}$

$=\frac{508.26}{1.11 \times 10^{-7}}=4.58 \times 10^{9} s^{-1}$

Collision time is given as:

$T=\frac{d}{v_{\text{ms }}}$

$=\frac{2 \times 10^{-10}}{508.26}=3.93 \times 10^{-13} s$

Time taken between successive collisions:

$T^{\prime}=\frac{l}{v_{\text{ms }}}$

$ \begin{aligned} & =\frac{1.11 \times 10^{-7} m}{508.26 m / s}=2.18 \times 10^{-10} s \\ & \quad \frac{T^{\prime}}{T}=\frac{2.18 \times 10^{-10}}{3.93 \times 10^{-13}}=500 \end{aligned} $

Hence, the time taken between successive collisions is 500 times the time taken for a collision.