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`color{blue}{star}` INTRODUCTION
`color{blue}{star}` MOLECULAR NATURE OF MATTER
`color{blue}{star}` BEHAVIOUR OF GASES


`color{blue} ✍️`Kinetic theory explains the behaviour of gases based on the idea that the gas consists of rapidly moving atoms or molecules.

`color{blue} ✍️`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.

`color{blue} ✍️`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.

`color{blue} ✍️`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.


`color{blue} ✍️`Richard Feynman, 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.

`color{blue} ✍️`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.

`color{blue} ✍️`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.

`color{blue} ✍️`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.

`color{blue} ✍️`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.

`color{blue} ✍️`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.

`color{blue} ✍️`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.

`color{blue} ✍️`A small number of atoms of each element combine to form a molecule of the compound. Gay Lussac’s law, also given in early 19th century, states: When gases combine chemically to yield another gas, their volumes are in the ratios of small integers.

`color{blue} ✍️`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.

`color{blue} ✍️`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 !

`color{blue} ✍️`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 `(10^-10 m)`. In solids, which are tightly packed, atoms are spaced about a few angstroms `(2 Å)` apart.

`color{blue} ✍️`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.

`color{blue} ✍️`The average distance a molecule can travel without colliding is called `color{lime} "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.

`color{blue} ✍️`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.

`color{blue} ✍️`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.

`color{blue} ✍️`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.

`color{blue} ✍️`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.

`color{blue} ✍️`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.


`color{blue} ✍️`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.

`color{blue} ✍️`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 Ch. 11) for a given sample of the gas.

`color{blue} {PV = KT}`

........... (13.1)

`color{blue} ✍️`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.

`color{blue} ✍️`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_B`.


`color {blue} {(P_1V_1)/(N_1T_1) = (P_2V_2)/(N_2T_2) =} " constant "color{blue} { = k_B}`


`color{blue} ✍️`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 same for all gases at a fixed temperature and pressure.

`color{blue} ✍️`The number in 22.4 litres of any gas is `6.02 × 10^23`. This is known as Avogadro number and is denoted by NA. The mass of 22.4 litres of any gas is equal to its molecular weight in grams at S.T.P (standard temperature 273 K and pressure 1 atm).

`color{blue} ✍️`This amount of substance is called a mole (see Chapter 2 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.

`color{blue} ✍️`The perfect gas equation can be written as

`color{blue} {PV = μ RT}`

........................ (13.3)

`color{blue} ✍️`where `μ` is the number of moles and `R = N_A k_B` is a universal constant. The temperature `T` is absolute temperature. Choosing kelvin scale for absolute temperature, `R = 8.314 J mol^-1K^-1`.


`color{blue} {mu = M/M_o = N/N_A}`


`color{blue} ✍️`where `M` is the mass of the gas containing N molecules, `M_0` is the molar mass and `N_A` the Avogadro’s number. Using Eqs. (13.4) and (13.3) can also be written as

`color{purple} { PV = k_BNT} ` or `color{purple} {P = k_B nT}`

`color{blue} ✍️`where n is the number density, i.e. number of molecules per unit volume. `k_B` is the Boltzmann constant introduced above. Its value in SI units is `1.38 × 10^-23 J K^-1`.

Another useful form of Eq. (13.3) is

`color{blue} {P =(rhoRT)/M_o}`


`color{blue} ✍️`where `ρ` is the mass density of the gas.

`color{blue} ✍️`A gas that satisfies Eq. (13.3) exactly at all pressures and temperatures is defined to be an `color{green} "ideal gas"`. An ideal gas is a simple theoretical model of a gas. No real gas is truly ideal. Fig. 13.1 shows departures from ideal gas behaviour for a real gas at three different temperatures.

`color{blue} ✍️`Notice that all curves approach the ideal gas behaviour for low pressures and high temperatures.

`color{blue} ✍️`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.

`color{blue} ✍️`If we fix `μ` and `T` in Eq. (13.3), we get

`color{blue} {PV = "constant"`


`color{blue} ✍️`i.e., keeping temperature constant, pressure of a given mass of gas varies inversely with volume.

`color{brown} bbul"Boyle’s law"`
`color{blue} ✍️`This is the famous `color{green} "Boyle’s law"`. Fig. 13.2 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.

`color{brown} bbul"Charles’ law"`
`color{blue} ✍️`Next, if you fix P, Eq. (13.1) shows that `V ∝ T` i.e., for a fixed pressure, the volume of a gas is proportional to its absolute temperature T (`color{green} "Charles’ law"`). See Fig. 13.3.

`color{blue} ✍️`Finally, consider a mixture of non-interacting ideal gases: `μ_1` moles of gas 1, `μ_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 :

`color{blue} {PV = ( μ_1 + μ_2 +… ) RT}`

.................... (13.7)


`color{blue} {P = mu_1 (RT)/V + mu_2 (RT)/V + ....}`


`color{blue} { = P_1 + P_2 + .....}`


`color{blue} ✍️`Clearly `color{purple} {P_1 = μ_1 R T//V}` is the pressure 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.

`color{blue} ✍️`Thus, the total pressure of a mixture of ideal gases is the sum of partial pressures. This is Dalton’s law of partial pressures.

`color{blue} ✍️`We next consider some examples which give us information about the volume occupied by the molecules and the volume of a single molecule.
Q 3230401312

The density of water is `1000 kg m^-3`. The density of water vapour at `100 °C` and 1 atm pressure is `0.6 kg 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.
Class 11 Chapter 13 Example 1

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 = (6 ×10^-4 )` 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. `6×10^-4`.
Q 3260401315

Estimate the volume of a water molecule using the data as The density of water is `1000 kg m^-3`. The density of water vapour at `100 °C` and 1 atm pressure is `0.6 kg m^-3`. The volume of a molecule multiplied by the total number gives ,what is called, molecular volume.
Class 11 Chapter 13 Example 2

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 kg 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)g = 18 g = 0.018` kg.

Since 1 mole contains about `6 × 10^23` molecules (Avogadro’s number), the mass of
a molecule of water is `(0.018)/(6 × 10^23) kg = 3 × 10^-26` kg. Therefore, a rough estimate of the volume of a water molecule is as follows :

Volume of a water molecule

`= (3 × 10^-26 kg)// (1000 kg m^-3)`

`= 3 × 10^-29 m^3`

`= (4/3) π ("Radius")^3`

Hence, Radius `≈ 2 ×10^-10 m = 2 Å`
Q 3270401316

What is the average distance between atoms (interatomic distance) in water? Use the data The density of water is `1000
kg m^-3`. The density of water vapour at `100 °C` and 1 atm pressure is `0.6 kg m^-3`. The volume of a molecule multiplied by the total number gives ,what is called, molecular volume and Volume of a water molecule is `(4//3)pi ("Radius")^3` , Radius is `2 Å`
Class 11 Chapter 13 Example 3

A given mass of water in vapour state has `1.67×10^3` times the volume of the same mass of water in liquid state (Ex. 13.1). This is also the increase in the amount of volume available for each molecule of water. When volume increases by 103 times the radius increases by `V^(1/3)` or 10 times, i.e., `10 × 2 Å = 20 Å`. So the average distance is `2 × 20 = 40 Å`.
Q 3280401317

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 `Ne = 20.2 u`, molecular mass of `O_2 = 32.0 u`.
Class 11 Chapter 13 Example 4

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_1V = μ_1 RT` and `P_2V = μ_2 RT`, i.e. `(P_1/P_2) = (μ_1 / μ_2)`.

Here 1 and 2 refer to neon and oxygen respectively.

Since `(P_1/P_2) = (3/2)` (given), `(μ_1/ μ_2) = 3/2`.

(i) By definition `μ_1 = (N_1/N_A )` and `μ_2 = (N_2/N_A)` where `N_1` and `N_2` are the number of molecules of 1 and 2, and `N_A` is the Avogadro’s number. Therefore, `(N_1/N_2) = (μ_1 / μ_2) = 3/2`.

(ii) We can also write `μ_1 = (m_1/M_1)` and `μ_2 = (m_2/M_2)` 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 `ρ_1` and `ρ_2` are the mass densities of 1 and 2 respectively, we have

`rho_1/rho_2 = (m_1//V)/(m_2//V) = m_1/m_2 = mu_1/mu_2 xx (M_1/M_2)`

`= 3/2 xx 20.2/32.0 = 0.947`