KINETIC THEORY-Notes

The properties of gases are comparatively easier to understand than those of solids and liquids. This simplicity arises mainly because, in a gas, the molecules are separated by large distances. As a result, the mutual interactions between molecules are negligible except during brief moments of collision. Due to this weak interaction, gases exhibit simple and predictable behaviour under ordinary conditions.

Experiments show that gases at low pressures and at temperatures much higher than those at which they liquefy or solidify obey a simple relation between pressure, volume, and temperature. For a given sample of gas, this relation is expressed as

\[ \begin{align} PV &= KT \tag{1} \end{align} \]

Here, \(T\) represents the absolute temperature measured on the kelvin scale, and \(K\) is a constant for the given sample of gas. However, the value of \(K\) changes when the amount of gas changes.

When the molecular nature of matter is taken into account, the constant \(K\) is found to be proportional to the number of molecules \(N\) present in the gas sample. Thus, we can write

\[ \begin{aligned} K &= Nk \end{aligned} \]

Experimental observations reveal that the proportionality constant \(k\) has the same value for all gases. This universal constant is known as the Boltzmann constant and is denoted by \(k_B\).

Substituting \(K = Nk_B\) into Eq. (1), we obtain

\[ \begin{align} PV &= Nk_B T \\ \frac{PV}{NT} &= k_B \tag{2} \end{align} \]

Equation (12.2) implies that if pressure, volume, and temperature are the same for different gases, then the number of molecules present must also be the same. This statement is known as Avogadro’s hypothesis, which asserts that equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules.

The number of molecules present in 22.4 litres of any gas at standard temperature and pressure (STP: 273 K and 1 atm) is \(6.02 \times 10^{23}\). This fixed number is called the Avogadro number, denoted by \(N_A\). The mass of 22.4 litres of a gas at STP, measured in grams, is numerically equal to its molecular mass. This quantity of substance is defined as one mole.

Avogadro originally proposed his hypothesis based on chemical reaction data. The kinetic theory of gases later provided a strong physical justification for this idea by relating macroscopic gas properties to molecular motion.

Using the concept of moles, the perfect gas equation can be written as

\[ \begin{align} PV &= \mu RT \tag{3} \end{align} \]

Here, \(\mu\) is the number of moles of the gas, and \(R = N_A k_B\) is the universal gas constant. On the kelvin scale of temperature, the value of \(R\) is

\[ \begin{aligned} R &= 8.314 \,\text{J mol}^{-1}\text{K}^{-1} \end{aligned} \]

The number of moles \(\mu\) can also be expressed in terms of molecular quantities as

\[ \begin{align} \mu &= \frac{N}{N_A} = \frac{M}{M_0} \tag{4} \end{align} \]

where \(M\) is the total mass of the gas containing \(N\) molecules and \(M_0\) is the molar mass. Substituting Eq. (12.4) into Eq. (12.3), the ideal gas equation can also be written in molecular form as

\[ \begin{aligned} PV &= Nk_B T \\ P &= k_B n T \end{aligned} \]

where \(n = \dfrac{N}{V}\) is the number density, defined as the number of molecules per unit volume. The value of the Boltzmann constant in SI units is

\[ \begin{aligned} k_B &= 1.38 \times 10^{-23} \,\text{J K}^{-1} \end{aligned} \]

These relations clearly show that the macroscopic properties of gases, such as pressure and temperature, are directly connected to the microscopic behaviour of molecules. Real gases approach ideal gas behaviour most closely at low pressures and high temperatures, where the assumptions of negligible molecular size and weak intermolecular forces remain valid.

Another useful form of Eq. (3) is \[P=\dfrac{\rho RT}{M_0}\] where \(\rho\) is the mass density of the gas.

A gas that satisfies the equation of state

\[ \begin{align} PV &= \mu RT \tag{3} \end{align} \]

exactly at all pressures and temperatures is defined as an ideal gas. An ideal gas is a simple theoretical model introduced to understand the behaviour of real gases under ordinary conditions. In practice, no real gas obeys this equation perfectly at all temperatures and pressures.

Experimental observations show that real gases deviate from ideal behaviour, especially at high pressures and low temperatures. However, as the pressure is reduced or the temperature is increased, these deviations become smaller. Under such conditions, gas molecules are far apart and the intermolecular forces become negligible. In the absence of significant molecular interactions, a real gas behaves approximately like an ideal gas.

Figure 12.1 (NCERT) illustrates the behaviour of a real gas at different temperatures. It is observed that all curves tend to approach the ideal gas behaviour at low pressures and high temperatures, confirming the validity of the ideal gas model under these conditions.

If the number of moles \(\mu\) and the temperature \(T\) are kept constant in Eq. (12.3), we obtain

\[ \begin{align} PV &= \text{constant} \tag{6} \end{align} \]

This result shows that, at constant temperature, the pressure of a fixed mass of gas varies inversely with its volume. This is known as Boyle’s law. Experimental pressure–volume curves confirm that Boyle’s law is obeyed closely at low pressures and high temperatures, where real gases behave nearly ideally.

Similarly, if pressure \(P\) is kept constant, Eq. (1) leads to

\[ \begin{aligned} V &\propto T \end{aligned} \]

Thus, for a fixed pressure, the volume of a gas is directly proportional to its absolute temperature. This relation highlights the role of temperature as a measure of the average kinetic energy of gas molecules.

Now consider a mixture of non-interacting ideal gases containing \(\mu_1, \mu_2, \ldots\) moles of different gases enclosed in a vessel of volume \(V\) at temperature \(T\) and pressure \(P\). For such a mixture, the equation of state becomes

\[ \begin{aligned} PV &= (\mu_1 + \mu_2 + \cdots)\,RT \end{aligned} \]

Dividing both sides by \(V\), we obtain

\[ \begin{align} P &= \frac{\mu_1 RT}{V} + \frac{\mu_2 RT}{V} + \cdots \end{align} \]

The quantity \(\dfrac{\mu_1 RT}{V}\) represents the pressure that gas 1 would exert if it alone occupied the same volume \(V\) at the same temperature \(T\). This pressure is called the partial pressure of gas 1 and is denoted by \(P_1\). Similarly, each gas in the mixture exerts its own partial pressure.

Hence, the total pressure of the mixture is equal to the sum of the partial pressures of the individual gases. This statement is known as Dalton’s law of partial pressures. It is valid for ideal gases and for real gases under conditions where intermolecular interactions are negligible.

Dalton’s law provides a simple and powerful method to analyse gas mixtures and further reinforces the idea that, in an ideal gas, each molecule behaves independently of others.

The kinetic theory of an ideal gas is a theoretical framework that explains the macroscopic properties of gases—such as pressure, temperature, and internal energy—on the basis of the microscopic motion of a large number of molecules. According to this theory, the behaviour of an ideal gas is the collective result of the continuous, random motion of its molecules and their elastic collisions with one another and with the walls of the container.

An ideal gas is a hypothetical gas whose molecules obey the assumptions of kinetic theory exactly. Though no real gas is perfectly ideal, many gases closely approximate ideal behaviour at low pressures and high temperatures.

Fundamental Assumptions of Kinetic Theory

The kinetic theory of an ideal gas is built upon the following assumptions:

• A gas consists of a very large number of identical molecules.
• The molecules are in continuous random motion in all directions.
• The size of each molecule is negligible compared to the average separation between molecules.
• Intermolecular forces are negligible except during collisions.
• Collisions between molecules and with the container walls are perfectly elastic.
• The duration of a collision is negligible compared to the time between successive collisions.
• Gas molecules obey Newton’s laws of motion.

These assumptions allow a simple mechanical treatment of gas behaviour.

Consider an ideal gas enclosed in a cubical container of side \(l\) and hence volume \(V = l^3\). Let each gas molecule have mass \(m\). The molecules move randomly in all directions with velocity components \((c_x, c_y, c_z)\).

Consider a molecule moving along the \(x\)-direction. When it collides elastically with a wall perpendicular to the \(x\)-axis, its momentum changes from \(mc_x\) to \(-mc_x\).

\[ \begin{aligned} \text{Change in momentum} &= (-mc_x) - (mc_x) \\ &= -2mc_x \end{aligned} \]

The time taken by the molecule to suffer successive collisions with the same wall is the time required to travel a distance \(2l\) along the \(x\)-direction.

\[ \begin{aligned} \Delta t &= \frac{2l}{c_x} \end{aligned} \]

Hence, the force exerted by one molecule on the wall is obtained by dividing the change in momentum by the time interval between successive collisions.

\[ \begin{aligned} F &= \frac{\text{change in momentum}}{\text{time}} \\ &= \frac{2mc_x}{2l/c_x} \\ &= \frac{mc_x^2}{l} \end{aligned} \]

If there are \(N\) molecules in the container, the total force exerted on the wall is

\[ \begin{aligned} F &= \frac{m}{l}\sum c_x^2 \end{aligned} \]

Pressure is defined as force per unit area. Since the area of the wall is \(A = l^2\),

\[ \begin{aligned} P &= \frac{F}{A} \\ &= \frac{m}{l^3}\sum c_x^2 \end{aligned} \]

Because molecular motion is completely random, the average of the squares of velocity components along the three mutually perpendicular directions are equal.

\[ \begin{aligned} \overline{c_x^2} = \overline{c_y^2} = \overline{c_z^2} = \frac{1}{3}\,\overline{c^2} \end{aligned} \]

Substituting this result in the expression for pressure, we obtain

\[ \begin{aligned} P &= \frac{m}{l^3}\,N\,\overline{c_x^2} \\ &= \frac{1}{3}\,\frac{Nm}{V}\,\overline{c^2} \end{aligned} \]

This is the kinetic theory expression for the pressure of an ideal gas. It shows that the macroscopic pressure of a gas originates from the microscopic motion of its molecules.

From the ideal gas equation, the relation between pressure, volume and temperature for \(N\) molecules of an ideal gas is

\[ \begin{aligned} PV &= Nk_B T \end{aligned} \]

According to the kinetic theory of gases, the pressure exerted by an ideal gas is given by

\[ \begin{aligned} P &= \frac{1}{3}\,\frac{Nm}{V}\,\overline{c^2} \end{aligned} \]

Equating the two expressions for pressure, we obtain

\[ \begin{aligned} \frac{1}{3}\,\frac{Nm}{V}\,\overline{c^2} &= \frac{Nk_B T}{V} \end{aligned} \]

Cancelling the common factors \(N\) and \(V\) from both sides, the above equation reduces to

\[ \begin{aligned} \frac{1}{3}\,m\,\overline{c^2} &= k_B T \end{aligned} \]

Multiplying both sides by \(\dfrac{3}{2}\), we obtain

\[ \begin{aligned} \frac{1}{2}\,m\,\overline{c^2} &= \frac{3}{2}\,k_B T \end{aligned} \]

The quantity \(\dfrac{1}{2}m\overline{c^2}\) represents the average translational kinetic energy of a gas molecule. Hence,

\[ \begin{aligned} \overline{E_k} &= \frac{3}{2}\,k_B T \end{aligned} \]

This result proves that the average translational kinetic energy of a gas molecule depends only on the absolute temperature and is independent of the nature of the gas. Therefore, temperature is a direct measure of the average kinetic energy of gas molecules.

The root mean square (rms) speed of gas molecules is defined as:

Using the kinetic energy relation:

where \(M\) is the molar mass of the gas.

Since intermolecular forces are absent in an ideal gas, its internal energy consists only of kinetic energy.

For one mole:

Thus, the internal energy of an ideal gas depends only on temperature and not on volume or pressure.

Monatomic Gases

A monatomic gas consists of single atoms, such as helium or neon. Since an atom has no internal structure, it can store energy only through translational motion.

A monatomic gas molecule has three translational degrees of freedom, corresponding to motion along the three mutually perpendicular directions in space. According to the law of equipartition of energy, each degree of freedom contributes an average energy of \(\frac{1}{2}k_BT\).

Hence, the average energy of one molecule at temperature \(T\) is

The total internal energy of one mole of a monatomic gas is therefore

The molar specific heat at constant volume, defined as the rate of change of internal energy with temperature, is

For an ideal gas, the general relation

holds. Hence, for a monatomic gas,

The ratio of specific heats is

Diatomic Gases

A diatomic gas molecule (such as \(\mathrm{H_2,\ N_2,\ O_2,\ \text{or }HCl}\)) consists of two atoms bonded together. At ordinary temperatures, such a molecule behaves like a rigid rotator.

A rigid diatomic molecule has:

• 3 translational degrees of freedom
• 2 rotational degrees of freedom



Thus, the total number of active degrees of freedom is 5.

Using the law of equipartition of energy, the internal energy of one mole of a rigid diatomic gas is

The molar specific heats are then

The ratio of specific heats is

If the diatomic molecule is not rigid and vibrational motion becomes active, one vibrational mode contributes two degrees of freedom (one kinetic and one potential). In this case, the total degrees of freedom become 7.

The internal energy then becomes

and the specific heats are

Polyatomic Gases

A polyatomic gas molecule generally has:

• 3 translational degrees of freedom
• 3 rotational degrees of freedom
• \(f\) vibrational modes

According to the law of equipartition of energy, each vibrational mode contributes energy equal to \(k_B\,T\) per molecule.

Thus, the internal energy of one mole of a polyatomic gas is

From this, the molar specific heats are obtained as

The ratio of specific heats is

Important Observations and Experimental Verification

The relation

is valid for all ideal gases, irrespective of whether they are monoatomic, diatomic, or polyatomic. Theoretical values of specific heats calculated by ignoring vibrational modes show good agreement with experimental values for many gases at ordinary temperatures. However, for some gases such as \(\mathrm{Cl_2,\ C_2H_6}\), and other complex polyatomic gases, experimental values are higher than theoretical predictions. This discrepancy indicates that vibrational modes become partially active, contributing additional energy. When these modes are included, the agreement between theory and experiment improves.

The law of equipartition of energy can be applied to understand the specific heat capacity of solids. In a solid, atoms are closely packed and are not free to move from one place to another. Instead, each atom vibrates about its mean equilibrium position. These vibrations are responsible for the internal energy of the solid.

Consider a solid consisting of Natoms. Each atom behaves like a harmonic oscillator and can vibrate independently along the three mutually perpendicular directions in space.

For an oscillation in one dimension, the energy consists of:

• one part due to kinetic energy, and
• one part due to potential energy.

According to the law of equipartition of energy, each quadratic term contributes an average energy of \(\frac{1}{2}k_BT\). Hence, for one-dimensional oscillation, the average energy per atom is

Since each atom vibrates in three dimensions, the average energy per atom becomes

For a solid containing Natoms, the total internal energy is

For one mole of a solid, the number of atoms \(N=N_A\), where \(N_A\) is Avogadro’s number. Thus,

where \(R=N_Ak_B\) is the universal gas constant.

In the case of solids, the change in volume with temperature is extremely small. Therefore, when heat is supplied at constant pressure, almost no work is done by the solid. Hence,

As a result, the molar specific heat capacity of a solid is given by

Differentiating the expression for internal energy,

This result implies that the molar specific heat capacity of a solid is approximately constant and equal to 3R, independent of temperature. This is known as the Dulong–Petit law, and it agrees well with experimental observations for many solids at ordinary temperatures.

Molecules in a gas move with very high speeds, comparable to the speed of sound. In spite of this, phenomena such as the slow spreading of a gas leaking from a cylinder, or the persistence of a smoke cloud for several hours, are commonly observed. These everyday observations indicate that gas molecules do not move freely in straight lines over long distances.

This behaviour arises because gas molecules, though extremely small, have a finite size. As a result, they frequently collide with one another. Each collision changes the direction of motion of a molecule, causing its path to be repeatedly deflected. Thus, instead of moving unhindered, a molecule follows a zig-zag path.

To describe this behaviour quantitatively, we introduce the concept of mean free path.

The mean free path is defined as the average distance travelled by a gas molecule between two successive collisions.

Derivation of Mean Free Path

Assume that gas molecules are spherical in shape, each having a diameter \(d\). Consider a single molecule moving with an average speed \(\langle v\rangle\) through a gas containing nmolecules per unit volume.

A collision occurs whenever the centre of another molecule comes within a distance dof the moving molecule. In a small time interval \(\Delta t\), the molecule travels a distance \(\langle v\rangle\Delta t\) and sweeps out a cylindrical volume given by

Any molecule whose centre lies within this volume will collide with the moving molecule. Since the number density of molecules is \(n\), the number of collisions suffered in time \(\Delta t\) is

Hence, the rate of collisions is

The average time between two successive collisions, called the mean free time \(\tau\), is therefore

The mean free path lis the average distance travelled during this time:

Correction Due to Relative Motion

In the above derivation, other molecules were assumed to be at rest. In reality, all molecules are in motion, and collisions depend on the relative velocity between molecules rather than the speed of a single molecule.

When this effect is taken into account, a more accurate treatment shows that the mean free path is given by

The density of water is 1000 kg m–3. The density of water vapour at 100 °C and 1 atm pressure is \(\mathrm{0.6\ kg\ m^{–3}}\). 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.

Solution

The density of water is given as \(1000\,\text{kg m}^{-3}\) and the density of water vapour at \(100^\circ\text{C}\) and \(1\,\text{atm}\) is \(0.6\,\text{kg m}^{-3}\). The molecular volume is taken as the actual volume occupied by the molecules, which can be approximated by the volume of water in the liquid state.

Let the mass of water vapour considered be \(1\,\text{kg}\).

\[ \begin{aligned} V_w &= \frac{\text{mass}}{\text{density of water}} \\ &= \frac{1}{1000} \\ &= 10^{-3}\,\text{m}^3 \end{aligned} \]

The total volume occupied by the same mass of water vapour is

\[ \begin{aligned} V_v &= \frac{\text{mass}}{\text{density of water vapour}} \\ &= \frac{1}{0.6} \\ &\approx 1.66\,\text{m}^3 \end{aligned} \]

The required ratio of molecular volume to the total volume of water vapour is therefore

\[ \begin{aligned} \frac{V_w}{V_v} &= \frac{10^{-3}}{1.66} \\ &\approx 6.0 \times 10^{-4} \end{aligned} \]

This very small value shows that the actual volume occupied by water molecules is negligible compared to the volume of water vapour, supporting a key assumption of the kinetic theory of gases.

Estimate the volume of a water molecule using the data in Example-1.

Solution

The molecular mass of water is obtained from its chemical formula \( \text{H}_2\text{O} \), which gives

\[ \begin{aligned} \text{Molecular mass of water} &= 2 + 16 \\ &= 18\,\text{g mol}^{-1} \end{aligned} \]

One mole of water contains \(6 \times 10^{23}\) molecules. Therefore, the mass of one water molecule is

\[ \begin{aligned} m &= \frac{18 \times 10^{-3}}{6 \times 10^{23}} \\ &= 3 \times 10^{-26}\,\text{kg} \end{aligned} \]

The density of water is given as \( \rho = 1000\,\text{kg m}^{-3} \). Using the relation between density, mass and volume,

\[ \begin{aligned} \rho &= \frac{m}{V} \\ V &= \frac{m}{\rho} \end{aligned} \]

Substituting the values,

\[ \begin{aligned} V &= \frac{3 \times 10^{-26}}{1000} \\ &= 3 \times 10^{-29}\,\text{m}^3 \end{aligned} \]

Assuming a water molecule to be spherical in shape, its volume can be written as

\[ \begin{aligned} \frac{4}{3}\pi r^3 &= 3 \times 10^{-29} \end{aligned} \]

Solving for the radius \(r\),

\[ \begin{aligned} r &= \left( \frac{3 \times 3 \times 10^{-29}}{4\pi} \right)^{1/3} \\ &\approx 1.93 \times 10^{-10}\,\text{m} \end{aligned} \]

Thus, the estimated radius of a water molecule is

\[ \begin{aligned} r &\approx 1.93\,\text{Å} \end{aligned} \]

This order of magnitude is consistent with typical molecular dimensions, validating the estimate.

What is the average distance between atoms (interatomic distance) in water? Use the data given in Examples-1 and 2.

Solution

From Example-1, the ratio of the volume of water vapour to the volume of liquid water for the same mass was found to be

\[ \begin{aligned} \frac{V_v}{V_w} &\approx 1.67 \times 10^{3} \end{aligned} \]

This means that when water changes from liquid to vapour, the volume increases by a factor of approximately \(1.67 \times 10^{3}\). Since the average separation between molecules scales with the cube root of volume, the ratio of intermolecular distances in vapour and liquid states is

\[ \begin{aligned} \frac{r_v}{r_w} &= \left(1.67 \times 10^{3}\right)^{1/3} \\ &\approx 10 \end{aligned} \]

From Example-2, the radius of a water molecule was estimated to be about \(2\,\text{Å}\), so the diameter of a water molecule is approximately

\[ \begin{aligned} d_w &= 2 \times 2\,\text{Å} \\ &= 4\,\text{Å} \end{aligned} \]

In liquid water, the average intermolecular distance is of the order of the molecular diameter. Hence, the average distance between molecules in water vapour becomes

\[ \begin{aligned} d_v &= 10 \times 4\,\text{Å} \\ &= 40\,\text{Å} \end{aligned} \]

Therefore, the average distance between water molecules (interatomic or intermolecular distance) under the given conditions is approximately \(40\,\text{Å}\).

A vessel contains two non reactive 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 \(\mathrm{O_2}\) = 32.0 u.

Solution

The vessel contains two non-reactive gases, neon and oxygen, at the same temperature and volume. The ratio of their partial pressures is given as

\[ \begin{aligned} \frac{P_1}{P_2} = \frac{3}{2} \end{aligned} \]

Using the ideal gas equation for each gas,

\[ \begin{aligned} P_1 V &= \mu_1 RT \\ P_2 V &= \mu_2 RT \end{aligned} \]

Dividing the two equations,

\[ \begin{aligned} \frac{P_1}{P_2} &= \frac{\mu_1}{\mu_2} \end{aligned} \]

Substituting the given ratio of partial pressures,

\[ \begin{aligned} \frac{\mu_1}{\mu_2} &= \frac{3}{2} \end{aligned} \]

Since the number of moles is related to the number of molecules by \(\mu = \frac{N}{N_A}\),

\[ \begin{aligned} \frac{N_1}{N_2} &= \frac{\mu_1}{\mu_2} \\ &= \frac{3}{2} \end{aligned} \]

Hence, the ratio of the number of molecules of neon to oxygen in the vessel is \(3:2\).

To find the ratio of mass densities, we use the relations \(\mu = \frac{m}{M}\) and \(\rho = \frac{m}{V}\), where \(M\) is the molar mass.

\[ \begin{aligned} \rho_1 &= \frac{m_1}{V}, \qquad \rho_2 = \frac{m_2}{V} \end{aligned} \]

Dividing the two expressions,

\[ \begin{aligned} \frac{\rho_1}{\rho_2} &= \frac{m_1}{m_2} \\ &= \frac{\mu_1 M_1}{\mu_2 M_2} \end{aligned} \]

Substituting the given values \(M_1 = 20.2\,\text{u}\) for neon and \(M_2 = 32.0\,\text{u}\) for oxygen,

\[ \begin{aligned} \frac{\rho_1}{\rho_2} &= \frac{3}{2} \times \frac{20.2}{32.0} \\ &\approx 0.95 \end{aligned} \]

Therefore, the ratio of the number of molecules of neon to oxygen is \(3:2\), and the ratio of their mass densities is approximately \(0.95:1\).

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

Solution

The flask contains argon and chlorine gases in the ratio \(2:1\) by mass and the temperature of the mixture is \(27^\circ\text{C}\). Since both gases are at the same temperature, their molecular motions are governed by the same thermal conditions.

The average kinetic energy of a gas molecule is given by

\[ \begin{aligned} \overline{K} &= \frac{1}{2} m v_{\text{rms}}^{2} \end{aligned} \]

For an ideal gas, the average kinetic energy per molecule depends only on temperature and is given by \(\frac{3}{2}k_B T\). Therefore, at the same temperature,

\[ \begin{aligned} \overline{K}_{\text{Ar}} &= \overline{K}_{\text{Cl}} \end{aligned} \]

Hence, the ratio of the average kinetic energy per molecule of argon to chlorine is

\[ \begin{aligned} \overline{K}_{\text{Ar}} : \overline{K}_{\text{Cl}} &= 1 : 1 \end{aligned} \]

Now, using the relation between rms speed and molecular mass,

\[ \begin{aligned} \frac{1}{2} m_{\text{Ar}} v_{\text{rms,Ar}}^{2} &= \frac{1}{2} m_{\text{Cl}} v_{\text{rms,Cl}}^{2} \end{aligned} \]

Rearranging the above expression,

\[ \begin{aligned} \frac{v_{\text{rms,Ar}}^{2}}{v_{\text{rms,Cl}}^{2}} &= \frac{m_{\text{Cl}}}{m_{\text{Ar}}} = \frac{M_{\text{Cl}}}{M_{\text{Ar}}} = \frac{70.9}{39.9} \end{aligned} \]

Taking the square root,

\[ \begin{aligned} \frac{v_{\text{rms,Ar}}}{v_{\text{rms,Cl}}} &= \sqrt{\frac{70.9}{39.9}} \approx 1.33 \end{aligned} \]

Thus, the ratio of average kinetic energy per molecule of argon to chlorine is \(1:1\), and the ratio of their root mean square speeds is approximately \(1.33:1\).

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.

Solution

Uranium has two isotopes of mass numbers 235 and 238. When present in uranium hexafluoride gas, both isotopic molecules are at the same temperature, so their average speeds depend only on their molecular masses.

The molecular mass of uranium hexafluoride containing \(^{235}\text{U}\) is

\[ \begin{aligned} \text{UF}_6 &= 235 + 6 \times 19 \\ &= 349 \end{aligned} \]

Similarly, the molecular mass of uranium hexafluoride containing \(^{238}\text{U}\) is

\[ \begin{aligned} \text{UF}_6 &= 238 + 6 \times 19 \\ &= 352 \end{aligned} \]

For gases at the same temperature, the average (or rms) speed is inversely proportional to the square root of molecular mass. Hence,

\[ \begin{aligned} \frac{v_{349}}{v_{352}} &= \sqrt{\frac{352}{349}} \\ &\approx 1.004 \end{aligned} \]

This shows that the uranium hexafluoride molecule containing the lighter isotope \(^{235}\text{U}\) has a slightly larger average speed.

The fractional difference in speeds is

\[ \begin{aligned} \frac{\Delta v}{v} &= \frac{1.004 - 1}{1.004} \times 100 \\ &\approx 0.4\% \end{aligned} \]

Thus, uranium hexafluoride containing \(^{235}\text{U}\) moves faster, and the percentage difference in their average speeds is approximately \(0.4\%\).

(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 ?

Solution

(a) When a molecule or an elastic ball strikes a massive wall or a bat held firmly at rest, it rebounds with the same speed because the wall or bat does not move and hence does not exchange kinetic energy with the ball. However, when the bat is moving towards the incoming ball, the situation changes. In an elastic collision, the relative speed of separation equals the relative speed of approach. Since the bat is moving towards the ball, the ball rebounds with a greater speed than before. Thus, the ball moves faster after the collision.

(b) When a gas in a cylinder is compressed by pushing the piston inward, the piston moves towards the gas molecules. Gas molecules colliding with the inward-moving piston rebound with higher speeds, similar to the ball rebounding faster from a moving bat as discussed in part (a). As a result, the average kinetic energy of the gas molecules increases. Since temperature is a measure of the average kinetic energy of molecules, the temperature of the gas rises during compression.

(c) When a compressed gas pushes the piston outward and expands, the piston moves away from the gas molecules. In this case, molecules colliding with the outward-moving piston rebound with reduced speeds. Consequently, the average kinetic energy of the gas molecules decreases. This leads to a fall in temperature, and the gas cools during expansion.

(d) Sachin Tendulkar’s use of a heavy cricket bat was indeed helpful. A heavier bat has greater momentum for the same swing speed. When the moving bat strikes the ball, the ball rebounds with a much higher speed due to the larger momentum transfer. This allows the ball to travel farther, giving the batsman an advantage while playing powerful shots.

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 °C ? \(\mathrm{(R = 8.31\ J\ mo1^{–1}\ K^{–1})}\).

Solution

The capacity of the cylinder is \(44.8\,\text{litres}\) and it contains helium gas at standard temperature and pressure. At STP, \(22.4\,\text{litres}\) of an ideal gas corresponds to one mole. Hence, the number of moles of helium present in the cylinder is

\[ \begin{aligned} n &= \frac{44.8}{22.4} \\ &= 2 \end{aligned} \]

The rise in temperature of the gas is given as \(\Delta T = 15^\circ\text{C} = 15\,\text{K}\). Helium is a monoatomic gas, so its molar specific heat at constant volume is

\[ \begin{aligned} C_V &= \frac{3}{2}R \end{aligned} \]

Since the cylinder is of fixed capacity, heating takes place at constant volume. Therefore, the heat required is

\[ \begin{aligned} Q &= n C_V \Delta T \\ &= 2 \times \frac{3}{2} R \times 15 \\ &= 2 \times 1.5 \times 8.31 \times 15 \\ &\approx 374\,\text{J} \end{aligned} \]

Thus, the amount of heat required to raise the temperature of the helium gas by \(15^\circ\text{C}\) is approximately \(374\,\text{J}\).

Estimate the mean free path for a water molecule in water vapour at 373 K. Use information from Exercises-1 and \(\mathrm{l = 2.9\ ×\ 10^{–7}\ m\ ≈\ 1500\ d}\)

Solution

The mean free path \(l\) of a gas molecule is related to its molecular diameter \(d\) by the expression

\[ \begin{aligned} l &\approx \frac{1}{\sqrt{2}\,\pi d^2 n} \end{aligned} \]

From the earlier exercise, the effective diameter of a water molecule is of the order of

\[ \begin{aligned} d &\approx 2 \times 10^{-10}\,\text{m} \end{aligned} \]

It is also given that, under the stated conditions of temperature \(T = 373\,\text{K}\), the mean free path of a water molecule in water vapour is approximately

\[ \begin{aligned} l &\approx 1500\,d \end{aligned} \]

Substituting the value of \(d\),

\[ \begin{aligned} l &= 1500 \times 2 \times 10^{-10} \\ &= 3.0 \times 10^{-7}\,\text{m} \end{aligned} \]

Thus, the mean free path of a water molecule in water vapour at \(373\,\text{K}\) is approximately

\[ \begin{aligned} l &\approx 2.9 \times 10^{-7}\,\text{m} \end{aligned} \]

This large value compared to the molecular size shows that water vapour is a highly dilute gas under these conditions.