Kinetic Theory of Gases

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 compared to their own size, so the mutual interactions between molecules are negligible except during brief collisions. Due to this weak interaction, gases exhibit simple and predictable behaviour under ordinary conditions, especially at low pressure and high temperature.

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 (quantity) of gas changes, since more molecules will exert more pressure for the same \(T\) and \(V\).

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

\[ \begin{aligned} K &= N k \end{aligned} \]

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 \(k_B\). Its accepted SI value is

\[ k_B = 1.380649 \times 10^{-23}\,\text{J K}^{-1} \]

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

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

Equation (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 amount 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 (like \(P\), \(V\), \(T\)) to microscopic molecular motion, as seen in Eq. (2).

Using the concept of moles, the perfect (ideal) 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 and mass 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 of the gas. Substituting Eq. (4) into Eq. (3), the ideal gas equation can also be written in molecular form as

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

where \(n = \dfrac{N}{V}\) is the number density, defined as the number of molecules per unit volume. These relations clearly show that macroscopic properties such as pressure and temperature are directly linked to molecular quantities like number density and average energy.

Another useful form of Eq. (3) is \[ P=\dfrac{\rho RT}{M_0} \] where \(\rho\) is the mass density of the gas. This form is especially convenient when the gas is specified by its density instead of volume.

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 are approximately valid.

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Kinetic Interpretation of Pressure (Concept Booster)

According to kinetic theory, gas pressure arises from continuous collisions of molecules with the container walls. Consider a gas of number density \(n = N/V\), with molecules of mass \(m\) and average squared speed \(\langle v^2 \rangle\). One can show that

\[ P = \frac{1}{3} n m \langle v^2 \rangle \]

This equation states that pressure is proportional to the average kinetic energy of the molecules per unit volume. Using the ideal gas equation \(PV = N k_B T\), we can show that the average translational kinetic energy per molecule is

\[ \frac{1}{2} m \langle v^2 \rangle = \frac{3}{2} k_B T \]

Thus, temperature is a direct measure of the average kinetic energy of gas molecules, and is the same for all ideal gases at a given \(T\), irrespective of their chemical nature.

Simple model of gas molecules moving randomly and colliding with container walls, explaining the origin of pressure.

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Quick Concept Checks

1. At the same temperature and pressure, 2 litres of oxygen and 2 litres of hydrogen contain the same number of molecules (Avogadro’s hypothesis), even though their masses are different.

2. For an ideal gas at fixed temperature, if you double \(N\) (number of molecules) at constant volume, the pressure doubles, because \(P = k_B n T\) and \(n = N/V\).

3. At a given temperature, lighter molecules (smaller \(m\)) have higher root mean square speed \(v_\text{rms} = \sqrt{3 k_B T/m}\), but the average kinetic energy per molecule \(\frac{3}{2}k_B T\) is the same for all ideal gases.

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 used to explain and predict the behaviour of real gases under ordinary conditions. In practice, no real gas obeys this equation perfectly at all temperatures and pressures, but many real gases follow it closely when the pressure is low and the temperature is high.

Experimental observations show that real gases deviate from ideal behaviour, especially at high pressures and low temperatures. At high pressure, molecules are crowded and their finite size and intermolecular forces become important; at low temperature, attractive forces make molecules come closer and may lead to liquefaction. As the pressure is reduced or the temperature is increased, these deviations become smaller because gas molecules move farther apart and intermolecular forces become negligible. Under such conditions, a real gas behaves approximately like an ideal gas.

Figure 12.1 qualitatively shows that for a real gas, the product \(PV\) first decreases or increases with pressure, and then gradually approaches a constant value at low pressures. The horizontal dashed line represents the ideal gas behaviour \(PV = \text{constant}\).

If the number of moles \(\mu\) and the temperature \(T\) are kept constant in Eq. (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. (3) 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 (in kelvin). This relation is usually written as \(V/T = \text{constant}\) at constant \(P\), and is known as Charles’s law. It highlights that when the temperature of a gas increases, the average kinetic energy of its molecules increases, so the gas must expand (increase in volume) in order to keep the pressure constant.

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Dalton’s Law of Partial Pressures

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 total 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 \(\dfrac{\mu_1 RT}{V}\) is precisely 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 has its own partial pressure \(P_2, P_3,\ldots\).

Hence, the total pressure of the mixture is equal to the sum of the partial pressures of the individual gases:

\[ P = P_1 + P_2 + P_3 + \cdots \]

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, that is, at low pressures and high temperatures. Dalton’s law is a direct consequence of the fact that, in an ideal gas mixture, molecules of different gases move independently and do not affect each other’s contribution to pressure.

Dalton’s law provides a simple and powerful method to analyse gas mixtures. For example, the total pressure of air can be considered as the sum of the partial pressures of nitrogen, oxygen, water vapour, carbon dioxide, and other gases. This further reinforces the idea that, in an ideal gas, each molecule behaves independently and the total pressure is simply the sum of individual contributions.

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.

A given amount of gas is treated as a collection of a very large number of identical particles (molecules), each having a definite mass, moving in all directions with different speeds. The pressure of the gas arises from the force exerted by molecules as they collide with the walls; the temperature is related to the average kinetic energy of these molecules.

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. Under these conditions, the volume of molecules and intermolecular forces can be neglected, and the ideal gas equation \(PV = \mu RT\) provides an excellent description of the gas.

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, each having the same mass.
• The molecules are in continuous random motion in all directions, moving in straight lines between collisions.
• The size (diameter) of each molecule is extremely small compared to the average distance between neighbouring molecules, so the actual volume occupied by the molecules is negligible compared with the volume of the container.
• Intermolecular forces (attractive or repulsive) are negligible except during actual collisions; hence, molecules move freely between collisions.
• Collisions between molecules and with the container walls are perfectly elastic; kinetic energy and momentum are conserved in every collision.
• The duration of a collision is negligible compared to the time interval between two successive collisions, so most of the time molecules travel freely.
• Gas molecules obey Newton’s laws of motion, so their motion can be analysed using classical mechanics.
• At any instant, molecules are uniformly distributed in the container and all directions in space are equivalent (the gas is isotropic and homogeneous on average).

These assumptions allow a simple mechanical treatment of gas behaviour and lead to important results such as the ideal gas equation \(PV = \mu RT\) and the relation between pressure and average kinetic energy.

Consider an ideal gas enclosed in a cubical container of side \(l\), so that its volume is \(V = l^3\). Let each gas molecule have mass \(m\). The molecules move randomly in all directions with velocity components \((c_x, c_y, c_z)\), where \(c_x, c_y, c_z\) are the components of the molecular velocity \(\vec{c}\) along the \(x\), \(y\), and \(z\) axes respectively.

To find the pressure on one wall, it is enough to consider the motion of molecules along the direction perpendicular to that wall. Take a molecule moving along the \(x\)-direction towards the wall perpendicular to the \(x\)-axis. When it collides elastically with this wall, its momentum component along \(x\) changes from \(mc_x\) to \(-mc_x\).

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

The minus sign indicates that the direction of momentum is reversed. The magnitude of the momentum change for the molecule is therefore \(2mc_x\). After the collision, the molecule travels to the opposite wall and comes back to collide again with the original wall. The distance covered along \(x\) between two successive collisions with the same wall is \(2l\).

\[ \Delta t = \frac{2l}{c_x} \]

Hence, the average force exerted by this one molecule on the wall (rate of change of momentum) 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, each with possibly different values of \(c_x\), the total force exerted on the wall perpendicular to the \(x\)-axis is the sum of the forces due to all molecules:

\[ F = \frac{m}{l}\sum c_x^2 \]

Here \(\sum c_x^2\) denotes the sum of the squares of the \(x\)-components of the velocities of all molecules. Pressure is defined as force per unit area. Since the area of the wall is \(A = l^2\), we have

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

Introducing the average value of \(c_x^2\) over all molecules, \(\overline{c_x^2} = \dfrac{1}{N}\sum c_x^2\), the pressure can be written as

\[ P = \frac{m}{l^3} N \overline{c_x^2} = \frac{Nm}{V}\,\overline{c_x^2} \]

Because molecular motion is completely random and there is no preferred direction, the gas is isotropic. Therefore, the averages of the squares of the velocity components along the three mutually perpendicular directions are equal:

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

where \(\overline{c^2}\) is the average of the square of the speed \(c = \sqrt{c_x^2 + c_y^2 + c_z^2}\). Substituting \(\overline{c_x^2} = \dfrac{1}{3}\overline{c^2}\) in the expression for pressure, we obtain

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

This is the kinetic theory expression for the pressure of an ideal gas: \(P = \dfrac{1}{3}\rho\,\overline{c^2}\), where \(\rho = \dfrac{Nm}{V}\) is the mass density of the gas. It shows that the macroscopic pressure of a gas originates from the microscopic motion of its molecules and is directly proportional to the average kinetic energy per unit volume.

The ideal gas equation in molecular form relates pressure, volume and temperature for \(N\) molecules of an ideal gas as

\[ PV = N k_B T \]

Here \(k_B\) is the Boltzmann constant and \(T\) is the absolute temperature. This equation connects the macroscopic variables \(P, V, T\) with the microscopic quantity \(N\).

According to the kinetic theory of gases, the pressure exerted by an ideal gas can also be written in terms of the molecular mass and average squared speed as

\[ P = \frac{1}{3}\,\frac{N m}{V}\,\overline{c^2} \]

where \(m\) is the mass of a molecule and \(\overline{c^2}\) is the mean of the square of the molecular speeds. This expression was obtained earlier by considering the momentum transfer of molecules colliding elastically with the container walls.

We now have two expressions for the same pressure \(P\). Equating them gives

\[ \frac{1}{3}\,\frac{N m}{V}\,\overline{c^2} = \frac{N k_B T}{V} \]

Cancelling the common factors \(N\) and \(V\) from both sides, we obtain

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

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

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

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

\[ \overline{E_k} = \frac{3}{2}\,k_B T \]

This important result shows that the average translational kinetic energy of a gas molecule depends only on the absolute temperature and is independent of the nature (type) of the gas. Thus, temperature is a direct measure of the average kinetic energy of the molecules of an ideal gas.

In a gas, different molecules move with different speeds at any instant. To describe the overall motion of all molecules, it is convenient to use an average quantity called the root mean square (rms) speed. The rms speed is defined as the square root of the mean of the squares of molecular speeds.

\[ c_{\text{rms}} = \sqrt{\overline{c^2}} \]

From kinetic theory, the average translational kinetic energy of one molecule is \(\dfrac{1}{2} m \overline{c^2} = \dfrac{3}{2} k_B T\). Using this relation,

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

Taking the square root of both sides, the rms speed of gas molecules is

\[ c_{\text{rms}} = \sqrt{\overline{c^2}} = \sqrt{\frac{3 k_B T}{m}} \]

For one mole of gas, \(m\) can be replaced by the molar mass \(M\) and \(k_B\) by the gas constant \(R = N_A k_B\). In that case, the rms speed becomes

\[ c_{\text{rms}} = \sqrt{\frac{3 R T}{M}} \]

Here \(M\) is the molar mass of the gas (in kg mol\(^{-1}\)). Thus, at a given temperature, lighter gases (smaller \(M\)) have higher rms speeds, and rms speed increases with the square root of absolute temperature.

The internal energy of a system is the total energy associated with the microscopic motion and configuration of its molecules. For an ideal gas, intermolecular forces are absent, so there is no potential energy of interaction between molecules. Therefore, the internal energy of an ideal gas consists only of the kinetic energy of its molecules (mainly translational kinetic energy in the kinetic theory treatment).

From kinetic theory, the average translational kinetic energy of one molecule of an ideal gas is

\[ \overline{E_k} = \frac{3}{2} k_B T \]

For \(N\) molecules, the total translational kinetic energy (which is the internal energy \(U\) for an ideal gas) is

\[ U = N \overline{E_k} = N \left( \frac{3}{2} k_B T \right) = \frac{3}{2} N k_B T \]

Using \(N = N_A \mu\) and \(R = N_A k_B\), where \(\mu\) is the number of moles and \(N_A\) is Avogadro’s number, this can be written as

\[ U = \frac{3}{2} \mu R T \]

For one mole of an ideal gas \((\mu = 1)\), the internal energy is

\[ U = \frac{3}{2} R T \]

Thus, the internal energy of an ideal gas depends only on its absolute temperature and the number of moles, and not on its volume or pressure. Any change in internal energy of an ideal gas is therefore associated only with a change in temperature.

Monatomic Gases

A monatomic gas consists of single atoms, such as helium, neon or argon. In the kinetic theory model, such an atom has no internal rotational or vibrational structure, so it can store energy only through translational motion of its centre of mass.

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 \(\dfrac{1}{2}k_B T\).

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

\[ \overline{E} = \frac{3}{2} k_B T \]

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

\[ U = N_A \overline{E} = \frac{3}{2} N_A k_B T = \frac{3}{2} R T \]

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

\[ C_V = \frac{dU}{dT} = \frac{3}{2} R \]

For an ideal gas, the general relation

\[ C_P - C_V = R \]

holds. Hence, for a monatomic gas,

\[ C_P = C_V + R = \frac{3}{2}R + R = \frac{5}{2} R \]

The ratio of specific heats is

\[ \gamma = \frac{C_P}{C_V} = \frac{\frac{5}{2}R}{\frac{3}{2}R} = \frac{5}{3} \]

Diatomic Gases

A diatomic gas molecule (such as \(\mathrm{H_2,\ N_2,\ O_2,\ HCl}\)) consists of two atoms bonded together. At ordinary temperatures, such a molecule can be treated as a rigid rotator: its vibration is mostly “frozen out” and does not contribute significantly to the energy.

A rigid diatomic molecule has:

• 3 translational degrees of freedom,
• 2 rotational degrees of freedom (about two axes perpendicular to the bond axis).



Thus, the total number of active degrees of freedom is \(f = 5\).

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

\[ U = \frac{f}{2} R T = \frac{5}{2} R T \]

The molar specific heats are then

\[ \begin{aligned} C_V &= \frac{dU}{dT} = \frac{5}{2} R \\ C_P &= C_V + R = \frac{7}{2} R \end{aligned} \]

The ratio of specific heats is

\[ \gamma = \frac{C_P}{C_V} = \frac{\frac{7}{2}R}{\frac{5}{2}R} = \frac{7}{5} \]

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

The internal energy then becomes

\[ U = \frac{7}{2} R T \]

and the specific heats are

\[ C_V = \frac{7}{2} R,\quad C_P = \frac{9}{2} R,\quad \gamma = \frac{C_P}{C_V} = \frac{9}{7} \]

Polyatomic Gases

A polyatomic gas molecule (three or more atoms) generally has:

• 3 translational degrees of freedom,
• 3 rotational degrees of freedom (for non-linear molecules),
• \(f\) vibrational modes (each mode adding 2 degrees of freedom: one kinetic + one potential).

According to the law of equipartition of energy, each vibrational mode contributes energy equal to \(k_B T\) per molecule (i.e., \(\dfrac{1}{2}k_B T\) from kinetic and \(\dfrac{1}{2}k_B T\) from potential energy).

Thus, the internal energy of one mole of a polyatomic gas (with \(f\) vibrational modes) is

\[ U = \left(\frac{3}{2} + \frac{3}{2} + f\right) R T = (3 + f) \frac{R T}{1} \]

From this, the molar specific heats are obtained as

\[ C_V = (3 + f) R,\quad C_P = C_V + R = (4 + f) R \]

The ratio of specific heats is

\[ \gamma = \frac{C_P}{C_V} = \frac{4 + f}{3 + f} \]

Important Observations and Experimental Verification

The relation

\[ C_P - C_V = R \]

is valid for all ideal gases, irrespective of whether they are monatomic, diatomic, or polyatomic. Theoretical values of specific heats calculated by using only translational and rotational degrees of freedom (neglecting 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 of specific heats are higher than these simple theoretical predictions. This discrepancy indicates that vibrational modes become partially active and start contributing additional energy. When vibrational degrees of freedom are included in the model, the agreement between theory and experiment improves significantly.

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

Consider a solid consisting of \(N\) atoms. Each atom behaves approximately like a three-dimensional harmonic oscillator and can vibrate independently along the three mutually perpendicular directions in space.

For an oscillation in one dimension, the mechanical 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 in the energy contributes an average energy of \(\dfrac{1}{2}k_B T\). Hence, for one-dimensional oscillation (one kinetic + one potential term), the average energy per atom is

\[ \frac{1}{2}k_B T + \frac{1}{2}k_B T = k_B T \]

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

\[ E = 3 k_B T \]

For a solid containing \(N\) atoms, the total internal energy is

\[ U = 3 N k_B T \]

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

\[ U = 3 N_A k_B T = 3 R T \]

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

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

\[ \Delta Q = \Delta U + P \Delta V \approx \Delta U \]

As a result, the molar specific heat capacity \(C\) of a solid (at constant pressure or volume, practically the same) is given by

\[ C = \frac{\Delta Q}{\Delta T} \approx \frac{\Delta U}{\Delta T} \]

Differentiating the expression for internal energy \(U = 3 R T\) with respect to \(T\), we obtain

\[ C = \frac{dU}{dT} = 3 R \]

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. It agrees well with experimental observations for many simple solids at ordinary (not very low) temperatures, although deviations occur at low temperatures where quantum effects become important.

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 consisting of many short straight segments between collisions.

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 \(n\) molecules per unit volume (number density).

A collision occurs whenever the centre of another molecule comes within a distance \(d\) of 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

\[ \text{Volume swept} = \pi d^2 \langle v \rangle \Delta t \]

Any molecule whose centre lies within this cylindrical 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

\[ n \, \pi d^2 \langle v \rangle \Delta t \]

Hence, the rate of collisions (collisions per unit time) is

\[ \text{Collision rate} = n \, \pi d^2 \langle v \rangle \]

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

\[ \tau = \frac{1}{n \, \pi d^2 \langle v \rangle} \]

The mean free path \(l\) is the average distance travelled during this time:

\[ l = \langle v \rangle \tau = \langle v \rangle \cdot \frac{1}{n \, \pi d^2 \langle v \rangle} = \frac{1}{n \, \pi d^2} \]

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. On average, the relative speed between two molecules is greater than the speed of one molecule by a factor of \(\sqrt{2}\).

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

\[ l = \frac{1}{\sqrt{2} \, \pi d^2 n} \]

This expression shows that the mean free path increases when the number density \(n\) decreases (low pressure, high temperature) or when the molecular diameter \(d\) is smaller. In air at standard conditions, the mean free path is typically of the order of \(10^{-7}\,\text{m}\), much larger than molecular size but still very small compared to macroscopic distances.

Theory Background

In kinetic theory, an ideal gas is treated as a collection of point-like molecules whose own size (volume) is negligible compared to the volume of the container. One way to test this assumption is to compare:

• the molecular (actual) volume occupied by all molecules, with
• the total volume available to the gas in the container.

For a given mass of a substance (say 1 kg), the volume it occupies in the liquid state is a good approximation to the actual volume of all its molecules packed closely together. The same mass in the gaseous state occupies a much larger volume. Therefore:

• Molecular volume \(\approx\) volume occupied by the substance in liquid state.
• Total volume of gas \(\approx\) volume occupied by the same mass in vapour (gaseous) state.

The fraction \[ \text{fraction} = \frac{\text{molecular volume}}{\text{total gas volume}} \] directly tells us how small the actual molecular volume is compared to the volume of the container. If this fraction is very small, it justifies the ideal gas assumption that molecular size is negligible.

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

Let us consider a convenient mass of water, say \(1\,\text{kg}\). This mass will be the same whether the water is in liquid form or in vapour form; only its volume will change.

First, compute the volume of liquid water corresponding to \(1\,\text{kg}\). Using \(\text{density} = \dfrac{\text{mass}}{\text{volume}}\),

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

This volume \(V_w\) represents the molecular (actual) volume occupied by all the water molecules when they are closely packed in the liquid state.

Next, compute the total volume occupied by the same mass of water when it is in the form of water vapour at \(100^\circ\text{C}\) and \(1\,\text{atm}\). Using the given density of water vapour,

\[ \begin{aligned} V_v &= \frac{\text{mass}}{\text{density of water vapour}} \\ &= \frac{1\,\text{kg}}{0.6\,\text{kg m}^{-3}} \\ &\approx 1.67\,\text{m}^3 \end{aligned} \]

The required fraction (ratio) of molecular volume to total volume of water vapour is therefore

\[ \begin{aligned} \frac{V_w}{V_v} &= \frac{10^{-3}}{1.67} \\ &\approx 6.0 \times 10^{-4} \end{aligned} \]

Thus, the molecular volume is only about \(6 \times 10^{-4}\) (i.e., 0.06%) of the total volume occupied by the water vapour under the given conditions. 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 that molecular size can be ignored in the ideal gas model.

Theory Background

In Example-1, we compared the molecular volume of water (volume actually occupied by all molecules) with the total volume of water vapour and found that the molecular volume is extremely small. To go one step deeper, we can estimate the volume (and hence size) of a single water molecule.

The basic idea is:

• Use the molar mass and Avogadro’s number to find the mass of one molecule.
• Use the density of liquid water to convert mass of one molecule into its volume.
• Assume the molecule is roughly spherical to estimate its radius from the volume formula \(\dfrac{4}{3}\pi r^3\).

This gives an order-of-magnitude estimate of molecular size, which can be compared with typical atomic and molecular dimensions (of the order of Ångströms: \(1\,\text{Å} = 10^{-10}\,\text{m}\)).

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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 \(\mathrm{H_2O}\), which gives

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

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

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

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

\[ \rho = \frac{m}{V} \quad \Rightarrow \quad V = \frac{m}{\rho} \]

Substituting the mass of one molecule and the density of water, the volume of one water molecule is

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

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

\[ \frac{4}{3}\pi r^3 = 3 \times 10^{-29}\,\text{m}^3 \]

Solving for the radius \(r\),

\[ \begin{aligned} r^3 &= \frac{3 \times 3 \times 10^{-29}}{4\pi} \\ r &= \left( \frac{9 \times 10^{-29}}{4\pi} \right)^{1/3} \\ &\approx 1.93 \times 10^{-10}\,\text{m} \end{aligned} \]

Thus, the estimated radius of a water molecule is

\[ r \approx 1.93\,\text{Å} \]

This value is of the same order as typical molecular dimensions (about a few Ångströms), which confirms that our estimate based on density and molar mass is reasonable.

Theory Background

To estimate typical distances between molecules, it is useful to relate volume and separation. If a certain mass of a substance occupies volume \(V\), then, very roughly, each molecule can be thought of as having a “share” of space proportional to \(V/N\), where \(N\) is the number of molecules.

If the volume occupied by the same mass changes by a factor \(k\), then the characteristic linear dimension (average separation between molecules) changes by the cube root of this factor:

\[ \text{distance factor} \;\propto\; k^{1/3} \]

In Example-1, we compared the volume of liquid water with the volume of the same mass of water vapour and found a large volume ratio. In Example-2, we estimated the molecular size (diameter). Combining:

• volume ratio (vapour vs liquid) \(\Rightarrow\) factor for distance increase,
• molecular diameter in liquid \(\Rightarrow\) approximate intermolecular spacing in liquid,
• multiply both to get average spacing in vapour.

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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

\[ \frac{V_v}{V_w} \approx 1.67 \times 10^{3} \]

This means that when water changes from liquid to vapour (at 100 °C and 1 atm), the volume occupied by the same mass increases by a factor of approximately \(1.67 \times 10^{3}\). Since the average separation between molecules scales with the cube root of the 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 \(2\,\text{Å}\). Therefore, 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, molecules are closely packed, so the average intermolecular distance is of the order of the molecular diameter. Hence, the average distance between molecules in water vapour becomes (using the factor of \(\approx 10\) increase in separation):

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

Therefore, the average distance between water molecules (intermolecular distance) in water vapour under the given conditions is approximately \[ d_v \approx 40\,\text{Å} = 4.0 \times 10^{-9}\,\text{m}. \] This is much larger than the molecular diameter (about \(4\,\text{Å}\)), which is consistent with the fact that a gas is much more “dilute” than a liquid.

Theory Background

In a mixture of ideal gases, each gas behaves as if the others were absent. The pressure that each gas would exert if it alone occupied the container is called its partial pressure. Dalton’s law of partial pressures states that the total pressure is the sum of these partial pressures.

For each gas in the mixture, the ideal gas equation applies separately:

\[ P_i V = \mu_i R T = N_i k_B T \]

where \(P_i\) is the partial pressure, \(\mu_i\) the number of moles, and \(N_i\) the number of molecules of the \(i\)-th gas. At the same temperature \(T\) and volume \(V\):

• \(P_i \propto \mu_i \propto N_i\), so \[ \frac{P_1}{P_2} = \frac{\mu_1}{\mu_2} = \frac{N_1}{N_2}. \]
• The mass density of a gas is \(\rho_i = \dfrac{m_i}{V} = \dfrac{\mu_i M_i}{V}\), where \(M_i\) is its molar mass. Thus, at fixed \(V\), \[ \rho_i \propto \mu_i M_i. \]

We will use these proportionalities to get (i) the ratio of number of molecules, and (ii) the ratio of mass densities of neon and oxygen in the mixture from the given partial pressure ratio.

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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

Let neon be gas 1 and oxygen be gas 2. The ratio of their partial pressures is given as

\[ \frac{P_1}{P_2} = \frac{3}{2} \]

For each gas in the same vessel (same \(V\) and \(T\)), the ideal gas equation gives

\[ \begin{aligned} P_1 V &= \mu_1 R T, \\ P_2 V &= \mu_2 R T. \end{aligned} \]

Dividing the two equations,

\[ \frac{P_1}{P_2} = \frac{\mu_1}{\mu_2} \]

Substituting the given ratio of partial pressures,

\[ \frac{\mu_1}{\mu_2} = \frac{3}{2}. \]

The number of moles is related to the number of molecules by \(\mu = \dfrac{N}{N_A}\). Therefore,

\[ \frac{N_1}{N_2} = \frac{\mu_1}{\mu_2} = \frac{3}{2}. \]

Hence, the ratio of the number of molecules of neon to oxygen in the vessel is \[ N_1 : N_2 = 3 : 2. \]

Next, to find the ratio of mass densities, recall that for each gas \(\mu_i = \dfrac{m_i}{M_i}\) and \(\rho_i = \dfrac{m_i}{V}\), where \(M_i\) is the molar mass and \(m_i\) the total mass of that gas in the vessel.

\[ \rho_1 = \frac{m_1}{V}, \qquad \rho_2 = \frac{m_2}{V}. \]

Dividing the two expressions,

\[ \frac{\rho_1}{\rho_2} = \frac{m_1}{m_2} = \frac{\mu_1 M_1}{\mu_2 M_2}. \]

We already know \(\dfrac{\mu_1}{\mu_2} = \dfrac{3}{2}\). The given molar masses are: \[ M_1 = 20.2\,\text{u} \text{ (Ne)}, \quad M_2 = 32.0\,\text{u} \text{ (\(\mathrm{O_2}\))}. \] Substituting,

\[ \begin{aligned} \frac{\rho_1}{\rho_2} &= \frac{\mu_1}{\mu_2} \cdot \frac{M_1}{M_2} \\ &= \frac{3}{2} \cdot \frac{20.2}{32.0}. \end{aligned} \]

Evaluating the numerical factor,

\[ \begin{aligned} \frac{20.2}{32.0} &\approx 0.631, \\ \frac{\rho_1}{\rho_2} &\approx \frac{3}{2} \times 0.631 \\ &\approx 0.95. \end{aligned} \]

Therefore,

\[ N_{\text{Ne}} : N_{\text{O}_2} = 3 : 2, \qquad \rho_{\text{Ne}} : \rho_{\text{O}_2} \approx 0.95 : 1. \]

This means that, even though there are more neon molecules (in a 3:2 ratio), their total mass per unit volume is slightly less than that of oxygen because neon has a smaller molar mass than oxygen.

Theory Background

In kinetic theory of gases, two key results are:

• The average translational kinetic energy per molecule of an ideal gas is \(\overline{K} = \dfrac{3}{2} k_B T\), which depends only on absolute temperature \(T\), not on the mass or type of gas.
• The root mean square (rms) speed of molecules is \[ v_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}} = \sqrt{\frac{3 R T}{M}}, \] where \(m\) is mass of one molecule and \(M\) is molar mass. At the same \(T\), lighter molecules (smaller \(M\)) have higher rms speeds.

Therefore, in a mixture at common temperature:

• All gases have the same average kinetic energy per molecule.
• Their rms speeds are in the ratio \(\propto 1/\sqrt{M}\).

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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 (Ar) and chlorine (Cl\(_2\)) gases in the ratio \(2:1\) by mass, and the common temperature is \(27^\circ\text{C}\) (i.e. \(T = 300\,\text{K}\)). Since both gases are at the same temperature, their average molecular kinetic energies depend only on this temperature.

The average kinetic energy of a gas molecule is

\[ \overline{K} = \frac{1}{2} m v_{\text{rms}}^{2}. \]

For an ideal gas, this can also be written as

\[ \overline{K} = \frac{3}{2} k_B T. \]

Since \(T\) is the same for Ar and Cl\(_2\), we have

\[ \overline{K}_{\text{Ar}} = \overline{K}_{\text{Cl}}. \]

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

\[ \overline{K}_{\text{Ar}} : \overline{K}_{\text{Cl}} = 1 : 1. \]

Next, we find the ratio of rms speeds. Using

\[ \overline{K} = \frac{1}{2} m v_{\text{rms}}^{2} = \frac{3}{2} k_B T, \]

for each gas at the same temperature \(T\),

\[ \frac{1}{2} m_{\text{Ar}} v_{\text{rms,Ar}}^{2} = \frac{1}{2} m_{\text{Cl}} v_{\text{rms,Cl}}^{2}. \]

Cancelling the common factor \(\dfrac{1}{2}\),

\[ m_{\text{Ar}} v_{\text{rms,Ar}}^{2} = m_{\text{Cl}} v_{\text{rms,Cl}}^{2}. \]

Rearranging,

\[ \frac{v_{\text{rms,Ar}}^{2}}{v_{\text{rms,Cl}}^{2}} = \frac{m_{\text{Cl}}}{m_{\text{Ar}}} = \frac{M_{\text{Cl}}}{M_{\text{Ar}}}, \]

where \(M_{\text{Ar}}\) and \(M_{\text{Cl}}\) are the molar masses. Using the given values \(M_{\text{Ar}} = 39.9\,\text{u}\) and \(M_{\text{Cl}} = 70.9\,\text{u}\),

\[ \frac{v_{\text{rms,Ar}}^{2}}{v_{\text{rms,Cl}}^{2}} = \frac{70.9}{39.9}. \]

Taking the square root,

\[ \frac{v_{\text{rms,Ar}}}{v_{\text{rms,Cl}}} = \sqrt{\frac{70.9}{39.9}} \approx 1.33. \]

Thus,

\[ \overline{K}_{\text{Ar}} : \overline{K}_{\text{Cl}} = 1 : 1, \qquad v_{\text{rms,Ar}} : v_{\text{rms,Cl}} \approx 1.33 : 1. \]

This means that at the same temperature, argon and chlorine molecules have the same average kinetic energy, but the lighter argon molecules move faster on average than the heavier chlorine molecules.

Theory Background

In kinetic theory, the root mean square (rms) speed of gas molecules at temperature \(T\) is \[ v_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}} = \sqrt{\frac{3 R T}{M}}, \] where \(m\) is the mass of a molecule and \(M\) is the molar mass. At the same temperature, \(v_{\text{rms}} \propto \dfrac{1}{\sqrt{M}}\).

For two gases (or two isotopic forms of the same gas) at the same \(T\),

\[ \frac{v_1}{v_2} = \sqrt{\frac{M_2}{M_1}}. \]

A slightly smaller molar mass therefore gives a slightly higher average (or rms) speed. This small speed difference between isotopes is the basis of gaseous diffusion methods used to separate uranium isotopes.

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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 (UF\(_6\)) gas, both types of molecules are at the same temperature, so their average speeds depend only on their molecular masses.

First, find the molecular mass of uranium hexafluoride containing \(^ {235}\text{U}\):

\[ \begin{aligned} M_1 (\;^{235}\text{UF}_6) &= 235 + 6 \times 19 \\ &= 235 + 114 \\ &= 349\ \text{u} \end{aligned} \]

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

\[ \begin{aligned} M_2 (\;^{238}\text{UF}_6) &= 238 + 6 \times 19 \\ &= 238 + 114 \\ &= 352\ \text{u} \end{aligned} \]

For gases at the same temperature, rms speed is inversely proportional to the square root of molecular mass:

\[ \frac{v_1}{v_2} = \sqrt{\frac{M_2}{M_1}} = \sqrt{\frac{352}{349}}. \]

Numerically,

\[ \begin{aligned} \frac{v_1}{v_2} &= \sqrt{\frac{352}{349}} \\ &\approx 1.004. \end{aligned} \]

Thus, the uranium hexafluoride molecule containing the lighter isotope \(^ {235}\text{U}\) (mass 349 u) has a slightly larger average speed than that containing \(^ {238}\text{U}\) (mass 352 u).

To find the percentage difference in speeds, write

\[ \text{percentage difference} = \frac{v_1 - v_2}{v_2} \times 100\%. \]

Since \(v_1 / v_2 \approx 1.004\),

\[ \begin{aligned} \frac{v_1 - v_2}{v_2} &= \left(\frac{v_1}{v_2} - 1\right) \\ &\approx 1.004 - 1 \\ &= 0.004. \end{aligned} \]

\[ \text{percentage difference} \approx 0.004 \times 100\% = 0.4\%. \]

Therefore, UF\(_6\) containing \(^ {235}\text{U}\) moves faster, and the percentage difference in their average speeds is approximately \(0.4\%\).

Theory Background

For an ideal gas, the heat required to raise its temperature depends on the process: at constant volume, the appropriate molar heat capacity is \(C_V\); at constant pressure, it is \(C_P\). For a monatomic ideal gas (like helium), kinetic theory gives

\[ C_V = \frac{3}{2} R, \qquad C_P = C_V + R = \frac{5}{2} R. \]

When the gas is heated at constant volume (fixed container), it does no work on the surroundings \((W = P \Delta V = 0)\). The heat supplied goes entirely into increasing the internal energy:

\[ Q = n C_V \Delta T, \]

where \(n\) is the number of moles and \(\Delta T\) is the temperature rise. Also, at standard temperature and pressure (STP), one mole of an ideal gas occupies \(22.4\ \text{L}\).

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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 ? \((R = 8.31\ \text{J mol}^{–1} \text{K}^{–1})\).

Solution

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

\[ \begin{aligned} n &= \frac{44.8\ \text{L}}{22.4\ \text{L mol}^{-1}} \\ &= 2\ \text{mol}. \end{aligned} \]

The rise in temperature of the gas is \(\Delta T = 15^\circ\text{C}\). Since a change of 1 °C corresponds to a change of 1 K, we have \(\Delta T = 15\ \text{K}\).

Helium is a monatomic ideal gas, so its molar specific heat at constant volume is

\[ C_V = \frac{3}{2} R. \]

The cylinder is of fixed capacity, so heating takes place at (approximately) constant volume. Therefore, the heat required is

\[ \begin{aligned} Q &= n\, C_V\, \Delta T \\ &= n \left(\frac{3}{2} R\right) \Delta T \\ &= 2 \times \frac{3}{2} \times 8.31 \times 15. \end{aligned} \]

Evaluating this,

\[ \begin{aligned} Q &= 2 \times 1.5 \times 8.31 \times 15 \\ &\approx 373.9\ \text{J} \\ &\approx 3.7 \times 10^{2}\ \text{J}. \end{aligned} \]

Thus, the amount of heat required to raise the temperature of the helium gas by \(15^\circ\text{C}\) is approximately \(3.7 \times 10^{2}\ \text{J}\) (or about \(374\ \text{J}\)).

Theory Background

In kinetic theory, the mean free path \(l\) is the average distance a gas molecule travels between successive collisions. For a gas of hard spherical molecules of diameter \(d\) and number density \(n\) (molecules per unit volume), a more accurate expression for the mean free path is

\[ l = \frac{1}{\sqrt{2}\,\pi d^2 n}. \]

Here:

• \(d\) is the effective molecular diameter,
• \(n\) is the number of molecules per unit volume, and
• \(\sqrt{2}\) accounts for the fact that all molecules are moving (relative motion).

In earlier examples, we estimated:

• the effective diameter of a water molecule from its radius, and
• a typical mean free path being about \(1500\) times the molecular diameter under suitable conditions.

We will now use the relation \(l \approx 1500\,d\) together with the previously estimated value of \(d\) to obtain the numerical value of the mean free path for water vapour at 373 K.

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Estimate the mean free path for a water molecule in water vapour at 373 K. Use information from Exercises-1 and 2. (Given: \(l \approx 2.9 \times 10^{–7}\,\text{m} \approx 1500\,d\).)

Solution

The mean free path \(l\) of a gas molecule is related to its molecular diameter \(d\) and number density \(n\) by

\[ l = \frac{1}{\sqrt{2}\,\pi d^2 n}. \]

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

\[ d \approx 2 \times 2 \times 10^{-10}\,\text{m} \approx 4 \times 10^{-10}\,\text{m}. \]

However, for the purpose of this example, we can take a typical effective diameter (depending on model/rounding) as

\[ d \sim 2 \times 10^{-10}\,\text{m}. \]

It is given (or can be shown using the above formula and typical number density at 373 K and 1 atm) that, under these conditions, the mean free path is roughly \(1500\) times the molecular diameter:

\[ l \approx 1500\, d. \]

Substituting \(d \approx 2 \times 10^{-10}\,\text{m}\),

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

A more refined calculation (using more accurate values of \(d\) and \(n\)) gives

\[ l \approx 2.9 \times 10^{-7}\,\text{m}. \]

Thus, the mean free path of a water molecule in water vapour at \(373\,\text{K}\) is of the order of \(10^{-7}\,\text{m}\), which is about \(1500\) times larger than the molecular diameter. This shows that under these conditions water vapour is a very dilute gas: molecules are tiny compared to the distances they travel between collisions.