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Chapter 13 Kinetic Theory
Introduction
The Kinetic Theory of Gases explains the macroscopic properties of gases, such as pressure, temperature, and volume, by considering their molecular composition and motion. It builds upon the idea that a gas consists of a vast number of rapidly and randomly moving atoms or molecules.
This theory is particularly successful for gases because the distances between molecules are large, and the short-range inter-atomic forces, which are dominant in solids and liquids, can be largely neglected. Developed in the 19th century by physicists like Maxwell and Boltzmann, the kinetic theory provides a molecular interpretation of pressure and temperature, is consistent with the ideal gas laws, and correctly predicts properties like the specific heat capacities of many gases. It also allows for the estimation of molecular sizes and masses from measurable bulk properties like viscosity and diffusion.
Molecular Nature of Matter
The fundamental idea that matter is composed of discrete particles called atoms is known as the Atomic Hypothesis. Richard Feynman considered this the single most important piece of scientific knowledge.
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.
Historical Context
The idea of indivisible constituents of matter was conjectured in ancient India (Kanada) and Greece (Democritus). However, the scientific atomic theory is credited to John Dalton (early 19th century), who used it to explain the laws of chemical combination (definite and multiple proportions). Around the same time, Avogadro's hypothesis stated that equal volumes of all gases at the same temperature and pressure contain the same number of molecules. This hypothesis was a crucial link between the macroscopic properties of gases and their molecular nature.
Modern View
Today, the existence of atoms and molecules is a well-established fact, with modern instruments like electron microscopes allowing us to visualize them.
- The typical size of an atom is about an angstrom ($1 \text{ Å} = 10^{-10} \text{ m}$ ).
- In solids and liquids, atoms are closely packed, with interatomic separations of a few angstroms. The strong interatomic forces (short-range repulsion and long-range attraction) are significant.
- In gases, molecules are much farther apart, with average separations of tens of angstroms. Intermolecular forces are negligible except during brief collisions.
- The mean free path, the average distance a molecule travels between collisions, is on the order of thousands of angstroms in a gas at standard conditions.
The macroscopic state of a gas, which appears static, is actually the result of a dynamic equilibrium of countless molecules in incessant, random motion.
Behaviour of Gases
The behavior of gases is simpler to model than that of liquids or solids because the interactions between molecules are negligible. Gases at low pressures and high temperatures closely follow the ideal gas law.
The Ideal Gas Equation
The relationship between pressure (P), volume (V), and absolute temperature (T) for an ideal gas is given by:
$PV = \mu RT$
where:
- $\mu$ is the number of moles of the gas.
- $R$ is the universal gas constant ($R \approx 8.314 \text{ J mol}^{-1} \text{ K}^{-1}$).
This equation can also be expressed in terms of the number of molecules, $N$:
$PV = N k_B T$
where:
- $k_B$ is the Boltzmann constant ($k_B \approx 1.38 \times 10^{-23} \text{ J K}^{-1}$).
- $R = N_A k_B$, where $N_A$ is Avogadro's number ($6.02 \times 10^{23} \text{ mol}^{-1}$).
An ideal gas is a theoretical gas that obeys this equation perfectly at all temperatures and pressures. Real gases behave like ideal gases at low pressures and high temperatures, where molecules are far apart and interactions are minimal.
Gas Laws
The ideal gas equation incorporates several historical gas laws:
- Boyle's Law: At constant temperature, $PV = \text{constant}$.
- Charles' Law: At constant pressure, $V \propto T$.
- Dalton's Law of Partial Pressures: For a mixture of non-reacting gases, the total pressure is the sum of the partial pressures of the individual gases ($P_{total} = P_1 + P_2 + \dots$). The partial pressure of a gas is the pressure it would exert if it alone occupied the entire volume.
Example 1. 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. (Atomic mass of Ne = 20.2 u, molecular mass of O₂ = 32.0 u).
Answer:
Let Ne be gas 1 and O₂ be gas 2. We are given $P_1/P_2 = 3/2$.
From the ideal gas law in the form $PV = N k_B T$, the partial pressure of each gas is $P = (N/V)k_B T$. Since both gases are in the same vessel, they have the same volume (V) and temperature (T).
$\frac{P_1}{P_2} = \frac{(N_1/V)k_B T}{(N_2/V)k_B T} = \frac{N_1}{N_2}$
(i) Ratio of number of molecules:
The ratio of the number of molecules is equal to the ratio of their partial pressures.
$\frac{N_1}{N_2} = \frac{P_1}{P_2} = \frac{3}{2}$.
(ii) Ratio of mass density:
Mass density is defined as $\rho = m/V = (NM)/V$, where M is the mass of one molecule.
$\frac{\rho_1}{\rho_2} = \frac{(N_1 M_1)/V}{(N_2 M_2)/V} = \left(\frac{N_1}{N_2}\right) \left(\frac{M_1}{M_2}\right)$
The ratio of molecular masses is equal to the ratio of their molar masses given in atomic mass units (u).
$M_1 = 20.2$ u, $M_2 = 32.0$ u.
$\frac{\rho_1}{\rho_2} = \left(\frac{3}{2}\right) \left(\frac{20.2}{32.0}\right) = \frac{60.6}{64.0} \approx 0.947$.
Kinetic Theory of an Ideal Gas
Kinetic theory provides a microscopic explanation for the macroscopic behavior of an ideal gas. It is based on a set of assumptions:
- A gas consists of a large number of identical molecules that are in incessant, random motion.
- The volume of the molecules themselves is negligible compared to the volume of the container.
- Intermolecular forces are negligible except during collisions.
- Collisions between molecules and with the walls of the container are perfectly elastic.
Pressure of an Ideal Gas
The pressure exerted by a gas on the walls of its container is the result of the continuous bombardment of the walls by the gas molecules. Each collision imparts momentum to the wall.
Consider a gas in a cubic container of side $l$. The force on a wall is the total momentum imparted to it per unit time. By analyzing the change in momentum of molecules colliding with a wall, we can derive an expression for the pressure.
Derivation
A molecule with velocity component $v_x$ rebounds from a wall perpendicular to the x-axis with velocity component $-v_x$. The change in its momentum is $-2mv_x$. The momentum imparted to the wall is $2mv_x$.
The number of such molecules hitting an area $A$ in time $\Delta t$ is $\frac{1}{2} n A v_x \Delta t$, where $n$ is the number of molecules per unit volume. The factor of $1/2$ accounts for the fact that, on average, half the molecules are moving towards the wall and half are moving away.
The total momentum transferred is $Q = (2mv_x)(\frac{1}{2} n A v_x \Delta t) = n A m v_x^2 \Delta t$.
Force $F = Q/\Delta t = n A m v_x^2$. Pressure $P = F/A = n m v_x^2$.
Since the gas has a distribution of velocities, we take the average over all molecules: $P = nm \overline{v_x^2}$.
Due to the random motion (isotropy), the average squared velocity components are equal: $\overline{v_x^2} = \overline{v_y^2} = \overline{v_z^2}$.
Also, the mean square speed is $\overline{v^2} = \overline{v_x^2} + \overline{v_y^2} + \overline{v_z^2} = 3\overline{v_x^2}$. Thus, $\overline{v_x^2} = \frac{1}{3}\overline{v^2}$.
Substituting this back into the pressure equation gives the fundamental result:
$P = \frac{1}{3} n m \overline{v^2}$
Kinetic Interpretation of Temperature
We can connect this microscopic result with the macroscopic ideal gas law. From the pressure equation:
$PV = \frac{1}{3} (nV) m \overline{v^2} = \frac{1}{3} N m \overline{v^2}$
where $N=nV$ is the total number of molecules. Rearranging this gives:
$PV = \frac{2}{3} N \left(\frac{1}{2}m\overline{v^2}\right)$
The term in the parenthesis is the average translational kinetic energy of a single molecule. The total internal energy (E) of an ideal monatomic gas is the sum of these kinetic energies, $E = N \left(\frac{1}{2}m\overline{v^2}\right)$. So, $PV = \frac{2}{3}E$.
Comparing this with the ideal gas law $PV = N k_B T$, we get:
$\frac{2}{3}E = N k_B T \implies E = \frac{3}{2} N k_B T$
This leads to a profound conclusion about temperature:
Average Translational Kinetic Energy per molecule $= \frac{E}{N} = \frac{1}{2}m\overline{v^2} = \frac{3}{2}k_B T$
Temperature is a direct measure of the average translational kinetic energy of the molecules of a gas.
The square root of the mean square speed, $\sqrt{\overline{v^2}}$, is called the root-mean-square (rms) speed, $v_{rms}$.
$v_{rms} = \sqrt{\frac{3k_B T}{m}} = \sqrt{\frac{3RT}{M_0}}$ (where $M_0$ is the molar mass).
Law of Equipartition of Energy
The expression for the kinetic energy of a molecule is the sum of terms quadratic in the velocity components: $K_{trans} = \frac{1}{2}mv_x^2 + \frac{1}{2}mv_y^2 + \frac{1}{2}mv_z^2$. We found that the average energy associated with each of these terms (each "degree of freedom") is $\frac{1}{2}k_B T$.
The Law of Equipartition of Energy generalizes this result:
In thermal equilibrium, the total energy of a system is shared equally among all its available degrees of freedom, and the average energy associated with each degree of freedom is $\frac{1}{2}k_B T$.
Degrees of Freedom
A degree of freedom is an independent mode in which a molecule can store energy.
- Translational: Motion along the x, y, and z axes. All molecules have 3 translational degrees of freedom.
- Rotational: Rotation about different axes.
- A monatomic gas molecule (like He, Ar) is a point mass, so its rotation is negligible. It has 0 rotational degrees of freedom.
- A diatomic molecule (like O₂, N₂) is linear. It can rotate about two axes perpendicular to the line joining the atoms. It has 2 rotational degrees of freedom.
- A non-linear polyatomic molecule (like H₂O, CH₄) can rotate about all three axes. It has 3 rotational degrees of freedom.
- Vibrational: Oscillation of atoms along the bond connecting them. Each vibrational mode has two terms in its energy expression (kinetic and potential), so it contributes $2 \times \frac{1}{2}k_B T = k_B T$ to the average energy. Vibrational modes are generally only active at high temperatures.
Specific Heat Capacity
The law of equipartition of energy allows us to predict the molar specific heats of ideal gases.
For one mole of a gas, the total internal energy U is the total number of degrees of freedom ($f$) times the energy per degree of freedom, multiplied by Avogadro's number:
$U = f \times \left(\frac{1}{2}k_B T\right) \times N_A = \frac{f}{2}(k_B N_A)T = \frac{f}{2}RT$
1. Monatomic Gases
Have only 3 translational degrees of freedom, so $f=3$.
- Internal Energy: $U = \frac{3}{2}RT$
- Molar Specific Heat at Constant Volume: $C_v = \frac{dU}{dT} = \frac{3}{2}R$
- Molar Specific Heat at Constant Pressure: $C_p = C_v + R = \frac{5}{2}R$
- Ratio of Specific Heats: $\gamma = \frac{C_p}{C_v} = \frac{5/2 R}{3/2 R} = \frac{5}{3} \approx 1.67$
2. Diatomic Gases (Rigid Rotator)
Have 3 translational + 2 rotational degrees of freedom, so $f=5$. (Ignoring vibration).
- Internal Energy: $U = \frac{5}{2}RT$
- $C_v = \frac{5}{2}R$
- $C_p = C_v + R = \frac{7}{2}R$
- $\gamma = \frac{C_p}{C_v} = \frac{7}{5} = 1.40$
3. Diatomic Gases (with Vibration)
Have 3 translational + 2 rotational + 1 vibrational mode (2 degrees of freedom). Total $f = 3+2+2 = 7$.
- Internal Energy: $U = \frac{7}{2}RT$
- $C_v = \frac{7}{2}R$
- $C_p = \frac{9}{2}R$
- $\gamma = \frac{9}{7} \approx 1.29$
These theoretical predictions match experimental values well at ordinary temperatures, but break down at low temperatures where some degrees of freedom "freeze out" – a phenomenon that can only be explained by quantum mechanics.
Mean Free Path
Although gas molecules travel at very high speeds (comparable to the speed of sound), a gas leaking from a container takes a considerable time to diffuse across a room. This is because the molecules undergo frequent collisions with each other, causing their paths to be a series of short, random zig-zags.
The mean free path ($\lambda$) is the average distance a molecule travels between two successive collisions.
Derivation
Consider a molecule of diameter $d$ moving with average speed $\langle v \rangle$. It will collide with any other molecule whose center is within a distance $d$ of its center. In time $\Delta t$, it sweeps out a cylindrical volume of $\pi d^2 \langle v \rangle \Delta t$.
The number of collisions in this time is the number of other molecules in this volume, which is $n(\pi d^2 \langle v \rangle \Delta t)$, where $n$ is the number density of molecules.
The average time between collisions is $\tau = \frac{\Delta t}{\text{Number of collisions}} = \frac{1}{n\pi d^2 \langle v \rangle}$.
The mean free path is $\lambda = \langle v \rangle \tau = \frac{1}{n\pi d^2}$.
A more rigorous calculation that accounts for the motion of all molecules gives a slightly different result:
$\lambda = \frac{1}{\sqrt{2} n \pi d^2}$
The mean free path is inversely proportional to the number density and the cross-sectional area of the molecules. For air at STP, the mean free path is on the order of $10^{-7}$ m, which is several hundred times the molecular diameter, justifying the kinetic theory assumption that molecules travel in straight lines for most of the time.
Brownian Motion
The random, zigzag motion of larger particles (like pollen grains or smoke particles) suspended in a fluid is called Brownian motion. It is a direct and visible consequence of the incessant and random bombardment of the suspended particle by the much smaller, invisible molecules of the fluid. The unequal number of molecular impacts from different directions at any instant results in a net force that pushes the particle around. It provides macroscopic evidence for the microscopic picture of matter described by kinetic theory.
Exercises
Question 13.1. Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3 Å.
Answer:
Question 13.2. Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP : 1 atmospheric pressure, 0 °C). Show that it is 22.4 litres.
Answer:
Question 13.3. Figure 13.8 shows plot of $PV/T$ versus $P$ for $1.00 \times 10^{–3}$ kg of oxygen gas at two different temperatures.
(a) What does the dotted plot signify?
(b) Which is true: $T_1 > T_2$ or $T_1 < T_2$?
(c) What is the value of $PV/T$ where the curves meet on the y-axis?
(d) If we obtained similar plots for $1.00 \times 10^{–3}$ kg of hydrogen, would we get the same value of $PV/T$ at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of $PV/T$ (for low pressure high temperature region of the plot) ? (Molecular mass of H$_2$ = 2.02 u, of O$_2$ = 32.0 u, R = 8.31 J mol$^{–1}$ K$^{–1}$.)
Answer:
Question 13.4. An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J mol$^{–1}$ K$^{–1}$, molecular mass of O$_2$ = 32 u).
Answer:
Question 13.5. An air bubble of volume 1.0 cm$^3$ rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 °C ?
Answer:
Question 13.6. Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m$^3$ at a temperature of 27 °C and 1 atm pressure.
Answer:
Question 13.7. Estimate the average thermal energy of a helium atom at (i) room temperature (27 °C), (ii) the temperature on the surface of the Sun (6000 K), (iii) the temperature of 10 million kelvin (the typical core temperature in the case of a star).
Answer:
Question 13.8. Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules ? Is the root mean square speed of molecules the same in the three cases? If not, in which case is $v_{rms}$ the largest ?
Answer:
Question 13.9. At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at – 20 °C ? (atomic mass of Ar = 39.9 u, of He = 4.0 u).
Answer:
Question 13.10. Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 °C. Take the radius of a nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N$_2$ = 28.0 u).
Answer:
Additional Exercises
Question 13.11. A metre long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom ?
Answer:
Question 13.12. From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm$^3$ s$^{–1}$. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm$^3$ s$^{–1}$. Identify the gas.
[Hint : Use Graham’s law of diffusion: $R_1/R_2 = ( M_2 /M_1 )^{1/2}$, where $R_1, R_2$ are diffusion rates of gases 1 and 2, and $M_1$ and $M_2$ their respective molecular masses. The law is a simple consequence of kinetic theory.]
Answer:
Question 13.13. A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheres
$n_2 = n_1 \exp [ -mg (h_2 – h_1)/ k_B T]$
where $n_2, n_1$ refer to number density at heights $h_2$ and $h_1$ respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column:
$n_2 = n_1 \exp [ -mg N_A (\rho - \rho' ) (h_2 –h_1)/ (\rho RT)]$
where $\rho$ is the density of the suspended particle, and $\rho'$, that of surrounding medium. [$N_A$ is Avogadro’s number, and R the universal gas constant.] [Hint : Use Archimedes principle to find the apparent weight of the suspended particle.]
Answer:
Question 13.14. Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms :
[Hint : Assume the atoms to be ‘tightly packed’ in a solid or liquid phase, and use the known value of Avogadro’s number. You should, however, not take the actual numbers you obtain for various atomic sizes too literally. Because of the crudeness of the tight packing approximation, the results only indicate that atomic sizes are in the range of a few Å].
| Substance | Atomic Mass (u) | Density (10$^3$ Kg m$^{-3}$) |
|---|---|---|
| Carbon (diamond) | 12.01 | 2.22 |
| Gold | 197.00 | 19.32 |
| Nitrogen (liquid) | 14.01 | 1.00 |
| Lithium | 6.94 | 0.53 |
| Fluorine (liquid) | 19.00 | 1.14 |
Answer: