Welcome to Lecture 103 of our comprehensive JEE Physics Master Series! Today, we delve into the microscopic core of Unit 9: Kinetic Theory of Gases by investigating the Fundamental Postulates of Kinetic Theory and the Rigorous Derivation of Gas Pressure. By treating macroscopic gas pressure as the time-averaged rate of momentum imparted by countless chaotic molecular collisions against container walls, the kinetic theory constructs a direct mathematical bridge between Newtonian mechanics and bulk thermal properties. We analyze the core kinetic assumptions, derive the master pressure formula $P = \frac{1}{3}\rho v_{\text{rms}}^2$, establish the exact relationship between gas pressure and translational kinetic energy density ($P = \frac{2}{3}E_v$), explore molecular collision dynamics, and solve representative JEE Advanced problem archetypes.
Detailed Concept Card: Kinetic Theory Postulates & The Origin of Gas Pressure
1. Foundational Postulates of the Kinetic Theory of Gases
The classical kinetic model of an ideal gas rests upon eight fundamental physical assumptions:
- Molecular Constitution & Identity: A given gas consists of an enormous number of identical, exceedingly small, spherical particles called molecules. All molecules of a given chemical species have identical mass ($m$) and structure.
- Point-Mass Approximation: The physical volume occupied by the molecules themselves is completely negligible compared to the total volume $V$ of the enclosing container ($V_{\text{molecules}} \ll V$).
- Continuous, Random, Chaotic Motion: Molecules are in constant, ceaseless, isotropic motion in all possible directions with a broad distribution of velocities, strictly obeying Newton’s laws of motion.
- Spatial Isotropy of Motion: In an equilibrium state, there is no preferred direction of molecular flight. The mean-square velocity components along the three mutually perpendicular Cartesian axes are identical: $$\mathbf{\overline{v_x^2} = \overline{v_y^2} = \overline{v_z^2} = \frac{1}{3}\overline{v^2}}$$
- Perfectly Elastic Collisions: Collisions of gas molecules with one another and with the container walls are instantaneous and perfectly elastic. Both total linear momentum and total translational kinetic energy are conserved during collisions; no mechanical energy is converted into heat or chemical excitation.
- Negligible Collision Duration: The time interval during which a collision takes place ($\tau_{\text{coll}} \sim 10^{-13}\text{ s}$) is negligible compared to the time interval spent between successive collisions ($\tau_{\text{flight}} \sim 10^{-9}\text{ s}$).
- Zero Intermolecular Forces: Molecules exert neither attractive nor repulsive forces on one another except during the brief collision contact. Consequently, the intermolecular potential energy is identically zero ($U_{\text{potential}} = 0$), and the total internal energy of an ideal gas consists entirely of translational kinetic energy ($U = K_{\text{trans}}$).
- Negligible Gravitational Influence: Due to their minuscule mass ($m \sim 10^{-26}\text{ kg}$) and high thermal velocities ($v \sim 500\text{ m/s}$), gravitational potential energy changes across laboratory-scale vessels are negligible compared to thermal kinetic energies.
2. Rigorous Derivation of Pressure Exerted by an Ideal Gas
Consider a cubical container of side length $L$ and total volume $V = L^3$, aligned with the coordinate axes, containing $N$ identical gas molecules each of mass $m$:
- Single Molecule Collision Dynamics:
Let a molecule move with velocity vector $\vec{v}_1 = (v_{x1}, v_{y1}, v_{z1})$. When it strikes a planar wall perpendicular to the $x$-axis (at $x = L$): – Because the collision is perfectly elastic, the normal component reverses ($v_{x1} \to -v_{x1}$), while tangential components $v_{y1}$ and $v_{z1}$ remain unaltered.
– Change in linear momentum of the molecule: $$\Delta p_{x1} = -m v_{x1} – (m v_{x1}) = -2 m v_{x1}$$ – By Newton’s Third Law, momentum transferred to the wall per collision is: $$\Delta p_{\text{wall}} = +2 m v_{x1}$$ - Time Between Successive Collisions:
After rebounding, the molecule travels across the container, reflects off the opposite wall at $x = 0$, and returns to the wall at $x = L$, covering a round-trip distance $2L$. The time interval between successive collisions on this specific wall is: $$\Delta t = \frac{2L}{v_{x1}}$$ - Average Force Exerted by One Molecule:
By Newton’s Second Law, the time-averaged force exerted by molecule 1 on the wall is: $$F_{1x} = \frac{\Delta p_{\text{wall}}}{\Delta t} = \frac{2 m v_{x1}}{2L / v_{x1}} = \frac{m v_{x1}^2}{L}$$ - Total Force Exerted by All $N$ Molecules:
Summing the contributions of all $N$ non-interacting molecules: $$F_{\text{total}, x} = \sum_{i=1}^N \frac{m v_{xi}^2}{L} = \frac{m}{L} \sum_{i=1}^N v_{xi}^2 = \frac{m N}{L} \overline{v_x^2}$$ where $\overline{v_x^2} = \frac{1}{N}\sum v_{xi}^2$ is the mean-square velocity along the $x$-direction. - Formulation of Pressure:
Pressure is force divided by wall area $A = L^2$: $$P = \frac{F_{\text{total}, x}}{A} = \frac{m N \overline{v_x^2}}{L \cdot L^2} = \frac{m N}{L^3} \overline{v_x^2} = \frac{m N}{V} \overline{v_x^2}$$ Using the isotropy condition $\overline{v_x^2} = \frac{1}{3}\overline{v^2} = \frac{1}{3}v_{\text{rms}}^2$: $$\mathbf{P = \frac{1}{3} \frac{m N}{V} \overline{v^2} = \frac{1}{3} \rho \, v_{\text{rms}}^2 = \frac{1}{3} n_0 m \, v_{\text{rms}}^2}$$ where $\rho = \frac{m N}{V}$ is the mass density and $n_0 = \frac{N}{V}$ is the number density of the gas.
3. Relation Between Gas Pressure & Kinetic Energy Density
The total translational kinetic energy $E_{\text{trans}}$ of all $N$ gas molecules enclosed within volume $V$ is:
$$E_{\text{trans}} = \sum_{i=1}^N \frac{1}{2} m v_i^2 = \frac{1}{2} m N \overline{v^2} = \frac{1}{2} M_{\text{total}} v_{\text{rms}}^2$$
The Translational Kinetic Energy Density ($E_v$), defined as kinetic energy per unit volume, is:
$$E_v = \frac{E_{\text{trans}}}{V} = \frac{1}{2}\left(\frac{m N}{V}\right) \overline{v^2} = \frac{1}{2} \rho \, v_{\text{rms}}^2$$
Comparing this with the pressure equation $P = \frac{1}{3}\rho v_{\text{rms}}^2$ reveals the fundamental connection:
$$\mathbf{P = \frac{2}{3} \left(\frac{1}{2}\rho v_{\text{rms}}^2\right) = \frac{2}{3} E_v \iff E_v = \frac{3}{2} P}$$
Fundamental Physical Theorem:
The pressure exerted by an ideal gas is numerically equal to exactly two-thirds of its translational kinetic energy density.
Total translational kinetic energy in volume $V$ is directly proportional to pressure and volume: $$\mathbf{E_{\text{trans}} = \frac{3}{2} P V = \frac{3}{2} n R T}$$
4. Root-Mean-Square Speed ($v_{\text{rms}}$) & Microscopic Temperature
Equating the macroscopic ideal gas equation $P V = N k_B T = n R T$ with the microscopic pressure formula $P V = \frac{1}{3} m N v_{\text{rms}}^2$:
$$\frac{1}{3} m N v_{\text{rms}}^2 = N k_B T \implies \mathbf{\frac{1}{2} m v_{\text{rms}}^2 = \frac{3}{2} k_B T}$$
This yields the profound microscopic interpretation: Absolute temperature $T$ is a direct measure of the average translational kinetic energy per molecule!
| Quantity | Mathematical Formula | Physical Significance |
|---|---|---|
| Mean Translational KE per Molecule | $$\overline{\epsilon}_{\text{trans}} = \frac{3}{2} k_B T$$ | Depends solely on absolute temperature; completely independent of gas atomicity, molecular mass, or volume. |
| Mean Translational KE per Mole | $$E_{\text{molar}} = N_A \left(\frac{3}{2} k_B T\right) = \frac{3}{2} R T$$ | Universal constant value for 1 mole of any ideal gas at temperature $T$. |
| Root-Mean-Square Speed ($v_{\text{rms}}$) | $$v_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}} = \sqrt{\frac{3 R T}{M}} = \sqrt{\frac{3 P}{\rho}}$$ | Molecules of lighter gases travel much faster than molecules of heavier gases at the same temperature ($v_{\text{rms}} \propto 1/\sqrt{M}$). |
Four Graded Solved Examples
Example 1 (JEE Main Level: Comprehensive Pressure and Kinetic Energy Accounting)
Problem Statement: A rigid container of volume $V = 5.0\text{ L} = 5.0 \times 10^{-3}\text{ m}^3$ contains $m_{\text{total}} = 8.0\text{ g}$ of Oxygen gas ($\text{O}_2$, molar mass $M = 32.0\text{ g/mol} = 0.032\text{ kg/mol}$) maintained at temperature $T = 300\text{ K}$. Taking the universal gas constant $R = 8.314\text{ J/(mol}\cdot\text{K)}$ and Avogadro’s number $N_A = 6.022 \times 10^{23}\text{ mol}^{-1}$, determine:
1. The mass density $\rho$ and number density $n_0$ of the gas.
2. The root-mean-square speed $v_{\text{rms}}$ of the Oxygen molecules.
3. The gas pressure $P$ calculated from kinetic theory.
4. The total translational kinetic energy $E_{\text{trans}}$ and the translational kinetic energy density $E_v$.
Solution:
Step 1: Calculate Mass Density and Number Density.
Total mass in SI units: $m_{\text{total}} = 8.0 \times 10^{-3}\text{ kg}$.
Mass density:
$$\rho = \frac{m_{\text{total}}}{V} = \frac{8.0 \times 10^{-3}\text{ kg}}{5.0 \times 10^{-3}\text{ m}^3} = \mathbf{1.60\text{ kg/m}^3}$$
Number of moles:
$$n = \frac{m_{\text{total}}}{M} = \frac{8.0\text{ g}}{32.0\text{ g/mol}} = 0.25\text{ mol}$$
Number density $n_0$:
$$n_0 = \frac{n N_A}{V} = \frac{(0.25\text{ mol})(6.022 \times 10^{23}\text{ molecules/mol})}{5.0 \times 10^{-3}\text{ m}^3} = \mathbf{3.011 \times 10^{25}\text{ molecules/m}^3}$$
Step 2: Root-Mean-Square Speed ($v_{\text{rms}}$).
$$v_{\text{rms}} = \sqrt{\frac{3 R T}{M}} = \sqrt{\frac{3 \times 8.314 \times 300}{0.032}} = \sqrt{\frac{7482.6}{0.032}} = \sqrt{233831.25} \approx \mathbf{483.56\text{ m/s}}$$
Step 3: Gas Pressure from Kinetic Theory.
$$P = \frac{1}{3} \rho v_{\text{rms}}^2 = \frac{1}{3}(1.60\text{ kg/m}^3)(233831.25\text{ m}^2/\text{s}^2) = \mathbf{124710\text{ Pa} \approx 1.247 \times 10^5\text{ Pa} \approx 1.23\text{ atm}}$$
Cross-check using ideal gas equation:
$$P = \frac{n R T}{V} = \frac{0.25 \times 8.314 \times 300}{5.0 \times 10^{-3}} = \frac{623.55}{0.005} = 124710\text{ Pa} \quad (\text{Exact match!})$$
Step 4: Translational Kinetic Energy & Density.
Total translational kinetic energy:
$$E_{\text{trans}} = \frac{3}{2} n R T = 1.5 \times (0.25 \times 8.314 \times 300) = 1.5 \times 623.55\text{ J} = \mathbf{935.33\text{ J}}$$
Kinetic energy density:
$$E_v = \frac{E_{\text{trans}}}{V} = \frac{935.33\text{ J}}{5.0 \times 10^{-3}\text{ m}^3} = \mathbf{187065\text{ J/m}^3}$$
Verify the fundamental theorem:
$$\frac{2}{3} E_v = \frac{2}{3}(187065\text{ J/m}^3) = 124710\text{ Pa} = P \quad (\text{Perfect verification!})$$
Example 2 (Molecular Momentum Transfer & Wall Collision Rate)
Problem Statement: A cubical box of side $L = 1.0\text{ m}$ (volume $V = 1.0\text{ m}^3$) contains $N = 10^{24}$ identical molecules of an ideal gas, each of mass $m = 5.0 \times 10^{-26}\text{ kg}$. The root-mean-square speed of the molecules is $v_{\text{rms}} = 600\text{ m/s}$.
1. Calculate the mean-square velocity component along the $x$-axis ($\overline{v_x^2}$) and the corresponding RMS velocity component $v_{x,\text{rms}}$.
2. Find the total pressure $P$ and total normal force $F$ exerted by the gas on one wall face.
3. If the gas is heated at constant volume such that $v_{\text{rms}}$ increases by $20\%$, find the percentage increase in the pressure exerted on the container walls.
Solution:
Step 1: Normal Velocity Component.
From spatial isotropy of molecular motion:
$$\overline{v_x^2} = \frac{1}{3}\overline{v^2} = \frac{1}{3}(600\text{ m/s})^2 = \frac{360000}{3} = \mathbf{120000\text{ m}^2/\text{s}^2}$$
RMS velocity component along the normal to the wall:
$$v_{x,\text{rms}} = \sqrt{\overline{v_x^2}} = \sqrt{120000} \approx \mathbf{346.41\text{ m/s}}$$
Step 2: Total Pressure and Force on One Face.
Total mass of gas in the container:
$$M_{\text{total}} = N m = (10^{24})(5.0 \times 10^{-26}\text{ kg}) = 0.050\text{ kg}$$
Mass density:
$$\rho = \frac{M_{\text{total}}}{V} = \frac{0.050\text{ kg}}{1.0\text{ m}^3} = 0.050\text{ kg/m}^3$$
Pressure:
$$P = \frac{1}{3} \rho v_{\text{rms}}^2 = \frac{1}{3}(0.050\text{ kg/m}^3)(360000\text{ m}^2/\text{s}^2) = \mathbf{6000\text{ Pa}}$$
Force on one face of area $A = L^2 = 1.0\text{ m}^2$:
$$F = P A = (6000\text{ Pa})(1.0\text{ m}^2) = \mathbf{6000\text{ N}}$$
Step 3: Effect of $20\%$ Increase in $v_{\text{rms}}$.
The new RMS speed is $v_{\text{rms}}’ = 1.20 \, v_{\text{rms}}$.
Since volume and mass density are held constant, pressure scales with the square of the speed:
$$P \propto v_{\text{rms}}^2 \implies P’ = (1.20)^2 P = 1.44 P$$
Percentage increase:
$$\text{Percentage Increase} = \frac{P’ – P}{P} \times 100\% = (1.44 – 1.00) \times 100\% = \mathbf{44\%}$$
Example 3 (JEE Advanced Level: Solid Angle Integration & Origin of the $1/3$ Factor)
Problem Statement: In kinetic theory, molecules strike a planar wall at various angles of incidence. Consider a stationary plane surface of area $A$ exposed to an ideal gas of number density $n_0$, where all molecules move with speed $v$. Derive the pressure formula by explicitly integrating over all angles of incidence $\theta$ (measured with respect to the surface normal) across the forward hemisphere, and rigorously prove the origin of the geometric factor $\frac{1}{3}$.
Solution:
Step 1: Solid Angle and Directional Distribution.
Because molecular directions are uniformly distributed over all $4\pi$ steradians of solid angle, the fraction of molecules traveling in directions within a differential solid angle $d\Omega$ is $\frac{d\Omega}{4\pi}$.
In spherical polar coordinates where $\theta$ is the angle with the inward normal to surface $A$:
$$d\Omega = 2\pi \sin\theta \, d\theta$$
The fraction of molecules moving toward the wall at angles between $\theta$ and $\theta + d\theta$ is:
$$\frac{d\Omega}{4\pi} = \frac{2\pi \sin\theta \, d\theta}{4\pi} = \frac{1}{2} \sin\theta \, d\theta$$
Step 2: Number of Collisions per Unit Time.
Molecules moving at angle $\theta$ have a normal velocity component toward the wall of $v_{\perp} = v \cos\theta$.
In time $dt$, all such molecules located within an oblique cylinder of base area $A$ and slant length $v \, dt$ (normal height $v \cos\theta \, dt$) will strike the wall. The volume of this cylinder is:
$$dV = A v \cos\theta \, dt$$
Number of molecules in this angular range striking area $A$ in time $dt$:
$$dN(\theta) = n_0 \left(\frac{1}{2} \sin\theta \, d\theta\right) (A v \cos\theta \, dt) = \frac{1}{2} n_0 A v \cos\theta \sin\theta \, d\theta \, dt$$
Step 3: Momentum Imparted and Total Pressure.
In a perfectly elastic collision, each molecule transfers normal momentum $\Delta p = 2 m v_{\perp} = 2 m v \cos\theta$.
Total momentum transferred by these molecules in time $dt$:
$$dp(\theta) = dN(\theta) \times (2 m v \cos\theta) = \left(\frac{1}{2} n_0 A v \cos\theta \sin\theta \, d\theta \, dt\right)(2 m v \cos\theta) = n_0 m A v^2 \cos^2\theta \sin\theta \, d\theta \, dt$$
To find the total force $F$, integrate $\theta$ over the entire forward hemisphere from $\theta = 0$ (normal incidence) to $\theta = \pi/2$ (glancing incidence):
$$F = \frac{1}{dt} \int_0^{\pi/2} dp(\theta) = n_0 m A v^2 \int_0^{\pi/2} \cos^2\theta \sin\theta \, d\theta$$
Let $u = \cos\theta \implies du = -\sin\theta \, d\theta$. When $\theta = 0, u = 1$; when $\theta = \pi/2, u = 0$:
$$\int_0^{\pi/2} \cos^2\theta \sin\theta \, d\theta = \int_0^1 u^2 du = \left[\frac{u^3}{3}\right]_0^1 = \mathbf{\frac{1}{3}}$$
Dividing force by area $A$ yields:
$$\mathbf{P = \frac{F}{A} = \frac{1}{3} n_0 m v^2 = \frac{1}{3} \rho v^2}$$
This confirms that the factor $\frac{1}{3}$ arises directly from the spherical geometry of 3D isotropic velocity space!
Example 4 (JEE Advanced Level: Gas Mixture in Thermal Equilibrium & Microscopic Dalton’s Law)
Problem Statement: A closed vessel of volume $V = 10.0\text{ L} = 10^{-2}\text{ m}^3$ contains a non-reacting mixture of $m_1 = 2.0\text{ g}$ of Helium ($\text{He}$, molar mass $M_1 = 4.0\text{ g/mol}$) and $m_2 = 14.0\text{ g}$ of Nitrogen ($\text{N}_2$, molar mass $M_2 = 28.0\text{ g/mol}$) in complete thermal equilibrium at temperature $T = 300\text{ K}$. Given $R = 8.314\text{ J/(mol}\cdot\text{K)}$:
1. Find the ratio of the root-mean-square speeds $\frac{v_{\text{rms, He}}}{v_{\text{rms, N}_2}}$.
2. Find the ratio of the average translational kinetic energy per molecule $\frac{\overline{\epsilon}_{\text{He}}}{\overline{\epsilon}_{\text{N}_2}}$.
3. Calculate the partial pressures $P_{\text{He}}, P_{\text{N}_2}$, and the total pressure $P_{\text{total}}$.
4. Show that the total pressure equals two-thirds of the sum of the individual kinetic energy densities ($P_{\text{total}} = \frac{2}{3}(E_{v, \text{He}} + E_{v, \text{N}_2})$).
Solution:
Step 1: Ratio of Root-Mean-Square Speeds.
Since both gases share the same equilibrium temperature $T$:
$$\frac{v_{\text{rms, He}}}{v_{\text{rms, N}_2}} = \frac{\sqrt{3RT / M_1}}{\sqrt{3RT / M_2}} = \sqrt{\frac{M_2}{M_1}} = \sqrt{\frac{28.0}{4.0}} = \sqrt{7} \approx \mathbf{2.646}$$
Helium molecules move roughly $2.65$ times faster than Nitrogen molecules.
Step 2: Ratio of Mean Translational Kinetic Energies per Molecule.
The mean translational kinetic energy per single molecule is $\overline{\epsilon} = \frac{3}{2} k_B T$, which depends strictly on temperature:
$$\frac{\overline{\epsilon}_{\text{He}}}{\overline{\epsilon}_{\text{N}_2}} = \frac{\frac{3}{2} k_B T}{\frac{3}{2} k_B T} = \mathbf{1}$$
Both species have identical average translational kinetic energies!
Step 3: Partial and Total Pressures.
Calculate moles of each gas:
$$n_{\text{He}} = \frac{2.0\text{ g}}{4.0\text{ g/mol}} = 0.50\text{ mol}$$
$$n_{\text{N}_2} = \frac{14.0\text{ g}}{28.0\text{ g/mol}} = 0.50\text{ mol}$$
Total moles: $n_{\text{total}} = 0.50 + 0.50 = 1.00\text{ mol}$.
Partial pressures:
$$P_{\text{He}} = \frac{n_{\text{He}} R T}{V} = \frac{(0.50\text{ mol})(8.314\text{ J/(mol}\cdot\text{K)})(300\text{ K})}{10^{-2}\text{ m}^3} = \frac{1247.1}{0.010} = \mathbf{124710\text{ Pa} \approx 1.247\text{ bar}}$$
$$P_{\text{N}_2} = \frac{n_{\text{N}_2} R T}{V} = \mathbf{124710\text{ Pa} \approx 1.247\text{ bar}}$$
Total pressure by Dalton’s Law:
$$P_{\text{total}} = P_{\text{He}} + P_{\text{N}_2} = 124710\text{ Pa} + 124710\text{ Pa} = \mathbf{249420\text{ Pa} \approx 2.494\text{ bar}}$$
Step 4: Microscopic Verification via Kinetic Energy Densities.
Kinetic energy density for each gas:
$$E_{v, \text{He}} = \frac{3}{2} P_{\text{He}} = 1.5 \times 124710\text{ Pa} = 187065\text{ J/m}^3$$
$$E_{v, \text{N}_2} = \frac{3}{2} P_{\text{N}_2} = 1.5 \times 124710\text{ Pa} = 187065\text{ J/m}^3$$
Total translational kinetic energy density:
$$E_{v, \text{total}} = E_{v, \text{He}} + E_{v, \text{N}_2} = 187065 + 187065 = 374130\text{ J/m}^3$$
Taking two-thirds:
$$\frac{2}{3} E_{v, \text{total}} = \frac{2}{3}(374130\text{ J/m}^3) = \mathbf{249420\text{ Pa} = P_{\text{total}}}$$
This rigorously proves that Dalton’s Law of Partial Pressures is the direct macroscopic expression of additivity of microscopic kinetic energy densities!
Practice Worksheet: 10 JEE Problems
Questions 1 to 5: JEE Main Level (Single Correct)
Q1. If $E_v$ represents the translational kinetic energy density (kinetic energy per unit volume) of an ideal gas, the pressure $P$ exerted by the gas is related to $E_v$ by:
(A) $P = E_v$
(B) $P = \frac{1}{2} E_v$
(C) $P = \frac{2}{3} E_v$
(D) $P = \frac{3}{2} E_v$
Q2. The root-mean-square speed of molecules of an ideal gas of density $\rho$ at pressure $P$ is given by:
(A) $\sqrt{\frac{P}{\rho}}$
(B) $\sqrt{\frac{3P}{\rho}}$
(C) $\sqrt{\frac{\rho}{3P}}$
(D) $\sqrt{\frac{2P}{3\rho}}$
Q3. The absolute temperature of a sample of an ideal gas is quadrupled ($T \to 4T$), while its volume is halved ($V \to V/2$). The root-mean-square speed of its molecules and its pressure will increase respectively by factors of:
(A) $2$ and $8$
(B) $4$ and $8$
(C) $2$ and $4$
(D) $4$ and $2$
Q4. The ratio of the root-mean-square speed of Helium molecules ($\text{He}$, molar mass $4\text{ g/mol}$) to that of Argon molecules ($\text{Ar}$, molar mass $40\text{ g/mol}$) at the same absolute temperature is:
(A) $10$
(B) $\sqrt{10}$
(C) $1/\sqrt{10}$
(D) $1/10$
Q5. An enclosure of volume $V$ contains $N$ molecules of an ideal gas, each of mass $m$. If the mean-square velocity of the molecules is $\overline{v^2}$, the pressure exerted on the container walls is:
(A) $\frac{m N \overline{v^2}}{V}$
(B) $\frac{1}{2} \frac{m N \overline{v^2}}{V}$
(C) $\frac{1}{3} \frac{m N \overline{v^2}}{V}$
(D) $\frac{2}{3} \frac{m N \overline{v^2}}{V}$
Questions 6 to 8: JEE Advanced Level (One or More Than One Correct)
Q6. Which of the following assumptions is/are fundamental postulates of the classical kinetic theory of ideal gases?
(A) The volume of the gas molecules is completely negligible compared to the total volume occupied by the gas.
(B) Collisions between molecules and with the walls of the container are perfectly elastic.
(C) The intermolecular forces of attraction or repulsion are zero except during the collision instant.
(D) The average translational kinetic energy of a molecule depends solely upon the absolute temperature of the gas.
Q7. Which of the following statements regarding the microscopic origin of gas pressure is/are correct?
(A) Pressure is the time-averaged rate of momentum transfer per unit area imparted by molecules rebounding from the container walls.
(B) If the collisions of molecules with the walls were inelastic such that molecules stuck to the wall on impact, the pressure exerted would decrease to half its original value.
(C) Gas pressure is isotropic because molecular velocities in thermal equilibrium are randomly and uniformly distributed in 3D space ($\overline{v_x^2} = \overline{v_y^2} = \overline{v_z^2} = \frac{1}{3}\overline{v^2}$).
(D) The pressure of an ideal gas does not depend on the geometrical shape of the container for a fixed volume, temperature, and quantity of gas.
Q8. A rigid container holds a thermal mixture of Helium gas ($\text{He}$, molar mass $4\text{ g/mol}$) and Oxygen gas ($\text{O}_2$, molar mass $32\text{ g/mol}$) in complete thermodynamic equilibrium at temperature $T$. Which of the following statements is/are correct?
(A) The average translational kinetic energy per molecule is identical for Helium and Oxygen.
(B) The ratio of the root-mean-square speeds of Helium to Oxygen is $v_{\text{rms, He}} / v_{\text{rms, O}_2} = 2\sqrt{2}$.
(C) During an elastic collision with the container wall, a single Oxygen molecule imparts a greater momentum than a single Helium molecule on average.
(D) Helium molecules strike the container walls with a higher collision frequency per unit area than Oxygen molecules of equal mole fraction.
Questions 9 to 10: Numerical / Integer Answer Type
Q9. A rigid vessel contains an ideal gas at pressure $P = 100\text{ kPa}$ and uniform mass density $\rho = 1.2\text{ kg/m}^3$. Calculate the root-mean-square speed $v_{\text{rms}}$ of the gas molecules (in $\text{m/s}$).
Q10. A cubical enclosure of side length $L = 2.0\text{ m}$ contains $N = 3.0 \times 10^{24}$ molecules of an ideal gas, each of mass $m = 4.0 \times 10^{-26}\text{ kg}$. The root-mean-square speed of the molecules is $v_{\text{rms}} = 500\text{ m/s}$. Calculate the total normal force exerted by the gas on ONE face of the cubical enclosure (in $\text{kN}$).
Detailed Solutions & Answer Key
Answer Key
| Question | Correct Option / Answer | Topic / Concept |
|---|---|---|
| Q1 | (C) $P = \frac{2}{3} E_v$ | Pressure and Kinetic Energy Density Relation |
| Q2 | (B) $\sqrt{3P/\rho}$ | RMS Speed in Terms of Pressure and Density |
| Q3 | (A) $2$ and $8$ | Scaling of $v_{\text{rms}}$ and Pressure with $T$ and $V$ |
| Q4 | (B) $\sqrt{10}$ | Ratio of RMS Speeds for He and Ar |
| Q5 | (C) $\frac{1}{3}\frac{m N \overline{v^2}}{V}$ | Master Pressure Equation of Kinetic Theory |
| Q6 | (A), (B), (C), (D) | Postulates of the Kinetic Theory of Gases |
| Q7 | (A), (B), (C), (D) | Microscopic Mechanism of Gas Pressure |
| Q8 | (A), (B), (C), (D) | Gas Mixture Collision Dynamics & Equipartition |
| Q9 | 500 | RMS Speed Calculation ($\sqrt{3 \times 10^5 / 1.2}\text{ m/s}$) |
| Q10 | 5 | Total Force on Cubical Enclosure Face ($5\text{ kN}$) |
Complete Step-by-Step Solutions
Solution 1:
From the kinetic derivation of gas pressure:
$$P = \frac{1}{3} \rho v_{\text{rms}}^2$$
The translational kinetic energy density is:
$$E_v = \frac{1}{2} \rho v_{\text{rms}}^2 \implies \rho v_{\text{rms}}^2 = 2 E_v$$
Substituting into the pressure formula:
$$P = \frac{1}{3}(2 E_v) = \mathbf{\frac{2}{3} E_v}$$
Correct Option: (C)
Solution 2:
From $P = \frac{1}{3} \rho v_{\text{rms}}^2$:
$$v_{\text{rms}}^2 = \frac{3P}{\rho} \implies \mathbf{v_{\text{rms}} = \sqrt{\frac{3P}{\rho}}}$$
Correct Option: (B)
Solution 3:
1. For root-mean-square speed:
$$v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \propto \sqrt{T}$$
When $T \to 4T$, $v_{\text{rms}}’ = \sqrt{4} v_{\text{rms}} = \mathbf{2 \, v_{\text{rms}}}$.
2. For gas pressure:
$$P = \frac{n R T}{V} \propto \frac{T}{V}$$
When $T \to 4T$ and $V \to V/2$:
$$P’ = \frac{4 T_0}{V_0 / 2} = \mathbf{8 P_0}$$
The speed increases by a factor of $2$, and the pressure increases by a factor of $8$.
Correct Option: (A)
Solution 4:
At the same temperature $T$:
$$v_{\text{rms}} \propto \frac{1}{\sqrt{M}}$$
Taking the ratio:
$$\frac{v_{\text{rms, He}}}{v_{\text{rms, Ar}}} = \sqrt{\frac{M_{\text{Ar}}}{M_{\text{He}}}} = \sqrt{\frac{40\text{ g/mol}}{4\text{ g/mol}}} = \mathbf{\sqrt{10}}$$
Correct Option: (B)
Solution 5:
In a volume $V$ containing $N$ molecules each of mass $m$ with mean-square velocity $\overline{v^2}$:
$$P = \frac{1}{3} \rho \overline{v^2} = \mathbf{\frac{1}{3} \frac{m N \overline{v^2}}{V}}$$
Correct Option: (C)
Solution 6:
(A) Correct: Gas molecules are treated as point particles whose volume is negligible compared to the vessel volume.
(B) Correct: Collisions conserve both momentum and translational kinetic energy.
(C) Correct: Zero intermolecular forces act at a distance; potential energy is negligible.
(D) Correct: Microscopic interpretation of temperature shows $\overline{\epsilon}_{\text{trans}} = \frac{3}{2} k_B T$, which depends solely on absolute temperature.
Correct Options: (A), (B), (C), (D)
Solution 7:
(A) Correct: Macroscopic pressure originates from the time-averaged normal momentum imparted per unit area during wall collisions.
(B) Correct: In an elastic collision, the molecule rebounds with equal and opposite normal velocity, imparting momentum $\Delta p = 2 m v_x$. If it sticks to the wall (completely inelastic collision), momentum transferred is only $\Delta p’ = m v_x$, which is exactly half, halving the pressure.
(C) Correct: Spatial isotropy ensures equal momentum transfer rate across all container orientations.
(D) Correct: For an ideal gas, $P = \frac{nRT}{V}$, which depends only on volume $V$, temperature $T$, and quantity $n$, independent of vessel shape.
Correct Options: (A), (B), (C), (D)
Solution 8:
(A) Correct: Mean translational kinetic energy per molecule is $\frac{3}{2} k_B T$, identical for all gases in thermal equilibrium.
(B) Correct: $\frac{v_{\text{rms, He}}}{v_{\text{rms, O}_2}} = \sqrt{\frac{M_{\text{O}_2}}{M_{\text{He}}}} = \sqrt{\frac{32}{4}} = \sqrt{8} = 2\sqrt{2}$.
(C) Correct: Momentum transferred per collision is $\Delta p = 2 m v \propto m \frac{1}{\sqrt{m}} = \sqrt{m}$. Since $M_{\text{O}_2} > M_{\text{He}}$, an Oxygen molecule imparts greater momentum per impact.
(D) Correct: Collision frequency per unit area is proportional to speed ($f \propto v \propto 1/\sqrt{m}$). Since Helium molecules move $2\sqrt{2}$ times faster, they strike the walls more frequently.
Correct Options: (A), (B), (C), (D)
Solution 9:
Given $P = 100\text{ kPa} = 10^5\text{ Pa}$ and $\rho = 1.2\text{ kg/m}^3$:
$$v_{\text{rms}} = \sqrt{\frac{3P}{\rho}} = \sqrt{\frac{3 \times 10^5\text{ Pa}}{1.2\text{ kg/m}^3}} = \sqrt{\frac{300000}{1.2}} = \sqrt{250000} = \mathbf{500\text{ m/s}}$$
Answer: 500
Solution 10:
1. Total volume of the cubical box:
$$V = L^3 = (2.0\text{ m})^3 = 8.0\text{ m}^3$$
2. Area of one face:
$$A = L^2 = (2.0\text{ m})^2 = 4.0\text{ m}^2$$
3. Total mass of gas:
$$M_{\text{total}} = N m = (3.0 \times 10^{24})(4.0 \times 10^{-26}\text{ kg}) = 0.12\text{ kg}$$
4. Mass density:
$$\rho = \frac{M_{\text{total}}}{V} = \frac{0.12\text{ kg}}{8.0\text{ m}^3} = 0.015\text{ kg/m}^3$$
5. Pressure exerted by the gas:
$$P = \frac{1}{3} \rho v_{\text{rms}}^2 = \frac{1}{3}(0.015\text{ kg/m}^3)(500\text{ m/s})^2 = 0.005 \times 250000 = 1250\text{ Pa}$$
6. Total force on one face:
$$F = P A = (1250\text{ Pa})(4.0\text{ m}^2) = 5000\text{ N} = \mathbf{5\text{ kN}}$$
Answer: 5