Table of Contents
How to Use This Page
Read each concept carefully, then check the formula, common mistake, and exam tip before moving to the next. This page completely covers the Kinetic Theory of Gases (KTG) for CBSE Class 11 Physics.
Key Concepts
Kinetic Theory of Gases
The microscopic behavior of macroscopic matter
1
Ideal Gas Equation
Formula
The macroscopic equation state for a perfect gas, combining Boyle’s, Charles’, and Avogadro’s laws. It relates Pressure ($P$), Volume ($V$), Temperature ($T$), and number of moles ($n$).
$$PV = nRT = N k_B T$$
$\text{Where } R = 8.314 \text{ J/mol K} \text{ and } k_B = 1.38 \times 10^{-23} \text{ J/K}$
2
Pressure of an Ideal Gas
Formula
The pressure exerted by a gas is due to the continuous bombardment of gas molecules on the walls of the container. It depends on density ($\rho$) and the mean square velocity.
$$P = \frac{1}{3} \rho v_{rms}^2 = \frac{1}{3} \left(\frac{M}{V}\right) v_{rms}^2$$
3
Root Mean Square (RMS) Velocity
Formula
The square root of the average of the squares of the velocities of the molecules. It is the most physically significant speed in KTG because it directly links to kinetic energy and temperature.
$$v_{rms} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3k_B T}{m}}$$
4
Average Kinetic Energy
Formula
The translational kinetic energy of a gas depends only on its absolute temperature. It does not depend on the pressure, volume, or the nature of the gas.
$$E_{\text{per molecule}} = \frac{3}{2} k_B T \quad | \quad E_{\text{per mole}} = \frac{3}{2} RT$$
5
Degrees of Freedom ($f$)
Definition
The number of independent ways a molecule can possess energy (translational, rotational, vibrational).
Monoatomic (e.g., He, Ar): $f = 3$
Diatomic (e.g., $O_2, H_2$): $f = 5$ (at room temp)
Monoatomic (e.g., He, Ar): $f = 3$
Diatomic (e.g., $O_2, H_2$): $f = 5$ (at room temp)
$$f = 3N – K \text{ (where } N \text{=atoms, } K \text{=constraints)}$$
6
Law of Equipartition of Energy
Formula
For a dynamical system in thermal equilibrium, the total energy is distributed equally amongst all the degrees of freedom. The energy associated with each molecule per degree of freedom is $\frac{1}{2} k_B T$.
$$E_{\text{total}} = f \times \frac{1}{2} k_B T \text{ (per molecule)}$$
7
Specific Heats from Degrees of Freedom
Formula
Using the Equipartition Law, we can predict the molar specific heats ($C_v$ and $C_p$) and their ratio ($\gamma$) for any ideal gas just by knowing its molecular structure.
$$C_v = \frac{f}{2} R \quad | \quad C_p = \left(\frac{f}{2} + 1\right) R \quad | \quad \gamma = \frac{C_p}{C_v} = 1 + \frac{2}{f}$$
8
Mean Free Path ($\lambda$)
Formula
The average distance a gas molecule travels between two successive collisions. It depends on the diameter of the molecule ($d$) and the number density ($n = N/V$).
$$\lambda = \frac{1}{\sqrt{2} \pi d^2 n} = \frac{k_B T}{\sqrt{2} \pi d^2 P}$$
Concept Deep Dive
01
Why Hydrogen Escapes Earth but Oxygen Doesn’t
A cosmic speed limit problemEarth’s escape velocity is $11.2 \text{ km/s}$. Since $v_{rms} = \sqrt{3RT/M}$, lighter gases have much higher velocities at the same temperature.
The molar mass of Hydrogen ($H_2$) is extremely small ($2 \text{ g/mol}$). At the high temperatures of the upper atmosphere, a significant fraction of Hydrogen molecules achieve speeds exceeding $11.2 \text{ km/s}$ and bleed off into space. Oxygen ($O_2$), being 16 times heavier, has an RMS speed that is $\frac{1}{4}$th that of Hydrogen. Almost no Oxygen molecules ever reach escape velocity, which is why Earth still has a breathable atmosphere today!
The molar mass of Hydrogen ($H_2$) is extremely small ($2 \text{ g/mol}$). At the high temperatures of the upper atmosphere, a significant fraction of Hydrogen molecules achieve speeds exceeding $11.2 \text{ km/s}$ and bleed off into space. Oxygen ($O_2$), being 16 times heavier, has an RMS speed that is $\frac{1}{4}$th that of Hydrogen. Almost no Oxygen molecules ever reach escape velocity, which is why Earth still has a breathable atmosphere today!
02
Pressure and Energy Connection
The $2/3$ RuleThere is a beautiful, direct relationship between the macroscopic pressure of a gas and its microscopic kinetic energy.
Let $E$ be the total translational kinetic energy, and let $E_{vol}$ be the Kinetic Energy per unit volume ($E/V$). From the pressure equation $P = \frac{1}{3}\rho v_{rms}^2$, we can algebraically show that the pressure exerted by an ideal gas is exactly two-thirds of its translational kinetic energy density.
Let $E$ be the total translational kinetic energy, and let $E_{vol}$ be the Kinetic Energy per unit volume ($E/V$). From the pressure equation $P = \frac{1}{3}\rho v_{rms}^2$, we can algebraically show that the pressure exerted by an ideal gas is exactly two-thirds of its translational kinetic energy density.
$$P = \frac{2}{3} E_{vol}$$
Compare & Contrast
✗ Ideal Gas
- Follows $PV=nRT$ strictly at all temperatures and pressures.
- Assumes intermolecular forces of attraction are zero.
- Assumes the volume occupied by the molecules themselves is zero (point masses).
- Cannot be liquefied, no matter how much pressure is applied.
✓ Real Gas
- Deviates from $PV=nRT$, especially at high pressure and low temperature.
- Intermolecular forces (Van der Waals forces) DO exist.
- Molecules have a finite, non-zero physical volume.
- Can be liquefied when cooled and compressed.
- Follows the Van der Waals equation: $(P + \frac{an^2}{V^2})(V – nb) = nRT$.
Remember
A Real Gas behaves almost exactly like an Ideal Gas under conditions of High Temperature and Low Pressure (because the molecules are moving too fast to attract each other, and they are spread so far apart their physical size doesn’t matter).
Common Mistakes to Avoid
Mistake 1
Confusing Molar Mass ($M$) and Molecular Mass ($m$):
If you use Universal Gas Constant ($R$), you MUST use Molar Mass ($M$ in kg/mol).
If you use Boltzmann’s Constant ($k_B$), you MUST use the mass of a single molecule ($m$ in kg).
Mixing them up ($v_{rms} = \sqrt{3RT/m}$) will result in answers that are off by a factor of Avogadro’s number ($10^{23}$)!
If you use Universal Gas Constant ($R$), you MUST use Molar Mass ($M$ in kg/mol).
If you use Boltzmann’s Constant ($k_B$), you MUST use the mass of a single molecule ($m$ in kg).
Mixing them up ($v_{rms} = \sqrt{3RT/m}$) will result in answers that are off by a factor of Avogadro’s number ($10^{23}$)!
Mistake 2
Using Grams instead of Kilograms: When calculating $v_{rms}$ using $\sqrt{3RT/M}$, students constantly put $M = 32$ for Oxygen. You MUST convert to standard SI units: $M = 32 \times 10^{-3} \text{ kg/mol}$. If you don’t, your velocity will come out 30 times too slow.
Mistake 3
Assuming Diatomic Gases Always Have $f=5$: At room temperature, a diatomic molecule has 3 translational + 2 rotational = 5 degrees of freedom. However, at very high temperatures, the atoms start vibrating vigorously, adding 2 more vibrational degrees of freedom, bringing $f = 7$. Always assume $f=5$ unless “high temperature” is specifically mentioned.
Exam Tips
Tip 1
The Velocity Ratios: Memorize the order and ratio of the three molecular speeds:
$v_{rms} : v_{avg} : v_{mp} = \sqrt{3} : \sqrt{8/\pi} : \sqrt{2} \approx 1.73 : 1.60 : 1.41$.
Root Mean Square is ALWAYS the fastest, Average is in the middle, and Most Probable is the slowest.
$v_{rms} : v_{avg} : v_{mp} = \sqrt{3} : \sqrt{8/\pi} : \sqrt{2} \approx 1.73 : 1.60 : 1.41$.
Root Mean Square is ALWAYS the fastest, Average is in the middle, and Most Probable is the slowest.
Tip 2
Mean Free Path proportionality: A frequent MCQ asks what happens to the mean free path ($\lambda$) if the volume is halved at constant temperature. Since $\lambda \propto 1/n$ and $n = N/V$, halving the volume doubles the density, which halves the mean free path.
Expected Exam Questions
SQ
Board Pattern Questions
Class 11 · Kinetic Theory · CBSE Exam1
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 $-20^\circ\text{C}$? (Atomic mass of Ar = 39.9 u, of He = 4.0 u). [2 marks]
Answer
$2523.7 \text{ K}$ or $2250.7^\circ\text{C}$
📝
Explanation
Given $v_{rms}(\text{Argon}) = v_{rms}(\text{Helium})$.
$\sqrt{\frac{3 R T_{\text{Ar}}}{M_{\text{Ar}}}} = \sqrt{\frac{3 R T_{\text{He}}}{M_{\text{He}}}}$
Squaring both sides and cancelling $3R$: $\frac{T_{\text{Ar}}}{M_{\text{Ar}}} = \frac{T_{\text{He}}}{M_{\text{He}}}$
First, convert Helium temp to Kelvin: $T_{\text{He}} = -20 + 273.15 = 253.15 \text{ K}$.
$T_{\text{Ar}} = T_{\text{He}} \times \left( \frac{M_{\text{Ar}}}{M_{\text{He}}} \right) = 253.15 \times \left( \frac{39.9}{4.0} \right)$
$T_{\text{Ar}} = 253.15 \times 9.975 = 2523.67 \text{ K}$.
2
Using the law of equipartition of energy, derive the ratio of specific heats ($\gamma = C_p/C_v$) for a diatomic gas at normal room temperature. [2 marks]
Answer
$\gamma = 1.4$
📝
Explanation
1. At room temperature, a diatomic gas has 5 degrees of freedom ($f = 5$): 3 translational and 2 rotational.
2. By equipartition of energy, internal energy per mole is $U = \frac{f}{2} RT = \frac{5}{2} RT$.
3. Molar specific heat at constant volume is $C_v = \frac{dU}{dT} = \frac{5}{2} R$.
4. By Mayer’s relation, $C_p = C_v + R = \frac{5}{2} R + R = \frac{7}{2} R$.
5. Ratio $\gamma = \frac{C_p}{C_v} = \frac{(7/2)R}{(5/2)R} = \frac{7}{5} = 1.4$.
3
Estimate the mean free path for a water molecule in water vapor at $373 \text{ K}$. Assume the collision diameter of the molecule is $2 \times 10^{-10} \text{ m}$ and the pressure is $1 \text{ atm}$ ($1.01 \times 10^5 \text{ Pa}$). [3 marks]
Answer
$\approx 2.87 \times 10^{-7} \text{ m}$
📝
Explanation
Mean free path formula: $\lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P}$
Given:
$k_B = 1.38 \times 10^{-23} \text{ J/K}$
$T = 373 \text{ K}$
$d = 2 \times 10^{-10} \text{ m} \implies d^2 = 4 \times 10^{-20} \text{ m}^2$
$P = 1.01 \times 10^5 \text{ Pa}$
$\lambda = \frac{1.38 \times 10^{-23} \times 373}{1.414 \times 3.14 \times 4 \times 10^{-20} \times 1.01 \times 10^5}$
$\lambda = \frac{514.7 \times 10^{-23}}{17.93 \times 10^{-15}} = 28.7 \times 10^{-8} \text{ m} = 2.87 \times 10^{-7} \text{ m}$.
Concept Map
Kinetic Theory connects to →
Thermodynamics
Thermal Properties (Specific Heats)
Mechanics (Momentum and Collisions)
Chemistry (States of Matter)
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