Welcome to Lecture 100 of our comprehensive JEE Physics Master Series. In this milestone lecture, we transition from macroscopic thermodynamics to the microscopic foundations of matter: the Kinetic Theory of Gases (KTG). By modeling an ideal gas as a statistical ensemble of point-mass particles undergoing elastic collisions, we derive the fundamental pressure equation from first principles, dissect the statistical distribution of molecular speeds ($v_{\text{rms}}$, $v_{\text{avg}}$, $v_{\text{mp}}$), and establish the Law of Equipartition of Energy across translational, rotational, and vibrational degrees of freedom.
1. Microscopic Derivation of Gas Pressure
Consider $N$ identical gas molecules, each of mass $m$, enclosed within a cubic vessel of side length $L$ and volume $V = L^3$. The velocity vector of the $i$-th molecule is $\vec{v}_i = v_{xi}\hat{i} + v_{yi}\hat{j} + v_{zi}\hat{k}$.
Under the core postulates of KTG (elastic impacts, negligible intermolecular potential energy, point particles):
1. Momentum Transfer: Upon colliding elastically with a planar wall perpendicular to the $x$-axis, the normal velocity reverses ($v_{xi} \to -v_{xi}$), giving a momentum change:
$$\Delta p_{xi} = -m v_{xi} – (m v_{xi}) = -2m v_{xi}$$
2. Collision Frequency: The time interval between successive collisions with the same face is $\Delta t = \frac{2L}{v_{xi}}$. Thus, the average force exerted on the wall by molecule $i$ is:
$$F_{xi} = \frac{|\Delta p_{xi}|}{\Delta t} = \frac{2m v_{xi}}{2L / v_{xi}} = \frac{m v_{xi}^2}{L}$$
3. Total Force and Pressure: Summing over all $N$ molecules, the net normal force is $F_x = \frac{m}{L} \sum_{i=1}^N v_{xi}^2$. The total pressure $P$ on area $A = L^2$ is:
$$P = \frac{F_x}{L^2} = \frac{m}{L^3} \sum_{i=1}^N v_{xi}^2 = \frac{m N}{V} \langle v_x^2 \rangle$$
By spatial isotropy of thermal motion, $\langle v_x^2 \rangle = \langle v_y^2 \rangle = \langle v_z^2 \rangle = \frac{1}{3}\langle v^2 \rangle = \frac{1}{3}v_{\text{rms}}^2$. Therefore, with total mass $M = N m$ and density $\rho = M/V$:
$$P = \frac{1}{3}\rho v_{\text{rms}}^2 = \frac{1}{3}\frac{N m}{V} v_{\text{rms}}^2$$
2. Molecular Speeds & Maxwell-Boltzmann Distribution
Combining $P V = \frac{1}{3} N m v_{\text{rms}}^2$ with the ideal gas law $P V = n R T = N k_B T$ (where $k_B = R/N_A$ is the Boltzmann constant and $M_{\text{mol}}$ is the molar mass), we obtain the key characteristic speeds:
1. Root-Mean-Square Speed ($v_{\text{rms}}$):
$$v_{\text{rms}} = \sqrt{\langle v^2 \rangle} = \sqrt{\frac{3 k_B T}{m}} = \sqrt{\frac{3 R T}{M_{\text{mol}}}}$$
2. Average / Mean Speed ($v_{\text{avg}}$): Derived from the first moment of the Maxwell-Boltzmann distribution:
$$v_{\text{avg}} = \langle v \rangle = \sqrt{\frac{8 k_B T}{\pi m}} = \sqrt{\frac{8 R T}{\pi M_{\text{mol}}}} \approx 0.921 \, v_{\text{rms}}$$
3. Most Probable Speed ($v_{\text{mp}}$): Peak of the Maxwell distribution function $\frac{dP(v)}{dv} = 0$:
$$v_{\text{mp}} = \sqrt{\frac{2 k_B T}{m}} = \sqrt{\frac{2 R T}{M_{\text{mol}}}} \approx 0.816 \, v_{\text{rms}}$$
Standard Ratio Hierarchy:
$$v_{\text{rms}} : v_{\text{avg}} : v_{\text{mp}} = \sqrt{3} : \sqrt{\frac{8}{\pi}} : \sqrt{2} \approx 1.224 : 1.128 : 1.000$$
Hence: $v_{\text{rms}} > v_{\text{avg}} > v_{\text{mp}}$.
3. Degrees of Freedom ($f$) & Law of Equipartition of Energy
The Law of Equipartition of Energy states that in thermal equilibrium at temperature $T$, every independent quadratic term in the system’s Hamiltonian energy function contributes an average energy of $\frac{1}{2}k_B T$ per molecule (or $\frac{1}{2}R T$ per mole).
For an ideal gas with $f$ active degrees of freedom:
- Internal Energy per mole: $U_m = \frac{f}{2} R T$
- Molar Heat Capacity at Constant Volume: $C_v = \left(\frac{\partial U_m}{\partial T}\right)_V = \frac{f}{2} R$
- Molar Heat Capacity at Constant Pressure: $C_p = C_v + R = \left(\frac{f}{2} + 1\right) R$
- Adiabatic Exponent: $\gamma = \frac{C_p}{C_v} = 1 + \frac{2}{f}$
| Gas Type | Translational | Rotational | Vibrational (High T) | Total ($f$) | $C_v$ | $C_p$ | $\gamma$ |
|---|---|---|---|---|---|---|---|
| Monoatomic (He, Ne, Ar) | 3 | 0 | 0 | 3 | $\frac{3}{2}R$ | $\frac{5}{2}R$ | $\frac{5}{3} \approx 1.67$ |
| Diatomic (Rigid) (O₂, N₂, H₂) | 3 | 2 | 0 | 5 | $\frac{5}{2}R$ | $\frac{7}{2}R$ | $\frac{7}{5} = 1.40$ |
| Diatomic (Non-rigid) | 3 | 2 | 2 (1 kinetic + 1 potential) | 7 | $\frac{7}{2}R$ | $\frac{9}{2}R$ | $\frac{9}{7} \approx 1.29$ |
| Non-linear Polyatomic (H₂O, NH₃) | 3 | 3 | $3N-6$ modes | 6 | $3R$ | $4R$ | $\frac{4}{3} \approx 1.33$ |
4. Solved Examples
Example 1 (Foundational – JEE Main):
At what temperature will the root-mean-square speed of hydrogen molecules ($M_{\text{mol}} = 2\text{ g/mol}$) equal the escape velocity from the Earth’s surface ($v_{\text{esc}} \approx 11.2\text{ km/s}$)? Take $R = 8.314\text{ J/(mol}\cdot\text{K)}$.
Solution:
Equating $v_{\text{rms}} = \sqrt{\frac{3 R T}{M_{\text{mol}}}}$ to $v_{\text{esc}}$:
$$v_{\text{rms}}^2 = \frac{3 R T}{M_{\text{mol}}} \implies T = \frac{M_{\text{mol}} v_{\text{esc}}^2}{3 R}$$
Converting to SI units: $M_{\text{mol}} = 2 \times 10^{-3}\text{ kg/mol}$, $v_{\text{esc}} = 1.12 \times 10^4\text{ m/s}$.
$$T = \frac{(2 \times 10^{-3}\text{ kg/mol})(1.12 \times 10^4\text{ m/s})^2}{3 \times 8.314\text{ J/(mol}\cdot\text{K)}} = \frac{2 \times 10^{-3} \times 1.2544 \times 10^8}{24.942} = \frac{2.5088 \times 10^5}{24.942} \approx \mathbf{10058\text{ K}} \approx \mathbf{1.01 \times 10^4\text{ K}}$$
Example 2 (Gas Mixture Thermodynamics – JEE Main/Advanced):
A gas mixture consists of $2\text{ moles}$ of helium (monoatomic) and $3\text{ moles}$ of oxygen (diatomic, treated as rigid rotors) at temperature $T$. Find (a) the equivalent molar heat capacity $C_{v,\text{mix}}$, and (b) the adiabatic exponent $\gamma_{\text{mix}}$ of the mixture.
Solution:
For helium: $n_1 = 2$, $f_1 = 3 \implies C_{v1} = \frac{3}{2}R$.
For oxygen: $n_2 = 3$, $f_2 = 5 \implies C_{v2} = \frac{5}{2}R$.
(a) Total internal energy $U = n_1 C_{v1} T + n_2 C_{v2} T = (n_1 + n_2) C_{v,\text{mix}} T$.
$$C_{v,\text{mix}} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} = \frac{2\left(\frac{3}{2}R\right) + 3\left(\frac{5}{2}R\right)}{2 + 3} = \frac{3R + 7.5R}{5} = \frac{10.5R}{5} = \mathbf{2.1 R} = \mathbf{\frac{21}{10}R}$$
(b) Molar heat capacity at constant pressure:
$$C_{p,\text{mix}} = C_{v,\text{mix}} + R = 2.1 R + R = 3.1 R$$
Adiabatic exponent:
$$\gamma_{\text{mix}} = \frac{C_{p,\text{mix}}}{C_{v,\text{mix}}} = \frac{3.1 R}{2.1 R} = \frac{31}{21} \approx \mathbf{1.476}$$
Example 3 (Calculus-Based Distribution – JEE Advanced):
A hypothetical two-dimensional gas of $N$ particles of mass $m$ is confined to an area $A$ in thermal equilibrium at temperature $T$. The number of molecules with speeds between $v$ and $v + dv$ is governed by the 2D Maxwell distribution:
$$dN(v) = C v e^{-\frac{m v^2}{2 k_B T}} dv$$
Determine: (a) the normalization constant $C$, (b) the most probable speed $v_{\text{mp}}$, and (c) the root-mean-square speed $v_{\text{rms}}$.
Solution:
(a) Normalizing over $v \in [0, \infty)$:
$$\int_0^\infty dN(v) = N \implies C \int_0^\infty v e^{-\frac{m v^2}{2 k_B T}} dv = N$$
Let $u = v^2 \implies du = 2v dv$. The integral becomes $\frac{1}{2} \int_0^\infty e^{-\frac{m u}{2 k_B T}} du = \frac{1}{2} \left(\frac{2 k_B T}{m}\right) = \frac{k_B T}{m}$.
$$C \left(\frac{k_B T}{m}\right) = N \implies C = \frac{N m}{k_B T}$$
(b) Most probable speed occurs at $\frac{d}{dv}(v e^{-\frac{m v^2}{2 k_B T}}) = 0$:
$$e^{-\frac{m v^2}{2 k_B T}} + v \left(-\frac{m v}{k_B T}\right) e^{-\frac{m v^2}{2 k_B T}} = 0 \implies 1 – \frac{m v^2}{k_B T} = 0 \implies v_{\text{mp}} = \sqrt{\frac{k_B T}{m}}$$
(c) Mean square speed:
$$\langle v^2 \rangle = \frac{1}{N} \int_0^\infty v^2 dN(v) = \frac{C}{N} \int_0^\infty v^3 e^{-\frac{m v^2}{2 k_B T}} dv = \frac{m}{k_B T} \int_0^\infty v^3 e^{-\frac{m v^2}{2 k_B T}} dv$$
Integrating by parts or standard gamma function gives $\langle v^2 \rangle = \frac{2 k_B T}{m}$.
$$v_{\text{rms}} = \sqrt{\langle v^2 \rangle} = \sqrt{\frac{2 k_B T}{m}}$$
Example 4 (Effusion & Kinetic Rates – JEE Advanced):
A vessel of volume $V$ contains an ideal gas at pressure $P_0$ and temperature $T$. Gas slowly effuses out through a pinhole of area $A$ into a vacuum. The rate of effusion of molecules is given by $\frac{dN}{dt} = -\frac{1}{4} n_{\text{vol}} v_{\text{avg}} A$, where $n_{\text{vol}} = \frac{N}{V}$. Assuming the temperature $T$ of the remaining gas is maintained strictly constant by thermal contact, find the pressure $P(t)$ inside the vessel as a function of time $t$.
Solution:
Since $T$ is constant, $v_{\text{avg}} = \sqrt{\frac{8 k_B T}{\pi m}}$ is independent of time.
Expressing effusion rate in terms of total molecule count $N(t)$:
$$\frac{dN}{dt} = -\frac{1}{4} \left(\frac{N(t)}{V}\right) v_{\text{avg}} A = -\left(\frac{A v_{\text{avg}}}{4V}\right) N(t)$$
Let the characteristic effusion time constant be $\tau = \frac{4V}{A v_{\text{avg}}}$.
$$\frac{dN}{N} = -\frac{dt}{\tau} \implies N(t) = N_0 e^{-t/\tau}$$
Since $P(t) V = N(t) k_B T$, pressure is directly proportional to molecule count $N(t)$ at constant $T$ and $V$:
$$P(t) = P_0 e^{-t/\tau} = P_0 \exp\left(-\frac{A \sqrt{8 k_B T / (\pi m)}}{4V} t\right) = \mathbf{P_0 \exp\left(-\frac{A t}{2V} \sqrt{\frac{2 k_B T}{\pi m}}\right)}$$
5. 10-Question Practice Worksheet (DPP)
Q1 (Single Choice): If the absolute temperature of a sample of gas is doubled, the root-mean-square speed of its molecules increases by a factor of:
(A) $2$
(B) $\sqrt{2}$
(C) $4$
(D) $\frac{1}{\sqrt{2}}$
Q2 (Single Choice): Four gas molecules have speeds $2.0\text{ km/s}$, $3.0\text{ km/s}$, $4.0\text{ km/s}$, and $5.0\text{ km/s}$. The ratio of their root-mean-square speed to their average speed ($v_{\text{rms}} / v_{\text{avg}}$) is:
(A) $1.000$
(B) $\frac{\sqrt{54}}{7} \approx 1.050$
(C) $\frac{\sqrt{50}}{7} \approx 1.010$
(D) $\frac{7}{\sqrt{54}}$
Q3 (Single Choice): A vessel contains equal masses of hydrogen ($M_{\text{mol}} = 2\text{ g/mol}$) and helium ($M_{\text{mol}} = 4\text{ g/mol}$) at the same temperature. The ratio of the average translational kinetic energy per molecule of hydrogen to that of helium is:
(A) $1 : 1$
(B) $1 : 2$
(C) $2 : 1$
(D) $\sqrt{2} : 1$
Q4 (Single Choice): The ratio of molar heat capacities $\gamma = C_p / C_v$ for a non-linear triatomic gas containing rigid molecules (ignoring vibrational modes) is:
(A) $1.40$
(B) $1.67$
(C) $1.33$
(D) $1.28$
Q5 (Single Choice): An ideal gas enclosed in a rigid container has initial pressure $P_0$. If the kinetic energy of all molecules is suddenly reduced to one-fourth of its initial value, the new pressure of the gas becomes:
(A) $4 P_0$
(B) $2 P_0$
(C) $P_0 / 2$
(D) $P_0 / 4$
Q6 (Single Choice): In a mixture of $1\text{ mole}$ of He and $1\text{ mole}$ of Ne, the total internal energy of the mixture at absolute temperature $T$ is:
(A) $3 R T$
(B) $\frac{5}{2} R T$
(C) $2 R T$
(D) $\frac{7}{2} R T$
Q7 (One or More than One Correct): Which of the following statements regarding the Maxwell-Boltzmann molecular speed distribution are strictly TRUE?
(A) The peak of the distribution shifts toward higher speeds as temperature increases.
(B) The area under the distribution curve $\int_0^\infty P(v) dv$ equals unity and is temperature-independent.
(C) For any ideal gas sample, $v_{\text{rms}} > v_{\text{avg}} > v_{\text{mp}}$.
(D) The fraction of molecules having speeds near $v = 0$ is exactly zero at all finite temperatures.
Q8 (One or More than One Correct): One mole of a rigid diatomic gas is heated so that its temperature rises by $\Delta T$. Which of the following quantities are independent of whether the process is isobaric, isochoric, or polytropic?
(A) Work done by the gas ($W$)
(B) Change in internal energy ($\Delta U$)
(C) Heat supplied to the gas ($Q$)
(D) Change in translational kinetic energy ($\Delta K_{\text{trans}}$)
Q9 (Numerical Value): An insulated container contains a mixture of $1\text{ mole}$ of oxygen ($O_2$, rigid diatomic) and $2\text{ moles}$ of argon ($Ar$, monoatomic) at absolute temperature $T$. The equivalent adiabatic exponent $\gamma_{\text{mix}}$ of the mixture can be written as a fraction $\frac{p}{q}$ in lowest terms. The value of $(p + q)$ is ________.
Q10 (Numerical Value): A container of volume $0.05\text{ m}^3$ contains an ideal gas at a pressure of $1.2 \times 10^5\text{ N/m}^2$. The total translational kinetic energy of all the gas molecules in the container in Joules is ________.
6. Answer Key & Detailed Step-by-Step Solutions
Answer Key:
Q1: (B) | Q2: (B) | Q3: (A) | Q4: (C) | Q5: (D) | Q6: (A) | Q7: (A, B, C, D) | Q8: (B, D) | Q9: 26 | Q10: 9000
Detailed Solutions
Solution 1:
Root-mean-square speed is given by $v_{\text{rms}} = \sqrt{\frac{3 R T}{M_{\text{mol}}}} \propto \sqrt{T}$.
When $T \to 2T$, $v_{\text{rms}}’ = \sqrt{2} v_{\text{rms}}$. Hence, (B) is correct.
Solution 2:
Given speeds: $v_1 = 2, v_2 = 3, v_3 = 4, v_4 = 5$ (in km/s).
1. Average speed: $v_{\text{avg}} = \frac{2 + 3 + 4 + 5}{4} = \frac{14}{4} = 3.5\text{ km/s} = \frac{7}{2}\text{ km/s}$.
2. Mean square speed: $\langle v^2 \rangle = \frac{2^2 + 3^2 + 4^2 + 5^2}{4} = \frac{4 + 9 + 16 + 25}{4} = \frac{54}{4} = 13.5\text{ (km/s)}^2$.
$$v_{\text{rms}} = \sqrt{13.5} = \frac{\sqrt{54}}{2}\text{ km/s}$$
$$\frac{v_{\text{rms}}}{v_{\text{avg}}} = \frac{\sqrt{54} / 2}{7 / 2} = \frac{\sqrt{54}}{7} \approx \frac{7.348}{7} \approx 1.050$$
Hence, (B) is correct.
Solution 3:
According to the Law of Equipartition of Energy, the average translational kinetic energy per molecule is $\langle K_{\text{trans}} \rangle = \frac{3}{2} k_B T$. Since both gases are at identical temperature $T$, this quantity depends strictly on temperature and is completely independent of the mass or molar mass of the molecule. Therefore, the ratio is $1 : 1$. Hence, (A) is correct.
Solution 4:
A non-linear triatomic molecule has 3 translational and 3 rotational degrees of freedom, giving $f = 3 + 3 = 6$.
Adiabatic exponent: $\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{6} = 1 + \frac{1}{3} = \frac{4}{3} \approx 1.33$. Hence, (C) is correct.
Solution 5:
From the kinetic theory pressure relation, $P = \frac{2}{3} \frac{E_{\text{trans}}}{V}$, where $E_{\text{trans}}$ is total translational kinetic energy.
For a rigid container, volume $V = \text{constant}$.
If $E_{\text{trans}} \to \frac{E_{\text{trans}}}{4}$, the pressure becomes $P = \frac{P_0}{4}$. Hence, (D) is correct.
Solution 6:
Both He and Ne are monoatomic gases with $f = 3$ degrees of freedom.
Internal energy for $n_1 = 1\text{ mole}$ of He: $U_1 = 1 \times \frac{3}{2} R T = \frac{3}{2} R T$.
Internal energy for $n_2 = 1\text{ mole}$ of Ne: $U_2 = 1 \times \frac{3}{2} R T = \frac{3}{2} R T$.
Total internal energy: $U = U_1 + U_2 = \frac{3}{2} R T + \frac{3}{2} R T = 3 R T$. Hence, (A) is correct.
Solution 7:
(A) True: $v_{\text{mp}} = \sqrt{2 k_B T / m} \propto \sqrt{T}$, so the distribution peak shifts rightward as $T$ increases.
(B) True: Total probability over all speeds is normalized to 1, regardless of temperature.
(C) True: $\sqrt{3} > \sqrt{8/\pi} > \sqrt{2} \implies 1.224 > 1.128 > 1.000$.
(D) True: Since $P(v) \propto v^2 e^{-m v^2 / 2 k_B T}$, as $v \to 0$, $P(v) \to 0$.
Hence, all statements (A, B, C, D) are correct.
Solution 8:
Internal energy $U$ and translational kinetic energy $K_{\text{trans}}$ are state functions depending solely on temperature change $\Delta T$:
$\Delta U = n C_v \Delta T = 1 \times \left(\frac{5}{2}R\right) \Delta T$
$\Delta K_{\text{trans}} = n \left(\frac{3}{2}R\right) \Delta T$
In contrast, work done $W$ and heat supplied $Q$ are path-dependent quantities that vary with the thermodynamic process. Hence, (B, D) are correct.
Solution 9:
For $O_2$ (rigid diatomic): $n_1 = 1$, $C_{v1} = \frac{5}{2}R$, $C_{p1} = \frac{7}{2}R$.
For $Ar$ (monoatomic): $n_2 = 2$, $C_{v2} = \frac{3}{2}R$, $C_{p2} = \frac{5}{2}R$.
Equivalent $C_{v,\text{mix}}$:
$$C_{v,\text{mix}} = \frac{1\left(\frac{5}{2}R\right) + 2\left(\frac{3}{2}R\right)}{1 + 2} = \frac{2.5 R + 3.0 R}{3} = \frac{5.5 R}{3} = \frac{11}{6} R$$
Equivalent $C_{p,\text{mix}} = C_{v,\text{mix}} + R = \frac{11}{6} R + R = \frac{17}{6} R$.
Adiabatic exponent:
$$\gamma_{\text{mix}} = \frac{C_{p,\text{mix}}}{C_{v,\text{mix}}} = \frac{17/6 R}{11/6 R} = \frac{17}{11}$$
Here $p = 17$ and $q = 11$, which are coprime. Thus, $p + q = 17 + 11 = \mathbf{28}$.
(Self-check: $17 + 11 = 28$).
Solution 10:
From the kinetic pressure relation:
$$P = \frac{2}{3} \frac{E_{\text{trans}}}{V} \implies E_{\text{trans}} = \frac{3}{2} P V$$
Substituting numerical values:
$$E_{\text{trans}} = \frac{3}{2} \times (1.2 \times 10^5\text{ N/m}^2) \times (0.05\text{ m}^3) = 1.5 \times 6000\text{ J} = \mathbf{9000\text{ J}}$$