Welcome to Lecture 104 of our comprehensive JEE Physics Master Series! Continuing in Unit 9: Kinetic Theory of Gases, we explore one of the most profound physical unifications in all of science: the Kinetic Interpretation of Temperature and the Root-Mean-Square (RMS) Speed of Gas Molecules. By equating the macroscopic ideal gas equation of state with the microscopic momentum transfer of chaotic molecular impacts, we establish that absolute temperature is nothing other than a direct measure of the average translational kinetic energy per molecule ($\overline{\epsilon}_{\text{trans}} = \frac{3}{2} k_B T$). We derive the master expressions for root-mean-square speed ($v_{\text{rms}}$), compare them with most probable ($v_{\text{mp}}$) and average ($\overline{v}$) speeds, explain why terrestrial and planetary atmospheres retain or lose specific gases based on gravitational escape velocity, analyze the thermal mixing of ideal gases, and solve standard JEE Advanced problem archetypes.
Detailed Concept Card: Kinetic Temperature & Molecular Speeds
1. Microscopic Interpretation of Temperature
From the macroscopic ideal gas equation of state:
$$P V = n R T = N k_B T$$
From the kinetic theory derivation of gas pressure for $N$ molecules of mass $m$ in volume $V$:
$$P = \frac{1}{3} \frac{m N}{V} v_{\text{rms}}^2 \implies P V = \frac{1}{3} m N v_{\text{rms}}^2 = \frac{2}{3} N \left(\frac{1}{2} m v_{\text{rms}}^2\right) = \frac{2}{3} N \overline{\epsilon}_{\text{trans}}$$
Equating both expressions for $PV$:
$$\frac{2}{3} N \overline{\epsilon}_{\text{trans}} = N k_B T \implies \mathbf{\overline{\epsilon}_{\text{trans}} = \frac{1}{2} m v_{\text{rms}}^2 = \frac{3}{2} k_B T}$$
The Core Thermodynamic Principle:
The absolute thermodynamic temperature $T$ of an ideal gas is directly proportional to the average translational kinetic energy of its molecules:
$$\mathbf{T = \frac{2}{3 k_B} \overline{\epsilon}_{\text{trans}}}$$
– Universality: The mean translational kinetic energy per molecule depends solely upon absolute temperature $T$. It is completely independent of molecular mass, chemical identity, gas atomicity, pressure, or volume.
– Absolute Zero ($T = 0\text{ K}$): The theoretical temperature at which all classical translational molecular motion ceases entirely ($v_{\text{rms}} = 0$).
2. Kinetic Energy Scales: Molecular, Molar, and Specific
Translational kinetic energy can be expressed across three standard physical scales:
| Scale | Quantity | Formula | Key Feature |
|---|---|---|---|
| Per Molecule | Average Translational KE ($\overline{\epsilon}$) | $$\overline{\epsilon} = \frac{3}{2} k_B T$$ | Identical for all gases at the same temperature (e.g., $\text{He}$ vs. $\text{UF}_6$). |
| Per Mole | Molar Translational KE ($E_{\text{molar}}$) | $$E_{\text{molar}} = N_A \left(\frac{3}{2} k_B T\right) = \frac{3}{2} R T$$ | Universal constant for 1 mole of any ideal gas ($\approx 3.74\text{ kJ/mol}$ at $300\text{ K}$). |
| Per Unit Mass (Specific) | Specific Translational KE ($e$) | $$e = \frac{E_{\text{molar}}}{M} = \frac{3}{2} \left(\frac{R}{M}\right) T = \frac{3}{2} r T$$ | Inversely proportional to molar mass $M$ ($e \propto 1/M$). Lighter gases possess far higher specific kinetic energy. |
3. Root-Mean-Square Speed ($v_{\text{rms}}$) Master Formulas
The root-mean-square speed is defined as the square root of the mean of the squares of molecular speeds:
$$v_{\text{rms}} = \sqrt{\overline{v^2}} = \sqrt{\frac{1}{N}\sum_{i=1}^N v_i^2}$$
Three equivalent formulations express $v_{\text{rms}}$ across microscopic, macroscopic, and hydrostatic variables:
$$\mathbf{v_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}} = \sqrt{\frac{3 R T}{M}} = \sqrt{\frac{3 P}{\rho}}}$$
- Temperature Dependence: $v_{\text{rms}} \propto \sqrt{T}$ (where $T$ is in Kelvin). To double the RMS speed, the absolute temperature must be quadrupled ($T \to 4T$).
- Molar Mass Dependence: At fixed temperature, $v_{\text{rms}} \propto \frac{1}{\sqrt{M}}$. Lighter molecules travel substantially faster than heavier molecules (e.g., $v_{\text{rms}}(\text{H}_2) = 4 \, v_{\text{rms}}(\text{O}_2)$ at the same temperature).
- Pressure Independence at Constant Temperature: If gas pressure is doubled at constant temperature, density doubles in direct proportion ($\rho \propto P$). The ratio $P/\rho = \frac{RT}{M}$ remains strictly constant, leaving $v_{\text{rms}}$ unchanged!
4. The Hierarchy of Molecular Speeds (Maxwell-Boltzmann Preview)
In a gas in thermal equilibrium, molecular speeds are distributed according to the Maxwell-Boltzmann distribution. Three distinct characteristic speeds describe the population:
- Most Probable Speed ($v_{\text{mp}}$): The speed possessed by the maximum number of molecules (peak of the distribution curve): $$v_{\text{mp}} = \sqrt{\frac{2 R T}{M}} = \sqrt{\frac{2 k_B T}{m}} \approx 1.414 \sqrt{\frac{R T}{M}}$$
- Average / Mean Speed ($\overline{v}$): The simple arithmetic mean of the speeds of all molecules: $$\overline{v} = \sqrt{\frac{8 R T}{\pi M}} = \sqrt{\frac{8 k_B T}{\pi m}} \approx 1.596 \sqrt{\frac{R T}{M}}$$
- Root-Mean-Square Speed ($v_{\text{rms}}$): Associated with the average kinetic energy: $$v_{\text{rms}} = \sqrt{\frac{3 R T}{M}} = \sqrt{\frac{3 k_B T}{m}} \approx 1.732 \sqrt{\frac{R T}{M}}$$
Universal Speed Hierarchy:
For any gas at any temperature:
$$\mathbf{v_{\text{mp}} < \overline{v} < v_{\text{rms}}}$$
$$\mathbf{v_{\text{mp}} : \overline{v} : v_{\text{rms}} = \sqrt{2} : \sqrt{\frac{8}{\pi}} : \sqrt{3} \approx 1 : 1.128 : 1.225}$$
$$v_{\text{mp}} \approx 0.816 \, v_{\text{rms}}, \quad \overline{v} \approx 0.921 \, v_{\text{rms}}$$
5. Planetary Atmosphere Retention & The Escape Velocity Criterion
For a celestial body to permanently retain an atmosphere of a given gas over astronomical timescales (billions of years), the thermal RMS speed of the gas molecules at the exosphere temperature must be substantially less than the gravitational escape velocity ($v_{\text{escape}} = \sqrt{2 g R}$):
$$\mathbf{v_{\text{rms}} \le \frac{1}{5} v_{\text{escape}} \quad \text{to} \quad \frac{1}{6} v_{\text{escape}}}$$
- If $v_{\text{rms}} > \frac{1}{5} v_{\text{escape}}$, a significant fraction of molecules in the high-energy Maxwellian tail possess speeds exceeding $v_{\text{escape}}$ and leak into space, completely depleting the atmosphere over geological time (Jeans escape).
- The Moon: With $v_{\text{escape}} \approx 2.38\text{ km/s}$ and daytime temperatures reaching $400\text{ K}$, $v_{\text{rms}}$ for $\text{O}_2$ and $\text{N}_2$ ($\sim 0.6\text{ km/s}$) exceeds $\frac{1}{5} v_{\text{escape}}$, explaining why the Moon is completely devoid of an atmosphere.
- Earth: With $v_{\text{escape}} \approx 11.2\text{ km/s}$, $\frac{1}{5} v_{\text{escape}} \approx 2.24\text{ km/s}$. For $\text{N}_2$ and $\text{O}_2$ at $300\text{ K}$, $v_{\text{rms}} \approx 0.5\text{ km/s} \ll 2.24\text{ km/s}$, so they are permanently retained. However, for $\text{H}_2$ ($v_{\text{rms}} \approx 1.93\text{ km/s}$) and $\text{He}$ ($v_{\text{rms}} \approx 1.37\text{ km/s}$), speeds easily exceed the threshold in the upper exosphere ($T \sim 1000\text{ K}$), explaining why free Hydrogen and Helium have escaped Earth’s atmosphere.
- Jupiter: With $v_{\text{escape}} \approx 60\text{ km/s}$ and low temperatures ($T \sim 150\text{ K}$), even light gases like $\text{H}_2$ and $\text{He}$ are firmly trapped.
Four Graded Solved Examples
Example 1 (JEE Main Level: Comprehensive Speed Calculations for Oxygen & Hydrogen)
Problem Statement: Given the universal gas constant $R = 8.314\text{ J/(mol}\cdot\text{K)}$:
1. Calculate the root-mean-square speed $v_{\text{rms}}$, average speed $\overline{v}$, and most probable speed $v_{\text{mp}}$ for Oxygen gas ($\text{O}_2$, molar mass $M = 32.0\text{ g/mol} = 0.032\text{ kg/mol}$) at temperature $T = 300\text{ K}$.
2. At what temperature will the root-mean-square speed of Hydrogen gas ($\text{H}_2$, molar mass $M = 2.0\text{ g/mol} = 0.002\text{ kg/mol}$) equal the root-mean-square speed of Oxygen gas at $300\text{ K}$?
3. At what temperature will the root-mean-square speed of Oxygen gas be doubled from its value at $300\text{ K}$?
Solution:
Step 1: Calculate Characteristic Speeds for $\text{O}_2$ at $300\text{ K}$.
Root-mean-square speed:
$$v_{\text{rms}} = \sqrt{\frac{3 R T}{M}} = \sqrt{\frac{3 \times 8.314 \times 300}{0.032}} = \sqrt{\frac{7482.6}{0.032}} = \sqrt{233831.25} \approx \mathbf{483.56\text{ m/s}}$$
Average speed:
$$\overline{v} = \sqrt{\frac{8 R T}{\pi M}} = \sqrt{\frac{8 \times 8.314 \times 300}{\pi \times 0.032}} = \sqrt{\frac{19953.6}{0.10053}} = \sqrt{198482.5} \approx \mathbf{445.51\text{ m/s}}$$
Most probable speed:
$$v_{\text{mp}} = \sqrt{\frac{2 R T}{M}} = \sqrt{\frac{2 \times 8.314 \times 300}{0.032}} = \sqrt{\frac{4988.4}{0.032}} = \sqrt{155887.5} \approx \mathbf{394.83\text{ m/s}}$$
Notice that $v_{\text{mp}} (394.8\text{ m/s}) < \overline{v} (445.5\text{ m/s}) < v_{\text{rms}} (483.6\text{ m/s})$, fully confirming the universal speed hierarchy.
Step 2: Temperature at which $v_{\text{rms}}(\text{H}_2) = v_{\text{rms}}(\text{O}_2 \text{ at } 300\text{ K})$.
Equating speeds:
$$\sqrt{\frac{3 R T_{\text{H}_2}}{M_{\text{H}_2}}} = \sqrt{\frac{3 R (300)}{M_{\text{O}_2}}} \implies \frac{T_{\text{H}_2}}{M_{\text{H}_2}} = \frac{300}{M_{\text{O}_2}}$$
$$T_{\text{H}_2} = 300 \times \left(\frac{M_{\text{H}_2}}{M_{\text{O}_2}}\right) = 300 \times \left(\frac{2.0}{32.0}\right) = \frac{300}{16} = \mathbf{18.75\text{ K}}$$
Because Hydrogen is $16$ times lighter than Oxygen, it requires a temperature of only $18.75\text{ K}$ (near liquid hydrogen temperatures) to match the thermal velocity of room-temperature Oxygen.
Step 3: Temperature for Doubling $v_{\text{rms}}(\text{O}_2)$.
Since $v_{\text{rms}} \propto \sqrt{T}$:
$$\frac{v_{\text{rms}}’}{v_{\text{rms}}} = \sqrt{\frac{T’}{T}} = 2 \implies \frac{T’}{T} = 4 \implies T’ = 4 \times 300\text{ K} = \mathbf{1200\text{ K}} = 926.85^\circ\text{C}$$
Example 2 (Planetary Atmosphere Retention & Lunar Escape Analysis)
Problem Statement: The Moon has a surface gravitational acceleration $g = 1.62\text{ m/s}^2$ and radius $R_M = 1.74 \times 10^6\text{ m}$. During lunar daytime, the surface temperature reaches $T_{\text{max}} = 400\text{ K}$. Given $R = 8.314\text{ J/(mol}\cdot\text{K)}$:
1. Calculate the gravitational escape velocity $v_{\text{esc}}$ from the Moon’s surface (in $\text{km/s}$).
2. Calculate the root-mean-square speeds of Hydrogen molecules ($\text{H}_2, M = 2.0\text{ g/mol}$) and Oxygen molecules ($\text{O}_2, M = 32.0\text{ g/mol}$) at $400\text{ K}$.
3. Applying the Jeans escape criterion ($v_{\text{rms}} > \frac{1}{5} v_{\text{esc}}$), explain why the Moon cannot retain an atmosphere of either gas.
Solution:
Step 1: Calculate Lunar Escape Velocity.
$$v_{\text{esc}} = \sqrt{2 g R_M} = \sqrt{2 \times (1.62\text{ m/s}^2) \times (1.74 \times 10^6\text{ m})} = \sqrt{5.6376 \times 10^6} \approx \mathbf{2374.4\text{ m/s} \approx 2.37\text{ km/s}}$$
The retention threshold is:
$$v_{\text{threshold}} = \frac{1}{5} v_{\text{esc}} = \frac{2374.4}{5} \approx \mathbf{474.9\text{ m/s}}$$
Step 2: Calculate Molecular Speeds at $400\text{ K}$.
For Hydrogen molecules ($\text{H}_2, M = 0.002\text{ kg/mol}$):
$$v_{\text{rms, H}_2} = \sqrt{\frac{3 \times 8.314 \times 400}{0.002}} = \sqrt{4988400} \approx \mathbf{2233.5\text{ m/s} \approx 2.23\text{ km/s}}$$
For Oxygen molecules ($\text{O}_2, M = 0.032\text{ kg/mol}$):
$$v_{\text{rms, O}_2} = \sqrt{\frac{3 \times 8.314 \times 400}{0.032}} = \sqrt{311775} \approx \mathbf{558.37\text{ m/s} \approx 0.56\text{ km/s}}$$
Step 3: Apply the Jeans Escape Criterion.
– For Hydrogen: $v_{\text{rms}} = 2233.5\text{ m/s}$ is almost equal to the escape velocity itself ($v_{\text{esc}} = 2374.4\text{ m/s}$). Hydrogen escapes almost immediately.
– For Oxygen: $v_{\text{rms}} = 558.4\text{ m/s} > v_{\text{threshold}} = 474.9\text{ m/s}$. Even for Oxygen, thermal agitation at $400\text{ K}$ elevates the molecular speed beyond one-fifth of escape velocity, causing the high-speed Maxwellian tail to evaporate into space over geological time.
Hence, the Moon is completely incapable of retaining any atmosphere.
Example 3 (JEE Advanced Level: Thermal Mixing of Two Ideal Gases & RMS Velocity Shifts)
Problem Statement: An insulated container is divided into two compartments by a thermal partition:
– Compartment 1 contains $n_1 = 2.0\text{ moles}$ of Argon gas ($\text{Ar}$, monoatomic, $M_1 = 40.0\text{ g/mol}$) at temperature $T_1 = 400\text{ K}$.
– Compartment 2 contains $n_2 = 3.0\text{ moles}$ of Helium gas ($\text{He}$, monoatomic, $M_2 = 4.0\text{ g/mol}$) at temperature $T_2 = 300\text{ K}$.
The partition is removed, allowing the gases to mix thoroughly and reach thermal equilibrium.
1. Determine the final equilibrium temperature $T_f$ of the mixture.
2. Calculate the ratio of the final mean translational kinetic energy of an Argon molecule to that of a Helium molecule ($\overline{\epsilon}_{\text{Ar}} / \overline{\epsilon}_{\text{He}}$).
3. Find the percentage change in the root-mean-square speed of the Argon molecules and Helium molecules as a result of the mixing.
Solution:
Step 1: Calculate Final Equilibrium Temperature $T_f$.
Because the container is insulated and no external work is performed ($Q = 0, W = 0$), total internal energy is conserved ($\Delta U = 0$).
For monoatomic gases ($C_v = \frac{3}{2} R$):
$$n_1 C_v (T_f – T_1) + n_2 C_v (T_f – T_2) = 0 \implies (n_1 + n_2) T_f = n_1 T_1 + n_2 T_2$$
$$T_f = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2} = \frac{(2.0)(400\text{ K}) + (3.0)(300\text{ K})}{2.0 + 3.0} = \frac{800 + 900}{5.0} = \frac{1700}{5.0} = \mathbf{340\text{ K}}$$
Step 2: Ratio of Final Mean Translational Kinetic Energies per Molecule.
At thermal equilibrium, both gases share the same temperature $T_f = 340\text{ K}$. Since $\overline{\epsilon}_{\text{trans}} = \frac{3}{2} k_B T$:
$$\frac{\overline{\epsilon}_{\text{Ar}}}{\overline{\epsilon}_{\text{He}}} = \frac{\frac{3}{2} k_B T_f}{\frac{3}{2} k_B T_f} = \mathbf{1}$$
The average translational kinetic energy per molecule is strictly equal for both gases.
Step 3: Percentage Change in Root-Mean-Square Speeds.
Since $v_{\text{rms}} \propto \sqrt{T}$:
– For Argon (cooled from $400\text{ K}$ to $340\text{ K}$):
$$\frac{v_{f, \text{Ar}}}{v_{i, \text{Ar}}} = \sqrt{\frac{T_f}{T_1}} = \sqrt{\frac{340}{400}} = \sqrt{0.85} \approx 0.9220$$
$$\text{Percentage Change for Ar} = (0.9220 – 1) \times 100\% = \mathbf{-7.80\%} \quad (\text{Speed decreases by } 7.8\%)$$
– For Helium (heated from $300\text{ K}$ to $340\text{ K}$):
$$\frac{v_{f, \text{He}}}{v_{i, \text{He}}} = \sqrt{\frac{T_f}{T_2}} = \sqrt{\frac{340}{300}} = \sqrt{\frac{17}{15}} \approx \sqrt{1.1333} \approx 1.0646$$
$$\text{Percentage Change for He} = (1.0646 – 1) \times 100\% = \mathbf{+6.46\%} \quad (\text{Speed increases by } 6.5\%)$$
Example 4 (JEE Advanced Level: Microscopic Mechanism of Compressive Heating by a Moving Piston)
Problem Statement: An ideal gas is enclosed in a cylinder with a frictionless piston moving inward with a constant small velocity $u$ ($u \ll v_{\text{rms}}$).
1. Consider a gas molecule moving with normal velocity component $v_x$ toward the advancing piston. Show that upon elastic collision with the piston, the molecule’s velocity component increases by $\Delta v_x = 2u$.
2. Determine the kinetic energy gained by the molecule during this single collision.
3. Summing over all collisions occurring in time $dt$, prove that the microscopic rate of kinetic energy gain by the gas molecules exactly equals the macroscopic rate of boundary work $P \left|\frac{dV}{dt}\right|$ performed on the gas.
Solution:
Step 1: Collision in the Frame of the Moving Piston.
Let the piston move in the $-x$ direction with speed $u$. In the reference frame of the piston:
– The incoming velocity of the molecule is $v_{x, \text{rel}} = v_x – (-u) = v_x + u$.
– Because the collision with the massive piston is perfectly elastic, the molecule reflects with reversed relative velocity:
$$v_{x, \text{rel}}’ = -(v_x + u)$$
Transforming back to the stationary laboratory frame:
$$v_x’ = v_{x, \text{rel}}’ + v_{\text{piston}} = -(v_x + u) – u = -(v_x + 2u)$$
The magnitude of the normal speed after rebound has increased from $v_x$ to $v_x + 2u$. Thus:
$$\mathbf{\Delta v_x = 2u}$$
Step 2: Kinetic Energy Gained per Collision.
The change in kinetic energy of the molecule is:
$$\Delta K = \frac{1}{2} m (v_x + 2u)^2 – \frac{1}{2} m v_x^2 = \frac{1}{2} m (v_x^2 + 4 v_x u + 4u^2) – \frac{1}{2} m v_x^2 = 2 m v_x u + 2 m u^2$$
Since $u \ll v_x$, the second-order term $2 m u^2$ is negligible:
$$\mathbf{\Delta K \approx 2 m v_x u}$$
Step 3: Total Rate of Kinetic Energy Increase.
The number of molecules with normal speed $v_x$ striking the piston of area $A$ in time $dt$ is:
$$dN = n_{0, x} A (v_x + u) dt \approx n_{0, x} A v_x dt$$
The total energy transferred to these molecules in time $dt$ is:
$$dE = dN \times \Delta K = (n_{0, x} A v_x dt)(2 m v_x u) = 2 n_{0, x} m v_x^2 A u \, dt$$
Summing over all molecules moving toward the piston:
$$\sum 2 n_{0, x} m v_x^2 = n_0 m \overline{v_x^2} = \frac{1}{3} n_0 m v_{\text{rms}}^2 = P$$
Therefore:
$$dE = P A u \, dt$$
Notice that in time $dt$, the piston moves inward by displacement $dx = u \, dt$, decreasing the container volume by $|dV| = A \, dx = A u \, dt$. Thus:
$$\mathbf{dE = P |dV|}$$
This rigorously proves that the macroscopic boundary work done ON a gas during compression ($dW = P |dV|$) is the exact macroscopic accumulation of microscopic velocity boosts ($\Delta v_x = 2u$) imparted by the advancing wall to rebounding gas molecules!
Practice Worksheet: 10 JEE Problems
Questions 1 to 5: JEE Main Level (Single Correct)
Q1. The average translational kinetic energy of a molecule of an ideal gas at temperature $T$ depends upon:
(A) Only the absolute temperature $T$ of the gas.
(B) Both the absolute temperature $T$ and the molar mass $M$ of the gas.
(C) The pressure $P$ and volume $V$ of the gas.
(D) The atomicity of the gas molecules.
Q2. For an ideal gas at a given temperature, the ratio of the most probable speed ($v_{\text{mp}}$), average speed ($\overline{v}$), and root-mean-square speed ($v_{\text{rms}}$) is:
(A) $1 : 1.225 : 1.128$
(B) $\sqrt{2} : \sqrt{\frac{8}{\pi}} : \sqrt{3}$
(C) $\sqrt{3} : \sqrt{\frac{8}{\pi}} : \sqrt{2}$
(D) $1 : 1 : 1$
Q3. At what temperature will the root-mean-square speed of Oxygen molecules ($\text{O}_2$) be double its value at $27^\circ\text{C}$?
(A) $54^\circ\text{C}$
(B) $327^\circ\text{C}$
(C) $927^\circ\text{C}$
(D) $1200^\circ\text{C}$
Q4. If the pressure of an ideal gas enclosed in a vessel is doubled while keeping its temperature strictly constant, the root-mean-square speed of its molecules will:
(A) Double ($2\times$)
(B) Increase by $\sqrt{2}$ times
(C) Remain completely unchanged
(D) Halve ($0.5\times$)
Q5. At the same temperature, the ratio of the root-mean-square speed of Hydrogen gas molecules ($\text{H}_2, M = 2\text{ g/mol}$) to that of Oxygen gas molecules ($\text{O}_2, M = 32\text{ g/mol}$) is:
(A) $16$
(B) $4$
(C) $1/4$
(D) $1/16$
Questions 6 to 8: JEE Advanced Level (One or More Than One Correct)
Q6. Which of the following relationships correctly describe the characteristic molecular speeds of an ideal gas of molar mass $M$ at temperature $T$?
(A) $v_{\text{rms}} = \sqrt{\frac{3 R T}{M}}$
(B) $v_{\text{mp}} = \sqrt{\frac{2}{3}} v_{\text{rms}}$
(C) $\overline{v} = \sqrt{\frac{8}{3\pi}} v_{\text{rms}}$
(D) For any gas at any non-zero temperature, $v_{\text{mp}} < \overline{v} < v_{\text{rms}}$
Q7. Which of the following statements regarding planetary atmospheres and escape velocity is/are correct?
(A) A planet loses a gas over geological time if the root-mean-square speed of the gas molecules at exospheric temperatures satisfies $v_{\text{rms}} \gtrsim \frac{1}{5} v_{\text{escape}}$.
(B) Lighter gases escape a planet’s gravity more readily than heavier gases at the same atmospheric temperature.
(C) The Moon has no atmosphere primarily because its escape velocity ($2.38\text{ km/s}$) is comparable to thermal molecular speeds.
(D) Giant planets like Jupiter retain abundant light gases (Hydrogen and Helium) due to high escape velocity and low surface temperatures.
Q8. Two non-reacting ideal gases, Gas 1 ($n_1\text{ moles}$, molar mass $M_1$) at temperature $T_1$ and Gas 2 ($n_2\text{ moles}$, molar mass $M_2$) at temperature $T_2$, are mixed inside a thermally insulated container. Which of the following statements is/are correct?
(A) The final equilibrium temperature is $T_f = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2}$ (assuming equal translational degrees of freedom).
(B) Total translational kinetic energy is conserved during the mixing process.
(C) At thermal equilibrium, the average translational kinetic energy per molecule is identical for both gases ($\overline{\epsilon}_1 = \overline{\epsilon}_2 = \frac{3}{2} k_B T_f$).
(D) If $M_1 < M_2$, the root-mean-square speed of Gas 1 molecules in the mixture is greater than that of Gas 2 molecules.
Questions 9 to 10: Numerical / Integer Answer Type
Q9. At what temperature (in $\text{K}$) will the root-mean-square speed of Nitrogen gas molecules ($\text{N}_2, M = 28\text{ g/mol}$) be equal to the root-mean-square speed of Oxygen gas molecules ($\text{O}_2, M = 32\text{ g/mol}$) at $400\text{ K}$?
Q10. An insulated chamber contains a mixture of $n_1 = 1.0\text{ mole}$ of an ideal gas at $T_1 = 300\text{ K}$ and $n_2 = 3.0\text{ moles}$ of another ideal gas at $T_2 = 500\text{ K}$. Assuming both gases possess only translational degrees of freedom, determine the final equilibrium temperature of the mixture $T_f$ (in $\text{K}$).
Detailed Solutions & Answer Key
Answer Key
| Question | Correct Option / Answer | Topic / Concept |
|---|---|---|
| Q1 | (A) Only absolute temperature $T$ | Translational Kinetic Energy Independence |
| Q2 | (B) $\sqrt{2} : \sqrt{8/\pi} : \sqrt{3}$ | Hierarchy of Maxwellian Speeds |
| Q3 | (C) $927^\circ\text{C}$ | Temperature Scaling for Doubling $v_{\text{rms}}$ |
| Q4 | (C) Remain completely unchanged | Pressure Independence of $v_{\text{rms}}$ at Constant $T$ |
| Q5 | (B) $4$ | Ratio of RMS Speeds for $\text{H}_2$ and $\text{O}_2$ |
| Q6 | (A), (B), (C), (D) | Characteristic Molecular Speed Formulations |
| Q7 | (A), (B), (C), (D) | Jeans Escape & Planetary Atmospheres |
| Q8 | (A), (B), (C), (D) | Thermal Mixing & Microscopic Equipartition |
| Q9 | 350 | Equal RMS Speed Temperature ($400 \times 28/32\text{ K}$) |
| Q10 | 450 | Thermal Equilibrium Temperature ($(300 + 1500)/4\text{ K}$) |
Complete Step-by-Step Solutions
Solution 1:
The average translational kinetic energy per molecule of any ideal gas is:
$$\overline{\epsilon}_{\text{trans}} = \frac{3}{2} k_B T$$
This quantity depends strictly and exclusively on the absolute temperature $T$. It is completely independent of molecular mass, chemical identity, gas atomicity, or pressure.
Correct Option: (A)
Solution 2:
The three characteristic speeds are:
$$v_{\text{mp}} = \sqrt{\frac{2RT}{M}}, \quad \overline{v} = \sqrt{\frac{8RT}{\pi M}}, \quad v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$$
Taking their ratio:
$$v_{\text{mp}} : \overline{v} : v_{\text{rms}} = \sqrt{2} : \sqrt{\frac{8}{\pi}} : \sqrt{3} \approx 1.414 : 1.596 : 1.732$$
Correct Option: (B)
Solution 3:
Initial temperature in Kelvin:
$$T_1 = 27 + 273.15 \approx 300\text{ K}$$
Since $v_{\text{rms}} \propto \sqrt{T}$:
$$\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = 2 \implies \frac{T_2}{T_1} = 4 \implies T_2 = 4 \times 300\text{ K} = 1200\text{ K}$$
Converting back to Celsius:
$$t_2 = 1200 – 273 = \mathbf{927^\circ\text{C}}$$
Correct Option: (C)
Solution 4:
From $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$, the RMS speed depends strictly on temperature $T$ and molar mass $M$. While $v_{\text{rms}} = \sqrt{\frac{3P}{\rho}}$, doubling pressure at constant temperature doubles the density in exact proportion ($\rho \propto P$). Thus, the ratio $\frac{P}{\rho}$ is constant, and $v_{\text{rms}}$ remains completely unchanged.
Correct Option: (C)
Solution 5:
At the same temperature $T$:
$$v_{\text{rms}} \propto \frac{1}{\sqrt{M}}$$
$$\frac{v_{\text{rms}}(\text{H}_2)}{v_{\text{rms}}(\text{O}_2)} = \sqrt{\frac{M_{\text{O}_2}}{M_{\text{H}_2}}} = \sqrt{\frac{32\text{ g/mol}}{2\text{ g/mol}}} = \sqrt{16} = \mathbf{4}$$
Correct Option: (B)
Solution 6:
(A) Correct: Standard definition $v_{\text{rms}} = \sqrt{3RT/M}$.
(B) Correct: $\frac{v_{\text{mp}}}{v_{\text{rms}}} = \frac{\sqrt{2RT/M}}{\sqrt{3RT/M}} = \sqrt{\frac{2}{3}}$.
(C) Correct: $\frac{\overline{v}}{v_{\text{rms}}} = \frac{\sqrt{8RT/\pi M}}{\sqrt{3RT/M}} = \sqrt{\frac{8}{3\pi}}$.
(D) Correct: Since $\sqrt{2} \approx 1.414 < \sqrt{8/\pi} \approx 1.596 < \sqrt{3} \approx 1.732$, $v_{\text{mp}} < \overline{v} < v_{\text{rms}}$ universally.
Correct Options: (A), (B), (C), (D)
Solution 7:
(A) Correct: Jeans escape occurs when thermal speed exceeds roughly one-fifth of escape velocity ($v_{\text{rms}} \gtrsim 0.2 v_{\text{esc}}$).
(B) Correct: Since $v_{\text{rms}} \propto 1/\sqrt{M}$, lighter gases have higher thermal speeds and evaporate into space faster.
(C) Correct: Low lunar surface gravity yields a low escape velocity ($2.38\text{ km/s}$), easily exceeded by thermal molecular speeds at $400\text{ K}$.
(D) Correct: High escape velocity ($60\text{ km/s}$) and low temperature on Jupiter ensure $v_{\text{rms}} \ll v_{\text{esc}}$ even for Hydrogen.
Correct Options: (A), (B), (C), (D)
Solution 8:
(A) Correct: For isolated mixing, conservation of internal energy gives $T_f = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2}$.
(B) Correct: Since the chamber is insulated and rigid ($Q = 0, W = 0$), total internal translational kinetic energy is conserved.
(C) Correct: In thermal equilibrium, temperature is uniform, so $\overline{\epsilon}_{\text{trans}} = \frac{3}{2} k_B T_f$ is identical for every molecule.
(D) Correct: Because $v_{\text{rms}} = \sqrt{3RT_f/M}$, the gas with smaller molar mass has a higher RMS speed.
Correct Options: (A), (B), (C), (D)
Solution 9:
Equating the root-mean-square speeds of $\text{N}_2$ and $\text{O}_2$:
$$v_{\text{rms}}(\text{N}_2) = v_{\text{rms}}(\text{O}_2)$$
$$\sqrt{\frac{3 R T_{\text{N}_2}}{M_{\text{N}_2}}} = \sqrt{\frac{3 R T_{\text{O}_2}}{M_{\text{O}_2}}} \implies \frac{T_{\text{N}_2}}{M_{\text{N}_2}} = \frac{T_{\text{O}_2}}{M_{\text{O}_2}}$$
Substitute $M_{\text{N}_2} = 28\text{ g/mol}$, $M_{\text{O}_2} = 32\text{ g/mol}$, and $T_{\text{O}_2} = 400\text{ K}$:
$$T_{\text{N}_2} = T_{\text{O}_2} \times \left(\frac{M_{\text{N}_2}}{M_{\text{O}_2}}\right) = 400 \times \left(\frac{28}{32}\right) = 400 \times \frac{7}{8} = 50 \times 7 = \mathbf{350\text{ K}}$$
Answer: 350
Solution 10:
By conservation of internal energy for translational ideal gases:
$$n_1 \left(\frac{3}{2} R T_1\right) + n_2 \left(\frac{3}{2} R T_2\right) = (n_1 + n_2) \left(\frac{3}{2} R T_f\right)$$
$$T_f = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2}$$
Substitute $n_1 = 1.0\text{ mol}, T_1 = 300\text{ K}, n_2 = 3.0\text{ mol}, T_2 = 500\text{ K}$:
$$T_f = \frac{(1.0 \times 300) + (3.0 \times 500)}{1.0 + 3.0} = \frac{300 + 1500}{4.0} = \frac{1800}{4.0} = \mathbf{450\text{ K}}$$
Answer: 450