Kinetic Theory of Gases — Easy Practice (Set 1)

Quick Recap — Kinetic Theory Basics

• Ideal gas equation: $PV = nRT$ (or $PV = NkT$), with $R = 8.31$ J/(mol K) and $k = R/N_A \approx 1.38\times10^{-23}$ J/K.
• Gas laws: Boyle ($PV$ constant), Charles ($V/T$ constant), Gay-Lussac ($P/T$ constant), all at fixed amount of gas.
• Pressure of a gas: $P = \tfrac13\rho\langle v^2\rangle$.
• Temperature and energy: the average translational kinetic energy per molecule is $\tfrac32 kT$.
• RMS speed: $v_{rms} = \sqrt{\dfrac{3RT}{M}} = \sqrt{\dfrac{3kT}{m}} = \sqrt{\dfrac{3P}{\rho}}$, so $v_{rms}\propto\sqrt{T}$ and $\propto 1/\sqrt{M}$.

Beyond-NCERT JEE Formulae

1. Three molecular speeds (Maxwell distribution)

The rms, average and most probable speeds:

$$v_{rms}=\sqrt{\dfrac{3RT}{M}}=\sqrt{\dfrac{3P}{\rho}}=\sqrt{\dfrac{3kT}{m}},\qquad \bar v=\sqrt{\dfrac{8RT}{\pi M}},\qquad v_{mp}=\sqrt{\dfrac{2RT}{M}}.$$

Their fixed ratio is $v_{mp}:\bar v:v_{rms}=\sqrt2:\sqrt{8/\pi}:\sqrt3=1:1.128:1.225$ (that is, $1.41:1.60:1.73$).

When to use. Reach for $v_{rms}$ in energy and pressure work, since it alone reproduces $\tfrac12 m\overline{v^2}$; $v_{mp}$ marks the peak of the distribution curve; $\bar v$ drives effusion and collision counting.

[JEE Tip] Every speed shares one $\sqrt{T/M}$ skeleton, so only the constant under the root changes and the order $v_{mp}<\bar v<v_{rms}$ never flips. Any question of the form "speed A at some $T_1$ equals speed B at some $T_2$" therefore collapses, after squaring, to matching a single (constant)$\times\dfrac{T}{M}$ group.

2. Pressure as an energy density

$$P=\dfrac13\rho v_{rms}^2=\dfrac13\dfrac{mN}{V}v_{rms}^2=\dfrac13 n m v_{rms}^2\quad(n=N/V).$$

Grouping the kinetic energy $\tfrac12 m v_{rms}^2$ gives $P=\dfrac23\dfrac{N}{V}\left(\dfrac12 m v_{rms}^2\right)=\dfrac23 E_{trans}$, where $E_{trans}$ is the translational kinetic energy per unit volume.

[JEE Tip] The reading $P=\tfrac23\times$(translational KE density) is the shortcut behind every "find the total translational KE" question: that total is simply $\dfrac32 PV$, needing neither $T$ nor the amount nor the identity of the gas.

3. Degrees of freedom, equipartition and gamma

Energy per molecule is $\dfrac f2 kT$, since each quadratic term carries $\tfrac12 kT$, and hence

$$C_V=\dfrac f2 R,\qquad C_P=\left(\dfrac f2+1\right)R,\qquad \gamma=1+\dfrac2f.$$

Monatomic gases take $f=3$ with $\gamma=\tfrac53\approx1.67$; rigid diatomic $f=5$ with $\gamma=\tfrac75=1.40$; non-linear polyatomic $f=6$ with $\gamma=\tfrac43\approx1.33$. A fully excited vibration adds $2$ more, one kinetic and one potential term, lifting a diatomic to $f=7$ and $\gamma=\tfrac97\approx1.29$.

Mixtures (common $T$, no reaction) use mole-weighted averages: $C_{V,mix}=\dfrac{n_1C_{V1}+n_2C_{V2}}{n_1+n_2}$, $f_{mix}=\dfrac{n_1f_1+n_2f_2}{n_1+n_2}$, and $M_{eff}=\dfrac{n_1M_1+n_2M_2}{n_1+n_2}$.

[JEE Tip] Never average $\gamma$ itself. Because $\dfrac{n}{\gamma-1}=\dfrac{nf}{2}$ is nothing but a count of DOF tokens, the fast route is $\dfrac{n_1+n_2}{\gamma_{mix}-1}=\dfrac{n_1}{\gamma_1-1}+\dfrac{n_2}{\gamma_2-1}$: add the tokens on the right, then divide the total moles by their sum.

4. Internal energy in three equal forms

$$U=\dfrac f2 nRT=\dfrac{nRT}{\gamma-1}=\dfrac{PV}{\gamma-1},$$

and the translational share is always $\dfrac32 nRT=\dfrac32 PV$.

[JEE Tip] The form $U=\dfrac{PV}{\gamma-1}$ reads internal energy straight off a state's pressure and volume, with no need to know $f$ or $T$ — invaluable when this chapter is stitched to thermodynamics.

5. Mean free path and collision frequency

$$\lambda=\dfrac{1}{\sqrt2\,\pi d^2 n}=\dfrac{kT}{\sqrt2\,\pi d^2 P}\quad\Rightarrow\quad \lambda\propto\dfrac{T}{P}\propto\dfrac1n,$$

with $n$ the number density. The collision frequency is $\nu=\dfrac{v_{rms}}{\lambda}=\sqrt2\,\pi d^2 n\, v_{rms}$ and the mean free time is $\tau=\dfrac1\nu$.

[JEE Tip] The $\sqrt2$, which comes from the targets moving too, is the single most-dropped factor; leaving it out makes $\lambda$ about $41$ percent too large. In a rigid vessel the number density is fixed, so $\lambda$ is fixed and only $\nu$ climbs on heating through $v_{rms}\propto\sqrt T$; the growth $\lambda\propto T$ holds only at constant $P$.

6. Speed of sound (cross-link)

Sound propagates as an adiabatic wave, so $v_{sound}=\sqrt{\dfrac{\gamma RT}{M}}$ and therefore $\dfrac{v_{sound}}{v_{rms}}=\sqrt{\dfrac{\gamma}{3}}$.

[JEE Tip] Since $\gamma<3$ for every gas, sound always trails $v_{rms}$; for air, with $\gamma=1.4$, the ratio is $\sqrt{1.4/3}\approx0.68$. Mind the swap: the sound speed carries $\gamma$ where the molecular speed carries $3$.

7. Graham's law of effusion

At a fixed temperature the effusion (or diffusion) rate is set by $\bar v\propto\dfrac{1}{\sqrt M}$, so $\dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1}}$.

[JEE Tip] Lighter gases leak faster — the very principle behind isotope separation; equivalently the times taken to release equal amounts scale as $\sqrt M$.

Solved Examples — Beyond-NCERT Formulae

Example 1 — rms vs average vs most probable speed (one gas). Find the three characteristic speeds of nitrogen ($M=0.028$ kg/mol) at $300$ K, taking $R=8.31$ J/mol.K, and confirm their ratio.

Step 1 (rms): $v_{rms}=\sqrt{\dfrac{3RT}{M}}=\sqrt{\dfrac{3(8.31)(300)}{0.028}}=\sqrt{2.671\times10^5}\approx517$ m/s.

Step 2 (average): $\bar v=\sqrt{\dfrac{8RT}{\pi M}}=\sqrt{\dfrac{8(8.31)(300)}{\pi(0.028)}}\approx476$ m/s.

Step 3 (most probable): $v_{mp}=\sqrt{\dfrac{2RT}{M}}=\sqrt{\dfrac{2(8.31)(300)}{0.028}}\approx422$ m/s.

Step 4 (ratio): $v_{mp}:\bar v:v_{rms}=422:476:517=1:1.128:1.225$, exactly $\sqrt2:\sqrt{8/\pi}:\sqrt3$. The three speeds differ only through their constants, and never in the order $v_{mp}<\bar v<v_{rms}$.

Example 2 — gamma of a mixture (DOF-token shortcut). A vessel holds $3$ mol of a monatomic gas ($\gamma_1=\tfrac53$) and $2$ mol of a rigid diatomic gas ($\gamma_2=\tfrac75$) at a common temperature. Find $\gamma_{mix}$.

Step 1 (tokens): $\dfrac{n_1}{\gamma_1-1}=\dfrac{3}{2/3}=4.5$ and $\dfrac{n_2}{\gamma_2-1}=\dfrac{2}{2/5}=5$, so their sum is $9.5$.

Step 2 (combine): with $n_1+n_2=5$, the relation $\dfrac{5}{\gamma_{mix}-1}=9.5$ gives $\gamma_{mix}-1=\dfrac{5}{9.5}=0.526$, hence $\gamma_{mix}\approx1.53$.

Step 3 (check): $C_{V,mix}=\dfrac{3(\tfrac32R)+2(\tfrac52R)}{5}=1.9R$, so $C_{P,mix}=2.9R$ and $\gamma_{mix}=\dfrac{2.9}{1.9}\approx1.53$. Both routes agree.

Example 3 — internal energy from $f$ and from $PV$. A rigid tank holds a rigid diatomic gas ($\gamma=\tfrac75$) at $P=2.0\times10^5$ Pa in a volume of $5.0$ L, that is $5.0\times10^{-3}$ m$^3$. Find its internal energy and split it into translational and rotational parts.

Step 1 (form $PV$): $PV=(2.0\times10^5)(5.0\times10^{-3})=1000$ J.

Step 2 (energy): $U=\dfrac{PV}{\gamma-1}=\dfrac{1000}{0.4}=2500$ J, which equals $\tfrac52 PV$ as it must for $f=5$.

Step 3 (split): the translational store is $\dfrac32 PV=1500$ J, leaving rotational $=2500-1500=1000$ J. The $3:2$ energy split mirrors the $3$ translational and $2$ rotational modes.

Example 4 — mean free path and collision frequency. Nitrogen molecules ($d=3.7\times10^{-10}$ m, $M=0.028$ kg/mol) sit at $300$ K and $1.0\times10^5$ Pa. Take $k=1.38\times10^{-23}$ J/K and find $\lambda$ and the collision frequency.

Step 1 (mean free path): $\lambda=\dfrac{kT}{\sqrt2\,\pi d^2 P}=\dfrac{(1.38\times10^{-23})(300)}{\sqrt2\,\pi(3.7\times10^{-10})^2(1.0\times10^5)}\approx6.8\times10^{-8}$ m.

Step 2 (frequency): with $v_{rms}\approx517$ m/s from Example 1, $\nu=\dfrac{v_{rms}}{\lambda}=\dfrac{517}{6.8\times10^{-8}}\approx7.6\times10^9$ s$^{-1}$.

Step 3 (picture): each molecule suffers about $7.6\times10^9$ collisions every second, yet travels only about $10^{-7}$ m between them. Dropping the $\sqrt2$ would wrongly inflate $\lambda$ and shrink $\nu$.

Example 5 — effect of temperature on the speeds. A gas starts at $300$ K. (a) By what factor does $v_{rms}$ change when it is heated to $675$ K? (b) To what temperature must it be taken to triple its most probable speed?

Step 1 (part a): since every speed obeys $v\propto\sqrt T$, the factor is $\sqrt{\dfrac{675}{300}}=\sqrt{2.25}=1.5$.

Step 2 (part b): tripling a speed needs $3^2=9$ times the absolute temperature, so $T=9(300)=2700$ K.

Step 3 (moral): because speeds track $\sqrt T$, even a modest change in speed demands a disproportionately large change in temperature.

Example 6 — speed of sound versus rms speed. For air ($\gamma=1.4$, $M=0.029$ kg/mol) at $300$ K, compare the speed of sound with the rms molecular speed. Take $R=8.31$ J/mol.K.

Step 1 (sound): $v_{sound}=\sqrt{\dfrac{\gamma RT}{M}}=\sqrt{\dfrac{1.4(8.31)(300)}{0.029}}\approx347$ m/s.

Step 2 (rms): $v_{rms}=\sqrt{\dfrac{3RT}{M}}=\sqrt{\dfrac{3(8.31)(300)}{0.029}}\approx508$ m/s.

Step 3 (ratio): $\dfrac{v_{sound}}{v_{rms}}=\dfrac{347}{508}=0.68=\sqrt{\dfrac{1.4}{3}}$, a number fixed purely by $\gamma$ and independent of $T$ and $M$. Sound is the slower of the two because $\gamma<3$ always.

Example 7 — Graham's law of effusion. Hydrogen ($M=2$ g/mol) and oxygen ($M=32$ g/mol) escape through the same fine pinhole at the same temperature. Compare their effusion rates.

Step 1 (ratio): $\dfrac{r_{H_2}}{r_{O_2}}=\sqrt{\dfrac{M_{O_2}}{M_{H_2}}}=\sqrt{\dfrac{32}{2}}=\sqrt{16}=4$.

Step 2 (meaning): hydrogen effuses $4$ times as fast, so oxygen needs $4$ times as long to release the same number of molecules. This $\dfrac{1}{\sqrt M}$ dependence is exactly what lets effusion separate light isotopes from heavy ones.