Q.Estimate the mean free path for a water molecule in water vapour at $373\ \text{K}$. Use information from Example 12.1 and Eq. (12.41) above (mean free path of an air molecule at STP, $l = 2.9 \times 10^{-7}\ \text{m}$).

Showing the 12 most recent of 39 on this concept.

1. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.The rms speed of oxygen molecule at some temperature is $150\ ms^{-1}$. Then the rms speed of hydrogen molecule at the same temperature is (A) $400\ ms^{-1}$ (B) $600\ ms^{-1}$ (C) $200\ ms^{-1}$ (D) $800\ ms^{-1}$

   ›Reveal solutionSolution

   Tests the molar-mass dependence of rms speed from kinetic theory. Answer: $600\ ms^{-1}$.

   Concept and Intuition

   At a given temperature, all gas molecules share the same average kinetic energy $\frac32k_BT$ (equipartition), regardless of their mass. This means lighter molecules must move faster on average to carry the same kinetic energy — specifically $v_{rms}=\sqrt{3RT/M}$, so $v_{rms}$ scales as $1/\sqrt{M}$.

   Step-by-Step Solution

   1. $v_{rms}=\sqrt{\dfrac{3RT}{M}}$, so at fixed $T$: $\dfrac{v_{H_2}}{v_{O_2}}=\sqrt{\dfrac{M_{O_2}}{M_{H_2}}}$.
   2. Molar masses: $M_{O_2}=32$ g/mol, $M_{H_2}=2$ g/mol. …

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2. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.The temperature at which the $rms$ speed of hydrogen molecules is equal to that of oxygen molecules at $47^\circ C$ is (A) 20 K (B) 80 K (C) 73 K (D) 3 K

   ›Reveal solutionSolution

   This tests the rms speed formula's dependence on temperature and molar mass. The answer is 20 K.

   Concept and Intuition

   The rms speed of a gas is $v_{rms} = \sqrt{\dfrac{3RT}{M}}$, where $M$ is the molar mass. For two different gases to have the same rms speed, the ratio $T/M$ must be the same for both — a lighter gas (like hydrogen) needs a proportionally lower temperature to match the same rms speed as a heavier gas (like oxygen) at a higher temperature.

   Step-by-Step Solution

   1. Convert oxygen's temperature to Kelvin: $47^\circ C = 47+273 = 320\ K$.
   2. Set rms speeds equal: $\sqrt{\dfrac{3RT_{H_2}}{M_{H_2}}} = \sqrt{\dfrac{3R(320)}{M_{O_2}}}$.
   3. Square and simplify: $\dfrac{T_{H_2}}{M_{H_2}} = \dfrac{320}{M_{O_2}}$.
   4. Molar masses: $M_{H_2}=2\ g/mol$, $M_{O_2}=32\ g/mol$. …

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3. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.The Mean free path of a gas molecule, whose diameter is $2 \times 10^{-10}$ m is $1.6 \times 10^{-7}$ m. Calculate the mean free path of another gas molecule whose diameter is $4 \times 10^{-10}$ m (under the same conditions) (A) $0.4 \times 10^{-7}$ m (B) $0.8 \times 10^{-7}$ m (C) $6.4 \times 10^{-7}$ m (D) $1.6 \times 10^{-7}$ m

   ›Reveal solutionSolution

   This tests the dependence of the kinetic-theory mean free path on molecular diameter: $\lambda \propto 1/d^2$ at fixed number density, so doubling the diameter quarters the mean free path.

   Concept and Intuition

   The mean free path of a gas molecule is the average distance travelled between successive collisions. Kinetic theory gives

   $\lambda = \frac{1}{\sqrt{2}\,\pi d^2 n},$

   where $d$ is the molecular diameter and $n$ is the number density of molecules. "Under the same conditions" means the same $n$ (same pressure and temperature, so same number density) — so the only thing that changes between the two gases is $d$, and $\lambda$ scales as $1/d^2$.

   Step-by-Step Solution

   1. Write the proportionality at fixed $n$: $\lambda \propto \dfrac{1}{d^2}$.
   2. So $\lambda_1 d_1^2 = \lambda_2 d_2^2$ (same constant of proportionality since $n$ is unchanged).
   3. Given $d_1 = 2\times10^{-10}$ m, $\lambda_1 = 1.6\times10^{-7}$ m, and $d_2 = 4\times10^{-10}$ m (twice $d_1$). …

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4. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.The approximate temperature at which the rms speed of Nitrogen gas molecule is $500\ ms^{-1}$. Gas constant $R=8.314\ J\ mol^{-1}K^{-1}$, Mass number of Nitrogen = 28 (A) 280 K (B) 300 K (C) 350 K (D) 250 K

   ›Reveal solutionSolution

   Invert the rms-speed formula $v_{rms}=\sqrt{3RT/M}$ to solve for temperature; with $M=28$ g/mol and $v_{rms}=500$ m/s, $T\approx280$ K.

   Concept and Intuition

   The rms speed of gas molecules links molecular mass, absolute temperature, and speed via kinetic theory: $v_{rms}=\sqrt{\dfrac{3RT}{M}}$, where $M$ is the molar mass in kg/mol. Rearranging lets us find the temperature at which a given rms speed occurs.

   Step-by-Step Solution

   1. Molar mass of $N_2$: $M = 28\times10^{-3}$ kg/mol.
   2. Square the speed: $v_{rms}^2 = 500^2 = 2.5\times10^5$ m²/s².
   3. $T = \dfrac{v_{rms}^2 M}{3R} = \dfrac{2.5\times10^5\times28\times10^{-3}}{3\times8.314}$.
   4. Numerator: $2.5\times10^5\times0.028 = 7000$. Denominator: $3\times8.314=24.942$. …

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5. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.The translational kinetic energy of the molecule of 22 grams of $\mathrm{CO_2}$ at $27°C$ is (Universal gas constant $R = 8.314\ J\ mol^{-1}\ °C^{-1}$) (A) 1870.6 J (B) 164.7 J (C) 2000 J (D) 2200 J

   ›Reveal solutionSolution

   This tests the formula for translational kinetic energy of an ideal gas, $\tfrac32 nRT$, applied regardless of the gas being diatomic (translational KE per mole is the same for all ideal gases). Answer: 1870.6 J.

   Concept and Intuition

   By the equipartition theorem, each molecule has 3 translational degrees of freedom (motion along x, y, z), each contributing $\tfrac12 k_BT$ of energy on average. For $n$ moles, the total translational kinetic energy is $\tfrac32 nRT$ — this formula applies to ANY ideal gas (monatomic, diatomic, or polyatomic), since translational motion doesn't depend on molecular structure; only rotational/vibrational contributions differ.

   Step-by-Step Solution

   1. Molar mass of $CO_2 = 12 + 2(16) = 44$ g/mol.
   2. Number of moles: $n = \dfrac{22}{44} = 0.5$ mol.
   3. Convert temperature to Kelvin: $T = 27 + 273 = 300$ K.
   4. Translational KE $= \dfrac{3}{2}nRT = \dfrac{3}{2}\times 0.5 \times 8.314 \times 300$. …

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6. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.If the mean free path of a nitrogen molecule in a vessel containing nitrogen at a pressure of 2.1 atm and a temperature of $27^{\circ}\text{C}$ is $\lambda$, then its mean free path at a pressure of 1.65 atm and a temperature of $57^{\circ}\text{C}$ is (A) $1.4\lambda$ (B) $2.1\lambda$ (C) $2.8\lambda$ (D) $3.5\lambda$

   ›Reveal solutionSolution

   Tests the pressure and temperature dependence of the mean free path of a gas molecule, $\lambda\propto T/P$.

   Concept and Intuition

   Mean free path depends on how "crowded" the gas is (number density $n=P/k_BT$) — raising pressure packs molecules closer (shorter $\lambda$), while raising temperature at fixed pressure spreads them out (longer $\lambda$). Combining both effects via $\lambda\propto T/P$ handles any simultaneous change.

   Step-by-Step Solution

   1. Mean free path formula: $\lambda = \dfrac{k_BT}{\sqrt2\,\pi d^2 P}$, so for the same gas (same molecular diameter $d$), $\lambda \propto \dfrac{T}{P}$.
   2. Convert temperatures to Kelvin: $T_1 = 27+273=300\,K$, $T_2=57+273=330\,K$.
   3. Given pressures: $P_1=2.1\,\text{atm}$, $P_2=1.65\,\text{atm}$.
   4. Ratio: $\dfrac{\lambda_2}{\lambda_1} = \dfrac{T_2}{T_1}\times\dfrac{P_1}{P_2} = \dfrac{330}{300}\times\dfrac{2.1}{1.65}$. …

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7. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.If the rms speed of 2 moles of a gas of mass 64 g at a temperature of $47^0\mathrm{C}$ is V, then the rms speed of 5 moles of another gas of mass 20 g at a temperature of $367^{\circ}\mathrm{C}$ is (A) 16V (B) 4V (C) 8V (D) 2V

   ›Reveal solutionSolution

   Comparing rms speeds via $v_{rms}=\sqrt{3RT/M}$ (with $M$ = mass per mole of each sample), the second gas's rms speed works out to exactly $4V$.

   Concept and Intuition

   The rms speed formula $v_{rms}=\sqrt{3RT/M}$ uses $M$ as the molar mass of the gas. Here we're given total mass and number of moles for each sample, so the molar mass for each gas is simply (total mass)/(number of moles) — these need not be looked up, since they're derivable from the given data.

   Step-by-Step Solution

   1. Gas 1: mass $=64$ g, moles $=2 \implies M_1 = 64/2 = 32\ \text{g/mol}$. Temperature $T_1 = 47^\circ\text C = 320\ \text K$. Its rms speed is $V = \sqrt{3RT_1/M_1}$.
   2. Gas 2: mass $=20$ g, moles $=5 \implies M_2 = 20/5 = 4\ \text{g/mol}$. Temperature $T_2 = 367^\circ\text C = 640\ \text K$. Its rms speed is $v_2=\sqrt{3RT_2/M_2}$.
   3. Take the ratio: …

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8. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.A thermally insulated vessel contains an ideal gas of molecular weight M of ratio of specific heats γ. It is moving with a speed V and is suddenly brought to rest. Assuming no heat is lost to the surroundings, temperature of gas increases by (A) $\left[\dfrac{(\gamma-1)MV^2}{2(\gamma+2)R}\right]K$ (B) $\left[\dfrac{(\gamma-1)MV^2}{2\gamma R}\right]K$ (C) $\left[\dfrac{\gamma MV^2}{2R}\right]K$ (D) $\left[\dfrac{(\gamma-1)MV^2}{2R}\right]K$

   ›Reveal solutionSolution

   The bulk kinetic energy of the moving, insulated gas converts entirely to internal energy when it stops; equating this to $C_v\Delta T$ gives $\Delta T=\dfrac{(\gamma-1)MV^2}{2R}$.

   Concept and Intuition

   When a container of gas moving with bulk speed $V$ is suddenly halted, the ordered (bulk) kinetic energy of the gas has nowhere to go except into the random (thermal) kinetic energy of the molecules — since the vessel is insulated, no heat escapes. This raises the gas's internal energy, and hence its temperature.

   Step-by-Step Solution

   1. Kinetic energy of the moving gas (mass $=M$ for one mole, molar mass $M$): $KE = \tfrac12MV^2$.
   2. This converts entirely into internal energy: $\Delta U = \tfrac12MV^2$.
   3. For one mole of an ideal gas, $\Delta U = C_v\Delta T$, and $C_v=\dfrac{R}{\gamma-1}$. …

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9. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.If 2 moles of an ideal monoatomic gas at a temperature of $27^\circ\mathrm{C}$ is mixed with 4 moles of another ideal monoatomic gas at a temperature of $327^\circ\mathrm{C}$, then the temperature of mixture of the two gases is (A) $300^\circ\mathrm{C}$ (B) $227^\circ\mathrm{C}$ (C) $233^\circ\mathrm{C}$ (D) $327^\circ\mathrm{C}$

   ›Reveal solutionSolution

   Mixing two monoatomic ideal gases at different temperatures conserves total internal energy; solving for the common final temperature gives $227^\circ\mathrm{C}$.

   Concept and Intuition

   When two gases are mixed in an insulated container (no heat exchange with surroundings, no work done), the total internal energy is conserved. For an ideal monoatomic gas, internal energy is $U=nC_vT=\tfrac32 nRT$, so equating total energy before and after mixing (with the same $C_v$ for both monoatomic gases) gives a mole-weighted average temperature.

   Step-by-Step Solution

   1. Convert to Kelvin: $T_1 = 27+273=300\ \mathrm{K}$, $T_2=327+273=600\ \mathrm{K}$.
   2. Energy conservation: $n_1C_vT_1+n_2C_vT_2=(n_1+n_2)C_vT_f$. Since both gases are monoatomic, $C_v$ cancels out. …

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10. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
    Q.The average translational kinetic energy of the oxygen molecules at a temperature of $127\,^{\circ}\mathrm{C}$ is (Boltzmann constant $=1.38\times10^{-23}\ \mathrm{J\,K^{-1}}$) (A) $4.07\times10^{-21}\ \mathrm{J}$ (B) $2.07\times10^{-21}\ \mathrm{J}$ (C) $8.28\times10^{-21}\ \mathrm{J}$ (D) $8.00\times10^{-21}\ \mathrm{J}$

    ›Reveal solutionSolution

    Average translational KE per molecule is $\tfrac32kT$; converting $127^\circ$C to $400$ K and substituting gives $8.28\times10^{-21}$ J, independent of which gas it is (kinetic theory result depends only on temperature).

    Concept and Intuition

    A key result of the kinetic theory of gases is that the average translational kinetic energy per molecule depends only on the absolute temperature, not on the identity or mass of the gas: $\overline{KE} = \tfrac32 k_BT$. This is why oxygen and any other ideal gas at the same temperature have the same average translational KE per molecule (though their average speeds differ, because speed also depends on mass).

    Step-by-Step Solution

    1. Convert temperature to Kelvin: $T = 127 + 273 = 400$ K.
    2. Average translational KE per molecule: $\overline{KE} = \dfrac32 k_B T$.
    3. Substitute: $\overline{KE} = 1.5\times(1.38\times10^{-23})\times400$. …

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11. AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
    Q.A closed vessel contains a gas at a pressure P. If 50% of the mass of the gas is removed and rms speed of the gas molecules is increased by 20%, then the pressure of the remaining gas is (A) $\dfrac{16P}{25}$ (B) $\dfrac{8P}{25}$ (C) $\dfrac{9P}{25}$ (D) $\dfrac{18P}{25}$

    ›Reveal solutionSolution

    Kinetic theory gives $P = \frac{1}{3}\rho v_{rms}^2$ where $\rho$ is the gas density. Halving the mass halves the density, and increasing $v_{rms}$ by 20% multiplies the speed-squared term by $1.2^2=1.44$; combining gives $P' = 0.72P = \frac{18P}{25}$.

    Concept and Intuition

    For a fixed volume (closed vessel), the pressure of an ideal gas from kinetic theory is

    $P = \frac{1}{3}\frac{M}{V}v_{rms}^2$

    where $M$ is the total mass of gas present and $V$ is the (fixed) container volume. Removing gas reduces $M$; changing the temperature (and hence the average molecular speed) changes $v_{rms}$. Both effects multiply into the pressure independently.

    Step-by-Step Solution

    1. Original pressure: $P = \dfrac{1}{3}\dfrac{M}{V}v_{rms}^2$.
    2. After removing 50% of the mass: $M' = \dfrac{M}{2}$.
    3. New rms speed: $v'_{rms} = 1.2\,v_{rms}$, so $(v'_{rms})^2 = 1.44\,v_{rms}^2$. …

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12. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
    Q.The mean translational kinetic energy of a gas molecule at a temperature of 47 °C is (Boltzman constant $= 1.38 \times 10^{-23} \mathrm{~J~K^{-1}}$) (A) $138 \times 10^{-4}$ J (B) $138 \times 10^{-4}$ eV (C) $414 \times 10^{-4}$ J (D) $414 \times 10^{-4}$ eV

    ›Reveal solutionSolution

    Compute $\tfrac32 kT$ at $T=320\,$K and convert joules to electron-volts; it comes out to $414\times10^{-4}$ eV.

    Concept and Intuition

    By the kinetic theory of gases, each translational degree of freedom of a molecule carries mean energy $\tfrac12 kT$ (equipartition theorem). With 3 translational degrees of freedom (x, y, z), the mean translational kinetic energy of a molecule is

    $\langle KE\rangle = \frac32 kT$

    independent of the gas's molar mass — it depends only on temperature.

    Step-by-Step Solution

    1. Convert temperature to kelvin: $T = 47+273 = 320\text{ K}$.
    2. Compute $kT$: $kT = (1.38\times10^{-23})(320) = 4.416\times10^{-21}\text{ J}$.
    3. Multiply by $\tfrac32$:

    $\langle KE\rangle = 1.5\times4.416\times10^{-21} = 6.624\times10^{-21}\text{ J}$

    4. Convert to eV using $1\text{ eV}=1.6\times10^{-19}\text{ J}$: $\langle KE\rangle = \frac{6.624\times10^{-21}}{1.6\times10^{-19}} \approx 0.0414\text{ eV} = 414\times10^{-4}\text{ eV}$ …

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