For a surface molecule (Note: more than one of the given options may be correct.) (a) the net force on it is zero. (b) there is a net downward force. (c) the potential energy is less than that of a molecule inside. (d) the potential energy is more than that of a molecule inside.

For a surface molecule (Note: more than one of the given options may be correct.) (a) the net force on it is z

NCERT Exemplar · Q6

Q. For a surface molecule (Note: more than one of the given options may be correct.)

(a) the net force on it is zero.

(b) there is a net downward force.

(d) the potential energy is more than that of a molecule inside.

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Concept understanding — Molecular Volume Fraction

Molecular Volume Fraction

Kinetic theory begins with a bold assumption: gas molecules are so small compared to the space they roam in that we can treat them as point particles with essentially zero volume. The molecular volume fraction is the number that checks whether that assumption is actually reasonable — it compares the volume genuinely occupied by the molecules themselves to the total volume the gas fills.

The Intuition

Picture a large empty hall with a handful of tennis balls rolling around freely. The balls take up hardly any of the hall's floor space — most of what you see is empty air. That is exactly the picture kinetic theory paints of an ordinary gas: the molecules are the tennis balls, and the container is the hall. The molecular volume fraction simply asks: what fraction of the hall's volume is actually tennis ball?

The Precise Statement

If a gas has $N$ molecules, each modelled as a sphere of radius $r$ (so each molecule has volume $\frac{4}{3}\pi r^{3}$ ), and the gas occupies a total volume $V$ , the molecular volume fraction is:

f = \frac{N \times \frac{4}{3}\pi r^{3}}{V}

This is a pure number (no units) — a ratio of two volumes.

A Worked Example (STP Oxygen)

Take one mole of oxygen gas at STP, where $N = N_A = 6.02 \times 10^{23}$ molecules and the molar volume is $V = 22.4\ \text{L} = 2.24 \times 10^{-2}\ \text{m}^3$ . Taking the molecular diameter to be about $3\ \text{Å} = 3 \times 10^{-10}\ \text{m}$ (radius $r = 1.5 \times 10^{-10}\ \text{m}$ ):

\text{Volume of one molecule} = \frac{4}{3}\pi r^{3} \approx 1.4 \times 10^{-29}\ \text{m}^3

\text{Total molecular volume} = N_A \times 1.4 \times 10^{-29} \approx 8.5 \times 10^{-6}\ \text{m}^3

f = \frac{8.5 \times 10^{-6}}{2.24 \times 10^{-2}} \approx 4 \times 10^{-4}

So the molecules themselves occupy only about 0.04% of the space the gas fills — the remaining 99.96% is empty. This tiny number is exactly why kinetic theory can safely ignore the size of individual molecules for an ordinary gas.

Note

This is also why the mean free path of a gas molecule (the average distance it travels between collisions) is far larger than the molecule's own diameter — there is enormous empty space to cross before the next collision.

Why It Matters: Ideal vs. Real Gas …

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A surface molecule has fewer neighbours pulling on it from above (only sparse vapour there vs. dense liquid below/sideways), so it feels a net inward (downward) force and carries higher potential energy than a molecule deep in the bulk. Correct options: (B) and (D).

Compare a molecule sitting deep inside the liquid to one sitting right at the free surface, in terms of the neighbours pulling on it from every direction.

A bulk (interior) molecule is surrounded by other liquid molecules on all sides — above, below, and all around — so the attractive pulls from every direction cancel out symmetrically. Net force = zero, and this is the lowest-energy (most stable) configuration, since the molecule is fully bonded to its neighbours in every direction.

A surface molecule has liquid neighbours below and to the sides, but only sparse vapor/air molecules above it (far weaker attraction from that side). The pulls from below and the sides no longer cancel — there is a net unbalanced force pulling the molecule inward, i.e., downward into the bulk. This rules out option (a) (net force is NOT zero) and confirms option (b) (there is a net downward force). …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
$1.1\ g\,L^{-1}$
›Reveal solutionSolution
This tests density of a gas mixture using $d = \dfrac{PM_{avg}}{RT}$ where $M_{avg}$ is the mixture's average molar mass; the answer is 1.515 g L $^{-1}$ .

Concept and Intuition

For an ideal gas mixture, the density can be found either by computing total mass and total volume separately, or directly via $d = \dfrac{P M_{avg}}{RT}$ , where $M_{avg}$ is the total mass divided by total moles. Both routes must agree since density is just mass per unit volume regardless of how many components are present.

Step-by-Step Solution

Moles of $N_2 = \dfrac{7\ g}{28\ g\,mol^{-1}} = 0.25$ mol.

Moles of $CO_2 = \dfrac{11\ g}{44\ g\,mol^{-1}} = 0.25$ mol.

Total moles $n = 0.25 + 0.25 = 0.5$ mol; total mass $= 7 + 11 = 18$ g.

Average molar mass $M_{avg} = \dfrac{18}{0.5} = 36\ g\,mol^{-1}$ . …
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TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
$100\%$
›Reveal solutionSolution
The average speed of molecules is proportional to $\sqrt{T}$ , so doubling speed quadruples the temperature. Halving the volume at constant $n$ then, via the ideal gas law, multiplies pressure by 8 — a 700% increase.

The key is to connect molecular speed to temperature, and then temperature and volume to pressure. For an ideal gas, the average speed of molecules is $v_{\text{avg}} \propto \sqrt{T}$ . Doubling the speed means the temperature becomes four times larger. Now, with the ideal gas law $PV = nRT$ , if $n$ is constant (the container is sealed), halving the volume and quadrupling the temperature together change the pressure by a factor of $4 \times 2 = 8$ . An eight-fold increase corresponds to a 700% rise, since the original pressure is 100% and the new pressure is 800% of it.

Let’s walk through it step by step.

Relate speed to temperature. For an ideal gas, the root-mean-square speed is $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$ , and the average speed $v_{\text{avg}} = \sqrt{\frac{8RT}{\pi M}}$ . Both are proportional to $\sqrt{T}$ . So if $v_{\text{avg}}$ doubles,

$\frac{v_{\text{avg,new}}}{v_{\text{avg,old}}} = 2 = \sqrt{\frac{T_{\text{new}}}{T_{\text{old}}}} \quad \Rightarrow \quad \frac{T_{\text{new}}}{T_{\text{old}}} = 4.$
Temperature quadruples.

Apply the ideal gas law. The container is sealed, so the number of moles $n$ is fixed. The ideal gas law $PV = nRT$ gives

$\frac{P_{\text{new}} V_{\text{new}}}{T_{\text{new}}} = \frac{P_{\text{old}} V_{\text{old}}}{T_{\text{old}}}.$
We know $V_{\text{new}} = \frac{1}{2} V_{\text{old}}$ and $T_{\text{new}} = 4 T_{\text{old}}$ . Substitute:
$\frac{P_{\text{new}} \cdot \frac{1}{2} V_{\text{old}}}{4 T_{\text{old}}} = \frac{P_{\text{old}} V_{\text{old}}}{T_{\text{old}}}.$

Solve for the new pressure. Cancel $V_{\text{old}}$ and $T_{\text{old}}$ (they are non-zero): $\frac{P_{\text{new}}}{8} = P_{\text{old}} \quad \Rightarrow \quad P_{\text{new}} = 8 P_{\text{old}}.$ …
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MHT-CET 2024Set pcm-2024-05-02-M1 markMCQ
$d$
›Reveal solutionSolution

Mean free path depends on number density, not directly on temperature; at constant volume the density is fixed, so $\lambda$ remains $1500\,d$ — option (A).

From kinetic theory (Class 12 NCERT/CBSE kinetic-theory chapter), the mean free path is
$\lambda=\frac{1}{\sqrt{2}\,\pi d^{2}\,n},$
where $d$ is the molecular diameter and $n=\dfrac{N}{V}$ is the number of molecules per unit volume.

The process here is at constant volume with the same gas sample, so the number of molecules $N$ and the volume $V$ are both unchanged — hence the number density $n$ is constant. Since $\lambda$ depends only on $d$ and $n$ (and not explicitly on temperature), raising the temperature from S.T.P. (273 K) to 373 K at fixed volume leaves the mean free path unchanged:

$\lambda_{373\,K}=\lambda_{S.T.P.}=1500\,d.$ …
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TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
$V_2^2$
›Reveal solutionSolution
The ratio $V_1/V_2$ is found by relating the mean square x‑component velocity of gas A to the mean square speed of gas B via the equipartition theorem. The result is $\sqrt{2/3}$ , so the correct option is (D).

The key idea is that temperature is a measure of the average kinetic energy per molecule, and this energy is shared equally among all independent degrees of freedom (equipartition). For a gas in thermal equilibrium at temperature $T$ , each translational degree of freedom contributes $\frac{1}{2} k_B T$ to the mean kinetic energy per molecule.

For gas B, the mean square speed $V_2^2$ involves all three velocity components:

$\frac{1}{2} (2m) \langle v_B^2 \rangle = \frac{3}{2} k_B T$
so $\langle v_B^2 \rangle = V_2^2 = \frac{3k_B T}{2m}$ .

For gas A, the mean square x‑component $V_1^2$ involves only one degree of freedom:

$\frac{1}{2} m \langle v_{A,x}^2 \rangle = \frac{1}{2} k_B T$
so $\langle v_{A,x}^2 \rangle = V_1^2 = \frac{k_B T}{m}$ .

Now we simply take the ratio:

$\frac{V_1}{V_2} = \sqrt{\frac{V_1^2}{V_2^2}} = \sqrt{\frac{k_B T / m}{3k_B T / (2m)}} = \sqrt{\frac{2}{3}}.$ …
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AP EAPCET 2022Set ap-2022-07-12-FN1 markMCQ
$Q. If the volume of a gas is reduced to half then the change in its mean free path is (A) Becomes half (B) Unchanged (C) Doubles (D) Depend on temperature$
›Reveal solutionSolution
Mean free path is inversely proportional to the number density of gas molecules; compressing the gas to half its volume doubles the density and halves the mean free path. Answer: becomes half.

Concept and Intuition

Mean free path is the average distance a molecule travels between successive collisions. It depends on how crowded the gas is — the more molecules packed into a given volume, the sooner a molecule bumps into another one, so $\lambda \propto 1/n$ , where $n=N/V$ is number density.

Step-by-Step Solution

Mean free path formula: $\lambda = \dfrac{1}{\sqrt2\,\pi d^2 n}$ , where $n=N/V$ (number of molecules per unit volume) and $d$ is molecular diameter (unchanged).

If volume $V$ is halved while the number of molecules $N$ stays fixed, the new number density is $n' = N/(V/2) = 2n$ .

New mean free path: $\lambda' = \dfrac{1}{\sqrt2\,\pi d^2(2n)} = \dfrac{\lambda}{2}$ . …
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