Q.Estimate the mean free path for a water molecule in water vapour at 373 K. Use information from Example 12.1 and Eq. (12.41) above (mean free path of an air molecule at STP, l=2.9×10−7 m).
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Kinetic Theory of Gases
Imagine you're sitting in a quiet room. The air around you feels still — but it isn't. Every second, billions of tiny particles (molecules of nitrogen, oxygen, and others) are zipping past you at hundreds of metres per second. They're constantly crashing into each other and into the walls, your skin, the furniture. You don't feel each individual hit because the molecules are so small and the collisions happen so fast. But collectively, those countless tiny impacts produce something you do feel: pressure.
That's the core intuition behind the kinetic theory of gases. It says: all the macroscopic properties of a gas — pressure, temperature, volume — can be explained by the motion of its molecules.
The Big Idea
Instead of treating a gas as a continuous, smooth substance (like a fluid), the kinetic theory treats it as a swarm of tiny, hard, perfectly elastic balls in constant, random motion. "Perfectly elastic" means that when two molecules collide, no kinetic energy is lost — they bounce off each other like ideal billiard balls, not like sticky clay.
From this simple picture, we can derive the gas laws (Boyle's, Charles's, Avogadro's) and even calculate things like the speed of sound in a gas.
The Five Assumptions (The Precise Statement)
For a gas to behave according to the kinetic theory in its simplest form, we make these assumptions:
-
A gas consists of a very large number of molecules.
The number is so huge that we can use statistics — individual molecules don't matter, only averages do.
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The molecules are in constant, random motion.
They move in straight lines until they hit something (another molecule or a wall). There's no preferred direction.
-
The molecules are point masses.
Their actual size is negligible compared to the distance between them. In other words, the volume of the molecules themselves is tiny compared to the volume of the container.
-
Collisions are perfectly elastic.
No kinetic energy is lost when molecules collide with each other or with the walls. Total energy of the system stays constant.
-
There are no intermolecular forces.
The molecules don't attract or repel each other except during collisions. Between collisions, they move freely.
These assumptions define an ideal gas. Real gases deviate from this behaviour at high pressure or low temperature, but the kinetic theory gives an excellent approximation for most everyday conditions.
How It Explains Pressure
Pressure is the force per unit area exerted by the gas on the walls of its container. In the kinetic picture:
- A molecule moving toward a wall hits it and bounces back.
- During the collision, the wall exerts a force on the molecule to reverse its momentum.
- By Newton's third law, the molecule exerts an equal and opposite force on the wall.
- Multiply that by the billions of collisions happening every second, and you get a steady, measurable pressure.
If you heat the gas, the molecules move faster. They hit the walls harder and more often — pressure increases. If you compress the gas into a smaller volume, molecules hit the walls more frequently — pressure increases again.
The Key Result: The Kinetic Equation
From these assumptions, we can derive a relationship between pressure P, volume V, and the average kinetic energy of the molecules. The result is:
PV=31Nmv2
Where:
- N = number of molecules
- m = mass of one molecule
- v2 = mean square speed of the molecules (average of the squares of their speeds)
Since the average kinetic energy of a molecule is K=21mv2, we can rewrite this as:
PV=32NK …
Concept: Molecular Volume Fraction — the mean free path scales inversely with the number density n of molecules. Since n∝P/T, we can compare the given air-at-STP value to water vapour at 373 K and 1 atm.
Step 1: At STP (273 K, 1 atm), air has lair=2.9×10−7 m. For water vapour at 373 K and 1 atm, the number density changes only with temperature (pressure same).
Step 2: Mean free path l∝1/n∝T (at constant P). So: …
The mean free path is l=1/(2nπd2) with n=P/kT. At fixed pressure l∝T, and taking the effective molecular size for water vapour as roughly that of air, scaling the given air value from 273 K to 373 K gives l≈4×10−7 m.
Reasoning
The mean free path depends on the number density n and the molecular diameter d:
l=2nπd21,n=kTP
At the same pressure, n∝1/T, so - treating the effective diameter of a water molecule as roughly the same as that of an air molecule (both a few angstrom, a fair estimate for an order-of-magnitude answer) - the mean free path scales directly with temperature:
lairlvapour=nvapournair=TairTvapour …
Rather than treating this as 'a different gas at a different condition,' notice only temperature is changing here (pressure stays at 1 atm), and l∝n−1∝T at fixed P (from PV=NkBT⇒n=P/kBT). That means you can scale the given air value directly by the temperature ratio, with no need to compute a molecular diameter or number density from scratch: $l_{water}=l_{air}\times(373/273)\approx4.0\times10 …
Showing the 12 most recent of 20 on this concept.
- KEAM 2026Set eng-2026-04174 marksMCQQ.The root mean square velocity of a gas molecule is 100 ms−1. The mass of the molecule is increased four times keeping the temperature constant. Then, the root mean square velocity is (A) 25 ms−1 (B) 50 ms−1 (C) 75 ms−1 (D) 2500 ms−1 (E) 125 ms−1
›Reveal solutionSolution
Since vrms∝1/m at constant T, increasing mass fourfold gives 50 ms−1.
The root-mean-square speed of gas molecules is:
vrms=m3kBT⇒vrms∝m1 (at constant T) …
- KEAM 2026Set eng-2026-04174 marksMCQQ.The mean free path of moving gas molecules is expressed as α∝nxdy where n is the number of molecules per unit volume and d is the size of the molecules. Then the values of x and y respectively, are (A) 2, -2 (B) -1, -2 (C) 1, 2 (D) 1, -2 (E) -2, -2
›Reveal solutionSolution
λ∝n−1d−2, giving x=−1 and y=−2.
The mean free path of a gas molecule is:
λ=2πnd21 …
- KEAM 2026Set eng-2026-04184 marksMCQQ.The average kinetic energy of a gas molecule is directly proportional to (A) its pressure (B) its volume (C) the square root of its absolute temperature (D) the square of its pressure (E) its absolute temperature
›Reveal solutionSolution
Kinetic theory gives average molecular KE =23kBT, so it is directly proportional to the absolute temperature T.
From kinetic theory, the average translational kinetic energy of a gas molecule is
⟨KE⟩=23kBT, …
- KEAM 2026Set eng-2026-04204 marksMCQQ.If the kinetic energy of 1 mole of a monoatomic gas is 24.942 kJ, then its temperature is (R=8.314 J kg−1K−1) (A) 200 K (B) 300 K (C) 400 K (D) 2000 K (E) 3000 K
›Reveal solutionSolution
Monoatomic gas: KE=23nRT; solve for T.
The translational kinetic energy of n moles of a monoatomic ideal gas is KE=23nRT. With n=1 and KE=24942 J: …
- KEAM 2026Set eng-2026-04204 marksMCQQ.If the rms speed of ideal gas molecules increases by 20% at constant volume, the percentage increase in their pressure is (A) 20 % (B) 40 % (C) 44 % (D) 10 % (E) 32 %
›Reveal solutionSolution
Since P∝vrms2 at constant volume, a factor of 1.2 on the speed gives 1.44 on the pressure, i.e. a 44% rise.
From kinetic theory the pressure of an ideal gas is
P=31VNmvrms2.
At constant volume, with fixed number and mass of molecules, everything except vrms2 is constant, so P∝vrms2.
If the rms speed increases by 20%, vrms′=1.20vrms, hence …
- KEAM 2026Set eng-2026-04214 marksMCQQ.Two perfect monoatomic gases at temperatures 300 K and 410 K are mixed without any loss of heat. If 1024 and 1023 are the number of molecules in the respective gases, then the temperature of the mixture is (A) 340 K (B) 310 K (C) 360 K (D) 350 K (E) 370 K
›Reveal solutionSolution
Molecule-number-weighted mean temperature: (1024⋅300+1023⋅410)/(1.1×1024)=310 K.
Both gases are monoatomic, so each molecule carries the same 23kT per unit temperature. With no heat loss, total internal energy is conserved and the equilibrium temperature is the number-weighted average:
T=N1+N2N1T1+N2T2. …
- KEAM 2026Set eng-2026-04214 marksMCQQ.If the temperature T of oxygen molecule is raised to 9T, then its root mean square speed v is increased to (A) 3v (B) 2v (C) 2v (D) 3v (E) 23v
›Reveal solutionSolution
Since vrms∝T, T→9T scales the speed by 9=3.
Root-mean-square speed vrms=m3kT∝T. …
- KEAM 2026Set pha-2026-0420F4 marksMCQQ.The temperature at which the rms speed of oxygen molecules becomes equal to the rms speed of hydrogen molecules at 300 K is: (A) 4800 K (B) 2400 K (C) 1200 K (D) 600 K (E) 300 K
›Reveal solutionSolution
vrms∝T/M; equating gives TO2=300×(32/2)=4800 K.
The rms speed is vrms=M3RT. Setting the rms speed of O2 (molar mass 32) at temperature T equal to that of H2 (molar mass 2) at 300 K: …
- KEAM 2025Set eng-2025-04274 marksMCQQ.The translational kinetic energy of an ideal gas containing N molecules at temperature T is (k - Boltzmann constant) (A) 25NkT (B) 21NkT (C) 23NkT (D) 27NkT (E) 29NkT
›Reveal solutionSolution
Average translational kinetic energy per molecule is 23kT; multiplying by N molecules gives 23NkT.
By the equipartition theorem, each of the three translational degrees of freedom contributes 21kT, so the average translational kinetic energy of one molecule is 23kT. F …
- KEAM 2025Set eng-2025-04274 marksMCQQ.For an ideal gas of molar mass M, the slope of the plot between the rms velocity (vrms along the y-axis) and the square root of absolute temperature (T along the x-axis) is (A) 3RM (B) M3R (C) 3MR (D) 3MR (E) M3R
›Reveal solutionSolution
Since vrms=3RT/M=3R/M⋅T, plotting vrms against T gives a straight line of slope 3R/M.
The rms speed of gas molecules is
vrms=M3RT=M3RT. …
- KEAM 2025Set eng-2025-04284 marksMCQQ.If the temperature of a gas is changed to 9 times the initial value, then the rms velocity of the gaseous molecule increases by (A) 9 times (B) 3 times (C) 3 times (D) 18 times (E) 12 times
›Reveal solutionSolution
vrms=3RT/M∝T; T→9T gives factor 9=3.
The rms speed of gas molecules is
vrms=M3RT∝T.
If the temperature becomes 9T, then …
- KEAM 2025Set eng-2025-04284 marksMCQQ.If the mean kinetic energy of one mole of helium gas at 400 K temperature is 5000 J, then that for one mole of neon gas at 800 K is (A) 5000 J (B) 50000 J (C) 10000 J (D) 2500 J (E) 500 J
›Reveal solutionSolution
Mean KE per mole =23RT∝T and is gas-independent; doubling T from 400 K to 800 K doubles 5000 J to 10000 J.
The mean (translational) kinetic energy of one mole of an ideal gas is
Eˉ=23RT, …
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