Kinetic Molecular Theory Of Gases

Q: What are the postulates of the kinetic molecular theory of gases?

A gas has a huge number of tiny, identical molecules whose own volume is negligible. They do not attract each other, move randomly in straight lines, and collide elastically with each other and the walls. Wall collisions cause pressure, speeds are distributed but the distribution is steady, and average kinetic energy is proportional to absolute temperature.

Q: How is the kinetic gas equation derived?

A molecule hitting a wall reverses its x-momentum, giving an impulse 2mu_x, and returns after 2l/u_x seconds. Its average force is m u_x squared over l. Summing over N molecules, and using the fact that u_x squared averages to one-third of u squared for random motion, gives pV = one-third m N u squared, the kinetic gas equation.

Q: What is the relation between kinetic energy and temperature of a gas?

Combining the kinetic gas equation with pV = RT gives the average translational kinetic energy as 3/2 RT per mole, or 3/2 kT per molecule. It depends only on absolute temperature, so all gases at the same temperature have the same average kinetic energy, 3.74 kJ per mole at 300 K.

Q: What is the ratio of most probable, average and rms speed?

The most probable speed is root of 2RT/M, the average speed root of 8RT/(pi M) and the rms speed root of 3RT/M. They are in the ratio 1 : 1.128 : 1.225 for every gas at every temperature, so the most probable speed is always the smallest and the rms speed the largest.

Q: How does temperature affect the Maxwell-Boltzmann distribution?

On heating, the peak of the Maxwell-Boltzmann curve moves to a higher speed and becomes lower, and the curve spreads out, while the area under it stays 1. A larger fraction of molecules moves at high speed, which is why reaction rates rise sharply with temperature.

Q: Why do lighter gas molecules move faster at the same temperature?

At the same temperature all gas molecules have the same average kinetic energy, one-half m u squared. A lighter molecule must move faster to carry the same energy. Speed varies as one over the square root of molar mass, so hydrogen moves four times faster than oxygen.

Q: Is kinetic theory of gases part of JEE Advanced chemistry?

Yes. The JEE Advanced syllabus lists kinetic theory of gases with average, root mean square and most probable velocities and their relation with temperature under States of Matter: Gases and Liquids. The chapter has been dropped from NCERT Class 11, JEE Main and NEET, so JEE Advanced is where it is examined.

Q: What types of kinetic theory questions are asked in JEE Advanced?

Typical questions compare speeds of two gases or two temperatures, test the shape of Maxwell-Boltzmann curves when temperature or molar mass changes, use dissociation to change molar mass, compute average kinetic energy per molecule, and find rms speed from pressure and density. Most reduce to u proportional to root T over M.

ChemistryStates Of MatterFor JEE aspirants

The kinetic molecular theory of gases explains the gas laws by picturing a gas as a huge number of tiny molecules in constant, random motion. Their collisions with the walls create pressure, and their average kinetic energy is fixed by the absolute temperature alone: $E_{k} = \frac{3}{2} R T$ per mole. From this come the three molecular speeds ( $u_{m p}$ , $u_{a v}$ , $u_{r m s}$ ) and the Maxwell-Boltzmann distribution. The kinetic molecular theory of gases, with the speeds and their temperature dependence, is examined in JEE Advanced.

On this page1Laws vs theory2Postulates3Pressure from collisions4KE and temperature5Three speeds6Maxwell curve7Gas laws explained8Examples

Key Formulas - Quick Reference

★ Must learn Kinetic gas equation: $p V = \frac{1}{3} m N \overline{u^{2}} = \frac{1}{3} m N u_{r m s}^{2}$ , and $p V = \frac{2}{3} E_{k}$ .
★ Must learn Average kinetic energy: $E_{k} = \frac{3}{2} R T$ per mole, $\frac{3}{2} k T$ per molecule; $k = \frac{R}{N _{A}} = 1.38 \times 1 0^{- 23}$ J K $^{- 1}$ .
★ Must learn $u_{r m s} = \frac{3 R T}{M} = \frac{3 p}{d}$ , with $M$ in kg mol $^{- 1}$ .
$u_{a v} = \frac{8 R T}{π M}$ and $u_{m p} = \frac{2 R T}{M}$ .
★ Must learn $u_{m p} : u_{a v} : u_{r m s} = 1 : 1.128 : 1.225$ (always $u_{m p} < u_{a v} < u_{r m s}$ ).
★ Must learn For any speed, $u \propto \frac{T}{M}$ : $\frac{u _{1}}{u _{2}} = \frac{T _{1} M _{2}}{T _{2} M _{1}}$ .
Definitions: $u_{a v} = \frac{u _{1} + u _{2} + \dots + u _{n}}{n}$ and $u_{r m s} = \frac{u _{1}^{2} + u _{2}^{2} + \dots + u _{n}^{2}}{n}$ .

1. From Laws to a Theory

Boyle's and Charles's laws are summaries of experiments: they say what happens. A theory is a mental model that says why. The kinetic molecular theory (KMT) builds a microscopic picture of a gas and then derives every gas law from it. Its success in matching experiment is the evidence that the picture is right.

2. Postulates of the Kinetic Molecular Theory

A gas consists of a very large number of identical molecules. They are so small and so far apart that their own volume is negligible compared with the volume of the container: they are treated as point masses.
There are no attractive forces between the molecules at ordinary temperature and pressure.
The molecules are in constant, random motion in straight lines, in all directions.
They collide with one another and with the walls. Pressure is the result of collisions with the walls.
All collisions are perfectly elastic: individual molecules may gain or lose energy, but the total kinetic energy is unchanged.
At any instant molecules have different speeds, which keep changing, but the distribution of speeds at a given temperature stays constant.
The average kinetic energy of the molecules is directly proportional to the absolute temperature.

Figure 1: The kinetic model. For

N_{2}

at STP the average spacing

(\frac{V}{N})^{1/3} = (\frac{22.7 \times 1 0 ^{- 3}}{6.022 \times 1 0 ^{23}})^{1/3} m \approx 3.3 nm

is about nine times the molecular size, so the molecules fill under

0.1%

of the volume.

Each postulate is chosen to match something we observe:

Figure 2: Every postulate is there to explain an observation. The two outlined in red, negligible molecular volume and no attraction, are exactly the ones that break down at high pressure and low temperature.

Postulates 1 and 2 are only approximations. At high pressure the molecular volume is no longer negligible, and at low temperature the attractions matter. This is why real gases deviate from ideal behaviour (see Deviation From Ideal Gas Behaviour).

3. Pressure from Molecular Collisions

Take $N$ molecules, each of mass $m$ , in a cube of side $l$ (volume $V = l^{3}$ ). Follow one molecule moving towards wall A with velocity component $u_{x}$ .

Figure 3: Each molecule hits wall A every

\frac{2 l}{u _{x}}

seconds and gives it

2 m u_{x}

each time, so its average force is

\frac{m u _{x}^{2}}{l}

. Adding all

N

molecules and using

\overline{u_{x}^{2}} = \frac{u ^{2}}{3}

gives

p = \frac{m N u ^{2}}{3 V}

On an elastic hit its $x$ -momentum changes from $+ m u_{x}$ to $- m u_{x}$ , so the wall receives an impulse $2 m u_{x}$ .
It travels to the opposite wall and back, a distance $2 l$ , before hitting A again: time between hits $= \frac{2 l}{u _{x}}$ .
Average force of this molecule on A $= \frac{2 m u _{x}}{2 l / u _{x}} = \frac{m u _{x}^{2}}{l}$ .
For all $N$ molecules: force $= \frac{m N u _{x}^{2}}{l}$ . Motion is random, so $\overline{u_{x}^{2}} = \overline{u_{y}^{2}} = \overline{u_{z}^{2}} = \frac{u ^{2}}{3}$ .
Pressure = force/area $= \frac{m N u ^{2}}{3 l \cdot l ^{2}} = \frac{m N u ^{2}}{3 V}$ .

p V = \frac{1}{3} m N \overline{u^{2}} = \frac{1}{3} m N u_{r m s}^{2}

This is the kinetic gas equation. Here $m N$ is the total mass of gas, so $u_{r m s} = \frac{3 p V}{m N} = \frac{3 p}{d}$ : the rms speed can be found from pressure and density alone.

Key idea

Pressure is momentum delivered to the walls per second per unit area:

p V = \frac{1}{3} m N u_{r m s}^{2}

4. Kinetic Energy and Temperature

The total translational kinetic energy is $E_{k} = \frac{1}{2} m N \overline{u^{2}}$ , so the kinetic gas equation becomes $p V = \frac{2}{3} E_{k}$ . For one mole, $p V = R T$ , so

E_{k} = \frac{3}{2} R T (per mole), \overline{ε} = \frac{3}{2} k T (per molecule)

At 300 K this is $3.74$ kJ mol $^{- 1}$ , or $6.21 \times 1 0^{- 21}$ J per molecule, for every gas. The average kinetic energy depends only on the absolute temperature, not on the nature, mass or pressure of the gas. Absolute temperature is simply a measure of this energy, and at 0 K the translational motion (in this classical model) would stop.

Figure 4: (a)

E_{k} = \frac{3}{2} R T

per mole is a straight line through the origin, identical for

He

N_{2}

C O_{2}

. (b) Speed grows only as

T

: at 300 K

u_{r m s}

517 m s^{- 1}

for

N_{2}

and

412 m s^{- 1}

for

C O_{2}

; quadrupling

T

only doubles the speed.

Exam Trick Energy follows $T$ , speed follows $T / M$ . At the same temperature

H_{2}

and

O_{2}

have equal average KE, but

H_{2}

moves

32/2 = 4

times faster. To double a speed you must quadruple the kelvin temperature.

Key idea

Average KE is $\frac{3}{2} R T$ per mole for every gas; temperature is a direct measure of molecular motion.

Quick Recall: tap to check

What is the average KE of one molecule of any gas at

27^{\circ} C

\frac{3}{2} k T = 1.5 \times 1.38 \times 1 0^{- 23} \times 300 = 6.21 \times 1 0^{- 21}

Why is

\overline{u_{x}^{2}} = \frac{1}{3} \overline{u^{2}}

u^{2} = u_{x}^{2} + u_{y}^{2} + u_{z}^{2}

, and random motion makes the three averages equal.

Which postulates fail for real gases?

Negligible molecular volume and no intermolecular attraction.

5. Three Molecular Speeds

Collisions keep changing each molecule's speed, so a gas can only be described by averages. Three are used:

Speed	Definition	Formula	$N_{2}$ at 300 K
Most probable, $u_{m p}$	speed of the largest number of molecules (peak of the curve)	$\frac{2 R T}{M}$	422 m s $^{- 1}$
Average, $u_{a v}$	arithmetic mean of all speeds	$\frac{8 R T}{π M}$	476 m s $^{- 1}$
Root mean square, $u_{r m s}$	square root of the mean of the squared speeds	$\frac{3 R T}{M}$	517 m s $^{- 1}$

The mean square speed $\overline{u^{2}}$ is the direct measure of kinetic energy, which is why $u_{r m s}$ appears in the kinetic gas equation. The formulas differ only in the number under the root, so the three speeds are always in the same ratio:

Figure 5: The three speeds differ only in the number under the root: 2,

\frac{8}{π} = 2.55

and 3. So their ratio is fixed for every gas at every temperature, and the order is always

u_{m p} < u_{a v} < u_{r m s}

u_{m p} : u_{a v} : u_{r m s} = 2 : \frac{8}{π} : 3 = 1 : 1.128 : 1.225

Exam Trick Remember the numbers under the root as 2, 2.55, 3 for mp, av, rms (

8/ π = 2.55

). Bigger number, bigger speed, so the order

u_{m p} < u_{a v} < u_{r m s}

writes itself. And always put

M

in kg mol

^{- 1}

M_{O_{2}} = 0.032

6. The Maxwell-Boltzmann Distribution

Maxwell and Boltzmann worked out how many molecules have each speed. The plot of the fraction of molecules against speed is the Maxwell-Boltzmann distribution:

Figure 6: Computed Maxwell-Boltzmann curve for

N_{2}

at 300 K. The long high-speed tail pulls the averages right of the peak:

u_{m p} = 422 < u_{a v} = 476 < u_{r m s} = 517 m s^{- 1}

. The area under the whole curve is 1 (all the molecules).

Very few molecules are very slow or very fast.
The peak is at $u_{m p}$ ; the curve is not symmetric but has a long tail on the high-speed side, which pulls $u_{a v}$ and $u_{r m s}$ to the right of the peak.
The total area under the curve is 1 (all the molecules), whatever the temperature or gas.
Individual speeds keep changing, but at a fixed temperature the curve itself does not change.

6.1 Effect of temperature

Figure 7: Heating

N_{2}

from 300 K to 900 K moves

u_{m p}

from 422 to

731 m s^{- 1}

(a factor

3

), lowers and broadens the peak, and raises the share of molecules faster than

800 m s^{- 1}

from

6.6%

49%

. This fast tail is why reaction rates climb steeply with temperature.

On heating, the peak shifts to a higher speed and becomes lower, and the curve broadens. More molecules move at high speed. Because the area stays 1, a taller peak and a wider spread cannot both happen.

6.2 Effect of molar mass

Figure 8: Same temperature, different molar mass. Because

u_{m p} = \frac{2 R T}{M}

, the peaks sit at 265, 422 and

1117 m s^{- 1}

for

C l_{2}

N_{2}

and He. Molecules of every gas have the same average KE; lighter ones need higher speeds to carry it.

Higher temperature (same gas)

Peak moves right ( $u_{m p} \propto T$ ), becomes lower, curve spreads. More fast molecules.

Lower molar mass (same T)

Peak moves right ( $u_{m p} \propto 1/ M$ ), becomes lower, curve spreads. The average KE stays the same.

JEE Advanced The distribution function is

f (u) = 4 π (\frac{M}{2 π R T})^{3/2} u^{2} e^{- M u^{2} /2 R T}

, where

f (u) d u

is the fraction of molecules with speeds between

u

and

u + d u

. Setting

\frac{df}{d u} = 0

gives

u_{m p} = \frac{2 R T}{M}

. Substituting back, the peak height

f (u_{m p}) = \frac{4}{π e} \frac{M}{2 R T} \propto \frac{M}{T}

, so heating a gas four-fold halves the peak height. The fraction of molecules with energy above

E

grows roughly as

e^{- E / R T}

, which is the origin of the Arrhenius factor in chemical kinetics.

Key idea

The Maxwell curve has fixed area: anything that raises speeds (higher

T

, lower

M

) moves the peak right and makes it lower and wider.

Quick Recall: tap to check

At the same temperature, which has the taller Maxwell peak,

C l_{2}

N_{2}

C l_{2}

: heavier molecules have a lower

u_{m p}

and a narrower, taller curve.

Ratio of

u_{r m s}

of He at 1200 K to

O_{2}

at 300 K?

\frac{1200 \times 32}{300 \times 4} = 32 = 5.66

Does heating raise the fraction of molecules at

u_{m p}

No. The peak height falls as

1/ T

; the molecules spread over more speeds.

7. The Gas Laws Follow from the Theory

Everything in the Gas Laws concept can be derived from $p V = \frac{2}{3} E_{k}$ and $E_{k} \propto T$ :

Law	How KMT explains it
Boyle	At fixed $T$ , $E_{k}$ is fixed, so $p V = \frac{2}{3} E_{k}$ is constant.
Charles	At fixed $p$ , $V = \frac{2 E _{k}}{3 p} \propto T$ .
Gay-Lussac	At fixed $V$ , $p = \frac{2 E _{k}}{3 V} \propto T$ : heated molecules hit harder and more often.
Avogadro	Equal $p$ , $V$ and $T$ give equal total $E_{k}$ ; equal KE per molecule then means equal $N$ .
Dalton	Molecules do not interact, so each gas delivers its own momentum to the walls independently.
Graham	Molecules escape through a hole at a rate $\propto u_{a v} \propto \frac{1}{M}$ .

Figure 9: Both laws come from

p V = \frac{2}{3} E_{k}

. Halving

V

at fixed

T

keeps

E_{k}

fixed, so

p

doubles. Doubling

T

at fixed

V

doubles

E_{k}

, so

p

doubles; the average speed rises only by

2

, but each hit is harder and hits come more often.

Because the theory reproduces every gas law and the measured speeds, it is accepted as a correct model of an ideal gas.

8. Solving Speed Problems

Figure 10: Most speed questions are ratios; the absolute formulas are needed only for a numerical speed. The single most common error is putting

M

in g mol

^{- 1}

, which makes

u

about 31.6 times too small.

8.1 The whole concept at a glance

Figure 11: The whole concept on one page. The chain to remember: collisions give pressure, pressure gives

E_{k} = \frac{3}{2} R T

, and

E_{k}

gives the speeds.

9. Solved Examples

Solved Example 1

Calculate

u_{m p}

u_{a v}

and

u_{r m s}

O_{2}

27^{\circ} C

Solution:

$T = 300$ K, $M = 0.032$ kg mol $^{- 1}$ , $\frac{R T}{M} = \frac{8.314 \times 300}{0.032} = 7.794 \times 1 0^{4}$ m $^{2}$ s $^{- 2}$ .

u_{m p} = 2 \times 7.794 \times 1 0^{4} = 395 m s^{- 1}, u_{a v} = \frac{8}{π} \times 7.794 \times 1 0^{4} = 446 m s^{- 1}

u_{r m s} = 3 \times 7.794 \times 1 0^{4} = 484 m s^{- 1}

Check: $395 : 446 : 484 = 1 : 1.128 : 1.225$ .

Solved Example 2

At what temperature will the rms speed of

S O_{2}

equal that of

O_{2}

at 300 K?

Solution:

Equal speeds need equal $\frac{T}{M}$ : $\frac{T}{64} = \frac{300}{32}$ , so $T = 600$ K.

Solved Example 3

Find the total translational kinetic energy of 2 mol of helium at

27^{\circ} C

, and the average KE of one He atom.

Solution:

$E_{k} = \frac{3}{2} n R T = 1.5 \times 2 \times 8.314 \times 300 = 7483$ J $= 7.48$ kJ.

Per atom: $\frac{3}{2} k T = 1.5 \times 1.38 \times 1 0^{- 23} \times 300 = 6.21 \times 1 0^{- 21}$ J. The same value holds for any gas at 300 K.

Solved Example 4

The density of a gas is 1.25 g L

^{- 1}

at 1 atm (101325 Pa). Find its rms speed without knowing its identity or temperature.

Solution:

$d = 1.25$ kg m $^{- 3}$ . $u_{r m s} = \frac{3 p}{d} = \frac{3 \times 101325}{1.25} = 2.432 \times 1 0^{5} = 493$ m s $^{- 1}$ .

Solved Example 5

The rms speed of

O_{2}

molecules at

T

K is

u

. If the absolute temperature is doubled and the

O_{2}

molecules dissociate into O atoms, the rms speed becomes:
(A)

u /2

(B)

u

(C)

2 u

(D)

2 u

Solution:

Answer: (D). $u \propto T / M$ . $T$ doubles and $M$ halves (32 to 16), so $\frac{T}{M}$ becomes four times larger and $u$ doubles.

Solved Example 6

When a gas is heated from 300 K to 600 K at constant volume, which statement about its Maxwell-Boltzmann curve is WRONG?
(A) The most probable speed increases by a factor

2

(B) The height of the peak increases
(C) The area under the curve is unchanged
(D) The fraction of fast molecules increases

Solution:

Answer: (B). The peak height is proportional to $M / T$ , so it falls by $2$ when $T$ doubles. The curve gets lower and wider; its area stays 1.

Solved Example 7

A container holds a mixture of

H_{2}

and

O_{2}

at 300 K. Find the ratio of (a) their average kinetic energies per molecule and (b) their rms speeds.

Solution:

(a) $1 : 1$ : average KE depends only on $T$ .

(b) $\frac{u _{H_{2}}}{u _{O_{2}}} = \frac{32}{2} = 4 : 1$ .

Practice Questions

Calculate the total number of electrons present in 1.4 g of dinitrogen gas. (Ex. 5.13)Answer: $\frac{1.4}{28} = 0.05$ mol $N_{2}$ $\times 14 \times 6.022 \times 1 0^{23} = 4.22 \times 1 0^{23}$ electrons.
How long would it take to distribute one Avogadro number of wheat grains at $1 0^{10}$ grains per second? (Ex. 5.14)Answer: $6.022 \times 1 0^{13}$ s $\approx 1.9 \times 1 0^{6}$ years: a sense of how many molecules a mole holds.
In terms of Charles's law, explain why $- 273^{\circ} C$ is the lowest possible temperature. (Ex. 5.21)Answer: $V_{t} = V_{0} (1 + \frac{t}{273.15})$ gives $V = 0$ at $- 273.15^{\circ} C$ ; lower $t$ would give a negative volume, which is impossible. In KMT terms, the average KE would reach zero.
rms speed of $H_{2}$ at $0^{\circ} C$ ?Answer: $\frac{3 \times 8.314 \times 273.15}{0.002} = 1.85 \times 1 0^{3}$ m s $^{- 1}$ .
At what temperature is the average speed of $N_{2}$ equal to 500 m s $^{- 1}$ ?Answer: $T = \frac{π M u ^{2}}{8 R} = \frac{π \times 0.028 \times 2.5 \times 1 0 ^{5}}{8 \times 8.314} = 331$ K.
By what factor does $u_{r m s}$ change if the pressure of a gas is doubled at constant temperature?Answer: No change: $u_{r m s} = 3 R T / M$ depends only on $T$ and $M$ (doubling $p$ also doubles $d$ ).
Average KE per molecule of $C O_{2}$ at 400 K?Answer: $\frac{3}{2} \times 1.38 \times 1 0^{- 23} \times 400 = 8.28 \times 1 0^{- 21}$ J.

Common Mistakes to Avoid

Watch out

Putting $M$ in g mol $^{- 1}$ in the speed formulas. It must be kg mol $^{- 1}$ , or the speed comes out about 31.6 times too small.
Thinking speed is proportional to $T$ . Speed is proportional to $T$ ; kinetic energy is proportional to $T$ .
Saying heavier gases have more kinetic energy at the same temperature. Average KE is the same for all gases; heavier molecules just move more slowly.
Saying the Maxwell peak rises on heating. It moves right and gets lower; the area is fixed at 1.
Using $\frac{3}{2} R T$ for one molecule. Per molecule it is $\frac{3}{2} k T$ , with $k = R / N_{A}$ .
Taking $u_{r m s}$ as the square of the mean speed. It is the square root of the mean of the squares, which is larger than $u_{a v}$ .
Swapping the order of the speeds. Always $u_{m p} < u_{a v} < u_{r m s}$ .
Forgetting that dissociation halves $M$ (for example $O_{2}$ to 2O), which raises the speed.

Frequently Asked Questions

What are the postulates of the kinetic molecular theory of gases?

A gas has a huge number of tiny, identical molecules whose own volume is negligible. They do not attract each other, move randomly in straight lines, and collide elastically with each other and the walls. Wall collisions cause pressure, speeds are distributed but the distribution is steady, and average kinetic energy is proportional to absolute temperature.

How is the kinetic gas equation derived?

A molecule hitting a wall reverses its x-momentum, giving an impulse 2mu_x, and returns after 2l/u_x seconds. Its average force is m u_x squared over l. Summing over N molecules, and using the fact that u_x squared averages to one-third of u squared for random motion, gives pV = one-third m N u squared, the kinetic gas equation.

What is the relation between kinetic energy and temperature of a gas?

Combining the kinetic gas equation with pV = RT gives the average translational kinetic energy as 3/2 RT per mole, or 3/2 kT per molecule. It depends only on absolute temperature, so all gases at the same temperature have the same average kinetic energy, 3.74 kJ per mole at 300 K.

What is the ratio of most probable, average and rms speed?

The most probable speed is root of 2RT/M, the average speed root of 8RT/(pi M) and the rms speed root of 3RT/M. They are in the ratio 1 : 1.128 : 1.225 for every gas at every temperature, so the most probable speed is always the smallest and the rms speed the largest.

How does temperature affect the Maxwell-Boltzmann distribution?

On heating, the peak of the Maxwell-Boltzmann curve moves to a higher speed and becomes lower, and the curve spreads out, while the area under it stays 1. A larger fraction of molecules moves at high speed, which is why reaction rates rise sharply with temperature.

Why do lighter gas molecules move faster at the same temperature?

At the same temperature all gas molecules have the same average kinetic energy, one-half m u squared. A lighter molecule must move faster to carry the same energy. Speed varies as one over the square root of molar mass, so hydrogen moves four times faster than oxygen.

Is kinetic theory of gases part of JEE Advanced chemistry?

Yes. The JEE Advanced syllabus lists kinetic theory of gases with average, root mean square and most probable velocities and their relation with temperature under States of Matter: Gases and Liquids. The chapter has been dropped from NCERT Class 11, JEE Main and NEET, so JEE Advanced is where it is examined.

What types of kinetic theory questions are asked in JEE Advanced?

Typical questions compare speeds of two gases or two temperatures, test the shape of Maxwell-Boltzmann curves when temperature or molar mass changes, use dissociation to change molar mass, compute average kinetic energy per molecule, and find rms speed from pressure and density. Most reduce to u proportional to root T over M.

Previous year questions on Kinetic Molecular Theory Of Gases

1 question from past papers, each with a step-by-step solution.

JEE Advanced 2024 Paper 1, Chemistry Section 1 Q1

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