JEE Mains Physics · Kinetic Theory

Pressure, Kinetic Energy and Temperature

Gas pressure comes from molecules bouncing off the walls, P = (1/3)ρv²rms, and the average translational kinetic energy of a molecule, (3/2)kT, depends on the temperature alone.

All Kinetic Theory notes JEE Mains Physics strategy

Why this matters

Seventeen PYQs, one of them asking for a number, and one from 2026. Four ask where pressure comes from and how PV is related to the kinetic energy, eight turn a temperature into a kinetic energy or back, and five compare two gases at one temperature. All seventeen rest on one fact: the average translational kinetic energy of a molecule is (3/2)kT, whatever the gas, the pressure or the volume.

Concept 1 of 3: Pressure from molecular impacts, and PV = (2/3)E

A molecule that hits a wall and bounces back elastically reverses the part of its velocity that points at the wall, so it hands the wall a little momentum. Billions of such hits each second add up to a steady force on every square metre: the pressure. Averaging over all directions gives P = ⅓ρ⟨v²⟩, and multiplying by V shows that PV is two-thirds of the gas's translational kinetic energy.

Definition

One elastic hit with velocity component u normal to the wall: momentum change $2mu$ .
$P = \dfrac{1}{3}\dfrac{N}{V}m\overline{v^{2}} = \dfrac{1}{3}\rho v_{rms}^{2}$ .
$PV = \dfrac{2}{3}E_{trans}$ , so $E_{trans} = \dfrac{3}{2}PV = \dfrac{3}{2}nRT$ . PV is NOT the translational energy itself.
Translational kinetic energy per unit volume $= \dfrac{3}{2}P$ .
Number of molecules from P, V and the mean energy per molecule $\overline{KE}$ : $N = \dfrac{3PV}{2\,\overline{KE}}$ .
Assumptions: molecules are points, exert no forces except in collisions, and collide elastically in a negligible time.

Kinetic-theory pressure

P = \frac{1}{3}\rho v_{rms}^{2} \qquad PV = \frac{2}{3}E_{trans}

Worked example

A gas at a pressure of

1.5 \times 10^{5}\ \text{Pa}

has a density of

1.8\ \text{kg/m}^{3}

. Find the rms speed of its molecules and the translational kinetic energy in each cubic metre.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 24 Jan 2023 · Q4Moderate

Example 1 · Kinetic Theory · Pressure, Kinetic Energy and Temperature

Given below are two statements: Statement I: The temperature of a gas is

-73^{\circ}C

. When the gas is heated to

527^{\circ}C

, the root mean square speed of the molecules is doubled. Statement II: The product of pressure and volume of an ideal gas will be equal to translational kinetic energy of the molecules. In the light of the above statements, choose the correct answer from the options given below:

Taking PV as the kinetic energy

PV is two-thirds of the translational kinetic energy, not equal to it. The energy is (3/2)PV.

Squaring the mean speed

Pressure depends on the mean of v², which is larger than the square of the mean speed. That is why the rms speed, not the average speed, appears in P = ⅓ρv²rms.

Drill 3 more on pressure from molecular impacts, and pv = (2/3)e

Concept 2 of 3: Average kinetic energy is (3/2)kT

Temperature is a measure of how much translational kinetic energy the molecules carry on average: (3/2)kT per molecule. So the energy is proportional to the kelvin temperature and to nothing else. Double the energy and the kelvin temperature doubles. To compare with an energy in electron-volts, set (3/2)kT equal to it.

Definition

Per molecule: $\overline{KE} = \dfrac{3}{2}kT$ ; per mole $\dfrac{3}{2}RT$ ; for n moles $\dfrac{3}{2}nRT$ .
$k = 1.38 \times 10^{-23}\ \text{J/K}$ ; $1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$ .
$\overline{KE} \propto T$ in kelvin. It does not depend on the pressure, the volume or the kind of gas.
A ratio of energies is a ratio of kelvin temperatures; subtract 273 only at the end.
An electron accelerated through a potential difference V gains $eV$ ; set $\dfrac{3}{2}kT = eV$ to find the matching temperature.
Total translational kinetic energy of a mass m of gas: $\dfrac{3}{2}\dfrac{m}{M}RT$ .

Mean translational kinetic energy

\overline{KE} = \frac{3}{2}kT \qquad E_{trans} = \frac{3}{2}nRT

Worked example

The average translational kinetic energy of a gas molecule is E at 27 °C. At what temperature is it 3E? What is E in electron-volts?

(k = 1.38 \times 10^{-23}\ \text{J/K})

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The same idea in a real exam question:

JEE Mains · 2024 · 9 April 2024 · Q93Moderate

Example 2 · Kinetic Theory · Pressure, Kinetic Energy and Temperature

The temperature of a gas is

-78^{\circ}C

and the average translational kinetic energy of its molecules is

K

. The temperature at which the average translational kinetic energy of the molecules of the same gas becomes

2\text{ }K

is:

Doubling the Celsius temperature

Twice the kinetic energy means twice the kelvin temperature. From 50 °C (323 K) that is 646 K, or 373 °C, not 100 °C.

Per molecule or per mole

(3/2)kT is for one molecule; (3/2)RT is for one mole. A numerical answer off by about 6 × 10²³ has mixed the two.

Drill 7 more on average kinetic energy is (3/2)kt

Concept 3 of 3: Same temperature, same mean kinetic energy

At one temperature every molecule, light or heavy, has the same average translational kinetic energy. A light molecule makes up for its small mass by moving fast, so its rms speed is higher, but its energy is the same. In a mixture the mass ratio of the gases changes nothing about the energy per molecule.

Definition

Equal T: $\tfrac{1}{2}m_1\overline{v_1^{2}} = \tfrac{1}{2}m_2\overline{v_2^{2}}$ , so $v_{rms} \propto \dfrac{1}{\sqrt{m}}$ .
In any mixture the translational kinetic energy per molecule is in the ratio 1 : 1, whatever the mass ratio.
Equal mean kinetic energy means equal temperature, whatever the size of the container.
If rotation is counted, the total energy per molecule is $\dfrac{f}{2}kT$ : a diatomic molecule then carries more than a monatomic one at the same T. Read "average kinetic energy" as translational unless the question says otherwise.
The average velocity of the molecules, a vector, is zero, and so is their average momentum; the average speed is not.

Equal temperatures

\tfrac{1}{2}m_1\overline{v_1^{2}} = \tfrac{1}{2}m_2\overline{v_2^{2}} = \tfrac{3}{2}kT

Worked example

A vessel at 350 K holds neon (M = 20 g/mol) and krypton (M = 84 g/mol) in the ratio 3 : 1 by mass. Find the ratio of their average translational kinetic energies per molecule, and of their rms speeds.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 4 Apr 2026 Shift 2 · Q8Moderate

Example 3 · Kinetic Theory · Pressure, Kinetic Energy and Temperature

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Assertion A: If the average kinetic energy of

H_{2}

and

O_{2}

molecules, kept in two different sized containers are same, then their temperatures will be same. Reason R: The r.m.s. speed of

H_{2}

and

O_{2}

molecules are same at same temperature. Choose the correct answer from the options given below

Letting the mass ratio decide the answer

A mixture's mass ratio is a distractor here. The energy per molecule depends on T alone, so it is 1 : 1 for any mix.

Equal energy is not equal speed

At one temperature the lighter molecule moves faster. Equal kinetic energies make the speeds unequal, in the ratio √(m₂/m₁).

Drill 4 more on same temperature, same mean kinetic energy

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Pressure from molecular impacts, and PV = (2/3)E
Kinetic-theory pressure
$P = \frac{1}{3}\rho v_{rms}^{2} \qquad PV = \frac{2}{3}E_{trans}$
Average kinetic energy is (3/2)kT
Mean translational kinetic energy
$\overline{KE} = \frac{3}{2}kT \qquad E_{trans} = \frac{3}{2}nRT$
Same temperature, same mean kinetic energy
Equal temperatures
$\tfrac{1}{2}m_1\overline{v_1^{2}} = \tfrac{1}{2}m_2\overline{v_2^{2}} = \tfrac{3}{2}kT$

Watch out for (6)

Taking PV as the kinetic energy
→ Pressure from molecular impacts, and PV = (2/3)E
Squaring the mean speed
→ Pressure from molecular impacts, and PV = (2/3)E
Doubling the Celsius temperature
→ Average kinetic energy is (3/2)kT
Per molecule or per mole
→ Average kinetic energy is (3/2)kT
Letting the mass ratio decide the answer
→ Same temperature, same mean kinetic energy
Equal energy is not equal speed
→ Same temperature, same mean kinetic energy

Test yourself on Kinetic Theory

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