Kinetic Theory of Gases Formula Sheet — JEE Main Physics

Every key Kinetic Theory of Gases formula, definition and theorem for JEE Main Physics in one place — with common examiner traps and worked examples. Free to read; blurt from memory, then check your gaps.

Syllabus — topics coveredNTA · 11 sub-topics

Equation of state of a perfect gas
Work done on compressing a gas
Kinetic theory assumptions
Concept of pressure
Kinetic interpretation of temperature
RMS speed of gas molecules
Degrees of freedom
Law of equipartition of energy
Applications to specific heat capacities
Mean free path
Avogadro's number

Ideal Gas Equation & Gas Laws

Ideal gas equation

P V = μ RT = N k_{B} T = \frac{m}{M} RT

μ =

moles,

N =

number of molecules,

m =

mass,

M =

molar mass,

T =

absolute temp (K)

Constants

▸ $R = 8.314 J mol^{- 1} K^{- 1}$ (universal gas constant).
▸ $k_{B} = \frac{R}{N _{A}} = 1.38 \times 1 0^{- 23} J K^{- 1}$ (Boltzmann).
▸ $N_{A} = 6.022 \times 1 0^{23} mol^{- 1}$ ; molar volume at STP ( $0^{\circ}$ C, $1$ atm) $= 22.4$ L.

Three P-V isotherms (hyperbolae PV=const) of an ideal gas at increasing temperatures — Each isotherm: $P V =$ const (Boyle's law); higher curve $=$ higher $T$ .

⚡ Shortcut · Combined gas law

For a fixed mass of gas

\frac{P _{1} V _{1}}{T _{1}} = \frac{P _{2} V _{2}}{T _{2}}

— covers Boyle/Charles/Gay-Lussac in one step. Density form:

\frac{P}{ρT} = \frac{R}{M} =

const.

Work done on compressing a gas

W_{o n} = - \int_{V_{1}}^{V_{2}} P d V; W_{o n, i so t h er ma l} = μ RT ln \frac{V _{1}}{V _{2}}

positive when the gas is compressed (

V_{2} < V_{1}

). Constant pressure:

W_{o n} = P (V_{1} - V_{2})

. Adiabatic:

W_{o n} = \frac{P _{2} V _{2} - P _{1} V _{1}}{γ - 1}

, which is MORE than isothermal for the same compression ratio, because P rises faster

🎯 Exam · Gas leaking or escaping

Mass left in a vessel

\propto \frac{P V}{T}

, so

\frac{m _{2}}{m _{1}} = \frac{P _{2} T _{1}}{P _{1} T _{2}}

(fixed V). Open vessel heated (P,V fixed):

m T =

const, so a fraction

1 - \frac{T _{1}}{T _{2}}

escapes. A leak at constant T leaves

v_{r m s}

UNCHANGED — only n and P fall.

Comparative: the gas laws

Law	Held const	Relation
Boyle	$T, μ$	$P V =$ const ( $P \propto 1/ V$ )
Charles	$P, μ$	$V / T =$ const ( $V \propto T$ )
Gay-Lussac	$V, μ$	$P / T =$ const ( $P \propto T$ )
Avogadro	$P, T$	equal $V \Rightarrow$ equal N
Dalton	$V, T$	$P = P_{1} + P_{2} + \dots$ (partials)

🚫 Examiner Trap · Ideal gas & gas laws

(1) T MUST be in

kelvin

in every gas law — using

^{\circ}

C is the #1 blunder. (2)

P V = μ RT

and

22.4

L/mol at STP hold only for an

ideal

gas — best at

low P

high T

. (3) Use

N k_{B}

for molecules but

μ R

for moles — don't mix N with

μ

. (4) Charles/Gay-Lussac are linear in

T (K)

, NOT in

^{\circ}

Kinetic Theory — Assumptions & Pressure

Assumptions

▸Large number of identical molecules in random motion; molecular size $≪$ inter-molecular distance.
▸No intermolecular forces except during $elastic$ collisions.
▸Collisions (with walls and each other) are perfectly elastic; collision duration negligible.
▸Between collisions molecules move in straight lines (Newton's first law).

Pressure of an ideal gas

P = \frac{1}{3} \frac{m N}{V} \overline{v^{2}} = \frac{1}{3} ρ \overline{v^{2}} = \frac{1}{3} ρ v_{r m s}^{2}

\overline{v^{2}} =

mean-square speed

= v_{r m s}^{2}

;

ρ = m N / V =

density;

N =

molecules

Pressure

\leftrightarrow

kinetic energy

P = \frac{2}{3} \frac{E}{V}

E =

total translational KE; pressure

= \frac{2}{3}

of translational KE per unit volume

A molecule strikes a wall with velocity +v_x and rebounds with -v_x, transferring momentum 2mv_x — Pressure $=$ rate of momentum transfer ( $2 m v_{x}$ per elastic bounce).

⚡ Shortcut · Pressure from rms speed

Since

\overline{v^{2}} = v_{r m s}^{2}

, just write

P = \frac{1}{3} ρ v_{r m s}^{2}

\Rightarrow v_{r m s} = 3 P / ρ

. The result is

independent of container shape

— only n and

\overline{v^{2}}

matter.

🚫 Examiner Trap · Pressure & mean-square speed

(1)

\overline{v^{2}}

is the

mean of the squares

, NOT

(\overset{v}{ˉ})^{2}

— and

\overline{v^{2}} \neq = v_{a v}^{2}

. (2) The

\frac{1}{3}

comes from averaging over 3 axes (

\overline{v_{x}^{2}} = \frac{1}{3} \overline{v^{2}}

) — don't drop it. (3) Pressure relates to

translational

KE only; rotation/vibration don't push the wall.

Kinetic Interpretation of Temperature

Mean translational KE per molecule

\overline{E} = \frac{1}{2} m \overline{v^{2}} = \frac{3}{2} k_{B} T

depends

only

on T — same for every gas at a given temperature

Total translational KE

E = \frac{3}{2} N k_{B} T = \frac{3}{2} μ RT

for N molecules /

μ

moles (translational part only)

Average kinetic energy increases linearly with absolute temperature through the origin — $\overline{E} \propto T$ — a kinetic definition of temperature.

Internal energy (ideal gas)

U = \frac{f}{2} μ RT

f =

degrees of freedom; ideal-gas

U

depends only on

T

, never on

P

V

⚡ Shortcut · Same KE, different speed

At equal T, He,

H_{2}

O_{2}

share the

same

\overline{E} = \frac{3}{2} k_{B} T

per molecule, but

v_{r m s} = 3 k_{B} T / m \propto 1/ m

— lighter

\Rightarrow

faster. Compare speeds via mass ratios alone.

🎯 Exam · Moving container stopped suddenly

Uniform motion of the container does NOT raise T (temperature measures RANDOM motion only). If an insulated container moving at v is stopped, its ordered KE becomes internal energy:

\frac{1}{2} M v^{2} = \frac{f}{2} R Δ T

per mole, so

Δ T = \frac{M v ^{2}}{f R}

(

M =

molar mass in kg/mol).

🚫 Examiner Trap · Temperature & internal energy

(1)

\overline{E} = \frac{3}{2} k_{B} T

translational

KE per molecule; total U uses

\frac{f}{2}

, not

\frac{3}{2}

. (2) Absolute zero

\Rightarrow

zero translational KE (classical) — T here is

kelvin

. (3)

\overline{E}

is independent of mass/pressure — only T. (4) Real-gas U also has intermolecular PE; ideal-gas U does not.

Molecular Speeds

Root-mean-square speed

v_{r m s} = \frac{3 RT}{M} = \frac{3 k _{B} T}{m} = \frac{3 P}{ρ}

M =

molar mass (kg/mol in SI),

m =

mass of one molecule

Average & most-probable

v_{a v} = \frac{8 RT}{π M}, v_{m p} = \frac{2 RT}{M}

v_{a v} =

arithmetic mean speed;

v_{m p} =

peak of the distribution

Maxwell speed distribution curve with most-probable, average and rms speeds marked in increasing order — Skewed distribution: $v_{m p} < v_{a v} < v_{r m s}$ (tail pulls $v_{r m s}$ highest).

Comparative: the three speeds

Speed	Formula ( $RT / M \times$ )	vs $v_{r m s}$
$v_{m p}$	$2 \approx 1.41$	$0.816 v_{r m s}$
$v_{a v}$	$8/ π \approx 1.60$	$0.921 v_{r m s}$
$v_{r m s}$	$3 \approx 1.73$	$1$

⚡ Shortcut · Memorise the ratio

v_{m p} : v_{a v} : v_{r m s} = 2 : 8/ π : 3 = 1 : 1.128 : 1.225

. All three

\propto T / M

, so quadruple T to double any speed.

Speeds in practice

▸Speed of sound $v_{s} = \frac{γ RT}{M}$ , so $\frac{v _{s}}{v _{r m s}} = \frac{γ}{3}$ (always less than $1$ ).
▸Temperature at which $v_{r m s}$ equals the escape speed: $T = \frac{M v _{e}^{2}}{3 R}$ (use $v_{e} = 11.2$ km/s for Earth).
▸Mixture: $v_{r m s}^{2} = \frac{\sum N _{i} v _{i}^{2}}{\sum N _{i}} = \frac{3 RT}{\sum μ _{i}} \sum \frac{μ _{i}}{M _{i}}$ — average $v^{2}$ weighted by number of molecules, NOT the average of the speeds.

🚫 Examiner Trap · Molecular speeds

(1) Put M in

kg/mol

(e.g. O

_{2} = 0.032

) — using g/mol gives v off by

1000

. (2)

v_{r m s} \propto T

, so doubling

T (K)

multiplies

v_{r m s}

2

, NOT 2. (3)

v_{r m s} > v_{a v} > v_{m p}

— don't swap the order. (4) Graham: diffusion rate

\propto v_{r m s} \propto 1/ M \propto 1/ ρ

— lighter gases diffuse faster.

Degrees of Freedom & Equipartition

Degrees of freedom

f

: The number of independent ways a molecule stores energy: translational (

3

), rotational (

2

linear,

3

non-linear), plus vibrational (each mode

= 2

quadratic terms).

Comparative: DOF by molecule type

Molecule	Trans	Rot	$f$ (room $T$ )
Monatomic (He)	3	0	3
Diatomic ( $O_{2}$ )	3	2	5
Diatomic + vib.	3	2	7 (high $T$ )
Polyatomic (non-linear)	3	3	6

Monatomic molecule with three translational axes and a diatomic dumbbell with three translational and two rotational axes — Monatomic ( $f = 3$ ) vs diatomic ( $f = 5$ ).

★ Remember · Law of equipartition

In equilibrium each

quadratic

degree of freedom carries average energy

\frac{1}{2} k_{B} T

per molecule (

\frac{1}{2} RT

per mole).

Internal energy

U = \frac{f}{2} μ RT

sum over all

active

degrees of freedom f

🚫 Examiner Trap · Degrees of freedom

(1) A

vibrational

mode counts as

TWO

quadratic terms (KE

+

PE)

\Rightarrow

energy

k_{B} T

, not

\frac{1}{2} k_{B} T

. (2) Monatomic gas has

no

rotational DOF (a point has no moment of inertia); a linear molecule has only

2

. (3) At room T vibration is usually

frozen

— use diatomic

f = 5

, not 7. (4) f is per molecule; multiply by

μ R /2

, not

μ R

Specific Heats of Gases

Molar specific heats

C_{V} = \frac{f}{2} R, C_{P} = (\frac{f}{2} + 1) R, γ = \frac{C _{P}}{C _{V}} = 1 + \frac{2}{f}

Mayer's relation

C_{P} - C_{V} = R

(per mole) holds for every ideal gas;

f =

active DOF

Gas	$f$	$C_{V}$	$C_{P}$	$γ$
Monatomic	3	$\frac{3}{2} R$	$\frac{5}{2} R$	$\frac{5}{3} \approx 1.67$
Diatomic (rigid)	5	$\frac{5}{2} R$	$\frac{7}{2} R$	$\frac{7}{5} = 1.40$
Polyatomic (non-linear)	6	$3 R$	$4 R$	$\frac{4}{3} \approx 1.33$

Trends & mixtures

▸Larger f $\Rightarrow$ larger $C_{V}, C_{P}$ but $smaller$ $γ$ .
▸Diatomic with vibration: $C_{V} = \frac{7}{2} R$ , $γ = \frac{9}{7}$ .
▸Mixture: $C_{V} = \frac{μ _{1} C _{V 1} + μ _{2} C _{V 2}}{μ _{1} + μ _{2}}$ (mole-weighted).

🎯 Exam · Solids (Dulong–Petit)

A solid atom has

6

quadratic DOF (

3

+

3

PE)

\Rightarrow

molar specific heat

C \approx 3 R \approx 25 J mol^{- 1} K^{- 1}

🚫 Examiner Trap · Specific heats of gases

(1)

C_{P} - C_{V} = R

per mole

; for

specific

(per kg) heats divide by M. (2)

C_{P} > C_{V}

always (extra R for expansion work). (3) Use the DOF

active

at that T — vibration frozen at room T

\Rightarrow

diatomic

C_{V} = \frac{5}{2} R

. (4) At very

low

T DOFs 'freeze out' (quantum) and

C_{V}

falls below these values.

Mean Free Path & Real Gases

Mean free path

ℓ = \frac{1}{2 πn d ^{2}} = \frac{k _{B} T}{2 π d ^{2} P}

n = N / V =

number density,

d =

molecular diameter; the

2

accounts for relative motion of molecules

Collision frequency

f_{co ll} = \frac{v _{a v}}{ℓ} = 2 πn d^{2} v_{a v}

collisions per second; collision time

τ = 1/ f_{co ll}

Zigzag path of a molecule between successive collisions with other molecules, one segment labelled mean free path — Average straight-line distance between successive collisions.

Dependence

▸ $ℓ \propto \frac{1}{n} \propto \frac{T}{P}$ — rises in a vacuum, falls at high pressure.
▸ $ℓ \propto \frac{1}{d ^{2}}$ — smaller molecules travel farther.
▸Air at STP: $ℓ \approx 7 \times 1 0^{- 8}$ m, about $200 \times$ the molecular diameter ( $d \approx 3.7 \times 1 0^{- 10}$ m).

Comparative: real-gas (van der Waals) corrections

Term	Corrects for	Effect
$a / V^{2}$ (add to P)	molecular attraction	real $P$ lower
$b$ (subtract from $V$ )	finite molecular size	free $V$ smaller

🚫 Examiner Trap · Mean free path & real gases

(1)

ℓ \propto T / P

— at fixed T, raising P

shortens

ℓ

; in a good vacuum

ℓ

can exceed the vessel size. (2) Keep the

2

factor (it comes from relative speed). (3) Real gases deviate most at

high P

low T

(van der Waals

(P + a / V^{2}) (V - b) = μ RT

). (4) Brownian motion is direct evidence of molecular motion, not a deviation.

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