JEE Mains Physics · Kinetic Theory

Ideal Gas Equation and Gas Laws

An ideal gas obeys PV = nRT = NkT with T in kelvin; for a fixed amount of gas PV/T stays constant, and when gas flows between vessels it is the total number of moles that stays fixed.

All Kinetic Theory notes JEE Mains Physics strategy

Why this matters

Twenty-seven PYQs, five of them asking for a number, and seven from 2026. Nine follow a fixed amount of gas from one state to another, thirteen count moles in a mixture or in joined vessels, and five read a gas-law graph or a pressure that changes through the gas. Convert every temperature to kelvin before the first ratio: a Celsius temperature in a ratio is the commonest slip in this chapter.

Concept 1 of 3: One fixed amount of gas: P₁V₁/T₁ = P₂V₂/T₂

For a fixed amount of gas, PV/T equals nR and so never changes. Hold any one of P, V and T, and the other two are tied: in a rigid sealed vessel the pressure rises in step with the kelvin temperature, and at constant temperature the volume grows as the pressure falls. A bubble rising through water is the classic case: the pressure on it drops from the water pressure at depth to the air pressure at the top.

Definition

$PV = nRT = NkT$ , with $R = 8.31\ \text{J mol}^{-1}\text{K}^{-1}$ and $k = R/N_A = 1.38 \times 10^{-23}\ \text{J/K}$ .
T is always in kelvin: $T = t + 273$ .
Fixed amount of gas: $\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}$ .
Rigid sealed vessel (V fixed): $P \propto T$ , so for a small change $\dfrac{\Delta P}{P} = \dfrac{\Delta T}{T}$ .
Under water at depth h: $P = P_0 + \rho gh$ ; at the surface only $P_0$ . Use the temperature at each level if the two differ.
Density of a gas: $\rho = \dfrac{PM}{RT}$ . A balloon of fixed volume displaces less air where $P/T$ is smaller, so it lifts less there.
Doubling the temperature means doubling T in kelvin; doubling the Celsius figure is a much smaller change.

Fixed amount of gas

PV = nRT = NkT \qquad \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

Worked example

A cylinder holds 40 cm³ of gas at 1.5 atm and 27 °C. The gas is compressed to 15 cm³ and its temperature rises to 102 °C. Find the new pressure.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 23 Jan 2026 Shift 2 · Q11Moderate

Example 1 · Kinetic Theory · Ideal Gas Equation and Gas Laws

An air bubble of volume

2.9{\text{ }cm}^{3}

rises from the bottom of a swimming pool of 5 m deep. At the bottom of the pool water temperature is

17^{\circ}C

. The volume of the bubble when it reaches the surface, where the water temperature is

27^{\circ}C

, is

\_\_\_\_

cm^{3}

. (

g = 10\text{ }m/s^{2}

, density of water

=10^{3}\text{ }kg/m^{3}

, and 1 atm pressure is

10^{5}\text{ }Pa

)

A Celsius temperature in a ratio

Going from 27 °C to 54 °C does not double anything: in kelvin it is 300 K to 327 K, a 9% rise. Convert first, every time.

Leaving out the air pressure at depth

The pressure on a bubble at depth h is P₀ + ρgh, not ρgh. Dropping P₀ gives a ratio that is far too large.

Drill 8 more on one fixed amount of gas: p₁v₁/t₁ = p₂v₂/t₂

Concept 2 of 3: Counting moles: mixtures and joined vessels

When gas can flow from one vessel to another, or several gases share one vessel, the quantity that is kept is the number of moles, n = PV/RT. Add up n for every part before the change, and set it equal to the total after. In a mixture each gas pushes on the walls as if it were alone, so the partial pressures add.

Definition

$n = \dfrac{m}{M} = \dfrac{N}{N_A} = \dfrac{PV}{RT}$ .
With the number density $n_V = N/V$ (molecules per m³): $P = n_VkT$ .
Dalton's law: $P = (n_1 + n_2 + \dots)\dfrac{RT}{V}$ ; each gas adds its own partial pressure.
Joined vessels reaching a common pressure P: $\sum \dfrac{P_iV_i}{T_i}$ before $= P\sum \dfrac{V_i}{T_i'}$ after.
Two known gases with total mass m: $n_1 + n_2 = \dfrac{PV}{RT}$ and $n_1M_1 + n_2M_2 = m$ . Two equations, two unknowns.
Same V and T: $P \propto n = m/M$ , so equal masses of a light gas and a heavy gas give the light gas the higher pressure.
A partition removed between two parts of the same kind of gas (same f) in an insulated container: the total internal energy $\tfrac{f}{2}\sum P_iV_i$ is kept, so $P = \dfrac{P_1V_1 + P_2V_2}{V_1 + V_2}$ .

Moles are conserved

n = \frac{PV}{RT} \qquad \sum_i \frac{P_iV_i}{T_i} = \text{constant}

Worked example

Two vessels of equal volume, joined by a thin tube, hold air at 120 kPa and 300 K. One vessel is then heated to 450 K while the other is kept at 300 K. Find the final pressure.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 6 Apr 2026 Shift 1 · Q10Moderate

Example 2 · Kinetic Theory · Ideal Gas Equation and Gas Laws

Two closed vessels of same volume are joined through a narrow tube and both vessels are filled with air of pressure 90 kPa and temperature 400 K . Keeping the temperature of one vessel constant at 400 K the second vessel temperature is raised to 500 K . The final pressure in the vessels is

\_\_\_\_

kPa .

Adding pressures instead of moles

When two vessels are joined, their pressures do not add. Their moles do. If the temperatures differ, write n = PV/RT for each part before you add.

k with moles, or R with molecules

P = n_V kT counts molecules per cubic metre; PV = nRT counts moles. Mixing them puts the answer off by a factor of 6 × 10²³.

Drill 12 more on counting moles: mixtures and joined vessels

Concept 3 of 3: Gas-law graphs and pressure that changes through the gas

Rearrange PV = nRT so that the two plotted quantities stand alone; whatever is left over is the slope. At constant volume, P = (nR/V)T is a straight line through absolute zero. At constant pressure, V = (nR/P)T, so a steeper line means a LOWER pressure. When the pressure is given as a function of volume, the temperature is T = PV/nR, so the hottest state is where the product PV is largest.

Definition

P against T at constant V: a straight line through 0 K, slope $\dfrac{nR}{V} = \dfrac{\rho R}{M}$ . For one gas, the steeper line is the denser sample.
P against t in °C: the lines, extended back, reach $P = 0$ at absolute zero, $-273\ ^{\circ}\text{C}$ .
V against T at constant P: slope $\dfrac{nR}{P}$ , so the steeper line has the lower pressure.
P against V at constant T: a hyperbola $PV =$ constant; the curve farther from the axes is the hotter one.
A given path $P(V)$ : $T = \dfrac{PV}{nR}$ ; find the hottest state from $\dfrac{d(PV)}{dV} = 0$ .
Pressure that varies through a gas (gravity, or a tube spun about one end): for a thin slice, $dP = \rho a\,dx$ with $\rho = \dfrac{PM}{RT}$ , so $\dfrac{dP}{P}$ is proportional to $dx$ and P varies exponentially. In an isothermal atmosphere $P = P_0e^{-Mgh/RT}$ ; in a spinning tube the acceleration $\omega^{2}x$ replaces g.

Hottest state on a path

T = \frac{PV}{nR} \qquad \frac{d(PV)}{dV} = 0

Worked example

One mole of an ideal gas follows the straight-line path

P = P_0 - aV

, with

P_0 = 2 \times 10^{5}\ \text{Pa}

and

a = 10^{7}\ \text{Pa/m}^{3}

. Find the highest temperature it reaches.

(R = 8.31)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 22 · Q15Moderate

Example 3 · Kinetic Theory · Ideal Gas Equation and Gas Laws

For an ideal gas the instantaneous change in pressure '

p

' with volume '

v

' is given by the equation

\frac{dp}{dv}= - ap

. If

p =p_{0}

v = 0

is the given boundary condition, then the maximum temperature one mole of gas can attain is: (Here

R

is the gas constant)

Reading a V–T slope as pressure

The slope of a V–T line is nR/P. A steeper line is a lower pressure, not a higher one.

Expecting a Celsius graph to pass through the origin

On a P–t graph with t in °C, the lines do not pass through zero. They meet the temperature axis at absolute zero, to the left of 0 °C.

Drill 4 more on gas-law graphs and pressure that changes through the gas

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)

One fixed amount of gas: P₁V₁/T₁ = P₂V₂/T₂
Fixed amount of gas
$PV = nRT = NkT \qquad \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$
Counting moles: mixtures and joined vessels
Moles are conserved
$n = \frac{PV}{RT} \qquad \sum_i \frac{P_iV_i}{T_i} = \text{constant}$
Gas-law graphs and pressure that changes through the gas
Hottest state on a path
$T = \frac{PV}{nR} \qquad \frac{d(PV)}{dV} = 0$

Watch out for (6)

A Celsius temperature in a ratio
→ One fixed amount of gas: P₁V₁/T₁ = P₂V₂/T₂
Leaving out the air pressure at depth
→ One fixed amount of gas: P₁V₁/T₁ = P₂V₂/T₂
Adding pressures instead of moles
→ Counting moles: mixtures and joined vessels
k with moles, or R with molecules
→ Counting moles: mixtures and joined vessels
Reading a V–T slope as pressure
→ Gas-law graphs and pressure that changes through the gas
Expecting a Celsius graph to pass through the origin
→ Gas-law graphs and pressure that changes through the gas

Test yourself on Kinetic Theory

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Kinetic Theory chapter test Drill all 27 questions