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JEE Mains Physics · Formula sheet

Kinetic Theory formulas

12 formulas, 1 reference table and 26 common traps for JEE Mains Physics Kinetic Theory, grouped by subtopic.

Full notes with worked examples

Ideal Gas Equation and Gas Laws

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

Fixed amount of gas

PV=nRT=NkTP1V1T1=P2V2T2PV = 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=PVRT∑iPiViTi=constantn = \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=PVnRd(PV)dV=0T = \frac{PV}{nR} \qquad \frac{d(PV)}{dV} = 0

Common traps

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.

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.

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.

Pressure, Kinetic Energy and Temperature

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Pressure from molecular impacts, and PV = (2/3)E

Kinetic-theory pressure

P=13ρvrms2PV=23EtransP = \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

KE‾=32kTEtrans=32nRT\overline{KE} = \frac{3}{2}kT \qquad E_{trans} = \frac{3}{2}nRT

Same temperature, same mean kinetic energy

Equal temperatures

12m1v12‾=12m2v22‾=32kT\tfrac{1}{2}m_1\overline{v_1^{2}} = \tfrac{1}{2}m_2\overline{v_2^{2}} = \tfrac{3}{2}kT

Common traps

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.

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.

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).

Molecular Speeds and Mean Free Path

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The rms speed scales as √(T/M)

rms speed

vrms=3RTMv2v1=T2M1T1M2v_{rms} = \sqrt{\frac{3RT}{M}} \qquad \frac{v_2}{v_1} = \sqrt{\frac{T_2M_1}{T_1M_2}}

The rms, average and most probable speeds

Three molecular speeds

vp=2RTMvˉ=8RTπMvrms=3RTMv_p = \sqrt{\frac{2RT}{M}} \qquad \bar{v} = \sqrt{\frac{8RT}{\pi M}} \qquad v_{rms} = \sqrt{\frac{3RT}{M}}

Mean free path and collision frequency

Mean free path

λ=12 πd2n=kT2 πd2PZ=vˉλ\lambda = \frac{1}{\sqrt{2}\,\pi d^{2}n} = \frac{kT}{\sqrt{2}\,\pi d^{2}P} \qquad Z = \frac{\bar{v}}{\lambda}

Common traps

Speed goes as √T, not T

Four times the kelvin temperature gives twice the speed, not four times. The same square root applies to the molar mass.

Celsius in the ratio

Equal T/M must use kelvin. With nitrogen at 77 °C, 77/28 would put helium at 11 °C instead of the true 50 K.

Using the rms speed when the average speed is asked

"Average speed" is √(8RT/πM); "rms speed" is √(3RT/M). They differ by about 8%, enough to land on a wrong option.

Molar mass in grams

With R = 8.31 J mol⁻¹ K⁻¹, M must be in kg/mol: 32 g/mol is 0.032 kg/mol. Grams make every speed about 32 times too small.

1/d instead of 1/d²

The mean free path goes as 1/d²: the target is an area, πd². Halving the diameter makes the path four times longer, not twice.

Heating at constant pressure versus in a sealed vessel

At constant pressure the gas spreads out and λ grows with T. In a sealed rigid vessel the molecules per cubic metre do not change, so λ stays the same and only the speed, and with it Z, goes up.

Degrees of Freedom and Specific Heats

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Cv, Cp and γ from the degrees of freedom

Heat capacities of an ideal gas

Cv=f2RCp=Cv+Rγ=1+2fC_v = \frac{f}{2}R \qquad C_p = C_v + R \qquad \gamma = 1 + \frac{2}{f}

Counting degrees of freedom

Gasf (trans + rot + vib)CvCpγ
Monatomic (He, Ne, Ar)3 (3 + 0 + 0)3R/25R/25/3 ≈ 1.67
Rigid diatomic (N₂, O₂ near room temperature)5 (3 + 2 + 0)5R/27R/27/5 = 1.40
Diatomic with one vibrational mode7 (3 + 2 + 2)7R/29R/29/7 ≈ 1.29
Rigid linear triatomic (CO₂)5 (3 + 2 + 0)5R/27R/27/5 = 1.40
Rigid non-linear (H₂O, NH₃, CH₄)6 (3 + 3 + 0)3R4R4/3 ≈ 1.33
Non-linear with v vibrational modes6 + 2v(3 + v)R(4 + v)R(4 + v)/(3 + v)
Each vibrational mode adds 2 to f, so it adds R to both Cv and Cp.
γ=1+2/f\gamma = 1 + 2/f: more degrees of freedom always means a smaller γ\gamma.

Common traps

Counting a vibrational mode once

A vibration stores kinetic and potential energy, so one mode adds 2 to f. Counting it as 1 gives the wrong Cv and γ.

Linear and non-linear triatomics

CO₂ is a straight line and has 5 degrees of freedom when rigid; H₂O is bent and has 6. The shape, not the number of atoms, decides the rotations.

Expecting γ to grow with f

γ = 1 + 2/f, so more degrees of freedom give a SMALLER γ. A gas with vibration has a lower γ than the same gas held rigid.

Cp − Cv = R is per mole

For specific heats per kilogram the difference is R/M. Check the units the question uses before subtracting.

Internal Energy and Gas Mixtures

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Internal energy U = n(f/2)RT

Internal energy of an ideal gas

U=nf2RT=f2PVΔU=nCvΔTU = n\frac{f}{2}RT = \frac{f}{2}PV \qquad \Delta U = nC_v\Delta T

A mixture as one equivalent gas, and the temperature after mixing

Equivalent gas

fmix=n1f1+n2f2n1+n2γmix=1+2fmixf_{mix} = \frac{n_1f_1 + n_2f_2}{n_1 + n_2} \qquad \gamma_{mix} = 1 + \frac{2}{f_{mix}}

Common traps

k for molecules, R for moles

Ten molecules carry an energy measured in kT; ten moles carry one measured in RT. The options often offer both with the same number in front.

"Neglect vibration" sets f = 5

A diatomic gas with vibration ignored is rigid: f = 5, not 7. Only an explicit vibrational mode adds 2.

Averaging γ directly

For equal moles of a monatomic and a rigid diatomic gas, the mean of 5/3 and 7/5 is about 1.53, but the true γ is 1.5. Average f or Cv first, then find γ.

Weighting temperatures by moles alone

When the gases have different f, each one's energy is n(f/2)RT, so the final temperature is weighted by nf, not by n.

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