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MHT-CET Physics · Kinetic Theory of Gases

Average Kinetic Energy, Equipartition and Specific Heats

At temperature T every molecule has average translational kinetic energy (3/2)kT whatever its mass, and each degree of freedom carries ½kT; counting f degrees of freedom gives C_v = (f/2)R, C_p = C_v + R and γ = 1 + 2/f.

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Why this matters

25 PYQs, 3 of them HARD. Ten use the average kinetic energy — it depends only on absolute temperature, and a container brought suddenly to rest turns its motion into heat. Fifteen count degrees of freedom to get C_v, C_p and γ, including a mixture of three gases and the same heat given at constant pressure and at constant volume. Two cards.

Concept 1 of 2: Average Kinetic Energy Depends Only on Temperature

The average translational kinetic energy of a molecule is (3/2)kT, the same for every gas at the same temperature — nitrogen and oxygen at one temperature have equal average kinetic energy though their speeds differ. So temperature measures that energy, and it goes as the kelvin temperature: halving the energy halves T in kelvin. An ideal monoatomic gas has no potential energy, so its internal energy is entirely kinetic. When an insulated container moving at speed V stops suddenly, the bulk kinetic energy ½MV² per mole becomes internal energy, n C_v ΔT, so ΔT = MV²/(fR).

Definition

  • Eˉ=32kT\bar{E} = \tfrac{3}{2}kT per molecule, independent of mass; Eˉ∝T\bar{E} \propto T (kelvin).
  • 399 °C ⇒ E; E/2 at 336 K = 63 °C. 27 °C ⇒ E; 327 °C ⇒ 2E.
  • Ideal gas: internal energy is all kinetic.
  • Container of molar mass M stopped: ΔT=MV2fR\Delta T = \dfrac{MV^2}{fR} (monoatomic 3R, rigid diatomic 5R).
  • Mean square x-velocity is one third of the mean square speed: ⟨vx2⟩=kTm\langle v_x^2 \rangle = \dfrac{kT}{m}.

Average kinetic energy

Eˉ=32kT,each degree of freedom 12kT\bar{E} = \frac{3}{2}kT, \qquad \text{each degree of freedom } \tfrac{1}{2}kT

Worked example

The average kinetic energy of a gas molecule at 27 °C is 6.2 × 10⁻²¹ J. What is it at 127 °C?
Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 22 April Shift I · Q30Moderate

Example 1 · Kinetic Theory of Gases · Average KE, Equipartition, and Specific Heats

The mean kinetic energy of the molecules of an ideal gas at 399∘C399^{\circ}C is ' E '. The temperature at which the mean kinetic energy of its molecules will be ' E/2E/2 ', is

Letting the heavier gas have more kinetic energy

At one temperature all gases have the same average kinetic energy. Molar mass changes the SPEED, not the energy.

Halving the Celsius temperature

E ∝ T in kelvin. E/2 at 399 °C (672 K) means 336 K, which is 63 °C — not 199.5 °C.
Drill 9 more on average kinetic energy depends only on temperature

Concept 2 of 2: Degrees of Freedom, C_v, C_p and γ

A monoatomic molecule moves in 3 directions: f = 3. A rigid diatomic molecule also rotates about 2 axes: f = 5. A diatomic molecule that vibrates adds 2 more: f = 7. Each degree of freedom holds ½RT per mole, so C_v = (f/2)R, C_p = C_v + R and γ = 1 + 2/f: 5/3, 7/5 and 9/7. Written with γ, C_v = R/(γ − 1) and C_p = γR/(γ − 1), so R/C_v = γ − 1 identifies the gas. For a mixture, add n·C_v over the components and divide by the total moles. Giving the same heat to a gas at constant pressure and at constant volume raises the temperature γ times more at constant volume.

Definition

  • f = 3 (monoatomic), 5 (rigid diatomic), 7 (diatomic with vibration).
  • Cv=f2RC_v = \dfrac{f}{2}R, Cp=Cv+RC_p = C_v + R, γ=1+2f\gamma = 1 + \dfrac{2}{f}.
  • Cv=Rγ−1C_v = \dfrac{R}{\gamma - 1}, Cp=γRγ−1C_p = \dfrac{\gamma R}{\gamma - 1}; RCv=0.4\dfrac{R}{C_v} = 0.4 ⇒ rigid diatomic.
  • Polyatomic with f vibrational modes: Cv=(3+f)RC_v = (3 + f)R, γ=4+f3+f\gamma = \dfrac{4 + f}{3 + f}.
  • Mixture: Cv,mix=∑niCv,i∑niC_{v,\text{mix}} = \dfrac{\sum n_iC_{v,i}}{\sum n_i} (4 H₂, 2 He, 1 H₂O ⇒ Cp=237RC_p = \dfrac{23}{7}R).
  • Same heat, constant P (A) and constant V (B): ΔTB=γ ΔTA\Delta T_B = \gamma\,\Delta T_A (49 K ⇒ 35 K).
  • Per unit mass: Cp−Cv=RMC_p - C_v = \dfrac{R}{M}, so ρ=PT(Cp−Cv)\rho = \dfrac{P}{T(C_p - C_v)}.

Equipartition

Cv=f2R,Cp=Cv+R,γ=1+2fC_v = \frac{f}{2}R, \qquad C_p = C_v + R, \qquad \gamma = 1 + \frac{2}{f}

Worked example

Heat needed to raise 14 g of nitrogen (M = 28) by 48 °C at constant pressure?
Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 15th May Shift 1 · Q28Hard

Example 2 · Kinetic Theory of Gases · Average KE, Equipartition, and Specific Heats

Four moles of hydrogen, two moles of helium and one mole of water vapour form an ideal gas mixture. CvC_{v} for hydrogen =52R= \frac{5}{2}R, CvC_{v} for helium =32R= \frac{3}{2}R, CvC_{v} for water vapour =3R= 3R. What is the molar specific heat at constant pressure of the mixture?

Averaging the specific heats of a mixture by gas, not by mole

Weight each C_v by its number of moles. Four moles of hydrogen count four times as much as one mole of water vapour.

Counting one vibrational degree of freedom

A vibrational mode stores both kinetic and potential energy, so it adds 2 to f. A vibrating diatomic has f = 7 and C_v = 7R/2.
Drill 14 more on degrees of freedom, c_v, c_p and γ

Summary — formulas & gotchas at a glance

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Formulas (2)

Watch out for (4)

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