Kinetic Theory

Physics · Class 11

Simulation · Physics · Class 11

Degrees of freedom decide a gas's specific heat

"Degrees of freedom decide a gas's specific heat" is a free, interactive Class 11 Physics simulation for NEET and JEE, built on the lesson "Specific heats of gases" in Kinetic Theory. Pick a molecule. The blocks count its squared terms: 3 for translation, 2 or 3 for rotation, and 2 for each vibration (kinetic and potential). Per mole each term adds ½R to C_v. More ways to hold energy mean a larger C_v and a γ closer to 1: 5/3, 7/5, 9/7, then (4 + f)/(3 + f). C_p − C_v = R for every ideal gas. JEE Main asked 28 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).

From the lesson Specific heats of gases in Kinetic Theory. Change the values and watch what happens.

Degrees of freedom decide a gas's specific heatPhysics · Class 11

In JEE Main

42 shifts analysed

28 questions from Kinetic Theory across the 42 JEE Main shifts Lumi has analysed (2017–2026).

See the past questions

The idea behind it

NCERT §12.6.1–§12.6.3

  • Monatomic gas: three translational degrees of freedom, so (3/2)k_BT per molecule and U = (3/2)k_BT × N_A = (3/2)RT per mole. Then C_v = dU/dT = (3/2)R.
  • For any ideal gas C_p − C_v = R, so a monatomic gas has C_p = (5/2)R and γ = C_p/C_v = 5/3.
  • Rigid diatomic gas (a dumbbell): 3 translational + 2 rotational = 5 degrees of freedom, U = (5/2)RT, C_v = (5/2)R, C_p = (7/2)R, γ = 7/5.
  • Diatomic gas that also vibrates: U = (5/2 k_BT + k_BT)N_A = (7/2)RT, so C_v = (7/2)R, C_p = (9/2)R, γ = 9/7.
  • Polyatomic gas with 3 translational, 3 rotational degrees of freedom and f vibrational modes: U = ((3/2)k_BT + (3/2)k_BT + f k_BT)N_A, so C_v = (3 + f)R, C_p = (4 + f)R and γ = (4 + f)/(3 + f).
  • C_p − C_v = R holds for every ideal gas, monatomic, diatomic or polyatomic.
  • Table 12.1 lists the predictions with vibration left out (J mol⁻¹ K⁻¹): a monatomic gas has C_v = 12.5, C_p = 20.8, γ = 1.67; a diatomic gas C_v = 20.8, C_p = 29.1, γ = 1.40; a triatomic gas C_v = 24.93, C_p = 33.24, γ = 1.33. In every row C_p − C_v = 8.31.
  • Table 12.2, measured values (same order): He 12.5, 20.8, 8.30, 1.66; Ne 12.7, 20.8, 8.12, 1.64; Ar 12.5, 20.8, 8.30, 1.67; H₂ 20.4, 28.8, 8.45, 1.41; O₂ 21.0, 29.3, 8.32, 1.40; N₂ 20.8, 29.1, 8.32, 1.40; H₂O 27.0, 35.4, 8.35, 1.31; CH₄ 27.1, 35.4, 8.36, 1.31.
  • Prediction and measurement agree well for these gases, so equipartition is well confirmed at ordinary temperatures. For gases such as Cl₂ and C₂H₆ the measured values are usually higher than Table 12.1, and including vibrational modes brings them closer.
  • Example 12.8: a fixed 44.8 L cylinder of helium at STP holds 2 mol (1 mol fills 22.4 L at 273 K and 1 atm). Fixed volume means C_v = (3/2)R applies, so heating it by 15.0 °C needs 2 × 1.5R × 15.0 = 45R = 45 × 8.31 = 374 J.

Questions

3 answers

What does the "Degrees of freedom decide a gas's specific heat" simulation show?
Pick a molecule. The blocks count its squared terms: 3 for translation, 2 or 3 for rotation, and 2 for each vibration (kinetic and potential). Per mole each term adds ½R to C_v.
Does JEE Main ask questions from Kinetic Theory?
Yes. JEE Main asked 28 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
Is this simulation free?
Yes. It is free and runs in the browser on a phone or a laptop. It belongs to the lesson "Specific heats of gases" in Kinetic Theory (Class 11 Physics).

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