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#molar specific heat

3 public questions tagged with this topic.

Isochoric Processes and Pressure-Temperature

What is the molar specific heat capacity at constant pressure for a solid if its molar specific heat is 24.4 J mol⁻¹ K⁻¹

**Pressure-temperature relation** at constant volume Gay-Lussac law P ∝ T, for V constant, P₁/T₁ = P₂/T₂, if T doubles from 300 K to 600 K P doubles, e.g., P₁=1 atm at 300 K P₂=2 atm at 600 K, no work done, ΔU = n C_v ΔT = Q. For solids, C ≈ 3R , but here C_v = 24.4 , and Δ V ≈ 0 , so C_p ≈ C_v .However, typically C_p - C_v = R , but for solids in PDF context, C is given directly.Since C = 24.4 is molar specific heat, C_p ≈ C_v = 24.4 J mol⁻¹ K⁻¹ (negligible Δ

Ref: NCERT > Physics Book > Thermodynamics > Isochoric Processes and Pressure-Temperature

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Collision Frequency and Mean Free Path Variation

The molar specific heat at constant pressure for a polyatomic gas with 2 vibrational modes is: (R = 8.31 J mol⁻¹ K⁻¹)

**Mean free path variation** λ ∝1/n ∝1/P at constant T, λ ∝ T/P, temperature increase increases λ because n decreases at constant P, but also v increases, overall λ ∝ T/P, for gas at 2 atm λ=4×10⁻⁷ m, at 4 atm λ=2×10⁻⁷ m halves when pressure doubles, as n doubles. Polyatomic gas: 3 translational + 3 rotational + 2 vibrational modes.Total degrees of freedom = 3 + 3 + 2 × 2 = 10.C_v = 5R, C_p = C_v + R = 6R = 6 × 8.31 = 49.86 J mol⁻¹ K⁻¹ . Substituting values gives 49.86 J mol⁻¹ K⁻¹, which matches expected kinetic theory

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Collision Frequency and Mean Free Path Variation

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Degrees of Freedom and Molar Specific Heat

A solid has a molar specific heat capacity of 24.9 J mol⁻¹ K⁻¹. What is the energy per degree of freedom per atom? (R =

**Molar specific heat** from equipartition C_v = f/2 R, C_p = f/2 R + R, γ = C_p/C_v =1+2/f, for f=3 γ=1.67, f=5 γ=1.4, f=6 γ=1.33, explaining specific heat variation with molecular structure, degrees of freedom determine heat capacity. C = f × (R)/(2), 24.9 = f × (8.31)/(2), f = (24.9 × 2)/(8.31) ≈ 6.Energy per degree of freedom = (1)/(2) k_B T, per mole = (1)/(2) R T.For 1 degree of freedom, energy per mole = (8.31)/(2) = 4.155 J mol⁻¹ K⁻¹ . Substituting values gives 4.16 J mol⁻¹ K⁻¹, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Degrees of Freedom and Molar Specific Heat

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