Skip to content

#solid heat capacity

2 public questions tagged with this topic.

Why does the specific heat capacity of a solid generally agree with 3R at ordinary temperatures?

**Isobaric process** constant pressure, work W = P ΔV = P(V₂ - V₁) = n R ΔT, for expansion ΔV positive W positive, for compression negative, heat Q = n C_p ΔT, ΔU = n C_v ΔT. Isothermal process constant temperature ΔU=0, work W = n R T ln(V₂/V₁) = n R T ln(P₁/P₂), Q = W, heat absorbed equals work done. The law of equipartition predicts that each atom in a solid has 3 degrees of freedom (vibrational), contributing (3)/(2) k_B T kinetic and potential energy per atom. For a mole, U = 3 R T , so C = (Δ U)/(Δ T) =

Ref: NCERT > Physics Book > Thermodynamics > Isobaric and Isothermal Processes Work Calculation

A solid has a molar specific heat capacity of 26.5 J mol⁻¹ K⁻¹. How many degrees of freedom per atom does it have? (R =

**Mean free path** λ = 1/(√2 n π d²) is average distance molecule travels between collisions, n number density (m⁻³), d molecular diameter (m), π≈3.14. Inversely proportional to n and d², larger n or d reduces λ. Rearranged d² = 1/(√2 n π λ), so d = √(1/(√2 n π λ)), enabling diameter estimation from measured λ and n. C = f × (R)/(2), 26.5 = f × (8.31)/(2).f = (26.5 × 2)/(8.31) ≈ 6.38 ≈ 6. Substituting values gives 6, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Mean Free Path and Molecular Diameter