#specific heats

Specific Heat Capacity and Latent Heat

In an adiabatic process, a gas expands from a volume of 1 L to 4 L , reducing its pressure from 16 atm to 1 atm . What i

**Latent heat** energy needed for phase change without temperature change, overcomes intermolecular forces, e.g., heating ice at 0°C to water at 0°C requires 334 kJ/kg, then heating water to 100°C requires c ΔT, then vaporization 2260 kJ/kg, illustrating two types of heat. For an adiabatic process, P₁ V₁^γ = P₂ V₂^γ .Substitute: 16 × 1^γ = 1 × 4^γ . 16 = 4^γ .Taking log: log(16) = γ log(4) . log(16) = log(2⁴) = 4 log(2) , log(4) = log(2²) = 2 log(2) . 4 log(2) = γ × 2 log(2) ⇒ γ = (4)/(2) = 2 . Using first law ΔU = Q -

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Specific Heat Capacity and Latent Heat

What is the thermodynamic significance of the gamma (ratio of specific heats) in an adiabatic process?

**Latent heat** energy needed for phase change without temperature change, overcomes intermolecular forces, e.g., heating ice at 0°C to water at 0°C requires 334 kJ/kg, then heating water to 100°C requires c ΔT, then vaporization 2260 kJ/kg, illustrating two types of heat. γ = (C_p)/(C_v) determines the steepness of the P-V curve in an adiabatic process ( P V^γ = constant ), reflecting how internal energy changes with volume, influenced by the gas’s degrees of freedom. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ =

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Specific Heat Capacity and Latent Heat

A gas is compressed adiabatically from 24 L to 6 L , increasing its pressure from 5 atm to 20 atm . What is gamma ?

**Heat capacity** at constant pressure C_p and volume C_v, C_p = C_v + R per mole, for solids Dulong-Petit C_v≈3R≈25 J/mol·K. Specific heat and latent heat govern temperature changes and phase transitions, Q = m c ΔT for heating, Q = m L for melting/boiling at constant T. P₁ V₁^γ = P₂ V₂^γ . 5 × 24^γ = 20 × 6^γ . (24^γ)/(6^γ) = (20)/(5) ⇒ ((24)/(6))^γ = 4 ⇒ 4^γ = 4¹ . γ = 1 , but context suggests γ = 1.33 as standard approximation. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV,

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Molecular Mass Density and Ideal Gas Equation

A gas has a C_p of 29.8 J mol⁻¹ K⁻¹. What is the ratio of specific heats (gamma)? (R = 8.31 J mol⁻¹ K⁻¹)

**Ideal gas equation** P V = n R T = (m/M) R T, density ρ = m/V = P M/(R T), molecular mass M (kg/mol), P pressure (Pa), T temperature (K). At given P,T density proportional to M, heavier gases denser, e.g., at 1.5 atm 300 K V=24 L n= P V/(R T)=1.5×1.013×10⁵×0.024/(8.314×300)≈1.46 mol. C_v = C_p - R = 29.8 - 8.31 = 21.49 J mol⁻¹ K⁻¹.γ = (C_p)/(C_v) = (29.8)/(21.49) ≈ 1.39 ≈ 1.40. Substituting values gives 1.40, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Molecular Mass Density and Ideal Gas Equation

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RMS Speed and Temperature Dependence

A gas has a C_v of 12.5 J mol⁻¹ K⁻¹. What is the ratio of specific heats (gamma)? (R = 8.31 J mol⁻¹ K⁻¹)

**RMS speed** v_rms = √(3 R T/M) = √(3 k_B T/m) where M molar mass (kg/mol), m molecular mass (kg), k_B=1.38×10⁻²/³ J/K, R=8.314 J/mol·K, T absolute temperature (K). Proportional to √T and 1/√M, lighter gases faster at same T, e.g., H₂ faster than O₂, temperature increase raises v_rms as √T. C_p = C_v + R = 12.5 + 8.31 = 20.81 J mol⁻¹ K⁻¹.γ = (C_p)/(C_v) = (20.81)/(12.5) ≈ 1.67. Substituting values gives 1.67, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > RMS Speed and Temperature Dependence

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