#C_p

Partial Pressures and Gas Mixtures

A gas has a C_p of 33.24 J mol⁻¹ K⁻¹. What is its C_v? (R = 8.31 J mol⁻¹ K⁻¹)

**Partial pressure concept** for mixture of non-reacting gases, total pressure sum of partials, each gas behaves independently, kinetic theory still holds with effective n = Σ n_i, mean free path depends on total n and cross-sections, mixture properties weighted by mole tions. C_p - C_v = R, C_v = C_p - R.C_v = 33.24 - 8.31 = 24.93 J mol⁻¹ K⁻¹. Substituting values gives 24.93 J mol⁻¹ K⁻¹, 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 > Partial Pressures and Gas Mixtures

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Molecular Mass Density and Ideal Gas Equation

A gas has a C_p of 35.4 J mol⁻¹ K⁻¹. What is its C_v? (R = 8.31 J mol⁻¹ K⁻¹)

**Molecular mass and density** relation ρ = P M/(R T) allows density calculation, ideal gas law also P = n k_B T where n number density, molecular mass determines mass per molecule, density increases with pressure and decreases with temperature, inverse T dependence. C_p - C_v = R, C_v = C_p - R.C_v = 35.4 - 8.31 = 27.09 J mol⁻¹ K⁻¹ ≈ 27.1 J mol⁻¹ K⁻¹. Substituting values gives 27.1 J mol⁻¹ K⁻¹, 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 > Molecular Mass Density and Ideal Gas Equation

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Internal Energy of Ideal Gases

A gas has a C_v of 21.0 J mol⁻¹ K⁻¹. What is its C_p? (R = 8.31 J mol⁻¹ K⁻¹)

**Internal energy of ideal gas** U = f/2 n R T depends only on temperature, f degrees of freedom, n moles, R=8.314 J/mol·K, for monatomic f=3 U=3/2 n R T, diatomic f=5 at moderate T U=5/2 n R T, independent of pressure or volume, only T matters for ideal gas. C_p = C_v + R.C_p = 21.0 + 8.31 = 29.31 J mol⁻¹ K⁻¹ ≈ 29.3 J mol⁻¹ K⁻¹. Substituting values gives 29.3 J mol⁻¹ K⁻¹, 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 > Internal Energy of Ideal Gases

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Degrees of Freedom and Molar Specific Heat

A gas has a C_v of 20.4 J mol⁻¹ K⁻¹. What is its C_p? (R = 8.31 J mol⁻¹ K⁻¹)

**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_p = C_v + R.C_p = 20.4 + 8.31 = 28.71 J mol⁻¹ K⁻¹ ≈ 28.7 J mol⁻¹ K⁻¹. Substituting values gives 28.7 J mol⁻¹ K⁻¹, 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 > Degrees of Freedom and Molar Specific Heat

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Gas Laws and Volume-Temperature Relations

A gas has a C_v of 17.1 J mol⁻¹ K⁻¹. What is its C_p? (R = 8.31 J mol⁻¹ K⁻¹)

**Ideal gas law** P V = n R T governs gas laws, at constant pressure V ∝ T, so temperature increase 300 K→600 K doubles volume 24→48 L. Charles' law quantitative prediction V₂ = V₁×(T₂/T₁), illustrating direct proportionality, absolute temperature must be in kelvin. C_p = C_v + R.C_p = 17.1 + 8.31 = 25.41 J mol⁻¹ K⁻¹ ≈ 25.4 J mol⁻¹ K⁻¹. Substituting values gives 25.4 J mol⁻¹ K⁻¹, 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 > Gas Laws and Volume-Temperature Relations

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Gas Laws and Volume-Temperature Relations

A gas has a C_v of 27.0 J mol⁻¹ K⁻¹. What is its C_p? (R = 8.31 J mol⁻¹ K⁻¹)

**Ideal gas law** P V = n R T governs gas laws, at constant pressure V ∝ T, so temperature increase 300 K→600 K doubles volume 24→48 L. Charles' law quantitative prediction V₂ = V₁×(T₂/T₁), illustrating direct proportionality, absolute temperature must be in kelvin. C_p = C_v + R.C_p = 27.0 + 8.31 = 35.31 J mol⁻¹ K⁻¹ ≈ 35.3 J mol⁻¹ K⁻¹. Substituting values gives 35.3 J mol⁻¹ K⁻¹, 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 > Gas Laws and Volume-Temperature Relations

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