Skip to content
Login Register

#partial pressure

30 public questions tagged with this topic.

Partial Pressures and Gas Mixtures

A gas mixture has 2 g of helium and 16 g of oxygen. What is the ratio of their partial pressures?

**Dalton's law of partial pressures** total pressure P_total = Σ P_i, P_i = X_i P_total, X_i mole tion, each gas exerts pressure as if alone, ideal gas mixture P_i V = n_i R T, partial pressure proportional to mole tion, e.g., air 79% N₂ 21% O₂ P_N₂=0.79 atm P_O₂=0.21 atm at 1 atm total. P = (μ RT)/(V), P_HeP_O₂ = μ_Heμ_O₂.μ_He = (2)/(4) = 0.5 mol, μ_O₂ = (16)/(32) = 0.5 mol.Ratio = (0.5)/(0.5) = 1:1. Substituting values gives 1:1, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Partial Pressures and Gas Mixtures

View Details
Partial Pressures and Gas Mixtures

A gas mixture contains 6 g of hydrogen and 48 g of oxygen. What is the ratio of their partial pressures?

**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. P = (μ RT)/(V), P_H₂P_O₂ = μ_H₂μ_O₂.μ_H₂ = (6)/(2) = 3 mol, μ_O₂ = (48)/(32) = 1.5 mol.Ratio = (3)/(1.5) = 2:1. Substituting values gives 2:1, 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

View Details
Partial Pressures and Gas Mixtures

A gas mixture contains 10 g of neon and 40 g of argon. What is the ratio of their partial pressures? (Atomic mass: Ne =

**Gas mixtures** ideal gas law applies to each component, P_total = Σ n_i R T/V, partial pressure P_i = n_i R T/V, mole tion X_i = n_i/n_total, P_i = X_i P_total, enabling calculation of individual pressures from composition, important for kinetic theory and chemistry. P = (μ RT)/(V), P_NeP_Ar = μ_Neμ_Ar.μ_Ne = (10)/(20.2) ≈ 0.495 mol, μ_Ar = (40)/(39.9) ≈ 1.0025 mol.Ratio = (0.495)/(1.0025) ≈ 0.494 ≈ 1:2. Substituting values gives 1:2, 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

View Details
RMS Speed and Temperature Dependence

A gas mixture contains 12 g of helium and 28 g of nitrogen. What is the ratio of their partial pressures?

**Maxwell-Boltzmann distribution** gives distribution of speeds, v_rms = √(3kT/m), most probable v_mp = √(2kT/m), average v_avg = √(8kT/πm), all ∝ √T, ratio v_rms:v_avg:v_mp =1.732:1.596:1.414, temperature raises all speeds proportionally. P = (μ RT)/(V), P_HeP_N₂ = μ_Heμ_N₂.μ_He = (12)/(4) = 3 mol, μ_N₂ = (28)/(28) = 1 mol.Ratio = (3)/(1) = 3:1. Substituting values gives 3:1, 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 > RMS Speed and Temperature Dependence

View Details
Internal Energy of Ideal Gases

A gas mixture contains 16 g of oxygen and 4 g of hydrogen. What is the ratio of their partial pressures?

**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. P = (μ RT)/(V), P_O₂P_H₂ = μ_O₂μ_H₂.μ_O₂ = (16)/(32) = 0.5 mol, μ_H₂ = (4)/(2) = 2 mol.Ratio = (0.5)/(2) = 1:4. Substituting values gives 1:4, 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

View Details
Internal Energy of Ideal Gases

A gas mixture contains 8 g of helium and 64 g of oxygen. What is the ratio of their partial pressures?

**Equipartition theorem** energy ½ k_B T per degree of freedom per molecule, f degrees give U = f/2 k_B T per molecule, f/2 R T per mole, internal energy function of T only for ideal gas, no intermolecular potential. P = (μ RT)/(V), P_HeP_O₂ = μ_Heμ_O₂.μ_He = (8)/(4) = 2 mol, μ_O₂ = (64)/(32) = 2 mol.Ratio = (2)/(2) = 1:1. Substituting values gives 1:1, 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

View Details
Degrees of Freedom and Molar Specific Heat

A gas mixture contains 4 g of helium and 16 g of oxygen. What is the ratio of their partial pressures?

**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. P = (μ RT)/(V), P_HeP_O₂ = μ_Heμ_O₂.μ_He = (4)/(4) = 1 mol, μ_O₂ = (16)/(32) = 0.5 mol.Ratio = (1)/(0.5) = 2:1. Substituting values gives 2:1, 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

View Details
Degrees of Freedom and Molar Specific Heat

A gas mixture contains 2 g of hydrogen and 32 g of oxygen. What is the ratio of their partial pressures?

**Specific heat relation** C_p - C_v = R for ideal gas per mole, Mayer's relation, due to work done at constant pressure, degrees of freedom include translational, rotational, vibrational, each quadratic term contributes ½ R to C_v. P = (μ RT)/(V), P_H₂P_O₂ = μ_H₂μ_O₂.μ_H₂ = (2)/(2) = 1 mol, μ_O₂ = (32)/(32) = 1 mol.Ratio = (1)/(1) = 1:1. Substituting values gives 1:1, 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

View Details
Gas Laws and Volume-Temperature Relations

A gas mixture contains 8 g of helium and 8 g of oxygen. What is the ratio of their partial pressures?

**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. P = (μ RT)/(V), P_HeP_O₂ = μ_Heμ_O₂.μ_He = (8)/(4) = 2 mol, μ_O₂ = (8)/(32) = 0.25 mol.Ratio = (2)/(0.25) = 8:1. Substituting values gives 8:1, 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

View Details