Dalton’s and Amagat’s Laws (Gas Mixtures)
1 / 10
A mixture of 2 moles of CO and 3 moles of CO₂ is at 500 K and a total volume of 0.2 m³. The total pressure of the mixture is:
5.195 bar
2.598 bar
20.78 bar
10.39 bar
Total moles n_total = 2 + 3 = 5. Using PV = nRT, P = (nRT)/V = (5 × 8.314 × 500)/0.2 = 20770 Pa = 20770 × 10⁻⁵ bar ≈ 10.39 bar.
2 / 10
The partial pressure of a gas in a mixture is 0.4 bar, and the total pressure is 2 bar. The partial volume of this gas, if the total volume is 0.1 m³, is:
0.02 m³
0.04 m³
0.08 m³
0.05 m³
Mole fraction yᵢ = Pᵢ/P_total = 0.4/2 = 0.2. By Amagat’s law, Vᵢ = yᵢ × V_total = 0.2 × 0.1 = 0.02 m³.
3 / 10
A gas mixture contains 1 kg of H₂ (molar mass 2 kg/kmol) and 2 kg of N₂ (molar mass 28 kg/kmol) at 1 bar and 300 K. The total volume of the mixture is:
172.6 m³
86.3 m³
43.15 m³
344.2 m³
Moles of H₂ = 1/2 = 0.5 kmol. Moles of N₂ = 2/28 = 1/14 ≈ 0.0714 kmol. Total moles n_total = 0.5 + 0.0714 = 0.5714 kmol. Using PV = nRT, V = (nRT)/P = (0.5714 × 8314 × 300)/(1 × 10⁵) ≈ 86.3 m³ (R = 8314 J/kmol·K).
4 / 10
A vessel contains a mixture of 40% N₂ and 60% O₂ by volume at 2 bar and 300 K. The partial pressure of O₂ is:
0.8 bar
1.2 bar
1.0 bar
1.6 bar
By Amagat’s law, volume fraction equals mole fraction for ideal gases at the same P and T. Thus, mole fraction of O₂ = 0.6. By Dalton’s law, P_O₂ = 0.6 × 2 = 1.2 bar.
5 / 10
A gas mixture at 400 K and 3 bar consists of 2 moles of CH₄ and 3 moles of Ar. The partial pressure of Ar is:
1.8 bar
2.4 bar
0.6 bar
Total moles n_total = 2 + 3 = 5. Mole fraction of Ar = 3/5. By Dalton’s law, P_Ar = (3/5) × 3 = 1.8 bar.
6 / 10
The mole fraction of a gas in a mixture is equal to:
The ratio of its partial pressure to the total pressure
The ratio of its partial volume to the total volume
Both A and B
The ratio of its mass to the total mass
For an ideal gas mixture, the mole fraction yᵢ = nᵢ/n_total. By Dalton’s law, yᵢ = Pᵢ/P_total (since Pᵢ = (nᵢ RT/V) and P_total = (n_total RT/V)). By Amagat’s law, yᵢ = Vᵢ/V_total (since Vᵢ = (nᵢ RT/P) and V_total = (n_total RT/P)). Thus, both A and B are correct.
7 / 10
A mixture of 1 mole of CO₂ and 4 moles of He is at 1 bar and 300 K. The partial volume of CO₂ according to Amagat’s law is:
0.00498 m³
0.0249 m³
0.0996 m³
0.01245 m³
By Amagat’s law, partial volume Vᵢ = (nᵢ/n_total) × V_total. Total moles n_total = 1 + 4 = 5. Total volume V_total = (n_total RT)/P = (5 × 8.314 × 300)/(1 × 10⁵) = 0.1245 m³. Mole fraction of CO₂ = 1/5. Thus, V_CO₂ = (1/5) × 0.1245 = 0.0249 m³.
8 / 10
A gas mixture contains 2 moles of N₂ and 3 moles of O₂ at 300 K and 2 bar total pressure. The partial pressure of N₂ is:
0.4 bar
By Dalton’s law, partial pressure Pᵢ = (nᵢ/n_total) × P_total. Total moles n_total = 2 + 3 = 5. Mole fraction of N₂ = 2/5. Thus, P_N₂ = (2/5) × 2 = 0.8 bar
9 / 10
Amagat’s law of partial volumes states that:
The total volume of a gas mixture is the sum of the partial pressures
The total volume of a gas mixture is the sum of the volumes each gas would occupy at the same pressure and temperature
The total pressure of a gas mixture is the sum of the partial volumes
The total volume is proportional to the total pressure
Amagat’s law states that for a mixture of non-reacting ideal gases at constant pressure and temperature, the total volume V = V₁ + V₂ + … + Vₙ, where Vᵢ is the volume the i-th gas would occupy alone at the total pressure P and temperature T.
10 / 10
Dalton’s law of partial pressures states that:
The total pressure of a gas mixture is the sum of the volumes of individual gases
The total pressure of a gas mixture is the sum of the partial pressures of individual gases
The total temperature of a gas mixture is the sum of individual gas temperatures
Dalton’s law states that for a mixture of non-reacting ideal gases at constant temperature and volume, the total pressure P = P₁ + P₂ + … + Pₙ, where Pᵢ is the partial pressure of the i-th gas.
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