What is the ratio of the effusion rates of N₂ and N₂O₄?
The ratio of the effusion rates of N₂ and N₂O₄ can be calculated using Graham's Law of Effusion. This law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. In this case, the molar mass of N₂ is 28 g/mol, and the molar mass of N₂O₄ is 92 g/mol. To find the ratio of their effusion rates, we can use the formula:
Effusion rate of N₂ / Effusion rate of N₂O₄ = √(Molar mass of N₂O₄ / Molar mass of N₂)
Plugging in the values, we get:
Effusion rate of N₂ / Effusion rate of N₂O₄ = √(92 g/mol / 28 g/mol)
Simplifying further, we have:
Effusion rate of N₂ / Effusion rate of N₂O₄ = √(3.286)
Calculating the square root, we find:
Effusion rate of N₂ / Effusion rate of N₂O₄ ≈ 1.81
Therefore, the ratio of the effusion rates of N₂ to N₂O₄ is approximately 1.81.
Graham's Law of Effusion
Graham's Law of Effusion states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. The effusion rate refers to how quickly a gas escapes through a tiny hole into a vacuum. In simple terms, lighter gas molecules effuse faster than heavier ones.
Calculating the Ratio of Effusion Rates
To calculate the ratio of the effusion rates of N₂ and N₂O₄, we first need to determine their molar masses. The molar mass of N₂ (Nitrogen gas) is 28 g/mol, and the molar mass of N₂O₄ (Dinitrogen tetroxide) is 92 g/mol.
By applying Graham's Law formula:
Effusion rate of N₂ / Effusion rate of N₂O₄ = √(Molar mass of N₂O₄ / Molar mass of N₂)
Effusion rate of N₂ / Effusion rate of N₂O₄ = √(92 g/mol / 28 g/mol)
Effusion rate of N₂ / Effusion rate of N₂O₄ = √(3.286)
Effusion rate of N₂ / Effusion rate of N₂O₄ ≈ 1.81
Therefore, the ratio of the effusion rates of N₂ to N₂O₄ is approximately 1.81. This means that Nitrogen gas (N₂) effuses 1.81 times faster than Dinitrogen tetroxide (N₂O₄) under the same conditions.