What is the order of increasing rate of effusion for the following gases at constant temperature? Put the slowest at the top, or first, and the fastest at the bottom, or last. Mwts: He = 4.0 g; C= 12.0 g; N = 14.0 g; O = 16.0 g; Ar = 39.95 g Ar, CO2, He, N2
The rate of effusion for a gas is inversely proportional to the square root of its molar mass. Therefore, the order of increasing rate of effusion for the given gases is:
1. CO2 (Molecular weight: 44.01 g/mol)
2. Ar (Molecular weight: 39.95 g/mol)
3. N2 (Molecular weight: 28.0 g/mol)
4. O (Molecular weight: 16.0 g/mol)
5. He (Molecular weight: 4.0 g/mol)
So, the slowest gas to effuse is CO2, followed by Ar, N2, O, and the fastest is He.
To determine the order of increasing rate of effusion for the given gases, we can use Graham's Law of Effusion, which states that the rate of effusion is inversely proportional to the square root of the molar mass of the gas.
1. Calculate the square root of the molar mass for each gas:
- Molar Mass of He (helium) = 4.0 g/mol, square root ≈ 2.0
- Molar Mass of C (carbon) = 12.0 g/mol, square root ≈ 3.5
- Molar Mass of N (nitrogen) = 14.0 g/mol, square root ≈ 3.7
- Molar Mass of O (oxygen) = 16.0 g/mol, square root ≈ 4.0
- Molar Mass of Ar (argon) = 39.95 g/mol, square root ≈ 6.3
2. Compare the square roots of the molar mass for each gas:
- The slowest gas is helium (He) with a square root value of 2.0.
- The next gas is carbon dioxide (CO2) with a square root value of 3.5.
- The next gas is nitrogen (N2) with a square root value of 3.7.
- The fastest gas is argon (Ar) with a square root value of 6.3.
Therefore, the order of increasing rate of effusion for the given gases, from slowest to fastest, is: He, CO2, N2, Ar.