Question

One way of separating oxygen isotopes is by gaseous diffusion of carbon monoxide. the gaseous diffusion process behaves like an effusion process. calculate the relative rates of effusion of 12c16o, 12c17o, and 12c18o. list

Answer 1

The relative rates of effusion of 12C16O, 12C17O, and 12C18O can be calculated using Graham's law. The relative rates are approximately 1, 1.03, and 1.04 respectively.

Step-by-step explanation:

The relative rates of effusion can be calculated using Graham's law, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. In this case, we are given three isotopes of carbon monoxide: 12C16O, 12C17O, and 12C18O. The molar mass of 12C16O is 28 g/mol, the molar mass of 12C17O is 29 g/mol, and the molar mass of 12C18O is 30 g/mol.

Using Graham's law, we can calculate the relative rates of effusion as follows:

Rate of effusion of 12C16O / Rate of effusion of 12C17O = √(Molar mass of 12C17O / Molar mass of 12C16O) = √(29 g/mol / 28 g/mol) ≈ 1.03

Rate of effusion of 12C16O / Rate of effusion of 12C18O = √(Molar mass of 12C18O / Molar mass of 12C16O) = √(30 g/mol / 28 g/mol) ≈ 1.04

Therefore, the relative rates of effusion of 12C16O, 12C17O, and 12C18O are approximately 1, 1.03, and 1.04 respectively.

Answer 2

Rates of effusion of gases is governed by the Graham's Law. This law relates the ratio of the rates of effusion to the ratio of the molar masses of the gases involved. With the assumption that pressure and temperature are similar for both gases, the Graham's law states that,
(rate of Gas A)/(rate of Gas B) = (molar mass of Gas B)^1/2 / (molar mass of Gas A)^1/2

Substituting the known values to the equation.

Part A:
rate of C12/ rate of O16 = sqrt (16/12) = 1.1543

The rate of C12 is 1.15 faster than O16.

Part B:
rate of C12/ rate of O17 = sqrt (17/12) = 1.19

The rate of C12 is 1.19 faster than O17.

Part C:
rate of C12/rate of O18 = sqrt (18/12) = 1.22
The rate of C12 is 1.22 faster than O18.