Argon exists in three isotopic forms, 36Ar,
38Ar, and 40Ar.
What is the relative rate of effusion of the
fastest to slowest, ufast/uslow?
The fastest will be Ar36; the slowest will be Ar40.
(rate 36/rate 40) = sqrt(40/36)
Grahams law, difusion rate is proportional to inverse sqrt of mass.
let the fastest rate be RATE
relative rate of second is RATE(sqrt(36/38)= .973 RATE
relative rate of last is
Rate(sqrt(36/40)=.949Rate
Please note that you didn't get too different answers here.
My response gives rate of fastest = 1.05 x slowest.
Bob P gave rate slowest = 0.949 x rate fastest.
Also note that 1/1.05(4) = 0.949
To calculate the relative rate of effusion between two gases, we can use Graham's law of effusion, 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 comparing the rates of effusion of 36Ar, 38Ar, and 40Ar. Since the molar mass of an isotope is proportional to its atomic mass, we can calculate the relative rates of effusion as follows:
1. Calculate the square root of the molar masses for each isotope:
- For 36Ar, the square root of its molar mass (36) is √36 = 6.
- For 38Ar, the square root of its molar mass (38) is √38 ≈ 6.16.
- For 40Ar, the square root of its molar mass (40) is √40 ≈ 6.32.
2. Calculate the relative rates of effusion by taking the ratio of the square roots:
- The relative rate of effusion between 36Ar and 38Ar is 6/6.16 ≈ 0.974.
- The relative rate of effusion between 36Ar and 40Ar is 6/6.32 ≈ 0.949.
- The relative rate of effusion between 38Ar and 40Ar is 6.16/6.32 ≈ 0.975.
Therefore, the relative rate of effusion of the fastest to the slowest is ufast/uslow ≈ 0.949.