Gas A has a density that is 1.20 times that of gas B. If gas A effuses through an orifice at a rate of 3.41g/L, at what rate will gas B effuse?
Hint: Ratio the molecular masses to the densities. Use this information with Graham's law of diffusion. Note that 3.41g/L is a rate. (3.11)
The part that confuses me is how to ratio the molecular masses after that I could use some steps just to make sure I get it right. thank you in advance.
The rate of effusion is inversely proportional to the square root of the molecular mass.
Let the rates be Ra, Rb.
Let the densities be Da, Db
Da / Db = 1/1.2
Based on the Law of effusion,
Ra/Rb = sqrt(Db) / sqrt(Da).
Db/Da = sqrt(1.2)/sqrt(1)
(3.41g/L) / Rb = sqrt(1/1.2)
(3.41g/L) / Rb = sqrt(1)/sqrt(1.2)
Solve for Rb
I think something is amiss here. If A has a higher density than B, then A must effuse slower than B (the answer GK gives of 3.73 supports that), but the answer of 3.11 doesn't support that. The only way I can come up with 3.11 is for the density of B to be 1.2 times that of A. Have I missed something? Perhaps I'm having trouble envisioning 3.41 g/L as a rate; g/L sounds like a density to me.
Now I am beginning to worry. I missed the units given. Could it be grams/minute instead of g/L?
Capacino, the ball is in your court.
To ratio the molecular masses of gas A and gas B, you need to divide the molecular mass of gas A by the molecular mass of gas B. The molecular mass can be found by looking up the molar mass of each gas in the periodic table.
Let's assume the molecular mass of gas A is MA and the molecular mass of gas B is MB.
Now, let's use Graham's law of diffusion, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
Mathematically, we can write this as:
Rate of effusion of gas A / Rate of effusion of gas B = √(MB / MA)
Since the rate of effusion of gas A is given as 3.41 g/L, we can write:
3.41 g/L (Rate of effusion of gas A) / Rate of effusion of gas B = √(MB / MA)
Now, let's solve for the rate of effusion of gas B.
1. Rearrange the equation to solve for the rate of effusion of gas B:
Rate of effusion of gas B = (3.41 g/L) / √(MB / MA)
2. Use the given information that the density of gas A is 1.20 times that of gas B:
Density of gas A / Density of gas B = 1.20
Since density is directly proportional to molar mass, we can write:
√(Molar mass of B / Molar mass of A) = 1.20
Square both sides of the equation:
(MB / MA) = 1.20^2
(MB / MA) = 1.44
3. Substitute this value into the equation for the rate of effusion of gas B:
Rate of effusion of gas B = (3.41 g/L) / √1.44
4. Simplify the expression:
Rate of effusion of gas B = (3.41 g/L) / 1.2
Rate of effusion of gas B ≈ 2.84 g/L
Therefore, the rate at which gas B will effuse is approximately 2.84 g/L.