if 465cm cube of so2 can diffuse through a porous partition in 30sec how long will (a) an equal volume (b) 630cm cube of hydrogen sulphide take to diffuse through the same partition
rate SO2 = volume/time = 465 cc/30 sec = ?
rate H2S = 465/time
MM = molar mass
(rate SO2/rate H2S) = sqrt(MM H2S/MMSO2)
Plug in and solve for time H2S
Do the same for part b but substitute 630 cc for volume H2S.
Post your work if you get stuck.
To determine the time required for the diffusion of different gases, we can use Graham's Law of Diffusion, which states that the rate of diffusion of a gas is inversely proportional to the square root of its molecular weight.
(a) Let's calculate the time for an equal volume, assuming the same conditions for diffusion:
Since the volume is the same, the only difference is the gas itself.
First, we need to determine the relative rates of diffusion for SO2 and H2S:
Molecular weight of SO2 = 64 g/mol
Molecular weight of H2S = 34 g/mol
Using Graham's Law, the ratio of the rates of diffusion is given by:
Rate of diffusion of SO2 / Rate of diffusion of H2S = √(Molecular weight of H2S / Molecular weight of SO2)
Let's substitute the molecular weights of SO2 and H2S to find the ratio:
Rate of diffusion of SO2 / Rate of diffusion of H2S = √(34 / 64) = 0.6667 (approximately)
Since the ratio is 0.6667, it means that H2S diffuses at approximately 0.6667 times the rate of SO2. Therefore, the time required for H2S to diffuse an equal volume would be approximately 1.5 times longer.
30 seconds * 1.5 ≈ 45 seconds
So, it would take around 45 seconds for an equal volume of H2S to diffuse through the same partition.
(b) Let's calculate the time for a volume of 630 cm³ of H2S:
Since the volume of H2S is given, we don't need to consider the ratio of rates of diffusion.
Using the ratio:
Rate of diffusion of SO2 / Rate of diffusion of H2S = √(34 / 64) = 0.6667 (approximately)
We can calculate the time required:
30 seconds * (630 cm³ / 465 cm³) * (1 / 0.6667) = 53.16 seconds (approximately)
So, it would take around 53.16 seconds for a volume of 630 cm³ of H2S to diffuse through the same partition.