Question

A certain volume of methane diffuses in 10 sec through a porous partition the time take by an equal volume of oxygen to diffuse under the same condition is

Answer 1

Answer:- 14.1 seconds

Solution:- It is based on Graham's law of diffusion rates. As per this law, "The rate of diffusion of a gas is inversely proportional to the square root of it's molar mass."

The formula for comparing the diffusion rates of two gases the formula is written as:

$\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}$

r stands for rate of diffusion and M stands for molar mass of the gas and the subscripts 1 and 2 are referring two different gases.

Let's say 1 is for methane and 2 is for oxygen. Molar mass of methane is 16 and molar mass of oxygen is 32.

Rate is how much volume diffused per unit time, so:

$\frac{\frac{V_1}{t_1}}{\frac{V_2}{t_2}}=\sqrt{\frac{M_2}{M_1}}$

Since the volumes are same, the formula could be written as:

$\frac{t_2}{t_1}=\sqrt{\frac{M_2}{M_1}}$

Let's plug in the values and solve this for the time taken to diffuse oxygen gas:

$\frac{t_2}{10sec}=\sqrt{\frac{32}{16}}$

$\frac{t_2}{10sec}=1.41$

$t_2=1.41*10sec$

$t_2=14.1sec$

So, oxygen gas will take 14.1 seconds to diffuse.

Answer 2

Explanation:

the answer of the quetion is 14seconds