Question

The ratio of the rate of diffusion of helium and methane under identical conditions of pressure and temperature will be [IIT $2005$]
A.$4$
B.$2$
C.$1$
D.$0.5$

Answer 1

Hint: Under the identical condition of pressure and temperature the rate of diffusion is inversely proportional to the square root of the molar mass of the gas. This is known as Graham’s law of diffusion. The ratio of the rate of diffusion of helium and methane gas is calculated by using this law.

Formula Used:The mathematical expression of the rate of diffusion can be expressed as:
$r\,\alpha \,\dfrac{1}{\sqrt{M}}$
And for two gases we can write this equation in the form of a ratio:
$\dfrac{{{r}_{1}}}{{{r}_{2}}}=\sqrt{\dfrac{{{M}_{2}}}{{{M}_{1}}}}$
Here ${{r}_{1}}\And {{r}_{2}}$ denotes the rate of diffusion of the gas $1$and $2$
And ${{M}_{1}}\And {{M}_{2}}$are the molar mass of the gas$1$and$2$

Complete step by step solution:When the gas molecules move from one place to another place along a concentration gradient is called diffusion. Generally, gas molecules are unaware of any concentration gradient; they just simply move randomly from the higher concentration region to the lower concentration region.
From Graham’s law of diffusion, we can write the relation between the rate of diffusion and the molar mass of gasses as:
$r\,\alpha \,\dfrac{1}{\sqrt{M}}$
Or, $r\,=\,\dfrac{k}{\sqrt{M}}$ [$k=$Proportionality constant]
For helium$(He)$gas,${{r}_{He}}=\dfrac{k}{\sqrt{{{M}_{He}}}}$
For methane$(C{{H}_{4}})$ gas,${{r}_{C{{H}_{4}}}}=\dfrac{k}{\sqrt{{{M}_{C{{H}_{4}}}}}}$
Now taking the ratio we get, $\dfrac{{{r}_{He}}}{{{r}_{C{{H}_{4}}}}}=\sqrt{\dfrac{{{M}_{C{{H}_{4}}}}}{{{M}_{He}}}}$ ………(i)
Now, ${{M}_{C{{H}_{4}}}}=$ (atomic weight of $C$)$+$ ($4\times $atomic weight of $H$)
$\therefore {{M}_{C{{H}_{4}}}}=(12+4\times 1)=16g/mol$
And ${{M}_{He}}=$ Atomic weight of $He$$=4g/mol$
From (i) we get,$\dfrac{{{r}_{He}}}{{{r}_{C{{H}_{4}}}}}=\sqrt{\dfrac{16g/mol}{4g/mol}}$
Or,$\dfrac{{{r}_{He}}}{{{r}_{C{{H}_{4}}}}}=2$
Therefore the ratio of the rate of diffusion of helium and methane under the identical condition of pressure and temperature is $2$.

Thus, option (B) is correct.

Note: There is a difference between the effusion and diffusion of gasses. Effusion occurs through gas particle movement via a tiny hole and here only a small hole allows gaseous molecules to move. But in the case of diffusion happens when the holes are larger than the mean free path or no holes are present.