Question

The ratio of rates of diffusion of $S{{O}_{2}}$,${{O}_{2}}$and $C{{H}_{4}}$is [BHU $1992$]
A.$1:\sqrt{2}:2$
B.$1:2:4$
C.$2:\sqrt{2}:1$
D.$1:2:\sqrt{2}$

Answer 1

Hint: Diffusion of gasses occurs when gas molecules move from one place to another along the concentration gradient. According to Graham’s law of diffusion, the rate of diffusion is inversely proportional to the molecular weight of the gas. By using Graham's law we can solve the given question.

Formula Used: The mathematical expression of Graham’s law of diffusion:
$\,r\,\,=\dfrac{k}{\sqrt{M}}$
Here $r\,=$Rate of diffusion
$M=$The molecular weight of the gas
$k=$proportionality constant

Complete step-by-step solution:Gaseous molecules tend to undergo diffusion as they possess kinetic energy. At higher temperatures when gaseous molecules have higher kinetic energy, the rate of diffusion becomes faster.
The rate of diffusion depends on different factors such as the amount of surface area, the distance of gaseous particles, the concentration gradient, etc.
Here in this problem, three different gasses$S{{O}_{2}}$,${{O}_{2}}$and $C{{H}_{4}}$ are given. To calculate the rate of diffusions we must have to calculate the molecular weight of these gasses.
The molecular weight of $S{{O}_{2}},\,\,\,{{M}_{S{{O}_{2}}}}=$(Atomic weight of $S$)$+$($2\times $Atomic weight of $O$)
$\therefore {{M}_{S{{O}_{2}}}}=32+16\times 2=64$
The molecular weight of ${{O}_{2}},\,\,{{M}_{{{O}_{2}}}}=$($2\times $Atomic weight of $O$)
$\therefore {{M}_{{{O}_{2}}}}=2\times 16=32$
The molecular weight of $C{{H}_{4}},{{M}_{C{{H}_{4}}}}=$(Atomic weight of $C$)$+\,$($4\times $Atomic weight of hydrogen)
$\therefore {{M}_{C{{H}_{4}}}}=12+4\times 1=16$
Putting these values in Graham’s law of diffusion we get,
${{r}_{S{{O}_{2}}}}:{{r}_{{{O}_{2}}}}:{{r}_{C{{H}_{4}}}}=\dfrac{1}{\sqrt{{{M}_{S{{O}_{2}}}}}}:\dfrac{1}{\sqrt{{{M}_{{{O}_{2}}}}}}:\dfrac{1}{\sqrt{{{M}_{C{{H}_{4}}}}}}$
Or,${{r}_{S{{O}_{2}}}}:{{r}_{{{O}_{2}}}}:{{r}_{C{{H}_{4}}}}=\dfrac{1}{\sqrt{64}}:\dfrac{1}{\sqrt{32}}:\dfrac{1}{\sqrt{16}}$
Or,${{r}_{S{{O}_{2}}}}:{{r}_{{{O}_{2}}}}:{{r}_{C{{H}_{4}}}}=\dfrac{1}{8}:\dfrac{1}{4\sqrt{2}}:\dfrac{1}{4}$
Or,${{r}_{S{{O}_{2}}}}:{{r}_{{{O}_{2}}}}:{{r}_{C{{H}_{4}}}}=\dfrac{1\times 2}{2}:\dfrac{1\times 2}{\sqrt{2}}:\dfrac{1\times 2}{1}=1:\sqrt{2}:2$
Therefore the ratio of the rate of diffusion of $S{{O}_{2}}$,${{O}_{2}}$and $C{{H}_{4}}$is $1:\sqrt{2}:2$

Thus, option (A) is correct.

Note: The rate of diffusion of a gas is also inversely proportional to time and the square root of the density of the gas. Graham’s law is used extensively in Physics, Chemistry, and many other branches. Therefore we must have to learn how to apply these laws to various problems.