Sigma Percentile
LEVELJEE Main

Animated Solution for Chemistry - States of Matter: When , and represent rate of diffusion, pressure and molecular mass, respectively, then the ratio of the rates of diffusion of two gases and , is given as

Select Answer:

Visualized Solution

\text{Visualizing Gas Diffusion}

  • Let's consider two gases, and , kept in separate containers.
  • They have pressures and , and molecular masses and .

\text{Graham's Law of Diffusion}

  • According to Graham's Law, the rate of diffusion is directly proportional to the pressure and inversely proportional to the square root of the molecular mass .

\text{Applying the Law to Gases A and B}

  • For Gas A:
  • For Gas B:

\text{Taking the Ratio}

  • Dividing the rate of Gas A by the rate of Gas B:

\text{Simplifying the Expression}

  • Rearranging the terms:

\text{Extensions of Graham's Law}

  • What if the gases were diffusing at different temperatures?
  • Think about how temperature affects the kinetic energy and thus the rate of diffusion.

The Sigma Insight: Gaseous State

Solution Diagram

Visualizing the Great Escape

Imagine you have two separate containers, one filled with Gas and the other with Gas . Both gases are restless, ready to diffuse out through a tiny pinhole into the open air. Gas is packed in at a pressure and has a molecular mass . Meanwhile, Gas is at a pressure with a molecular mass . The core question we need to answer is: how fast do they escape relative to each other?
To compare their rates of diffusion, we need a powerful theoretical tool. This is where Graham's Law of Diffusion comes into play.

The Master Equation

Graham's Law
Graham's Law tells us that the rate at which a gas diffuses, denoted by , is directly proportional to its pressure . This makes intuitive sense—higher pressure means the gas molecules are colliding with the walls (and the pinhole) more frequently, effectively pushing the gas out faster.
However, the rate is also inversely proportional to the square root of its molecular mass . Heavier gases are sluggish; they move slower at a given temperature compared to lighter gases. Mathematically, we express this beautiful relationship as:
Now, let's apply this raw setup to our specific gases. For Gas , the rate will be proportional to divided by the square root of :
Similarly, for Gas , the rate will be proportional to divided by the square root of :

The Final Calculation

The question asks for the ratio of their rates of diffusion, . So, let's divide the first expression by the second. We must be careful not to make a silly mistake while rearranging the fractions.
Let's simplify this complex fraction. The pressure terms and group together cleanly as . The square root of in the denominator of the bottom fraction flips up to the numerator, giving us the square root of .
In mathematics, a square root can also be written as a power of . Rewriting our expression in this format yields:
And there we have it! This exact expression matches option (b) perfectly.

The Way Forward

While we've solved the problem, it's always good to think a bit deeper. What if the two gases were kept at different temperatures? How would that change our formula? Remember, an increase in temperature increases the kinetic energy of the gas molecules, which in turn increases the rate of diffusion. The generalized form of Graham's Law includes temperature in the denominator inside the square root: . Keep this generalized form in mind for tackling more advanced problems in the future!

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