Introduction to Graham's Law
Graham's Law of Diffusion states that the rate of diffusion or effusion of a gas is inversely proportional to the square root of its molar mass (or density) at constant temperature and pressure. Mathematically, it is expressed as:
$$Rate \propto \frac{1}{\sqrt{M}}$$
Where $M$ is the molar mass of the gas. For two gases, the ratio of their rates is:
$$\frac{Rate_1}{Rate_2} = \sqrt{\frac{M_2}{M_1}}$$
Applications
- Separation of Isotopes: Used in gas centrifuges for uranium enrichment.
- Gas Identification: Determining the molar mass of unknown gases.
- Leak Detection: Identifying leaks in gas systems based on effusion rates.
Problem Solving: HCl and NH3 Diffusion
Problem: A 2-meter tube has $HCl$ ($M_1 = 36.5$ g/mol) at one end and $NH_3$ ($M_2 = 17$ g/mol) at the other. Where does $NH_4Cl$ (ammonium chloride) appear?
Step 1: Understand the chemistry. When $NH_3$ and $HCl$ meet, they react: $NH_3(g) + HCl(g) \rightarrow NH_4Cl(s)$. The white smoke appears where they meet.
Step 2: Set up the ratio. Let $x$ be the distance from the $HCl$ end. The distance from the $NH_3$ end is $(2-x)$. Since $Rate = \frac{distance}{time}$, and time is equal for both:
$$\frac{Distance_{NH_3}}{Distance_{HCl}} = \sqrt{\frac{M_{HCl}}{M_{NH_3}}}$$
$$\frac{2-x}{x} = \sqrt{\frac{36.5}{17}} \approx \sqrt{2.147} \approx 1.465$$
Step 3: Calculate $x$. $2-x = 1.465x$ $2 = 2.465x$ $x = 0.811$ meters.
Conclusion: The ammonium chloride will first appear at approximately 0.81 meters from the $HCl$ end.