The density of the given gas at constant pressure and temperature is `rho` and its rate of diffusion isr. If density of the gas becone `rho//3` then rate of diffusion becomes

Options

A. `sqrt2 r`
B. r
C. `r//sqrt3`
D. `sqrt3 r`

Answer

To solve the problem, we will use Graham's law of effusion, which states that the rate of diffusion of a gas is inversely proportional to the square root of its density.

### Step-by-Step Solution:

1. **Understanding Graham's Law**: According to Graham's law, the rate of diffusion (R) is inversely proportional to the square root of the density (ρ) of the gas. Mathematically, this can be expressed as:
\[ R \propto \frac{1}{\sqrt{\rho}} \]

2. **Setting Up the Initial Conditions**: Let the initial density of the gas be \( \rho \) and the initial rate of diffusion be \( R \).
\[ R_1 = R \quad \text{and} \quad \rho_1 = \rho \]

3. **Changing the Density**: The problem states that the density of the gas becomes \( \frac{\rho}{3} \). We denote the new rate of diffusion as \( R_2 \) and the new density as \( \rho_2 \):
\[ \rho_2 = \frac{\rho}{3} \]

4. **Applying Graham's Law**: According to Graham's law, we can write the relationship between the initial and final rates of diffusion and densities:
\[ \frac{R_1}{R_2} = \frac{\sqrt{\rho_2}}{\sqrt{\rho_1}} \]

5. **Substituting Known Values**: Substitute \( R_1 = R \), \( \rho_1 = \rho \), and \( \rho_2 = \frac{\rho}{3} \):
\[ \frac{R}{R_2} = \frac{\sqrt{\frac{\rho}{3}}}{\sqrt{\rho}} \]

6. **Simplifying the Equation**: This simplifies to:
\[ \frac{R}{R_2} = \frac{\sqrt{\rho}}{\sqrt{3} \cdot \sqrt{\rho}} = \frac{1}{\sqrt{3}} \]

7. **Finding \( R_2 \)**: Rearranging the equation gives:
\[ R_2 = R \cdot \sqrt{3} \]

8. **Final Result**: Thus, the new rate of diffusion when the density becomes \( \frac{\rho}{3} \) is:
\[ R_2 = R \sqrt{3} \]