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 the Initial Conditions**:
- Let the initial density of the gas be \( \rho \).
- The initial rate of diffusion is given as \( r \).

2. **Applying Graham's Law**:
- According to Graham's law, the rate of diffusion \( r \) is inversely proportional to the square root of the density \( \rho \):
\[ r \propto \frac{1}{\sqrt{\rho}} \]
- This can be expressed as:
\[ r = k \cdot \frac{1}{\sqrt{\rho}} \]
where \( k \) is a proportionality constant.

3. **Changing the Density**:
- Now, the density of the gas changes to \( \frac{\rho}{3} \).
- We need to find the new rate of diffusion, which we will denote as \( r' \).

4. **Calculating the New Rate of Diffusion**:
- Using the same relationship, the new rate of diffusion \( r' \) can be expressed as:
\[ r' = k \cdot \frac{1}{\sqrt{\frac{\rho}{3}}} \]
- Simplifying this expression:
\[ r' = k \cdot \frac{1}{\sqrt{\rho}} \cdot \sqrt{3} = r \cdot \sqrt{3} \]

5. **Final Result**:
- Therefore, the new rate of diffusion when the density becomes \( \frac{\rho}{3} \) is:
\[ r' = r \cdot \sqrt{3} \]

### Conclusion:
The rate of diffusion becomes \( r \cdot \sqrt{3} \) when the density of the gas decreases to \( \frac{\rho}{3} \).