At constant volume and temperature conditions, the rates of diffusion `r_(A)` and `r_(B)` of gases A and B having densities `P_(A)` and `P_(B)` are related by the expression :

Options

A. `r_(A)=r_(B).(p_(A)//p_(B))^(2)`
B. `r_(A)=r_(B)(p_(A)//p_(B))^(1//2)`
C. `r_(A)=(r_(B).p_(A)//p_(B))^(1//2)`
D. `r_(A)=r_(B)(p_(A)//p_(B))^(1//2)`

Explanation

To solve the problem regarding the rates of diffusion of gases A and B at constant volume and temperature, we can follow these steps:

### Step 1: Understand the relationship between diffusion rates and density
The rates of diffusion of two gases are inversely proportional to the square root of their densities. This relationship can be expressed mathematically as:
\[ \frac{r_A}{r_B} = \sqrt{\frac{P_B}{P_A}} \]
where \( r_A \) and \( r_B \) are the rates of diffusion of gases A and B, and \( P_A \) and \( P_B \) are their respective densities.

### Step 2: Relate density to molar mass
Density (\( P \)) of a gas can be expressed in terms of its molar mass (\( M \)) and the ideal gas law. The formula for density is:
\[ P = \frac{M}{RT} \]
where \( R \) is the universal gas constant and \( T \) is the temperature in Kelvin.

### Step 3: Substitute density in the diffusion rate equation
Substituting the expression for density into the diffusion rate equation gives:
\[ \frac{r_A}{r_B} = \sqrt{\frac{M_B}{M_A}} \]
This shows that the rates of diffusion are inversely proportional to the square root of the molar masses of the gases.

### Step 4: Conclusion
At constant volume and temperature, the rates of diffusion of gases A and B can be expressed as:
\[ \frac{r_A}{r_B} = \sqrt{\frac{M_B}{M_A}} \]
This means that the gas with a lower molar mass will diffuse faster than the gas with a higher molar mass.

### Final Answer
The expression relating the rates of diffusion \( r_A \) and \( r_B \) of gases A and B is:
\[ \frac{r_A}{r_B} = \sqrt{\frac{M_B}{M_A}} \]