The rate of effusion of two gases `'a'` and `'b'` under identical conditions of temperature and pressure are in the ratio of `2:1` What is the ratio of `rms` velocity of their molecules if `T_(a)` and `T_(b)` are in the ratio of `2:1`?

Options

A. `2 : 1`
B. `sqrt2 : 1`
C. `2sqrt2 : 1`
D. `1 : sqrt2`

Explanation

To solve the problem, we need to find the ratio of the root mean square (rms) velocities of two gases, A and B, given that the rate of effusion of these gases is in the ratio of 2:1 and their temperatures are in the ratio of 2:1.

### Step-by-Step Solution:

1. **Understanding Graham's Law of Effusion**:
Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, this can be expressed as:
\[ \frac{\text{Rate}_A}{\text{Rate}_B} = \sqrt{\frac{M_B}{M_A}} \]
where \(M_A\) and \(M_B\) are the molar masses of gases A and B, respectively.

2. **Given Information**:
We are given that:
\[ \frac{\text{Rate}_A}{\text{Rate}_B} = \frac{2}{1} \]
This implies:
\[ \frac{2}{1} = \sqrt{\frac{M_B}{M_A}} \]

3. **Squaring Both Sides**:
To eliminate the square root, we square both sides:
\[ \left(\frac{2}{1}\right)^2 = \frac{M_B}{M_A} \]
This gives:
\[ \frac{4}{1} = \frac{M_B}{M_A} \implies M_B = 4M_A \]

4. **Root Mean Square Velocity Formula**:
The rms velocity (\(V_{\text{rms}}\)) of a gas is given by the formula:
\[ V_{\text{rms}} = \sqrt{\frac{3RT}{M}} \]
where \(R\) is the universal gas constant and \(T\) is the temperature.

5. **Finding the Ratio of rms Velocities**:
For gases A and B, the rms velocities can be expressed as:
\[ V_{\text{rms}, A} = \sqrt{\frac{3RT_A}{M_A}}, \quad V_{\text{rms}, B} = \sqrt{\frac{3RT_B}{M_B}} \]
Therefore, the ratio of rms velocities is:
\[ \frac{V_{\text{rms}, A}}{V_{\text{rms}, B}} = \frac{\sqrt{\frac{3RT_A}{M_A}}}{\sqrt{\frac{3RT_B}{M_B}}} \]
This simplifies to:
\[ \frac{V_{\text{rms}, A}}{V_{\text{rms}, B}} = \sqrt{\frac{T_A}{T_B}} \cdot \sqrt{\frac{M_B}{M_A}} \]

6. **Substituting Known Ratios**:
We know:
- \(\frac{T_A}{T_B} = \frac{2}{1}\)
- \(\frac{M_B}{M_A} = \frac{4}{1}\)

Substituting these values into the equation gives:
\[ \frac{V_{\text{rms}, A}}{V_{\text{rms}, B}} = \sqrt{\frac{2}{1}} \cdot \sqrt{\frac{4}{1}} = \sqrt{2} \cdot 2 = 2\sqrt{2} \]

7. **Final Ratio**:
Thus, the ratio of the rms velocities of gases A and B is:
\[ \frac{V_{\text{rms}, A}}{V_{\text{rms}, B}} = 2\sqrt{2} : 1 \]

### Final Answer:
The ratio of the rms velocity of molecules of gas A to gas B is \(2\sqrt{2} : 1\).