Let A and B the two gases and given: `(T_(A))/(M_(A))=4.(T_(B))/(M_(B))`, where T is the temperature and M is the molecular mass. If `C_(A)` and `C_(B)` are the rms speed, then the ratio `(C_(A))/(C_(B))` will be equal to ………..

Answer

To solve the problem, we need to find the ratio of the root mean square (rms) speeds of two gases A and B, given the relationship between their temperatures and molecular masses.

### Step-by-Step Solution

1. **Understand the formula for rms speed**:
The rms speed \( C \) of a gas is given by the formula:
\[ C = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the universal gas constant, \( T \) is the temperature, and \( M \) is the molecular mass.

2. **Write the expressions for \( C_A \) and \( C_B \)**:
For gas A:
\[ C_A = \sqrt{\frac{3RT_A}{M_A}} \]
For gas B:
\[ C_B = \sqrt{\frac{3RT_B}{M_B}} \]

3. **Find the ratio \( \frac{C_A}{C_B} \)**:
\[ \frac{C_A}{C_B} = \frac{\sqrt{\frac{3RT_A}{M_A}}}{\sqrt{\frac{3RT_B}{M_B}}} \]
This simplifies to:
\[ \frac{C_A}{C_B} = \sqrt{\frac{T_A}{M_A} \cdot \frac{M_B}{T_B}} \]

4. **Use the given relationship**:
We are given:
\[ \frac{T_A}{M_A} = 4 \cdot \frac{T_B}{M_B} \]
Substitute this into the ratio:
\[ \frac{C_A}{C_B} = \sqrt{\left(4 \cdot \frac{T_B}{M_B}\right) \cdot \frac{M_B}{T_B}} \]

5. **Simplify the expression**:
The \( T_B \) terms cancel out:
\[ \frac{C_A}{C_B} = \sqrt{4} = 2 \]

### Final Answer:
Thus, the ratio \( \frac{C_A}{C_B} \) is equal to **2**.

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