Speed of sound wave in a gas `V_(1)` and rms speed of molecules of the gas at the same temperature is `v_(2)`.

Options

A. `v_(1)=v_(2)`
B. `v_(1)ltv_(2)`
C. `v_(1)gtv_(2)`
D. `v_(1)lev_(2)`

Answer

To solve the problem of comparing the speed of sound in a gas \( V_1 \) and the root mean square (rms) speed of the molecules of the gas \( V_2 \) at the same temperature, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formulas**:
- The speed of sound in a gas is given by the formula:
\[ V_1 = \sqrt{\frac{\gamma RT}{M}} \]
where:
- \( \gamma \) = ratio of specific heats (\( \frac{C_p}{C_v} \))
- \( R \) = universal gas constant
- \( T \) = absolute temperature
- \( M \) = molar mass of the gas

- The rms speed of the molecules in a gas is given by the formula:
\[ V_2 = \sqrt{\frac{3RT}{M}} \]

2. **Set Up the Ratio**:
- To compare \( V_1 \) and \( V_2 \), we can set up the ratio:
\[ \frac{V_1}{V_2} = \frac{\sqrt{\frac{\gamma RT}{M}}}{\sqrt{\frac{3RT}{M}}} \]

3. **Simplify the Ratio**:
- The \( RT/M \) terms cancel out in the ratio:
\[ \frac{V_1}{V_2} = \sqrt{\frac{\gamma}{3}} \]

4. **Analyze the Value of \( \gamma \)**:
- The value of \( \gamma \) for most gases is less than 3 (for example, for diatomic gases like nitrogen and oxygen, \( \gamma \) is approximately 1.4).
- Therefore, since \( \gamma < 3 \), we can conclude:
\[ \frac{V_1}{V_2} < 1 \]

5. **Conclusion**:
- Since \( \frac{V_1}{V_2} < 1 \), it follows that:
\[ V_1 < V_2 \]
- Thus, the speed of sound in the gas \( V_1 \) is less than the rms speed of the molecules \( V_2 \).

### Final Answer:
The speed of sound \( V_1 \) is less than the rms speed of the molecules \( V_2 \).