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, we need to compare the speed of sound in a gas \( V_1 \) with the root mean square (RMS) speed of the gas molecules \( V_2 \).

### Step 1: Write down the formulas for \( V_1 \) and \( V_2 \)

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

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

### Step 2: Compare \( V_1 \) and \( V_2 \)

To compare \( V_1 \) and \( V_2 \), we can express both speeds in terms of \( \sqrt{\frac{RT}{M}} \):

1. The expression for \( V_1 \) can be rewritten as:
\[ V_1 = \sqrt{\gamma} \cdot \sqrt{\frac{RT}{M}} \]

2. The expression for \( V_2 \) can be rewritten as:
\[ V_2 = \sqrt{3} \cdot \sqrt{\frac{RT}{M}} \]

### Step 3: Analyze the relationship between \( \gamma \) and 3

Since \( \gamma \) (the adiabatic index) is always less than 3 for any real gas (as it depends on the degrees of freedom of the gas molecules), we can conclude:
\[ \sqrt{\gamma} < \sqrt{3} \]

### Step 4: Conclude the comparison

From the above analysis, we have:
\[ V_1 = \sqrt{\gamma} \cdot \sqrt{\frac{RT}{M}} < \sqrt{3} \cdot \sqrt{\frac{RT}{M}} = V_2 \]

Thus, we conclude that:
\[ V_1 < V_2 \]

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