Question

In a mixture of gases, the average number of degrees of freedom per molecule is 6. The rms speed of the molecule of the gas is \(c\), then the velocity of sound in the gas is

• A. \(\dfrac{c}{\sqrt{3}}\)
• B. \(\dfrac{c}{\sqrt{2}}\)
• C. \(\dfrac{2c}{3}\)
• D. \(\dfrac{3c}{3}\)

Hint:

The speed of sound is always less than the rms speed of the molecules because \(\gamma\) is always less than 3. Specifically, \(v_s = c_{rms} \sqrt{\gamma/3}\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We need to relate the root mean square (rms) speed of gas molecules to the speed of sound. Both depend on the temperature and molar mass of the gas, but they differ by a factor involving the adiabatic index (\(\gamma\)).

Step 2: Key Formula or Approach:
1. RMS speed (\(c\)) = \(\sqrt{\frac{3RT}{M}}\) 2. Velocity of sound (\(v_s\)) = \(\sqrt{\frac{\gamma RT}{M}}\) 3. Ratio: \(v_s = c \sqrt{\frac{\gamma}{3}}\) 4. \(\gamma = 1 + \frac{2}{f}\), where \(f\) is the degrees of freedom.

Step 3: Detailed Explanation:
Given \(f = 6\): \[ \gamma = 1 + \frac{2}{6} = 1 + \frac{1}{3} = \frac{4}{3} \] Now, substitute \(\gamma\) into the sound velocity relation: \[ v_s = c \sqrt{\frac{4/3}{3}} = c \sqrt{\frac{4}{9}} = \frac{2c}{3} \]

Step 4: Final Answer
The velocity of sound in the gas is \(\dfrac{2c}{3}\).