Question:medium

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

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The speed of sound is always less than the rms speed of the molecules because \(\gamma\) is always less than 3. Specifically, \(v_s = c_{rms} \sqrt{\gamma/3}\).
Updated On: Apr 15, 2026
  • \(\dfrac{c}{\sqrt{3}}\)
  • \(\dfrac{c}{\sqrt{2}}\)
  • \(\dfrac{2c}{3}\)
  • \(\dfrac{3c}{3}\)
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The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept: 
The root mean square (rms) speed of gas molecules and the velocity of sound in the gas are both related to the temperature and the molar mass of the gas. The velocity of sound also depends on the adiabatic index ($\gamma$). 
Step 2: Key Formula or Approach: 
The formula for rms speed is: \[ v_{\text{rms}} = c = \sqrt{\frac{3RT}{M}} \] The formula for the velocity of sound in a gas is: \[ v_s = \sqrt{\frac{\gamma RT}{M}} \] The adiabatic index $\gamma$ is related to the degree of freedom ($f$) by: \[ \gamma = 1 + \frac{2}{f} \] Step 3: Detailed Explanation: 
Given the average degrees of freedom $f = 6$: \[ \gamma = 1 + \frac{2}{6} = 1 + \frac{1}{3} = \frac{4}{3} \] From the rms speed equation, we can express $\sqrt{\frac{RT}{M}}$ in terms of $c$: \[ \sqrt{\frac{RT}{M}} = \frac{c}{\sqrt{3}} \] Substitute this into the velocity of sound equation: \[ v_s = \sqrt{\gamma} \cdot \sqrt{\frac{RT}{M}} = \sqrt{\frac{4}{3}} \cdot \left(\frac{c}{\sqrt{3}}\right) \] \[ v_s = \frac{2}{\sqrt{3}} \cdot \frac{c}{\sqrt{3}} = \frac{2c}{3} \] Step 4: Final Answer: 
The velocity of sound in the gas is $\frac{2c}{3}$.

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