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

Hint:

The speed of sound in a gas is given by: \[ v_{{sound}} = \sqrt{\frac{\gamma RT}{M}} \] where \( \gamma \) depends on the degrees of freedom \( f \) as: \[ \gamma = 1 + \frac{2}{f} \]

Answer 1

The Correct Option is C

Step 1: {Formula for RMS speed and speed of sound} 
The root mean square (RMS) speed of gas molecules is given by: \[ v_{{rms}} = \sqrt{\frac{3RT}{M}} \] The velocity of sound in a gas is: \[ v_{{sound}} = \sqrt{\frac{\gamma RT}{M}} \] where \( \gamma \) is the adiabatic index. 
Step 2: {Relation between \( v_{{sound}} \) and \( v_{{rms}} \)} 
Dividing the equations: \[ \frac{v_{{sound}}}{v_{{rms}}} = \sqrt{\frac{\gamma}{3}} \] 
Step 3: {Finding \( \gamma \) using degrees of freedom} 
For a mixture of gases with an average degree of freedom \( f = 6 \): \[ \gamma = 1 + \frac{2}{f} = 1 + \frac{2}{6} = \frac{4}{3} \] 
Step 4: {Compute velocity of sound} 
\[ v_{{sound}} = \sqrt{\frac{4/3}{3}} v_{{rms}} = \frac{2}{3} v_{{rms}} \] Since \( v_{{rms}} = c \), we get: \[ v_{{sound}} = \frac{2c}{3} \] Thus, the correct answer is \( \frac{2c}{3} \).