Question 20: Thermodynamics & KTG

Question

In a mixture of gases, the average number of degree of freedoms per molecule is 6 . The rms speed of the molecule of the gas is $ c $, then the velocity of sound in the gas is

Accepted Answer

In a mixture of gases where the average number of degrees of freedom per molecule is 6, the velocity of sound in the gas can be determined using the following relationships:
The root mean square (rms) speed $ v_{\mathrm{rms}} $ is given by:
$   v_{\mathrm{rms}} = \sqrt{\frac{3 R T}{M}}   $
The velocity of sound $ v_{	ext{sound}} $ is given by:
$   v_{	ext{sound}} = \sqrt{\frac{\gamma R T}{M}}   $
The ratio of the velocity of sound to the rms speed is:
$ \frac{v_{	ext{sound}}}{v_{\mathrm{rms}}} = \sqrt{\frac{\gamma}{3}} $
Given that the degrees of freedom $ f = 6 $, we calculate the adiabatic index $ \gamma $ as follows:
$ \gamma = 1 + \frac{2}{f} = 1 + \frac{2}{6} = \frac{4}{3} $
Substituting $ \gamma = \frac{4}{3} $ into the equation for the velocity of sound:
$ v_{	ext{sound}} = \sqrt{\frac{4/3}{3}} \, v_{\mathrm{rms}} = \frac{2}{3} v_{\mathrm{rms}} $
Therefore, the velocity of sound in the gas is:
$ \frac{2c}{3} $

Answer

$ \frac{c}{\sqrt{3}} $

Answer

$ \frac{c}{\sqrt{2}} $

Answer

$ \frac{2 c}{3} $

Answer

$ \frac{3 c}{3} $

Thermodynamics & KTG

In a mixture of gases, the average number of degree of freedoms 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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