Question

Under standard conditions the gas density is 1.3 $\frac{mg}{c{m}^3}$ and the velocity of sound propagation in it is 330 $\frac{m}{s}$, then the number of degrees of freedom of gas is.

Answer 1

In standard condition,
Pressure of gas is P = ${10}^5\frac{N}{m^2}$
$V_{rms}=\sqrt{\frac{3P}{\rho }}$
here $\rho =1.3\frac{Kg}{m^3}$ and $V_{sound}=330\frac{m}{sec}$
$V_{rms}=\sqrt{\frac{3\times (10{)}^5}{1.3}}=480\frac{m}{sec}$
Using $\frac{V_{rms}}{V_{sound}}=\sqrt{\frac{3}{\gamma }}$
$\frac{480}{330}=\sqrt{\frac{3}{\gamma }}=\gamma =1.4$
Also $\frac{V_{rms}}{V_{sound}}=\sqrt{\frac{3}{\gamma }}=\sqrt{\frac{3f}{f+2}}$
$\frac{3}{\gamma }=\frac{3f}{f+2}$
f + 2 = $\gamma$f = 1.4 f
2 = 1.4 f - f = 0.4 f
f = $\frac{2}{0.4}=5$

Aliter,
Directly use $V_{sound}=\sqrt{\frac{\gamma P}{\rho }}$, and find $\gamma$, then find f, using $\gamma =\frac{f+2}{f}$