Question

Under standard conditions, the density of a gas is $1.3 \, mg \, cm^{- 3}$ and the velocity of propagation of sound, in it, is $330 \, m \, s^{- 1}$ . The number of degrees of freedom of gas is

Answer 1

$V_{s o u n d}=\sqrt{\frac{\gamma P}{\rho }}$ and $\gamma =1+\frac{2}{f}$
$\therefore $ $1+\frac{2}{f}=\frac{v_{s o u n d}^{2} \times \rho }{P}$
$1+\frac{2}{f}=\frac{(330 \times(1.3)}{1.01 \times 10^{5}}$
$1+\frac{2}{f}=1.40$
$f=5$