Under standard conditions, the density of a gas is \(\frac{1400}{1089} \mathrm{~kg} \mathrm{-}…
Under standard conditions, the density of a gas is \(\frac{1400}{1089} \mathrm{~kg} \mathrm{-} \mathrm{m}^{-3}\) and the speed of sound propagation in it is \(330 \mathrm{~ms}^{-1}\), then the number of degrees of freedom of the gas molecules is
2
7
5
3
Solution
Given, density of gas, \(\rho=\frac{1400}{1089} \mathrm{~kg} / \mathrm{m}^3\) speed of sound, \(v=330 \mathrm{~m} / \mathrm{s}\) and under standard condition, Pressure of gas, \(p=1 \times 10^5 \mathrm{~N} / \mathrm{m}^2\)
If \(\gamma\) be the ratio of \(C_p\) and \(C_V\) of a gas, then the speed of sound in gas is given by
\(\begin{aligned}
& \qquad v=\sqrt{\frac{\gamma P}{\rho}} \text { or } \frac{\gamma P}{\rho}=v^2 \\
& \quad \gamma=\frac{v^2 \rho}{p}=\frac{330 \times 330}{10^5} \times \frac{1400}{1089} \\
& \quad \gamma=1.4 \\
& \text { So, } \quad \gamma=\frac{C_p}{C_V}=1.4
\end{aligned} \quad\left[\begin{array}{l}
\because \gamma=1+\frac{2}{f} \\
\text { Since, for diatomic gas, the volume of }
\end{array}\right] \begin{aligned}
& \gamma \text { is 1.4. } \\
& \text { Hence, the degree of freedom for }
\end{aligned}\)
So, \(\gamma=\frac{C_p}{C_V}=1.4\)
Since, for diatomic gas, the volume of \(\left[\begin{array}{l}
\because \gamma=1+\frac{2}{f} \\
1.4=1+\frac{2}{f} \\
\therefore f=5
\end{array}\right]\)\(\gamma\) is 1.4.
Hence, the degree of freedom for
diatomic gas is equal to 5.