Under standard conditions, the density of a gas is ( (1400)/(1089)...

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 ($$\begin{aligned} & \qquad v=\sqrt{\frac{\gamma P}{\rho}} \text { or } \frac{\gamma P}{\rho}=v^2 \\ & \quad \gamma=\frac{v^2 \rho}{p}=\frac{330 \times 330}{10^5} \times \frac{1400}{1089}\\ & \quad \gamma=1.4 \\ & \text { So, } \quad \gamma=\frac{C_p}{C_V}=1.4 \end{aligned}$\$quad\left[$\begin{array}{l} \because \ggamma=1+\frac{2}{f}\\ \text { Since, for diatomic gas, the volume of } \end{array}$\right]$\begin{aligned} \text{ and } \gamma \text { is 1.4. }\\ \text{ and } \text { Hence, the degree of freedom for } \end{aligned}$) So, (\gamma=\frac{C_p}{C_V}=$1.4) Since, for diatomic gas, the volume of \$(\left[\begin{array}{l} \because \ggamma=1+\frac{2}{f}\\ 1.4=1+\frac{2}{f}\\ \therefore f=5 \end{array}$$\right])(\gamma) is 1.4. Hence, the degree of freedom for diatomic gas is equal to 5.