Use the formula $v=\sqrt{\frac{\gamma P}{\rho}}$ to explain why the speed of sound in air:
$(a)$ is independent of pressure,
$(b)$ increases with temperature,
$(c)$ increases with humidity.

(N/A) $v=\sqrt{\frac{\gamma P}{\rho}} \dots (i)$
$(a)$ Density $\rho = \frac{M}{V}$,where $M$ is the mass and $V$ is the volume. Substituting this into $(i)$,we get $v = \sqrt{\frac{\gamma PV}{M}}$. For an ideal gas,$PV = RT$. Thus,$v = \sqrt{\frac{\gamma RT}{M}}$. Since $R, T, M,$ and $\gamma$ are constant at a given temperature,$v$ is independent of pressure.
$(b)$ From the relation $v = \sqrt{\frac{\gamma RT}{M}}$,we see that $v \propto \sqrt{T}$. As the temperature $T$ increases,the speed of sound $v$ increases.
$(c)$ Let $\rho_m$ and $\rho_d$ be the densities of moist and dry air. Since water vapor has a lower molecular mass than dry air (mostly $N_2$ and $O_2$),the density of moist air is less than that of dry air $(\rho_m < \rho_d)$. From $v = \sqrt{\frac{\gamma P}{\rho}}$,since $v \propto \frac{1}{\sqrt{\rho}}$,a lower density results in a higher speed of sound. Thus,the speed of sound increases with humidity.