(a) According to Boyle’s law, we know that PV = constant at a constant temperature ... (i) Now volume of gas, $V=\frac{M}{\rho}$ ... (ii) ∴ From (i) and (ii), we get $\frac{P M}{\rho}$ = constant since mass of a gas remains constant ∴$\frac{P}{\rho}$ =constant g is always constant for a given gas or air ∴ From $v=\sqrt{\frac{\gamma P}{\rho}}$ equation, we conclude that velocity of sound in air always remain constant if its temperature is constant. (b) Effect of temperature : We know that $P V = R T$ or $P = \frac{R T}{V}$ Also $v=\sqrt{\frac{\gamma P}{\rho}}$ $=\sqrt{\frac{\gamma R T}{\rho V}}=\sqrt{\frac{\gamma R T}{M}}$ where $M = \rho V$ = molecular weight of the gas As $\gamma$, R and M are constants, so $v \propto \sqrt{T}$ i.e. velocity of sound in a gas is directly proportional tothe square root of its temperature, hence we conclude that the velocity of sound in air increases with increase in temperature. (c) Effect of humidity : The presence of water vapours in air changes the density of air, thus the velocity of sound changeswith humidity of air. Let $\rho_{m}=$ density of moist air. $\rho_{d}=$ density of dry air. $v_{m}=$ velocity of sound in moist air. $v_{d}=$ velocity of sound in dry air. From formula $v=\sqrt{\frac{\gamma P}{\rho}}$, we get $v_m=\sqrt{\frac{\gamma P}{\rho_m}}$... (i) and $v_d=\sqrt{\frac{\gamma P}{\rho_d}}$ ... (ii) Dividing (i) by (ii), we get $\frac{v_{m}}{v_{d}}=\sqrt{\frac{\rho_{d}}{\rho_{m}}}$ ... (iii) Also we know that density of water vapours is less than the density of dry air i.e. dry air is heavier thanwater vapours as the molecular mass of water is less than that of $\mathrm{N}_2 (28)$ and $\mathrm{O}_2 (32)$, so $\rho_{m} < \rho_{d}$ or $\frac{\rho_{d}}{\rho_{m}} > 1$ ... (iv) ∴ From (iii) and (iv), we get $\frac{v_{m}}{v_{d}} > 1$ or $v_{m} > v_{d}$ i.e. velocity of sound in air increases with humidity, i.e. velocity of sound in moist air is greater than velocity of sound in dry air. That is why sound travels faster on rainy day than on a dry day.