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Discuss the effect of (i) Density (ii) humidity on the velocity of sound in gases? |
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Answer» Solution :(i) Effect of density: Let us consider two gases with DIFFERENT denisties having same temperature and pressure. Then the SPEED of sound in the two gases are `v_(1)=sqrt((gamma_(1)P)/(rho_(1)))""......(1)` and `v_(2)=sqrt((gamma_(2)P)/(rho_(2)))"".....(2)` Taking ratio of equation (1) and (2), we get `(v_(1))/(v_(2))=sqrt((gamma_(1)P)/(rho_(1)))/(sqrt((gamma_(2)P)/(rho_(2))))=sqrt((gamma_(1)P_(2))/(gamma_(2)rho_(1)))` For gases having same value of `gamma`, `(v_(1))/(v_(2))=sqrt((rho_(2))/(rho_(2)))"".....(3)` Thus the velocity of sound in a gas is inversely proportional to the square root of the density of the gas. (II) Effect of moisture (humidity): We know that density of moist air is 0.625 of that of dry air that means the presence of moisture in air (INCREASE in humidity) DECREASES its density. Hence speed of sound increases with rise in humidity. `v=sqrt((gammaP)/(rho))` Let `rho_(1),v_(1)andrho_(2),v_(2)` be the density and speeds of sound in dry air and moist air, respectively. Then `(v_(1))/(v_(2))=sqrt((gamma_(1)P)/(rho_(1)))/(sqrt((gamma_(2)P)/(rho_(2))))=sqrt((rho_(2))/(rho_(1)))` if `gamma_(1)=gamma_(2)` Since P is the total atmospheric pressure, it can be shown that `(rho_(2))/(rho_(1))=(P)/(p_(1)+0.625p_(2))` where `p_(1)andp_(2)` are the partial pressures of dry air and water vapour respectively. Then `v_(1)=v_(2)sqrt((P)/(p_(1)+0.625p_(2)))` |
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