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Question
Use the formula `v = sqrt((gamma P)/rho)` to explain why the speed of sound in air increases with humidity.
Solution 1
Let `"v"_"m" and "v"_"d"` be the speeds of sound in moist air and dry air respectively.
Let `rho_"m"` and `rho_"d"` be the densities of moist air and dry air respectively.
Take the relation:
`"v" = sqrt((gamma "P")/rho)`
Hence, the speed of sound in moist air is:
`"v"_"m" = sqrt((gamma "P")/rho_"m")` ....(i)
And the speed of sound in dry air is:
`"v"_"d" = sqrt((gamma "P")/rho_"d")` ...(ii)
On dividing equations (i) and (ii), we get:
`"v"_"m"/"v"_"d" = sqrt((gamma"P")/rho_"m" xx rho_"d"/(gamma"P")) = sqrt(rho_"d"/rho_"m")`
However, the presence of water vapour reduces the density of air, i.e.,
`rho_"d" < rho_"m"`
`:. v_"m" > v_"d"`
Hence, the speed of sound in moist air is greater than it is in dry air. Thus, in a gaseous medium, the speed of sound increases with humidity.
Solution 2
We are given that `"v" = sqrt((gamma "p")/rho)`
We know PV = nRT (For n moles of ideal gas)
`=> "PV" = "m"/"M" "RT"`
where m is the total mass and M is the molecular mass of the gas
`:. "P" = "m"/"M" * "RT"/"M" = (rho "RT")/"M"`
`=> "P"/rho = "RT"/"M"`
Increase in humidity decrease the effective density of air. Therefore the velocity `("v" prop 1/sqrtrho)` increase.
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