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Use the formula v=γPρ to explain why the speed of sound in air increases with humidity.

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प्रश्न

Use the formula `v = sqrt((gamma P)/rho)` to explain why the speed of sound in air increases with humidity.

संख्यात्मक
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उत्तर १

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.

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उत्तर २

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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अध्याय 14: Waves - Exercises [पृष्ठ ३८७]

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एनसीईआरटी Physics Part 1 and 2 [English] Class 11
अध्याय 14 Waves
Exercises | Q 4.3 | पृष्ठ ३८७

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