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Question
Use the formula `v = sqrt((gamma P)/rho)` to explain why the speed of sound in air increases with temperature.
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Solution 1
Take the relation:
`v = sqrt((gammaP)/rho)` ...(i)
For one mole of an ideal gas, the gas equation can be written as:
PV = RT
`"P" = ("RT")/"V"` .....(ii)
Substituting equation (ii) in equation (i), we get:
`"v" = sqrt((gamma"RT")/("V"rho)) = sqrt((gamma"RT")/"M")` ....(iv)
Where,
Mass, M = ρV is a constant
γ and R are also constants
We conclude from equation (iv) that `v prop sqrt"T"`.
Hence, the speed of sound in a gas is directly proportional to the square root of the temperature of the gaseous medium, i.e., the speed of the sound increases with an increase in the temperature of the gaseous medium and vice versa.
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"`
Since `"P"/rho = "RT"/"M"`, therefore, `v = sqrt((gamma "P")/rho)`
= `sqrt((gamma "RT")/"M")`
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