Answer to Question #100908 in Optics for daniel

Question

Solve the question one in exercise one
You stand on the edge of a canyon 420 m and clap your hands; the echo comes back to you 2.56 sec later. What is the temperature of the air? (γ_air=1.40 and M=28.8gm)
Solution
The speed of sound in a gas is v=√(β/ρ)=√(γRT/M)
⇒T=(Mv^2)/γR=[(2.88×10^(-2) kg)(328 m/s^2 )]⁄([(1.4)(8.31)]=266K=-7℃)

Answer 1

First of all let's calculate the velocity of the sound wave$=2*d/t$

where distance is taken as (d+d) because sound wave travels from speaker to canyon and then it returns back to the speaker and t is the time taken by the sound wave to cover the (d+d) distance

velocity of sound wave $=$ $2*420/2.56=328m/s$

velocity of the sound wave =$\sqrt{\smash[b]{{\gamma}RT/M}}$

where $\gamma$ $=1.4$ , $R$$=8.314$, and $M=28.8*10^{-3}kg$

putting these in the above equation ,

$328= \sqrt{\smash[b]{{1.4*8.314*T/(28.8*10^{-3})}}}$

on calculating we get,

$T=266.2K$

In $^0C$ ,

$T$ $=266.2-273=-7.2^0C$