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Question

The absolute temperature of air in a region linearly increases from T1 to T2 in a space of width d. Find the time taken by a sound wave to go through the region in terms ofT1,T2, d and the speed V of sound at 273K. Evaluate this time for T1=280K,T2=310K,d=33mandV=330ms-1.


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Solution

Step1: Given data

The initial temperature of the air T1=280K

Final temperature T2=310K

Width of spaced=33m

Speed of wave V=330ms-1

Let us suppose the time taken by a sound wave to go through the region is t.

Step2: Formula Used

Using interpolation,

The temperature variation at a given position x is,

T=T1+T2-T1d×x.............(1)

Speed of sound at any temperature is given by-

V=γRTM

Where γ is the ratio of specific heat at constant pressure to the specific heat at constant volume,

M is the molar mass of gas,

R is the universal gas constant.

Step 3:Calculating the relation between VandT

It is understood that,

VT. So,

V at 273K

V=γR×273M

V at temperature TK

VT=γR×TM

Thus the ratio will be-

VTV=T273VT=VT273..................(2)

Where VT is the speed of the sound wave at any temperature T.

Step 4: Calculating the required time

We know that, Speed=distancetime

VT=dxdt.

So,

dt=dxVT............(3)

Substitute value of VT from equation (2) to equation (3)

dt=dxV273T

On integrating,

0tdt=273V0ddxT

Now putting the value of T from equation (1)-

0tdt=273V0ddxT1+T2-T1d×xt=273V0ddxT1+T2-T1d×x

Using the property rule, X-12dx=2X12 here, X=T1+T2-T1dx applying the limits also put the given values we get-

t=2273VT1+T2-T1dx0dT2-T1d

t=273V×2dT2-T1×T2-T1t=273330×2×33310-280×310-280t=0.05×2.2×0.8736t=0.09609sec

Hence, the evaluated time will be 0.09609sec.


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