Question

The speed of sound in a container of gas at 20.0°C is v. If the temperature of the gas is increased to 80.0°C, the new speed will be?

Answer 1

Answer : The new speed will be, 1.098v

Explanation :

The formula used for speed of sound in a container of gas is:

$c=\sqrt{\frac{\gamma RT}{M}}$

where,

c = speed of sound

R = gas constant

$\gamma$ = adiabatic constant

T = absolute temperature

M = molecular mass of gas

As, $c\propto \sqrt{T}$

or, we can say that

$\frac{c_1}{c_2}=\sqrt{\frac{T_1}{T_2}}$            .............(1)

Given:

$T_1$ = $20.0^oC=273+20.0=293K$

$T_1$ = $80.0^oC=273+80.0=353K$

$c_1$ = v

Now put all the given values in the above formula 1, we get:

$\frac{v}{c_2}=\sqrt{\frac{293}{353}}$

$c_2=1.098v$

Thus, the new speed will be, 1.098v

Answer 2

The rise in temperature from 20 to 80 degree Celsius results in the speed of sound to be doubled.

The speed of sound has been the transmission of the sound energy in the system from one end to another. The speed of sound (c) in gas has been given by:

$c=\sqrt{\dfrac{\gamma RT}{M} }$

Where, the adiabatic constant for the reaction, has been $\gamma$

The gas constant for the reaction has been R

The temperature of the gas has been T

The molar mass of the gas has been M

Computation for New speed of sound

The relationship between the speed of sound and temperature of gas has been given as:

$\dfrac{c_1}{c_2} =\sqrt{\dfrac{T_1}{T_2} }$

The initial speed of sound has been, $c=v$

The initial temperature of gas has been, $T_1=20.0\;^\circ \rm C$

The final temperature of gas has been, $T_2=80\;^\circ \rm C$

Substituting the values for final speed of sound, $c_2$ as follows:

$\dfrac{v}{c_2}=\sqrt{\dfrac{20}{80} } \\v^2\;\times\;80=c_2^2\;\times\;20\\c_2=2v$

The final speed of sound in the container of gas been doubled with the rise in temperature.

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