Question

The velocity of sound at $27^\circ C$ is $340{\text{ m/s}}{\text{.}}$ What is the velocity of sound in air at $127^\circ C{\text{ ?}}$

Answer 1

Hint
We are provided with a temperature and the velocity of sound at that temperature. We have given another temperature, we have to find the velocity of the sound at the given temperature. Use the relation between the temperature and the velocity of the sound and relate the given values to find the answer.

Complete step by step answer
Sound is produced through the vibration of particles. While moving back and forth the particles bump into each other and produce sound. These particles result in a formation of mechanical waves known as sound waves. Sound is a sequence of waves of pressure that propagates through compressible media like air liquid and solid. During the propagation of sound waves the waves are refracted, reflected and altered by the medium by which it travels.
The speed of sound in a medium depends on the temperature of the medium whether it is solid or liquid or air. In any medium if we increase the temperature the speed of sound will increase. This happens because air molecules have more energy at higher temperatures, which means their vibration increases when the temperature increases. This makes the sound waves to travel faster in high temperature because the sound waves are propelled by collisions between the molecules.
The kinetic energy of the sound wave is proportional to the temperature
The relation between the velocity of the sound and temperature is given by the equation,
${V^2} \propto T$
Given,
When the temperature is $27^\circ C$ the velocity of sound is $340{\text{ m/s}}{\text{.}}$
When the temperature is $127^\circ C$ the velocity of sound =?
Temperature one, ${T_1} = 27^\circ C$
$ \Rightarrow 27^\circ C = (27 + 273)K$
$ \Rightarrow 300K$
Temperature two, ${T_2} = 127^\circ C$
$ \Rightarrow 127^\circ C = (127 + 273)K$
$ \Rightarrow 400K$
The velocity at $300K$is ${V_1} = 340{\text{ m/s}}$
The velocity at $400K$is ${V_2} = ?$
The relation between the velocity of the sound and temperature is
${V^2} \propto T$
Then,
$ \Rightarrow \dfrac{{{V_1}^2}}{{{V_2}^2}} = \dfrac{{{T_1}}}{{{T_2}}}$
$ \Rightarrow \dfrac{{{{340}^2}}}{{{V_2}^2}} = \dfrac{{300}}{{400}}$
$ \Rightarrow \dfrac{{115600}}{{{V_2}^2}} = \dfrac{3}{4}$
$ \Rightarrow {V_2}^2 = 115600 \times \dfrac{4}{3}$
$ \Rightarrow {V_2}^2 = 115600 \times \dfrac{4}{3}$
$ \Rightarrow {V_2}^2 = 154133.3333{\text{ m/s}}$
$ \Rightarrow {V_2} = \sqrt {154133.3333{\text{ }}} {\text{m/s}}$
$ \Rightarrow {V_2} = 392.5{\text{ m/s}}$
When the temperature is $127^\circ C$ the velocity of sound is $392.5{\text{ m/s}}$.

Note
We are provided with the temperature in Celsius we have to convert it into kelvin. Because kelvin is the standard unit of temperature. To change the temperature from Celsius to kelvin, add $273$ to the given temperature in Celsius, because $0$degree Celsius is equal to $273$ kelvin.