Question

The velocity of sound in air at N.T.P is $330{\text{ m }}{{\text{s}}^{ - 1}}$ . What will be its value when temperature is doubled and pressure is halved?
${\text{(i) 65 m }}{{\text{s}}^{ - 1}}$
${\text{(ii) 330 m }}{{\text{s}}^{ - 1}}$
${\text{(iii) 330 }}\sqrt 2 {\text{ m }}{{\text{s}}^{ - 1}}$
${\text{(iv) }}\dfrac{{330}}{{\sqrt 2 }}{\text{ m }}{{\text{s}}^{ - 1}}$

Answer 1

The speed of sound under N.T.P is given as $330{\text{ m }}{{\text{s}}^{ - 1}}$ , we will use the Newton’s formula for finding the velocity of sound and then find the dependencies of velocity of sound. Also we will double the temperature and reduce the pressure, thus we will find the ratio of them.
$v{\text{ = }}\sqrt {\dfrac{{\gamma RT}}{M}}$
Where, $\gamma$ is adiabatic constant, $R$ is gas constant, $T$ is the temperature of a medium and $M$ is the molecular mass of a given gas or medium.

Complete step by step answer:
According to Newton the speed or velocity of sound in any medium can be calculated by using the formula as:
$v{\text{ = }}\sqrt {\dfrac{{\gamma RT}}{M}}$ ____________ $(1)$
From the above formula we can observe that velocity of sound does not depend on pressure of gas but it depend on temperature of gas as,
$v{\text{ }} \propto {\text{ }}\sqrt T$
Let ${v_1}$ be the velocity of sound at N.T.P which is given as $330{\text{ m }}{{\text{s}}^{ - 1}}$ . We can write equation $(1)$ for velocity ${v_1}$ as:
${v_1}{\text{ = }}\sqrt {\dfrac{{\gamma RT}}{M}}$ _______________ $(2)$
Now when temperature gets doubled, equation $(1)$ can be written as,
$v{\text{ = }}\sqrt {\dfrac{{\gamma RT}}{M}}$
$v{\text{ = }}\sqrt {\dfrac{{\gamma R{\text{ }} \times {\text{ 2}}T}}{M}}$
$v{\text{ = }}\sqrt 2 {\text{ }} \times {\text{ }}\sqrt {\dfrac{{\gamma RT}}{M}}$
Now using the value from equation $(2)$ we can get the speed of sound as,
$v{\text{ = }}\sqrt 2 {\text{ }} \times {\text{ }}{v_1}$
We know that ${v_1}$ is the velocity of sound at N.T.P which is given as $330{\text{ m }}{{\text{s}}^{ - 1}}$ . Therefore,
$v{\text{ = }}\left( {\sqrt 2 {\text{ }} \times {\text{ 330}}} \right){\text{ m }}{{\text{s}}^{ - 1}}$
Hence the correct option is ${\text{(iii) 330 }}\sqrt 2 {\text{ m }}{{\text{s}}^{ - 1}}$ .

Note:
The velocity of sound does depend on pressure of medium or gas. This is because it has no effect of change in pressure on velocity of sound. We can also find the speed of sound by directly using the relation $v{\text{ }} \propto {\text{ }}\sqrt T$ . The answer would be the same in both situations. There is no need to substitute the value of $R$ as it would be replaced at the end.