Question

The speed of sound in gas at NTP is \[300\,{\text{m}}\,{{\text{s}}^{ - 1}}\] . If the pressure is increased four times without a change in temperature the velocity of sound will be?
A. \[150\,{\text{m}}\,{{\text{s}}^{ - 1}}\]
B. \[300\,{\text{m}}\,{{\text{s}}^{ - 1}}\]
C. \[300\,{\text{m}}\,{{\text{s}}^{ - 1}}\]
D. \[1200\,{\text{m}}\,{{\text{s}}^{ - 1}}\]

Answer 1

Hint: To solve this, first use, the formula of the velocity of sound which also includes atomicity of the gas. First the ratio between the pressure and density of the gas and then substitute in the expression for velocity. It is important to note that \[\gamma \] which represents atomicity is a constant term for a specific gas.

Complete step by step answer:
The speed of a gas molecule is proportional to the temperature, which is inversely proportional to the gas's molar mass. In other words, as the temperature of a gas sample increases, the molecules accelerate and as a result the root mean square molecular velocities rise.

Humidity has no impact on sound speed nor does air pressure alone. In an ideal approximation to petrol, air pressure has no effect at all. This is because pressure and density both contribute equally to the velocity of the sound and the two effects cancel out in an ideal gas, leaving only the temperature effect.

The formula which give the formula of sound is:
\[v = \sqrt {\dfrac{{\gamma P}}{\rho }} \] …… (1)
Where,
\[\gamma \] indicates atomicity of the gaseous medium.
\[P\] indicates pressure of gas.
\[\rho \] indicates density of gas.

Again, we have a formula of density:
\[\rho = \dfrac{{PM}}{{RT}}\] …… (2)
Here, temperature is constant.
So, rearranging the equation (2), we get:
$\dfrac{\rho }{P} = \dfrac{M}{{RT}} \\
\dfrac{P}{\rho } = \dfrac{{RT}}{M} \\$
Since, the fraction \[\dfrac{{RT}}{M}\] is constant, so \[\dfrac{P}{\rho }\] is also constant.
Mathematically, it can be written as:
\[\dfrac{P}{\rho } = \dfrac{{RT}}{M} = \]constant …… (3)

Using the equation (3) in equation (1):
\[v = \sqrt {\dfrac{{\gamma P}}{\rho }} \]

\[v = \sqrt {\dfrac{{\gamma RT}}{M}} \] …… (4)
In the equation (4), all the quantities are constant, so it can be concluded that velocity of the sound won’t depend on the pressure. Whether the pressure is increased or decreased, it won’t affect the speed of sound. Hence the speed of the sound remains \[300\,{\text{m}}\,{{\text{s}}^{ - 1}}\].

The correct option is B.

Note: While solving this problem, it is important to remember that inside an ideal gas approximation, air pressure has no effect at all. This is because both pressure and density relate similarly to sound velocity, and the two effects balance out in an ideal gas, leaving only the temperature influence. Owing to lower temperatures, sound normally travels more slowly at higher altitudes.