Question

The velocity of sound in a gas is $30{\text{ }}m/s$ at $27^\circ C$ . What is the velocity of the sound in the same gas at $127^\circ C$ ?​
(A) $20m/s$
(B) $40m/s$
(C) $20\sqrt 3 m/s$
(D) $60m/s$​

Answer 1

For a gas, $\dfrac{{PV}}{T} = nR$ (where $P$ is the pressure of the gas, $V$ is the volume of the gas, $T$ is the temperature of the gas, $R$ is the gas constant and $n$ is the number of moles of the gas), which means that $\dfrac{{PV}}{T}$ is always constant. The velocity of sound is directly proportional to the root of the temperature of the gaseous medium. With the help of these properties, the velocity of sound in the gas at temperature $127^\circ C$ can be easily predicted.

Formula used :
$v = \sqrt {\dfrac{{\gamma RT}}{M}}$ where $v$ is the velocity of sound in the gas, $\gamma$ is the atomicity of the gas, $R$ is the gas constant, $T$ is the temperature of the gas and $M$ is the molar mass of the gas.

Complete step by step solution:
Converting the given temperatures from Celsius to Kelvin, we have
$27^\circ C = 273 + 27K = 300K$
$127^\circ C = \left( {127 + 273} \right)K = 400K$
According to the given question ,
The velocity of sound in a gas at $27^\circ C$ is ${v_{300K}} = 30m/s$ .
From the formula $v = \sqrt {\dfrac{{\gamma RT}}{M}}$ ,
We can infer that $v \propto \sqrt T$ ___________ (a)
Therefore, using equation (a) in context of the different velocities, and plugging in values, we get
${v_{300K}} \propto \sqrt {300K}$ _______________ (i)
${v_{400K}} \propto \sqrt {400K}$ _______________ (ii)
Now, dividing equation (ii) by equation (i), we have
$\Rightarrow \dfrac{{{v_{400K}}}}{{{v_{300K}}}} = \dfrac{{\sqrt {400K} }}{{\sqrt {300K} }}$
Putting value of ${v_{300K}}$ in the above equation, we get
$\Rightarrow \dfrac{{{v_{400K}}}}{{30m/s}} = \dfrac{{\sqrt {400K} }}{{\sqrt {300K} }} \\\ \Rightarrow \dfrac{{{v_{400K}}}}{{30m/s}} = \dfrac{{20}}{{10\sqrt 3 }} \\\ \Rightarrow {v_{400K}} = \dfrac{{2 \times 30m/s}}{{\sqrt 3 }} \\\$
Multiplying the denominator and numerator of Right Hand Side (RHS) with $\sqrt 3$ , we get
$\Rightarrow {v_{400K}} = \dfrac{{2 \times 30 \times \sqrt 3 m/s}}{3} \\\ \Rightarrow {v_{400K}} = 20\sqrt 3 m/s \\\$
Hence, option (C) is the correct answer.

Note:
When two same proportionality equations with different values are divided, their proportionality constants are cancelled out. Thus the proportionality symbol changes to the equal to symbol. Also, remember to convert Celsius values to kelvin before using them in the equations.