Question

The density of air at NTP is $1.293kg{m^{ - 3}}$and density of mercury at ${0^ \circ }C{\text{ is 13}}{\text{.6}} \times {\text{1}}{{\text{0}}^3}kg{m^{ - 3}}.{\text{ If }}{C_p} = 0.2417{\text{ }}calk{g^{ - 10}}{C^{ - 1}}{\text{and }}{{\text{C}}_v} = 0.17115$, the speed of sound in air at ${100^ \circ }C$will be $\left( {g = 9.8Nk{g^{ - 1}}} \right)$
A. $260m{s^{ - 1}}$
B. $322m{s^{ - 1}}$
C. $350.2m{s^{ - 1}}$
D. $369.4m{s^{ - 1}}$

Answer 1

Hint: NTP stands for Normal temperature and pressure, the values of these are here fixed as$T = {25^ \circ }C$, and $P = 1{\text{ }}bar$. Additionally, the speed of light is directly proportional to the square root of temperature, T.

Formula Used:
$\gamma = \dfrac{{{C_p}}}{{{C_v}}}$
$v \propto \sqrt T$

Complete step by step answer:
At Normal temperature and pressure (NTP),
\eqalign{ & T = {25^ \circ }C{\text{ }} = 298.15K \cr & P = 1{\text{ }}bar = {10^5}Pa \cr}
Give that Density of air at NTP,
$\rho {\text{ = 1}}{\text{.293}}kg{m^{ - 3}}$
We know that:
$\gamma = \dfrac{{{C_p}}}{{{C_v}}}$
Substituting values of Cp and Cv in the above equation we get:
$\gamma = \dfrac{{0.2417}}{{0.1715}} = 1.4$
At NTP, the speed of sound in air is:
\eqalign{ & {v_{{{25}^ \circ }C}} = \sqrt {\dfrac{{\lambda P}}{\rho }} \cr & = \sqrt {\dfrac{{1.4 \times {{10}^5}}}{{1.293}}} = 330.15m{s^{ - 1}} \cr}
Let the speed of sound in air at $100^o$C be represented by ${v_{{{100}^ \circ }C}}$
We know that
$v \propto \sqrt T$
Hence the ratio of speed at $100^0$C and that at $25^0$C will be:
\eqalign{ & \dfrac{{{v_{{{100}^ \circ }C}}}}{{{v_{{{25}^ \circ }C}}}} = \sqrt {\dfrac{{373.15}}{{298.15}}} \cr & \Rightarrow \dfrac{{{v_{{{100}^ \circ }C}}}}{{{v_{{{25}^ \circ }C}}}} = 1.118 \cr}
So, value of speed at $100^0$C is
\eqalign{ & \Rightarrow {v_{{{100}^ \circ }C}} = 1.118 \times 330.15 \cr & \Rightarrow {v_{{{100}^ \circ }C}} = 369.4m{s^{ - 1}} \cr}
Therefore, the correct option is D. i.e., $369.4m{s^{ - 1}}$

Additional Information:
For an ideal gas the relationship between Cp and Cv is given by:
Cp-Cv=R.
where R is the gas constant.
The absolute temperature of a given gas is proportional to the square of the rms speed of its molecules.
Thus, the absolute temperature of a given sample of gas is proportional to the total translational kinetic energy of its molecules.

Note: The density of air is also known as the atmospheric density. It is the ratio of mass upon volume of earth’s atmosphere. Just like the physical quantity of air pressure, air density also tends to decrease with increase in altitude. Additionally, it is also dependent on other quantities like atmospheric pressure, humidity and temperature.