Question

At NTP the density of a gas is $1.3kg/{m^3}$and the velocity of sound propagation in the gas is 330m/s. The degree of freedom of the gas molecule is:
1) 3
2) 5
3) 6
4) 7

Answer 1

Hint:- Here at NTP the pressure P of gas is$1.015 \times {10^5}KN/{m^2}$. We have been given a density of gas$(\rho = 1.3kg/{m^3})$, velocity of sound “v = 330m/s”. Apply the formula for speed of sound$v = \sqrt {\dfrac{B}{\rho }} $; where:
B = Bulk Modulus = $\gamma P$($\gamma $= Adiabatic constant, P = Pressure), $\rho $= Density of gas, v = velocity of sound. Then equate$\gamma = 1 + \dfrac{2}{f}$; where: f = degree of freedom. Put the given value and solve for the unknown.

Complete step-by-step solution
The speed of the sound wave is given by:
$v = \sqrt {\dfrac{B}{\rho }} $;
Put$B = \gamma P$;
$v = \sqrt {\dfrac{{\gamma P}}{\rho }} $;
To remove the square roots take square on both sides of the equation;
${v^2} = \dfrac{{\gamma P}}{\rho }$;
Take pressure and the density on LHS and solve for$\gamma $.
$\gamma = \dfrac{{{v^2}\rho }}{P}$;
Put the necessary value in the above equation and solve,
$\gamma = \dfrac{{330 \times 330 \times 1.3}}{{1.015 \times {{10}^5}}}$;
The value comes out to be:
$\gamma = 1.4$;
For finding the degree of freedom put the value of $\gamma $in the below given equation:
$\gamma = 1 + \dfrac{2}{f}$;
Solve for degree of freedom f.
$1.4 = 1 + \dfrac{2}{f}$
Simplify the equation
\[1.4f = f + 2\]
\[0.4f = 2\];
Take “0.4” to RHS:
\[f = \dfrac{2}{{0.4}}\];
\[f = \dfrac{{2 \times 10}}{4}\];
The degree of freedom is:
\[f = 5\];

Final Answer: Option “2” is correct. The degree of freedom of the gas molecule is 5.

Note:- Here we need to find the relation between the speed of the sound, pressure and the density. Put the necessary given values and solve for the adiabatic constant$\gamma $. After finding the value of adiabatic constant put the value in the equation$\gamma = 1 + \dfrac{2}{f}$and solve for the degree of freedom f.