Question

How many degrees of freedom have the gas molecules, if under standard conditions the gas density is $\rho = 1.3\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}$ and velocity of sound propagation is $v = 330\;{\rm{m/s}}$?

Answer 1

The above problem can be resolved using the concepts and applications of the degree of freedom for a gas molecule. The degree of freedom for a gas molecule can be obtained by the mathematical relation of the root mean square velocity of the gas molecule and its random velocity. This formula can be given in terms of the degree of freedom upon subsequent substitution, the final value is obtained.

Complete step by step answer:
Given:
The density of gas is, $\rho = 1.3\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}$.
The velocity of sound propagation is,$v = 330\;{\rm{m/s}}$.
The mathematical relation for the root mean square velocity and the velocity of sound propagation is given as,
$\dfrac{{{v_{rms}}}}{v} = \sqrt {\dfrac{3}{\gamma }}$…… (1)
Here, $\gamma$ is the adiabatic index and its value is 1.4.
The mathematical relation for the root mean square velocity and the velocity of sound propagation is also given as:
$\dfrac{{{v_{rms}}}}{v} = \sqrt {\dfrac{{3f}}{{f + 2}}}$ …… (2)
Here, f, is the number of degrees of freedom.
And, ${v_{rms}}$ is the root mean square velocity of gas molecule and its value is,
${v_{rms}} = \sqrt {\dfrac{{3P}}{\rho }}$ ……. (3)
Here, P is the value of pressure of gas at standard conditions, and its value is $P = {10^5}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}$.
Solve the above expression by substituting the values as,

{v_{rms}} = \sqrt {\dfrac{{3 \times {{10}^5}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}}}{{1.3\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}}}} \\\ {v_{rms}} = 480.38\;{\rm{m/s}} \end{array}$$ Then the number of degree of freedom is by comparing the equation 1 and 2 as, $$\begin{array}{l} \sqrt {\dfrac{3}{\gamma }} = \sqrt {\dfrac{{3f}}{{f + 2}}} \\\ 3f + 6 = 3f\gamma \\\ 3f + 6 = 3f\left( {1.4} \right)\\\ f = 5 \end{array}$$ Therefore, the number of degrees of freedom is 5. **Note:** To resolve the above problem, it is important to learn the degree of molecules. Degree and its values for some known gaseous substances should be known. These considerations will help to identify the value of the velocity of gas also.