Question

At standard temperature and pressure, the density of a gas is $1.3kgm^{-3}$ and the speed of sound in gas is $300 ms^{-1}$. Then what will be the degree of freedom of the gas?
A. 5
B. 6
C. 7
D. 8

Answer 1

Hint: Degree of freedom of a system is the number of independent parameters that defines its configuration or state. Here we will use the following relation
$v = \sqrt {\dfrac{{\gamma P}}{\rho }} $ Where
 v= speed of sound
 P= pressure of gas
 $\rho $= density of gas and
 $\gamma $ is the ratio of specific heats at constant pressure and constant volume.

Step by step solution:
1. It is given that density of gas that is $\rho = 1.3kg{m^ - }^3$ and speed of sound in gas that is $v = 330m{s^{ - 1}}$.
In question it is given that pressure and temperature are at standard conditions hence we will take the value of pressure as, $P = 1.013 \times {10^5}$ Pa.
1. Using the above relation,
$ \Rightarrow $ $v = \sqrt {\dfrac{{\gamma P}}{\rho }} $
$\therefore $ $\gamma = \dfrac{{\rho {v^2}}}{P}$
$ \Rightarrow $ $\gamma = \dfrac{{1.3 \times {{(330)}^2}}}{{1.013 \times {{10}^5}}} = 1.397$
$ \Rightarrow $ $\gamma \approx 1.4$

2. Now we have to find out the degree of freedom using the value which has been calculated above for that we have
$\lambda = 1 + \dfrac{2}{f}$
Where f is the degree of freedom
$ \Rightarrow $ $1.4 = 1 + \dfrac{2}{f}$
$ \Rightarrow $ $0.4 = \dfrac{2}{f}$
Now after rearranging it we get,
$f = 5$
Which means that the degree of freedom of gas is 5. Hence option A is correct.

Additional information:
Molecules have different types of configuration according to which they have different degree of freedoms for example:
A) A molecule of monoatomic gas has only 3 (translational) degrees of freedom, i.e. f=3.
B) A molecule of a diatomic gas has 5 degrees of freedom (3-translational and 2-rotational) at ordinary atmospheric temperatures (because the vibrational modes are not excited). Therefore, f=5.
C) A molecule of a triatomic or polyatomic gas has 6 degrees of freedom (3 translational and 3 rotational) that is f=6.

Note: We could also find the value of $\lambda $ here by using the formula $P = \dfrac{1}{3}\rho \overline {{\upsilon ^2}} $
Where $\overline {{\upsilon ^2}} = \dfrac{{\sum {{\upsilon ^2}} }}{N}$ is the mean square speed .