Question

How many degrees of freedom have the gas molecules if at STP the gas density is 1.3 $kg{{m}^{-3}}$ and the velocity of sound in the gas is 330m/s?

Answer 1

Hint: Use the formula for the speed of sound in a gas i.e. $v=\sqrt{\dfrac{\gamma P}{\rho }}$. With this formula find the value of $\gamma $. Then find f using the relation between f and $\gamma $ for a gas, which is $\dfrac{1}{\gamma -1}=\dfrac{f}{2}$.
Formula used:
$v=\sqrt{\dfrac{\gamma P}{\rho }}$
$\dfrac{1}{\gamma -1}=\dfrac{f}{2}$

Complete answer:
Let us first understand what is meant by degrees of freedom.
The term degree of freedom of the system, tells us about the number of free or independent motions of the system. In other words, the degree of freedom is the number equal to the types of independent motions that the system can have.
When we say that an atom or a molecule is in motion, we know that it possesses some energy. The total energy of the atom or molecule is divided into different types. The total number of types (or forms) of energies possessed by the atom or the molecule is called the degree of freedom.
The independent motions can be translation and rotational or we can also say that a molecule possesses translation and rotational energies.
The number of degrees of freedom is denoted by f.
To find the degrees of freedom of the given gas molecules we will use the formula for the speed of sound in a gas, i.e. $v=\sqrt{\dfrac{\gamma P}{\rho }}$ ….. (i),
Where v is the speed of the sound, P and $\rho $ are the pressure and density of the gas respectively. $\gamma $ is the ratio of the specific heats of the gas molecules at constant pressure to constant volume.
For a gas, $\dfrac{1}{\gamma -1}=\dfrac{f}{2}$
$\Rightarrow f=\dfrac{2}{\left( \gamma -1 \right)}$ ….. (ii).
Let us find the value with help of equation (i) and then find f with equation (ii).
It is given that v=330m/s and $\rho =1.3kg{{m}^{-3}}$.
At STP, the pressure of the gas is P=1atm=$1.013\times {{10}^{5}}Pa$.
Substitute the values in equation (i).
 $\Rightarrow 330=\sqrt{\dfrac{\gamma \left( 1.013\times {{10}^{5}} \right)}{1.3}}$
$\Rightarrow {{330}^{2}}=\dfrac{\gamma \left( 1.013\times {{10}^{5}} \right)}{1.3}$
$\Rightarrow \gamma =\dfrac{{{330}^{2}}\times 1.3}{\left( 1.013\times {{10}^{5}} \right)}=1.4$.
Now, substitute the value of $\gamma $ in equation (ii).
$\Rightarrow f=\dfrac{2}{\left( 1.4-1 \right)}=\dfrac{2}{0.4}=5$
Therefore, the number of degrees of freedom of the gas molecules is 5.

Note:
Note that the number of degrees of freedom of the molecules of a gas are independent of the pressure and density of the gas. It depends on the structure of the molecules, whether they are monoatomic, diatomic, triatomic and so on.
Also note that f is just a number and has no units.