Speed of sound in gases

Let $v$ be the speed of sound. Let $\gamma$ be the Adiabatic Index. Let $P$ be the pressure. Let $\rho$ be the density. Then, it is known that $v = \sqrt {\dfrac {\gamma \cdot P}{\rho}}.$ Thus, $\gamma = \dfrac {\rho v^2}{P}.$ But we also know that if $f$ is the number of degrees of freedom, $\gamma = 1 + \dfrac {2}{f} \implies f = \dfrac {2}{\gamma - 1}.$ So we solve: $\begin{aligned} f &= \dfrac {2}{\gamma - 1} \\&= \dfrac {2}{\dfrac {\rho v^2}{P} - 1}. \end{aligned}$ Substituting $\rho = 1.3 \, \text{kg/m}^3$ , $v = 330 \, \text{m/s}$ , and $P = 1.0105 \times 10^{5} \, \text{Pa}$ , we get $f = \boxed {5}.$

It should be clear that this gas is very near the values of Oxygen gas. One way of figuring this out more directly would be to note that, in order for this to be the case, we can, thermodynamically, derive the Gamma value of the gas to be $\gamma \approx \frac{7}{5}$ , which would be the value necessary for a diatomic molecule (and, hence, it must have 5 degrees of freedom).

Assuming that this gas is ideal and the postulates of classical thermodynamics/kinetic theory, we have: $c=\sqrt{\left(\frac{\partial p}{\partial \rho}\right)_q}$ For some density $\rho$ , pressure $p$ , and constant heat exchange $q$ (with the means taken to be adiabatic), and speed of transmission, $c$ .

This is, then: $c=\sqrt{\frac{C_p p}{C_v \rho}}$ We solve for $\gamma=\frac{C_p}{C_v}$ to get: $\frac{\rho}{p}c^2=\gamma\approx1.4009$ Hence, we have that our gas (if ideal) must be a diatomic molecule; from this, we note that diatomic molecules have one rotational symmetry in 3 dimensions, giving us $6-1=5$ degrees of freedom.

Please feel free to ask for a more in-depth explanation/motivation, if necessary.

As we now that V=√(gamma×P)/d As By putting value we gamma=1.4 For which degree of freedom is 5....

We can write $v = \sqrt{(\gamma * P/\rho)}$

Squaring and putting in the given values, we find $\gamma = 1.401$

We know that $\gamma = C_p/C_v$

$C_v = i*R/2$

$C_p = C_v + R = (i+2) * R/2$

$\gamma = (i+2)/i$

$=> \gamma = 1 + 2/i$

$=> i = 2 / (\gamma - 1)$

$=> i = 4.99$

Therefore, we can approximate $i$ to be $5$

$i = \boxed{5}$

Since we have v = (YP/d) Y=Cp/Cv; P is pressure and d is density. On substituting we get Y = 1.4. Also Cp = R + Cv [Mayer's relation]. Therefore we we Cv = R/0.8. Also Cv is fR/2. f is degree of freedom. Therefore f = 5..

Let velocity of sound be 'v' m/s, density be 'd' kg/cubic.m, pressure be 'P' Pa, ratio of heat capacities be 'g' and degree of freedom be 'i'.

Given, v = 330 m/s, d = 1.3 Kg/cubic.m and P = 101050 Pa.

So,

v = sqrt((P * g)/d) ..................(relation for velocity of sound varying with pressure and density)

330 * 330 * 1.3 = 101050 * g

g = 1.4

Now,

g = 1 + (2/i) .........................(relation between g and i)

1.4 = 1 + (2/i)

(1/i) = 0.2

i = 5 (Ans)

Velocity of sound = sqrt(gamma x P/density). Gamma is1.4 so the gas is diatomic and has 5 degreees of freedom