Answer

Verified

404.1k+ views

**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.

Recently Updated Pages

Cryolite and fluorspar are mixed with Al2O3 during class 11 chemistry CBSE

Select the smallest atom A F B Cl C Br D I class 11 chemistry CBSE

The best reagent to convert pent 3 en 2 ol and pent class 11 chemistry CBSE

Reverse process of sublimation is aFusion bCondensation class 11 chemistry CBSE

The best and latest technique for isolation purification class 11 chemistry CBSE

Hydrochloric acid is a Strong acid b Weak acid c Strong class 11 chemistry CBSE

Trending doubts

Give 10 examples for herbs , shrubs , climbers , creepers

Difference Between Plant Cell and Animal Cell

Write a letter to the principal requesting him to grant class 10 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Name 10 Living and Non living things class 9 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

List some examples of Rabi and Kharif crops class 8 biology CBSE

Write the 6 fundamental rights of India and explain in detail