How many degrees of freedom have the gas molecules, if under standard conditions the gas density is `1.3 kg m^(-3)` and the velocity of sound propagation in it is `C=330 m s^(-1)`.

Options

A. 2
B. 3
C. 4
D. 5

Answer

D

Explanation

`v=sqrt((gammaP)/rho)" "thereforev^(2)=sqrt((gammaP)/rho)`
`thereforegamma=(v^(2)rho)/P=(330xx330xx1.3)/10^(5)" "thereforegamma=1.4`
If f is the number of degrees of freedom of the gas, then
`gamma=1+2/f`
`therefore1.4-1=2/f" "thereforef=2/0.4=5`