Under standard conditions the density of a gas is `(1400)/(1089)` kg `m^(-3)` and the speed of sound progagation in it is 330 `ms^(-1)` then the number of degrees of freedom of the gas molecules is

Options

A. 2
B. 7
C. 5
D. 3

Answer

C

Explanation

Given density of gas `p = (1400)/(1089) kg//m^(3)`
speed of sound v= 330 `m//s`
and under standard condition
Pressure of gas `p = 1 xx 10^(5) N//m^(2)`
If `gamma` by the ratio of `C_(p)` and `C_(v )` of gas then the speed of sound in gas is given by
`v= sqrt((gamma p )/( p ) ) " or " (gammap )/( p ) = v^(2)`
`gamma =1.4`
` gamma = (c_(p))/( c_(v )) = 1.4 `
since for diatomic gas the volume of `gamma` is 1.4
Hence the degree of freedom for diatomic gas is equal to 5.