Under standard conditions, the density of a gas is 1.3 mg cm- 3 and the velocity of propagation of sound, in it, is 330 m s- 1 . The number of degrees of freedom of gas is

Question

Under standard conditions, the density of a gas is 1.3 mg cm- 3 and the velocity of propagation of sound, in it, is 330 m s- 1 . The number of degrees of freedom of gas is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Vs o u n d=√(&gamma; P/ρ ) and &gamma; =1+(2/f) ∴ 1+(2/f)=(vs o u n d2 × ρ /P) 1+(2/f)=((330 ×(1.3)/1.01 × 105) 1+(2/f)=1.40 f=5

Q. Under standard conditions, the density of a gas is $1.3 m g c m^{- 3}$ and the velocity of propagation of sound, in it, is $330 m s^{- 1}$ . The number of degrees of freedom of gas is

$V_{so u n d} = \frac{γ P}{ρ}$ and $γ = 1 + \frac{2}{f}$
$∴$ $1 + \frac{2}{f} = \frac{v _{so u n d}^{2} \times ρ}{P}$
$1 + \frac{2}{f} = \frac{( 330 \times ( 1.3 )}{1.01 \times 1 0 ^{5}}$
$1 + \frac{2}{f} = 1.40$
$f = 5$

Q. Under standard conditions, the density of a gas is 1.3mgcm−3 and the velocity of propagation of sound, in it, is 330ms−1 . The number of degrees of freedom of gas is

Vsound​=ργP​​ and γ=1+f2​ ∴ 1+f2​=Pvsound2​×ρ​ 1+f2​=1.01×105(330×(1.3)​ 1+f2​=1.40 f=5

Q. Under standard conditions, the density of a gas is $1.3 m g c m^{- 3}$ and the velocity of propagation of sound, in it, is $330 m s^{- 1}$ . The number of degrees of freedom of gas is

$V_{so u n d} = \frac{γ P}{ρ}$ and $γ = 1 + \frac{2}{f}$
$∴$ $1 + \frac{2}{f} = \frac{v _{so u n d}^{2} \times ρ}{P}$
$1 + \frac{2}{f} = \frac{( 330 \times ( 1.3 )}{1.01 \times 1 0 ^{5}}$
$1 + \frac{2}{f} = 1.40$
$f = 5$