Under standard conditions,the density of a ga | vedclass

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(C) Given: Density of gas,$ho = \frac{1400}{1089} \ kg/m^3$,speed of sound,$v = 330 \ m/s$,and under standard conditions,Pressure of gas,$P = 1 	im

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$5$

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(C) Given: Density of gas,$ho = \frac{1400}{1089} \ kg/m^3$,speed of sound,$v = 330 \ m/s$,and under standard conditions,Pressure of gas,$P = 1 	imes 10^5 \ N/m^2$.The speed of sound in a gas is given by the formula: $v = \sqrt{\frac{\gamma P}{ho}}$.Squaring both sides,we get: $v^2 = \frac{\gamma P}{ho} \implies \gamma = \frac{v^2 ho}{P}$.Substituting the values: $\gamma = \frac{(330)^2 	imes (1400/1089)}{10^5} = \frac{108900 	imes 1400}{10^5 	imes 1089} = \frac{108900}{1089} 	imes \frac{1400}{100000} = 100 	imes 0.014 = 1.4$.We know that the adiabatic index $\gamma$ is related to the degrees of freedom $f$ by the formula: $\gamma = 1 + \frac{2}{f}$.Substituting $\gamma = 1.4$: $1.4 = 1 + \frac{2}{f} \implies 0.4 = \frac{2}{f} \implies f = \frac{2}{0.4} = 5$.Thus,the number of degrees of freedom is $5$.

Under standard conditions,the density of a gas is $\frac{1400}{1089} \ kg \ m^{-3}$ and the speed of sound propagation in it is $330 \ ms^{-1}$. The number of degrees of freedom of the gas molecules is:

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