InterviewSolution
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InterviewSolution
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How many degress of freedom have the gas molecules, if under standard conditions the gas density is `rho = 1.3 kg//m^3` and velocity of sound propagation o it is `v = 330m//s` ? |
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Answer» `v = sqrt((gamma P)/(rho)) implies gamma (v^(2) rho)/(P)` If `r` = degrees of freedom of rotational motion, them `gamma + 1 (2)/(r )` `:.r = (2)/((v^(2) rho)/(p) -1) = (2)/((330^(2) xx 1.3)/(1.015 xx 10^(5) - 1)) = 5` |
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