InterviewSolution
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InterviewSolution
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As adiabatic vessel contains `n_(1) = 3` mole of diatomic gas. Moment of inertia of each molecule is `I = 2.76 xx 10^(-46) kg m^(2)` and root-mean-square angular velocity is `omega_(0) = 5 xx 10^(12) rad//s`. Another adiabatic vessel contains `n_(2) = 5` mole of a monatomic gas at a temperature `470 K`. Assume gases to be ideal, calculate root-mean-square angular velocity of diatomic molecules when the two vessels are connected by a thin tube of negligible volume. Boltzmann constant `k = 1.38 xx 10^(-23) J//:"molecule"`. |
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Answer» We know according to law of equipartition of energy, each gas molecule has `1//2 kT` energy associated with each of its degrees of freedom. As a diatomic gas molecule has two rotational degrees of freedom, final temperature the system is `T_(f) = (f_(1) n_(1) T_(1) + f_(2) n_(2) T_(2))/(f_(1) n_(1) + f_(2) n_(2))` Here for diatomic gas, `f_(1) = 5, n = 3` and `T_(1) = 250 K` For monatomic gas, `f_(2) = 3, n_(2) = 5` and `T_(2) = 470 K` Thus from Eq. (i), `T_(f) = (5 xx 3 xx 250 + 3 xx 5 xx 470)/(5 xx 3 + 3 xx 5) = 360 ` Thus final mixuture of the two gases is at temperature `360 `. If final rms angular velocity of diatomic gas molecules is `omega_(rms f)`, according to law of equipartition of energy, we have `(1)/(2) I omega_(rms f)^(2) = kT` or `omega_(rms f) = sqrt((2 kT)/(I))` `= sqrt((2 xx 1.38 xx 10^(-23) xx 360)/(2.76 xx 10^(-46)))` `omega_(rms f) = 6 xx 10^(12) rad//s` |
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