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The kinetic molecular theory attributes an average kinetic energy of (3)/(2) kT to each particle. What rms speed would a mist particle of mass 10^(-12) g have at room temperature (27^(@)C) according to the kinetic molecular theory ? |
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Answer» Solution :KE per MOLECULE `= (3)/(2)kT` `= (3)/(2).(R)/(N).T`. If the mass of ONE molecule is m then KE of this molecule `= (1)/(2) MC^(2)`, where c is the rms SPEED. `therefore (1)/(2)mC^(2) = (3)/(2) (R)/(N).T` or `C = sqrt(3 xx (R)/(N) xx (T)/(m))` `= sqrt((3 xx 8.314 xx 10^(7) xx 300)/(6.022 xx 10^(23) xx 10^(-12))) = 0.35 cm//s` |
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