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Let vecv,v_(rms)andv_(p) respectively denote the mean speed, the root mean square speed and the most probable speed of the molecules in an ideal monatomic gas at absolute temperature T. The mass of a molecule is m. Then |
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Answer» no molecule can have speed greater than `v_(RMS)` `v_(p)=sqrt((2RT)/(M))` `therefore v_(rms):barv:v_(p)=sqrt(3):sqrt(2.5):sqrt(2)` So, `v_(p)ltbarvltv_(rms)and(v_(rms))/(v_(p))=sqrt((3)/(2))orv_(rms)^(2)=(3)/(2)v_(p)^(2)` `therefore` Average kinetic energy `=(1)/(2)mv_(rms)^(2)=(1)/(2)mxx(3)/(2)v_(p)^(2)=(3)/(4)mv_(p)^(2)` |
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