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CBSE Class 11-science Answered

10th?
Asked by smanishkumar2002 | 04 Aug, 2018, 05:38: AM
answered-by-expert Expert Answer
Q.
The kinetic molecular theory attributes an average translational kinetic energy of 32RTN to each particle. What rms speed would a mist particle of mass 10−12 g have a root temperature (27∘C) according to kinetic theory of gases.
 
 
Solution:
 
Let the kinetic energy of particle be,
 
straight K. straight E space equals 1 half mμ squared straight mu equals space rms space velocity straight K. straight E space per space molecule space equals fraction numerator 3 RT over denominator 2 straight N end fraction 1 half mμ squared equals fraction numerator 3 RT over denominator 2 straight N end fraction straight mu space equals square root of fraction numerator 3 RT over denominator Nm end fraction end root straight m equals space 10 to the power of negative 12 end exponent space straight g straight R space equals 8.314 cross times 10 to the power of 7 space erg straight T space equals 300 space straight K straight mu space equals space square root of fraction numerator 3 cross times 8.314 cross times 300 over denominator 6.023 cross times 10 to the power of 23 cross times 10 to the power of negative 12 end exponent end fraction end root space space space equals 0.35 space cm divided by sec
The rms velocity is 0.35 am/sec
Answered by Varsha | 05 Aug, 2018, 04:26: PM

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