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Calculate the mean free path of air molecules at STP. The diameter of N_(2) and O_(2) is about 3xx10^(-10)m

Answer»

SOLUTION :`P=1 atm =1.01xx10^(5)Pa,k_(B)=1.38xx10^(-23)JK^(-1),T=273K`
from ideal fgas law,`N=P/(KT)`
`n=(1.01xx10^(5))/(1.38xx10^(-23)xx273)=(1.01xx10^(5)xx10^(23))/(376.74)=2.68xx10^(-3)xx10^(28)`
`n=2.68xx10^(25)" molecules"//m^(3)`
Mean free path of the air molecule,
`gamma=1/(sqrt2pind^(2))=1/(1.414xx3.14xx2.68xx10^(25)xx(3xx10^(-10))^(2))`
`1/(1.0709xx10^(-18)xx10^(25))=0.9338xx1^(-7)`
` gamma =9.3xx10^(-8)m`


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