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Find out the temperature at which the rm...

Find out the temperature at which the rms speed of nitrogen molecules will be equal to the escape velocity from the earth's gravity. Given,mass of a nitrogen atom =`23.24 times 10^-24 g`, average radius of the earth=6390 km,g=`980 cm.s^-2`, Boltzman constant `=1.37 times 10^-16 erg.^@C^-1`

Text Solution

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rms speed of a molecule `=sqrt((3RT)/M)`,
escape velocity `=sqrt(2gR_1)`, where `R_1`= radius of the earth
`sqrt((3RT)/M)=sqrt(2gR_1)`
or,`T=2/3(gMR_1)/R`
Now, R=Nk and M=mN
where m=mass of a nitrogen molecule
`=2 times (23.24 times 10^-24)g`,
N=number of nitrogen molecules
So, `T=2/3.*(gMNR_1)/(Nk)=2/3.(gmR_1)/k`
`=2/3times(980times(2times23.24times10^-24)times(6390times10^5))/(1.37times10^-16)`
`=1.42 times 10^5 K`
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