Question

calculate the temperature at which the root mean square speed of hydrogen and oxygen molecules will be equal to their escape velocities from the earth's gravitational field. the radius of the earth is 6400 km.

take $N_{A}$ = 6 * $10^{26}$ $mol^{-1}$ ,, g= 9.8 $ms^{-2}$ , $k_{b}$ = 1.38 * $10^{-23}$ $JK^{-1}$

Answer 1

root mean velocity = √3kT/m
where k is boltzman constant , T is temperature and m is mass of one molecule

mass of one molecule of H2=2/6.023 x 10^23
= 2 × 1.66 × 10^-24 g
= 3.22 × 10^-27 kg

in the same way
mass of one molecule of O2 = 16 x 1.66 x 10^-27 kg

now,
escape velocity = √2gR
where g is acceleration due to gravity , and R is the radius of earth .

a/c to question ,

√(3kT/m) =√2gR
take square both sides

3kT/m = 2gR

T = 2gRm/3k = (2/3) (Rgm/k)

now, temperature when RMS velocity equal to escape velocity ,

T = (2/3)( 6400 x 10^3 x 10 x 3.22 x 10^-27/1.38 x 10^-23 }

=(2/3){ 6400 x 3.22/1.38} K
=9955.5 K

in the same way temperature at which RMS velocity of oxygen = escape velocity .

T = (2/3){6400 x 10^3 x 10 x 16 x 1.66 x 10^-27/1.38 x 10^-23}

= (2/3){6400 x 16x 1.66 /1.38} K

=82117.87 K