Question

at what temperature the RMS speed of oxygen molecule becomes just sufficient for escaping from the earth atmosphere?(given mass of oxygen molecule(m)=
${2.76}^{ - 26} kg$
Boltzmann constant=
$1.38 \times {10}^{ - 23} j k {}^{ - 1}$
$1)2.508 \times {10}^{4}$
$2) 1.254 \times {10}^{4}$
$3)5.016 \times {10}^{4}$
$4)8.360 \times {10}^{4}$

​

Answer 1

Answer:

Since, V

esc

=11.2 km/s

R.M.S. =

M

3KT

M

3KT

=11200

T=

3×1.38×10

−23

11200

2

×2.76×10

−26

T=8.360×10

4

K