A gas molecule of mass M at the surface of the Earth has kinetic energy equivalent to 0° C. If it were to go up straight without colliding with any other molecules, how high it would rise ? Assume that the height attained is much less than radius of the earth. (kB is Boltzmann constant)

Question

A gas molecule of mass M at the surface of the Earth has kinetic energy equivalent to 0° C. If it were to go up straight without colliding with any other molecules, how high it would rise ? Assume that the height attained is much less than radius of the earth. (kB is Boltzmann constant) (A) 0 (B) (273 kB/2 Mg) (C) (546 kB/3 Mg) (D) (819 kB/2 Mg)

Tardigrade Admin · Accepted Answer

(3/2) KT =(1/2) mv 2 v 2=(3 KT / m ) v 2= u 2+2 as O =(3 KT / m )-2 gs S =(3 K /2) ((273)/ mg ) S =(819 K /2 mg )

$0$

$\frac{273 k _{B}}{2 M g}$

$\frac{546 k _{B}}{3 M g}$

$\frac{819 k _{B}}{2 M g}$

$\frac{3}{2} K T = \frac{1}{2} m v^{2}$
$v^{2} = \frac{3 K T}{m}$
$v^{2} = u^{2} + 2$ as
$O = \frac{3 K T}{m} - 2 g s$
$S = \frac{3 K}{2} \frac{( 273 )}{m g}$
$S = \frac{819 K}{2 m g}$

Q. A gas molecule of mass M at the surface of the Earth has kinetic energy equivalent to 0∘C. If it were to go up straight without colliding with any other molecules, how high it would rise ? Assume that the height attained is much less than radius of the earth. (kB​ is Boltzmann constant)

0

2Mg273kB​​

3Mg546kB​​

2Mg819kB​​

23​KT=21​mv2 v2=m3KT​ v2=u2+2 as O=m3KT​−2gs S=23K​mg(273)​ S=2mg819K​

$0$

$\frac{273 k _{B}}{2 M g}$

$\frac{546 k _{B}}{3 M g}$

$\frac{819 k _{B}}{2 M g}$

$\frac{3}{2} K T = \frac{1}{2} m v^{2}$
$v^{2} = \frac{3 K T}{m}$
$v^{2} = u^{2} + 2$ as
$O = \frac{3 K T}{m} - 2 g s$
$S = \frac{3 K}{2} \frac{( 273 )}{m g}$
$S = \frac{819 K}{2 m g}$