At what temperature does the rms speed of oxygen molecules equal the escape velocity of - (i) Earth, (ii) the Moon? The gravitational acceleration of the Moon is 0.6g.

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D

Putting the values of R and gravitational acceleration in the expression of $v_{e}$ ,

$v_{e}^{E a r t h} = 11200 m / s$

$v_{e}^{M o o n} = 2380 m / s$

The molar mass of $O_{2}$ is 32gms, or 0.032kgs. Now from kinetic theory,

$v_{r m s} = \sqrt{\frac{3 R T}{M}}$

$\Rightarrow T = [\frac{M v_{r m s}^{2}}{3 R}] .$

(i) For $v_{r m s}$ of $O_{2}$ to be $v_{e}^{E a r t h -}$

$T = [\frac{0.032 \times (11200)^{2}}{3 \times 8.314}] K$

$= 1.61 \times 10^{5} K$ .

(ii) For $v_{r m s}$ of $O_{2}$ to be $v_{e}^{M o o n -}$

$T = [\frac{0.032 \times (2380)^{2}}{3 \times 8.314}] K$

$= 7.27 \times 10^{3} K$ .

Suggest Corrections

Similar questions

At what temperature does the rms speed of oxygen molecules equal the escape velocity of - (i) Earth, (ii) the Moon? Assume gravitational acceleration at the moon is 0.6g.

Q. Compute the temperature at which the rms speed is equal to the speed of escape from the surface of the earth for hydrogen and for oxygen. The temperature of the upper atmosphere is about 1000 K. Would you expect to find a lot of hydrogen there or a lot of oxygen.

Q. At what temperature will the rms speed of oxygen molecules becomes just sufficient to escape from the earth's atmosphere?(Given: Mass of oxygen molecule

(m) = 2.76 \times 10^{- 26} k g

, Boltzmann constant i.e.

k_{B} = 1.38 \times 10^{- 23} J K^{- 1}

)

Q. At what absolute temperature T , is the root mean square speed of a hydrogen molecule equal to its escape velocity from the surface of the moon? The radius of moon is R, g is the acceleration due to gravity on moon's surface, m is the mass of hydrogen molecule and "k" is the Boltzmann constant

Join BYJU'S Learning Program