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.
Putting the values of R and gravitational acceleration in the expression of ve,
vEarthe=11200 m/s
vMoone=2380 m/s
The molar mass of O2 is 32gms, or 0.032kgs. Now from kinetic theory,
vrms=√3RTM
⇒T=[Mv2rms3R].
(i) For vrms of O2 to be vEarth−e
T=[0.032×(11200)23×8.314]K
=1.61×105K.
(ii) For vrms of O2 to be vMoon−e
T=[0.032×(2380)23×8.314]K
=7.27×103K.