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Question

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.


A

5.6×105K;7.0×105K

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B

2.4×106K;7.0×103K

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C

3.5×104K;700K

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D

1.6×105K;7.0×103K

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Solution

The correct option is D

1.6×105K;7.0×103K


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 vEarthe

T=[0.032×(11200)23×8.314]K

=1.61×105K.

(ii) For vrms of O2 to be vMoone

T=[0.032×(2380)23×8.314]K

=7.27×103K.


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