At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth's atmosphere? |
(Given : [NEET - 2018] |
Mass of oxygen molecule \[\text{(m)=2}\text{.76 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{--26}}}\text{kg}\] |
Boltzmann's constant \[{{\text{k}}_{\text{B}}}\text{=1}\text{.38 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{--23}}}\text{J}{{\text{K}}^{\text{--1}}}\text{)}\] |
A) \[\text{5}\text{.016 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{4}}}\text{ K}\]
B) \[\text{8}\text{.360 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{4}}}\text{ K}\]
C) \[\text{2}\text{.508 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{4}}}\text{ K}\]
D) \[\text{1}\text{.254 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{4}}}\text{ K}\]
Correct Answer: B
Solution :
[b] \[{{\text{V}}_{\text{escape}}}\text{=11200 m/s}\] |
Say at temperature T it attains \[{{\text{V}}_{\text{escape}}}\] |
So, \[\sqrt{\frac{\text{3}{{\text{k}}_{\text{B}}}\text{T}}{{{\text{m}}_{{{\text{O}}_{\text{2}}}}}}}=11200m/s\] |
On solving, |
\[\text{T=8}\text{.360 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{4}}}\text{ K}\] |
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