Find the temperature at which average speed of oxygen molecule be sufficient so as to escape from the earth
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At what atmospheric temperature would oxygen molecules move at a root mean square speed higher than the escape velocity for Earth?
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910
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5 ANSWERS

Harsh Raman
Answered May 13, 2018
Taking,
Mass of Oxygen molecule, M = 2.76x10^-26 kg
Boltzmann constant, k= 1.38x10^-23 J/K
Escape velocity, v= 11.2x10^3 m/s
Rms velocity = (3kT/M)^1/2
Rms velocity = (3 x 1.38x10^-23 x T/2.76 x 10^-26)^1/2 = 11.2x10^3
Squaring both sides,
3 x 1.38x10^-23 J/K x T/2.76 x 10^-26 kg = 1.25x10^8 m/s
T=1.25x10^8 m/s x 2.76x10^-26 kg /3 x 1.38x10^-23 J/K
Solving the above equation you’d get,
T = 8.3 x 10^4 K
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Answer
910
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5 ANSWERS

Harsh Raman
Answered May 13, 2018
Taking,
Mass of Oxygen molecule, M = 2.76x10^-26 kg
Boltzmann constant, k= 1.38x10^-23 J/K
Escape velocity, v= 11.2x10^3 m/s
Rms velocity = (3kT/M)^1/2
Rms velocity = (3 x 1.38x10^-23 x T/2.76 x 10^-26)^1/2 = 11.2x10^3
Squaring both sides,
3 x 1.38x10^-23 J/K x T/2.76 x 10^-26 kg = 1.25x10^8 m/s
T=1.25x10^8 m/s x 2.76x10^-26 kg /3 x 1.38x10^-23 J/K
Solving the above equation you’d get,
T = 8.3 x 10^4 K
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