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At what temperautre will the rms velocity of oxygen moleucle be sufficient, so as to escape from the earth ?Escape velocity from the earth is 11.0 km/s and the mass of 1 moleucle of oxygen 
(Boltzman constant k =
)
(Boltzman constant k =
) - a)
- b)
- c)
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
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Understanding RMS Velocity
The root mean square (RMS) velocity of a gas is a measure of the average velocity of its molecules. It can be calculated using the formula:
RMS velocity (v_rms) = √(3kT/m)
Where:
- k = Boltzmann constant = 1.38 x 10^-23 J/K
- T = Temperature in Kelvin
- m = Mass of one molecule of the gas
Escape Velocity Requirement
The escape velocity from the Earth is given as 11.0 km/s, which is equivalent to 11,000 m/s. For oxygen to escape Earth's gravity, its RMS velocity must be at least equal to this escape velocity.
Mass of Oxygen Molecule
The mass of one molecule of oxygen is given as:
- m = 5.34 x 10^-26 kg
Setting Up the Equation
To find the temperature at which the RMS velocity equals the escape velocity, we set:
v_rms = 11,000 m/s
Substituting into the RMS velocity formula gives:
11,000 = √(3kT/m)
Isolating Temperature
Squaring both sides:
(11,000)^2 = (3kT/m)
Now solving for T:
T = (m * (11,000)^2) / (3k)
Calculating Temperature
Substituting the values:
T = (5.34 x 10^-26 kg * (11,000 m/s)^2) / (3 * 1.38 x 10^-23 J/K)
Performing the calculation yields:
T ≈ 156,000 K, which can be rounded to 1.56 x 10^5 K.
Conclusion
Thus, the temperature at which the RMS velocity of an oxygen molecule is sufficient to escape from Earth is approximately:
Option C: 1.56 x 10^5 K
The root mean square (RMS) velocity of a gas is a measure of the average velocity of its molecules. It can be calculated using the formula:
RMS velocity (v_rms) = √(3kT/m)
Where:
- k = Boltzmann constant = 1.38 x 10^-23 J/K
- T = Temperature in Kelvin
- m = Mass of one molecule of the gas
Escape Velocity Requirement
The escape velocity from the Earth is given as 11.0 km/s, which is equivalent to 11,000 m/s. For oxygen to escape Earth's gravity, its RMS velocity must be at least equal to this escape velocity.
Mass of Oxygen Molecule
The mass of one molecule of oxygen is given as:
- m = 5.34 x 10^-26 kg
Setting Up the Equation
To find the temperature at which the RMS velocity equals the escape velocity, we set:
v_rms = 11,000 m/s
Substituting into the RMS velocity formula gives:
11,000 = √(3kT/m)
Isolating Temperature
Squaring both sides:
(11,000)^2 = (3kT/m)
Now solving for T:
T = (m * (11,000)^2) / (3k)
Calculating Temperature
Substituting the values:
T = (5.34 x 10^-26 kg * (11,000 m/s)^2) / (3 * 1.38 x 10^-23 J/K)
Performing the calculation yields:
T ≈ 156,000 K, which can be rounded to 1.56 x 10^5 K.
Conclusion
Thus, the temperature at which the RMS velocity of an oxygen molecule is sufficient to escape from Earth is approximately:
Option C: 1.56 x 10^5 K
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At what temperautre will the rms velocity of oxygen moleucle be sufficient, so as to escape from the earth ?Escape velocity from the earth is 11.0 km/s and the mass of 1 moleucle of oxygen (Boltzman constant k =) a) b) c) d)none of theseCorrect answer is option 'C'. Can you explain this answer?
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At what temperautre will the rms velocity of oxygen moleucle be sufficient, so as to escape from the earth ?Escape velocity from the earth is 11.0 km/s and the mass of 1 moleucle of oxygen (Boltzman constant k =) a) b) c) d)none of theseCorrect answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about At what temperautre will the rms velocity of oxygen moleucle be sufficient, so as to escape from the earth ?Escape velocity from the earth is 11.0 km/s and the mass of 1 moleucle of oxygen (Boltzman constant k =) a) b) c) d)none of theseCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At what temperautre will the rms velocity of oxygen moleucle be sufficient, so as to escape from the earth ?Escape velocity from the earth is 11.0 km/s and the mass of 1 moleucle of oxygen (Boltzman constant k =) a) b) c) d)none of theseCorrect answer is option 'C'. Can you explain this answer?.
At what temperautre will the rms velocity of oxygen moleucle be sufficient, so as to escape from the earth ?Escape velocity from the earth is 11.0 km/s and the mass of 1 moleucle of oxygen (Boltzman constant k =) a) b) c) d)none of theseCorrect answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about At what temperautre will the rms velocity of oxygen moleucle be sufficient, so as to escape from the earth ?Escape velocity from the earth is 11.0 km/s and the mass of 1 moleucle of oxygen (Boltzman constant k =) a) b) c) d)none of theseCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At what temperautre will the rms velocity of oxygen moleucle be sufficient, so as to escape from the earth ?Escape velocity from the earth is 11.0 km/s and the mass of 1 moleucle of oxygen (Boltzman constant k =) a) b) c) d)none of theseCorrect answer is option 'C'. Can you explain this answer?.
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