If it has enough kinetic energy, a moleculeat the surface of the Earth can escape the Earth’s gravitation.The acceleration of gravity is 9.8 m/s2,and the Boltzmanns’ constant is 1.38066×10−23 J/K. Using energy conservation, a) determine theminimum kinetic energy needed to escape interms of the mass of the molecule m , the free-fall acceleration at the surface g, and the radius of the Earth R

b) Calculate the temperature for which the min- imum escape energy is 11 times the average kinetic energy of an oxygen molecule.

K₁ + U₁ = K₂ + U₂ where K₁ and K₂ are the initial and final kinetic energies of the molecule and U₁ and U₂ are the initial and final potential energies of the molecule.

K₁ + -GMm/R = 0 + 0

K₁ = GMm/R since GM = gR²,

K₁ = mgR

So its minimum kinetic energy is mgR

b. The average kinetic energy energy of a gas molecule is K = 3/2kT. If K₁ = 11K, then

mgR = 11 ×3/2kT = 33/2kT.

T = 2mgR/k

For oxygen molecule, m = 32 g/6.02 × 10²³ = 5.32 × 10⁻²⁶ kg/molecule

T = 2 × 5.32 × 10⁻²⁶ kg × 9.8 m/s² 6.4 × 10⁶ m/1.38 10⁻²³ J/K = 483.6 × 10³ K

tallinn tallinn · Answer 2 · 2022-04-01T11:01:18+02:00

The average kinetic energy of a gas molecule is mgR 483.6 × 10³ K

Calculation of Average kinetic energy

a. Now we are using the law of conservation of energy,

Then K₁ + U₁ = K₂ + U₂ where K₁ and K₂ are the initial and also final kinetic energies of the molecule and U₁ and then U₂ are the initial and then final potential energies of the molecule.

Now K₁ + -GMm/R = 0 + 0

After that K₁ = GMm/R since GM = gR²,

Then K₁ = mgR

Hence, its minimum kinetic energy is mgR

b. When The average kinetic energy of a gas molecule is K = 3/2kT. If K₁ = 11K, then

Then mgR = 11 ×3/2kT = 33/2kT.

After that T = 2mgR/k

Then For oxygen molecule, m = 32 g/6.02 × 10²³ = 5.32 × 10⁻²⁶ kg/molecule

Therefore, T = 2 × 5.32 × 10⁻²⁶ kg × 9.8 m/s² 6.4 × 10⁶ m/1.38 10⁻²³ J/K = 483.6 × 10³ K

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