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Calculate the temperature for which the minimum escape energy is 15 times the average kinetic energy of an oxygen molecule. Answer in units of K.

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onyebuchinnaji

The temperature for which the minimum escape energy is 15 times the average kinetic energy of an oxygen molecule is 10,684.4 K.

Conservation of energy

The temperature for which the minimum escape energy is 15 times the average kinetic energy of an oxygen molecule is determined by applying principle of conservation of energy as shown below;

E = K + U

E = K - GmM/R

0 = K - GmM/R

K = mgR

When the minimum escape energy is 15 times average kinetic energy;

15K = mgR

[tex]15 \times \frac{3}{2} kT = mgR[/tex]

where;

  • m is mass of oxygen per atom = (0.032 kg)/(6.02 x 10²³) = 5.316 x 10⁻²⁶ kg/atom
  • g is acceleration due gravity = 9.8 m/s²
  • R is radius of Earth = 6.371 x 10⁶ m.
  • k is Boltzmanns’ constant =   1.38066 x 10⁻²³ J/K.

[tex]T = \frac{2mgR}{15 \times 3k} \\\\T = \frac{(2)(5.316 \times 10^{-26})(9.8)(6.371 \times 10^6)}{45(1.38066 \times 10^{-23})} \\\\T = 10,684.4 \ K[/tex]

Thus, the temperature for which the minimum escape energy is 15 times the average kinetic energy of an oxygen molecule is 10,684.4 K.

Learn more about minimum escape energy here: https://brainly.com/question/16008094

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