The temperature at which the root-mean-square speed of carbon dioxide molecules becomes just sufficient for escaping Earth's atmosphere is approximately
(Mass of carbon dioxide molecule $= 7 \times 10^{-26}\ \text{kg}$ , Boltzmann's constant $k_B = 14 \times 10^{-24}\ J/K$ , escape velocity from Earth's surface $= 11\ \text{km/s}$ )

Question

The temperature at which the root-mean-square speed of carbon dioxide molecules becomes just sufficient for escaping Earth's atmosphere is approximately
(Mass of carbon dioxide molecule

= 7 \times 10^{-26}\ \text{kg}

, Boltzmann's constant

k_B = 14 \times 10^{-24}\ J/K

, escape velocity from Earth's surface

= 11\ \text{km/s}

)

Quantrex Academy · Accepted Answer

Correct answer: A. 2.016 × 10^5\ K

The root-mean-square speed of gas molecules is given by:

v_{rms} = \sqrt{\dfrac{3 k_B T}{m}}

For the molecules to escape Earth's atmosphere, their rms speed must be equal to the escape velocity

v_e

:

\sqrt{\dfrac{3 k_B T}{m}} = v_e

Squaring both sides and solving for

T

:

T = \dfrac{m v_e^2}{3 k_B}

Substituting the given values (

m = 7 \times 10^{-26}\ \text{kg}

,

v_e = 11 \times 10^3\ \text{m/s}

,

k_B = 14 \times 10^{-24}\ \text{J/K}

):

T = \dfrac{7 \times 10^{-26} \times (11 \times 10^3)^2}{3 \times 14 \times 10^{-24}}

T = \dfrac{7 \times 121 \times 10^{-20}}{42 \times 10^{-24}}

T = \dfrac{121}{6} \times 10^4

T \approx 20.166 \times 10^4\ \text{K}

T \approx 2.016 \times 10^5\ \text{K}

Answer:

2.016 \times 10^5\ \text{K}

The temperature at which the root-mean-square speed of carbon...

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