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 10⁻²⁶ kg , Boltzmann's constant k_B = 14 10⁻²⁴ J/K , escape velocity from Earth's surface = 11 km/s )

Step-by-step solution

The root-mean-square speed of gas molecules is given by: v_ rms = 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 : 3 k_B T m = v_e Squaring both sides and solving for T : T = m v_e^2 3 k_B Substituting the given values ( m = 7 10⁻²⁶ kg , v_e = 11 10^3 m/s , k_B = 14 10⁻²⁴ J/K ): T = 7 10⁻²⁶ (11 10^3)^2 3 14 10⁻²⁴ T = 7 121 10⁻²⁰ 42 10⁻²⁴ T = 121 6 10^4 T 20.166 10^4 K T 2.016 10^5 K Answer: 2.016 10^5 K