Answer
a. $1.61\times10^5 K$.
b. $2.01\times10^4 K$.
c. See answer.
Work Step by Step
The rms speed is given by equation 13–9, $v_{rms}=\sqrt{3kT/m}$. Set this equal to the escape speed and solve for the temperature.
a. Use oxygen.
$$T=\frac{mv_{rms}^2}{3k}=\frac{2(15.99)(1.66\times10^{-27}kg)(1.12\times10^4 m/s)^2}{3(1.38\times10^{-23}J/K)}=1.61\times10^5 K$$
b. Now use helium atoms:
$$T=\frac{mv_{rms}^2}{3k}=\frac{(4.00)(1.66\times10^{-27}kg)(1.12\times10^4 m/s)^2}{3(1.38\times10^{-23}J/K)}=2.01\times10^4 K$$
c. Our calculated “escape temperature” is very high for oxygen molecules, so they remain in the atmosphere. However, helium can escape at a much lower temperature. During the Earth’s formation process, the atmospheric temperature was high enough for helium atoms to escape.
