Question

# At what temperature does the rms speed of (a)$$H_2$$ ( molecular hydrogen and ( b) $$O_2$$ ( molecular oxygen) equal the escape speed from Earth? At what temperature does the rms speed of (c) $$H_2$$ and (d) $$O_2$$ equal the escape speed from the Moon ( where the gravitational acceleration at the surface has magnitude $$0.16\ g$$? Considering the answers to parts (a) and (b), should there be much (e) hydrogen and (f) oxygen high in Earth's upper atmosphere, where the temperature is about $$1000\ K$$?

Solution

## The rms speed of molecules in a gas is given by $$v_{rms}=\sqrt{3RT/M}$$, where $$T$$ is the temperature and $$M$$ is the molar mass of the gas. The speed required for escape from Earth's gravitational pull is $$v=\sqrt{2gr_e}$$, where $$g$$ is the acceleration due to gravity at Earth's surface and $$r_e (=6.37\times 10^6\ m)$$ is the radius of Earth. To derive this expression, take the zero of gravitational potential energy to be at infinity. Then, the gravitational potential energy of a particle with mass $$m$$ at Earth's surface is $$U=-GMm/r_e^2=-mgr_e$$,where $$g=GM/r_e^2$$ was used. If $$v$$ is the speed of the particle, then its original energy is $$E=-mgr_e+\dfrac{1}{2}mv^2$$. If the particle is just able to travel far away, its kinetic energy must tend toward zero as its distance from Earth becomes large without bound. This means $$E=0$$ and $$v=\sqrt{2gr_e}$$. We equate the expressions for the speeds to obtain $$\sqrt{3RT/M}=\sqrt{2gr_e}$$. The solution for $$T$$ is $$T=2gr_eM/3R$$.(a)The molar mass of hydrogen is $$2.02\times 10^{-3}\ kg/mol$$, so for that gas$$T=\dfrac{2(9.8\ m/s^2)(6.37\times 10^6\ m)(2.02\times 10^{-3}\ kg/mol)}{3(8.31\ J/mol. K)}=1.0\times 10^4\ K$$.The solution for $$T$$ is $$T=2gr_eM/3R$$.(b) The molar mass of oxygen is $$32.0\times 10^{-3}\ kg/mol$$, so for that gas$$T=\dfrac{2(9.8\ m/s^2)(6.37\times 10^6\ m)(3.20\times 10^{-3}\ kg/mol)}{3(8.31\ J/mol. K)}=1.6\times 10^5\ K$$.(c) Now, $$T=2g_mr_mM/3R$$, where $$r_m=1.74\times 10^6\ m$$ is the radius of the Moon and $$g_m=0.16\ g$$ is the acceleration due to gravity at the Moon's surface. For hydrogen, the temperature is $$T=\dfrac{2(0.16)(9.8\ m/s^2)(1.74\times 10^6\ m)(2.02\times 10^{-3}\ kg/mol)}{3(8.31\ J/mol. K)}=4.4\times 10^2\ K$$.(d) For oxygen, the temperature is $$T=\dfrac{2(0.16)(9.8\ m/s^2)(1.74\times 10^6\ m)(32.0\times 10^{-3}\ kg/mol)}{3(8.31\ J/mol. K)}=7.0\times 10^3\ K$$.(e) The temperature high in Earth's atmosphere is great enough for a significant number of hydrogen atoms in the tail of the Maxwellian distribution to escape. As a result, the atmosphere is depleted of hydrogen.(f) On the other hand, very few oxygen atoms escape. So there should be much oxygen high in Earth's upper atmosphere.Physics

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