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 $1000 K$ ?

Given:- Radius of earth $= 6400 km$ and Radius of moon $= 1740 km$ .

Accepted Answer

The root-mean-square speed of molecules in a gas

$V_{rms} = \sqrt{\frac{3 R T}{M}}$ .... $(1)$

Where, $T$ is temperature of gas, $M$ is the molar mass of the gas.

The speed required to escape from gravitation field is $V = \sqrt{2 g R_{e}}$ ...... $(2)$

Where, $R_{e}$ is radius of earth $(6.4 km)$ , and $g$ is acceleration due to gravity.

From $(1)$ & $(2)$ , we will get,

$\sqrt{\frac{3 R T}{M}} = \sqrt{2 g R_{e}}$

Squaring on both sides.

$T = \frac{2 g R_{e} M}{3 R}$ ...... $(3)$

$(a)$ The molar mass of hydrogen, $M_{H} = 2 \times 10^{- 3} kg {mol}^{- 1}$ , Then the required temperature for the hydrogen gas will be,

$T_{H} = \frac{2 g R M_{H}}{3 R}$

Here, $g = 9.8 m s^{- 2}$ ,universal gas constant, $R = 8.31 J mol^{- 1} K^{- 1}$ , radius, $R = 6 .4 \times 10^{6} m$

$T_{H} = \frac{2 (9 .8 m s^{- 2}) (6 .4 \times 10^{6} m) (2 \times 10^{- 3} kg m^{- 1})}{(3 \times 8 .31 J {mol}^{- 1} K^{- 1})}$

$T_{H} = 1 .1 \times 10^{4} K$ .

$(b)$ Similarly, for oxygen gas $M_{O} = 32 \times 10^{- 3} kg {mol}^{- 1}$

$T_{O} = \frac{2 (9 .8 m s^{- 2}) (6 .4 \times 10^{6} m) (32 \times 10^{- 3} kg m^{- 1})}{(3 \times 8 .31 J (mol - K)^{- 1})}$

$\Rightarrow T_{O} = 1 .6 \times 10^{5} K$ .

On the surface of the moon the required temperature of the gas will be,

$T = \frac{2 g_{m} R_{m} M_{H}}{3 R}$

where $R_{m}$ Radius of moon, $R_{m} = 1 .74 \times 10^{6} m$

Acceleration due to gravity $g_{m} = 0 .16 g$

(c)The temperature for hydrogen gas,

$T_{H} = \frac{2 (0 .16) (9.8 m s^{- 2}) (1 .74 \times 10^{6} m) (2 \times 10^{- 3} kg m^{- 1})}{3 (8 .31 J (mol - K)^{- 1})}$

$\Rightarrow T_{H} = 4 .4 \times 10^{2} K$ .

(d)The temperature for oxygen gas,

$T_{O} = \frac{2 (0 .16) (9 .8 m s^{- 2}) (1 .74 \times 10^{6} m) (32 \times 10^{- 3} Kg {mol}^{- 1})}{3 (8 .31 J (mol - K)^{- 1})}$

$\Rightarrow T_{O} = 7 \times 10^{3} K$ .

In earth's atmosphere temperature is very high for hydrogen, as a result of this atmosphere is depleted of hydrogen.

Comparing the oxygen and hydrogen, very few oxygen atoms escape so oxygen will be there in earth's upper atmosphere.

Important Questions on The Kinetic Theory of Gases

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