SATHEE: Heat and Thermodynamics 4 Question 8

Heat and Thermodynamics 4 Question 8

9. The temperature, at which the root mean square velocity of hydrogen molecules equals their escape velocity from the earth, is closest to :

[Boltzmann constant $k_{B} = 1.38 \times 10^{- 23} J / K,$ Avogadro number $N_{A} = 6.02 \times 10^{26} / k g$ , Radius of earth $= 6.4 \times 10^{6} m$ , Gravitational acceleration on earth $= 10 m s^{- 2}$ ]

(2019 Main, 8 Apri, II)

(a) $10^{4} K$

(b) $650 K$

(d) $800 K$

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Solution:

Root mean square velocity of hydrogen molecule is given as

$v_{r m s} = \sqrt{\frac{3 k_{B} T}{m}}$

Escape velocity of hydrogen molecule from the earth is given as

$v_{e} = \sqrt{2 g R_{e}}$

Given,

$\begin{aligned} v_{r m s} & = v_{e} \\ \sqrt{\frac{3 k_{B} T}{m}} & = \sqrt{2 g R_{e}} \Rightarrow T = \frac{2 g R_{e} m}{3 \times k_{B}} \end{aligned}$

Substituting the given values, we get

$T = \frac{2 \times 10 \times 6.4 \times 10^{6} \times 2}{3 \times 1.38 \times 10^{- 23} \times 6.02 \times 10^{26}} \approx 10^{4} K$

Alternate Solution

At rms speed, average thermal kinetic energy of a hydrogen gas molecule is $= \frac{3}{2} k_{B} T$

and if $v_{e} =$ escape velocity of the gas molecule from the earth, then its kinetic energy is $K E = \frac{1}{2} m v_{e}^{2}$ , where $m$ is the mass of the gas molecule.

Equating the above thermal and kinetic energies, we have

$\frac{3}{2} k_{B} T = \frac{1}{2} m v_{e}^{2} or T = \frac{m v_{e}^{2}}{3 k_{B}}$

Here,

$\begin{aligned} k_{B} = 1.38 \times 10^{- 23} J K^{- 1}, \\ v_{e} = 11.2 \times 10^{3} m s^{- 1}, \\ m = \frac{2}{6.02 \times 10^{26}} = 0.332 \times 10^{- 26} k g \end{aligned}$