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 $k _B=1.38 \times 10^{-23} J / K, \quad$ Avogadro number $N _A=6.02 \times 10^{26} / kg$, Radius of earth $=6.4 \times 10^{6} m$, Gravitational acceleration on earth $=10 ms^{-2}$ ]
(2019 Main, 8 Apri, II)
(a) $10^{4} K$
(b) $650 K$
(c) $3 \times 10^{5} K$
(d) $800 K$
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Solution:
- Root mean square velocity of hydrogen molecule is given as
$$ v _{rms}=\sqrt{\frac{3 k _B T}{m}} $$
Escape velocity of hydrogen molecule from the earth is given as
$$ v _e=\sqrt{2 g R _e} $$
Given,
$$ \begin{aligned} v _{rms} & =v _e \\ \sqrt{\frac{3 k _B T}{m}} & =\sqrt{2 g R _e} \Rightarrow T=\frac{2 g R _e m}{3 \times k _B} \end{aligned} $$
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 $KE=\frac{1}{2} m v _e^{2}$, where $m$ is the mass of the gas molecule.
Equating the above thermal and kinetic energies, we have
$$ \frac{3}{2} k _B T=\frac{1}{2} m v _e^{2} \text { or } T=\frac{m v _e^{2}}{3 k _B} $$
Here,
$$ \begin{aligned} & k _B=1.38 \times 10^{-23} JK^{-1}, \\ & v _e=11.2 \times 10^{3} ms^{-1}, \\ & m=\frac{2}{6.02 \times 10^{26}}=0.332 \times 10^{-26} kg \end{aligned} $$
Substituting these value in Eq. (i), we get
$T=\frac{0.332 \times 10^{-26} \times\left(11.2 \times 10^{3}\right)^{2}}{3 \times 1.38 \times 10^{-23}} \approx 10^{4} K$
