Question

The temperature, at which the root mean square velocity o f 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 $N_A=6.02\times 10^{26}/kg$
Radius of Earth : $6.4\times 10^{6}m$
Gravitational acceleration on Earth = $10m{s}^{-2}$]

• A. $650K$
• B. $3\times 10^{5}K$
• C. $10^{4}K$
• D. $800K$

Answer 1

Answer: (C)

$v_{rms}=\sqrt{\frac{3RT}{m}}{v}_{escape}=\sqrt{2g{R}_e}$
$v_{rms}=v_{escape}$
$\frac{3RT}{m}=2g{R}_e$
$\frac{3\times 1.38\times 10^{-23}\times 6.02\times 10^{26}}{2}\times T$
$=2\times 10\times 6.4\times 10^6$
$T=\frac{4\times 10\times 6.4\times 10^6}{3\times 1.38\times 6.02\times 10^3}=10\times 10^3=10^{4}k$