Question

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 $N_A=6.02\times {10}^{26}/kg$
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

The correct option is C ${10}^{4}K$

$v_{rms}=\sqrt{\frac{2RT}{m}}$ $v_{escape}=\sqrt{2g{R}_e}$
$v_{rms}=v_{escape}$
$\frac{3RT}{m}=2g{R}_e$
$\frac{3\times 1.38\times {10}^{-23}\times 6.02\times {10}^{26}}{2}\times T$
$=2\times 10\times 6.4\times {10}^6$
$T=\frac{4\times 10\times 6.4\times {10}^6}{3\times 1.38\times 6.02\times {10}^3}=10\times {10}^3={10}^{4}k$
Note : Question gives avogadro Number
$N_A=6.02\times {10}^{26}$ / kg but we take $N_A=6.02\times {10}^{26}$ / kmol.