| [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) 650 K
B) \[3\times {{10}^{5}}K\]
C) \[{{10}^{4}}\] K
D) 800 K
Correct Answer: C
Solution :
\[{{\text{v}}_{rms}}=\sqrt{\frac{3RT}{m}}\] \[{{\text{v}}_{escape}}=\sqrt{2g{{R}_{e}}}\] \[{{\text{v}}_{rms}}={{v}_{escape}}\] \[\frac{3RT}{m}=2g{{R}_{e}}\] \[\frac{3\times 1.38\times {{10}^{-23}}\times 6.02\times {{10}^{26}}}{2}\times T\] \[=2\times 10\times 6.4\times {{10}^{6}}\] \[T=\frac{4\times 10\times 6.4\times {{10}^{6}}}{3\times 1.38\times 6.02\times {{10}^{3}}}=10\times {{10}^{3}}={{10}^{4}}k\] Note: Question gives avogadro Number \[{{N}_{A}}=6.02\times {{10}^{26}}/kg\] but we take \[{{N}_{A}}=6.02\times {{10}^{26}}/kmol.\]
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