A) 101 K
B) 102 K
C) 103 K
D) 104 K
Correct Answer: D
Solution :
Key Idea: For gas molecules to escape from earths atmosphere their root mean square velocity must be equal to escape velocity. As we know gas molecules cannot escape from earths atmosphere because their root mean square velocity is less than escape velocity at earths surface. If we fill this requirement, then gas molecules can escape from earths atmosphere. i.e., \[{{v}_{rms}}={{v}_{es}}\] or \[\sqrt{\frac{3RT}{M}}=\sqrt{2\,g{{R}_{e}}}\] or \[T=\frac{2Mg{{R}_{e}}}{3R}\] ... (i) Given, \[M=2\times {{10}^{-3}}kg,\,g=9.8\,m/{{s}^{2}}\] \[{{R}_{e}}=6.4\times {{10}^{6}}m,\,\,R=8.31\,J/mol-K\] Substituting in Eq. (i), we have \[T=\frac{2\times 2\times {{10}^{-3}}\times 9.8\times 6.4\times {{10}^{6}}}{3\times 8.31}\] \[={{10}^{4}}K\]
You need to login to perform this action.
You will be redirected in
3 sec
