A) 0
B) \[\frac{273{{k}_{B}}}{2Mg}\]
C) \[\frac{546{{k}_{B}}}{3Mg}\]
D) \[\frac{819{{k}_{B}}}{2Mg}\]
Correct Answer: D
Solution :
Kinetic energy of each molecule, \[K.E.=\frac{3}{2}{{K}_{B}}T\] In the given problem, Temperature, T = 0°C = 273 K Height attained by the gas molecule, h = ? \[K.E.=\frac{3}{2}{{K}_{B}}(273)=\frac{819{{K}_{B}}}{2}\] \[K.E.=P.E.\] \[\Rightarrow \]\[\frac{819{{K}_{B}}}{2}=Mgh\]or\[h=\frac{819{{K}_{B}}}{2Mg}\]You need to login to perform this action.
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