A) 844 K
B) 64 K
C) \[273{}^\circ C\]
D) 273 K
Correct Answer: C
Solution :
The root mean square velocity \[{{v}_{rms}}=\sqrt{\frac{3RT}{M}}\] where R is gas constant, T the absolute temperature and M the molecular weight. Given, \[{{v}_{He}}={{v}_{H}},\]\[{{T}_{H}}=273\,K,\]\[{{M}_{H}}=2,\] \[{{M}_{He}}=4\] \[\therefore \] \[\frac{{{v}_{He}}}{{{v}_{H}}}=\sqrt{\frac{{{T}_{H}}}{{{T}_{He}}}\times \frac{{{M}_{He}}}{{{M}_{H}}}}\] \[1=\sqrt{\frac{273}{{{T}_{He}}}\times \frac{4}{2}}\] \[\Rightarrow \] \[{{T}_{He}}=546\,K\] In \[{}^\circ C,\] \[{{T}_{He}}=(546-273){}^\circ C=273{}^\circ C\]
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