A) \[844\text{ }K\]
B) \[64\text{ }K\]
C) \[{{273}^{o}}C\]
D) \[273\text{ }K\]
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, \[{{v}_{He}}={{v}_{H}},\,\,\,\,\,{{T}_{H}}=273K,\,\,\,\,\,\,\,\,{{M}_{H}}=2,\,\,\,\,{{M}_{He}}=4\] \[\therefore \] \[\frac{{{v}_{He}}}{{{v}_{H}}}=\sqrt{\frac{{{T}_{H}}}{{{T}_{He}}}\times \frac{{{M}_{He}}}{{{M}_{H}}}}\] \[1=\sqrt{\frac{273}{{{T}_{He}}}\times \frac{4}{2}}\] \[\Rightarrow \] \[{{T}_{He}}=546K\] In \[^{o}C,\] \[{{T}_{He}}={{(546-273)}^{o}}C={{273}^{o}}C\]
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