A) \[-216{}^\circ C\]
B) \[-235{}^\circ C\]
C) \[-254{}^\circ C\]
D) \[-264{}^\circ C\]
Correct Answer: C
Solution :
Root mean square velocity of gas \[{{v}_{rms}}=\frac{3RT}{M},\] where M= molecular weight and \[{{v}_{rms}}\] is equal for both gases. \[\therefore \] \[\frac{{{T}_{H}}}{{{M}_{H}}}=\frac{{{T}_{O}}}{{{M}_{O}}}\] \[\Rightarrow \] \[\frac{{{T}_{H}}}{2}=\frac{273+31}{32}\] \[\Rightarrow \] \[{{T}_{H}}=\frac{304\times 2}{32}=19K\] \[{{T}_{H}}=19-273=-{{254}^{o}}C\]
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