A) \[{{1092}^{o}}C\]
B) \[{{1492}^{o}}C\]
C) 273 K
D) \[{{819}^{o}}C\]
Correct Answer: D
Solution :
Given, \[{{v}_{{{O}_{2}}}}=\frac{1}{2}\,{{v}_{{{H}_{2}}}}\] \[\therefore \] \[\sqrt{\frac{3RT}{32}}=\frac{1}{2}\sqrt{\frac{3R\times 273}{2}}\] \[\frac{T}{32}=\frac{273}{8}\] \[\therefore \] \[T=4\times 273\] \[T=1092\,\,K=1092-273\] \[T={{819}^{o}}C\]
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