A) 1200 K
B) 150 K
C) 600 K
D) 300 K
Correct Answer: B
Solution :
\[\frac{{{V}_{av}}C{{H}_{4}}}{{{V}_{ab}}{{O}_{2}}}=\sqrt{\frac{{{T}_{C{{H}_{4}}}}}{{{T}_{{{O}_{2}}}}}.\frac{{{M}_{{{O}_{2}}}}}{{{M}_{C{{H}_{4}}}}}}=1\] \[\frac{{{T}_{C{{H}_{4}}}}}{300}.\frac{32}{16}=1\]; \[{{T}_{C{{H}_{4}}}}={{150}^{o}}K\]You need to login to perform this action.
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