A) \[6.17\times {{10}^{-21}}J\]
B) \[6.17\times {{10}^{-20}}J\]
C) \[61.7\times {{10}^{-21}}J\]
D) \[7.16\times {{10}^{-20}}J\]
Correct Answer: A
Solution :
Here: Temperature\[(T)={{25}^{o}}C=298\,\,K\]. Average kinetic energy permolecule\[=\frac{3RT}{2n}=\frac{3\times 8.314\times 298}{2\times (6.02\times {{10}^{23}})}\]\[=6.17\times {{10}^{-21}}J\](where\[R=8.314\])
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