A) \[1.2\times {{10}^{2}}J\]
B) \[3.4\times {{10}^{3}}J\]
C) \[1.66\times {{10}^{4}}J\]
D) \[2.97\times {{10}^{4}}J\]
Correct Answer: B
Solution :
Key Idea: Average kinetic energy per molecule is equal to product of mass of \[1\,\,g\] molecule and square of mean square velocity. The kinetic energy of \[1\,\,g-\]mol is \[E=\frac{1}{2}M\,{{\bar{v}}^{2}}=\frac{1}{2}M\left( \frac{3RT}{M} \right)\] \[\left[ \because \,\,\bar{v}=\sqrt{\frac{3RT}{M}} \right]\] where \[R\] is gas constant. Putting the numerical values, we have \[E=\frac{3}{2}\times 8.31\times 2.73=3.4\times {{10}^{3}}J\]
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