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 1g-mol is \[E=\frac{1}{2}M{{\bar{v}}^{2}}=\frac{1}{2}M\left( \frac{3RT}{M} \right)\] \[\left[ \because \,\bar{v}\,=\sqrt{\frac{3RT}{M}} \right]\] \[E=\frac{3}{2}RT\] where R is gas constant. Putting the numerical values, we have \[E=\frac{3}{2}\times 8.31\times 273=3.4\times {{10}^{3}}J\]
You need to login to perform this action.
You will be redirected in
3 sec
