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Theo kinetic energy for 14 grams of nitrogen gas at 127 degreeC1s nearly (mol. mass of nitrogen= 28 and gas constant = 8.31 J/mol/K)

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1.0 J

4.15 J

2492.2 J

3.3 J

answer is C.

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Detailed Solution

$\begin{array}{l} K E = \frac{3}{2} \times 1.38 \times 10^{- 23} \times \frac{14}{28} \times 6.02 \times 10^{23} \times 400 \\ = 2492.28 \end{array}$

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