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Question: The kinetic energy for 14 grams of nitrogen gas at \(127^{o}C\) is nearly (mol. mass of nitrogen = 2...

The kinetic energy for 14 grams of nitrogen gas at 127oC127^{o}C is nearly (mol. mass of nitrogen = 28 and gas constant = 8.31JK−1mol−1)JK^{- 1}mol^{- 1})

A

1.0 J

B

4.15 J

C

2493 J

D

3.3 J

Answer

2493 J

Explanation

Solution

K.E.=32RTmol−1= \frac{3}{2}RTmol^{- 1}

or K.E.=32nRT=32×1428×8.31×400J=2493J= \frac{3}{2}nRT = \frac{3}{2} \times \frac{14}{28} \times 8.31 \times 400J = 2493J