Question

The kinetic energy for 14 grams of nitrogen gas at ${127}^{0}C$ is nearly (mol. mass of nitrogen = 28 and gas constant = 8.31 $J{K}^{-1}mo{l}^{-1}$)

• A. $1230J$
• B. $4015J$
• C. $2493J$
• D. $990J$

Answer 1

Answer: (C)

$2493J$

$KE=\frac{3}{2}nRT$ $n=\frac{14}{28}=0.5$

$=\frac{3}{2}\times 0.5\times 8.314\times (127+273)$

$=2494.2J$