Question

The average kinetic energy of one molecule of an ideal gas ${27}^{\circ }C$ and $1atm$ pressure is:

• A. $900cal{K}^{-1}mo{l}^{-1}$
• B. $6.21\times {10}^{-21}J{K}^{-1}molecul{e}^{-1}$
• C. $336.7J{K}^{-1}molecul{e}^{-1}$
• D. $3741.3J{K}^{-1}mo{l}^{-1}$

Answer 1

The correct option is B $6.21\times {10}^{-21}J{K}^{-1}molecul{e}^{-1}$

$KE=\frac{3}{2}\frac{R}{N}T$
$=\frac{3}{2}\times \frac{8.314}{6.023\times {10}^{23}}\times 300$
$=6.21\times {10}^{-21}J{K}^{-1}molecul{e}^{-1}$.