Question

The average K.E. of an ideal gas per molecule in SI units at ${25}^{o}C$ is :

• A. $6.17\times {10}^{-21}$ kJ
• B. $6.17\times {10}^{-21}$ J
• C. $6.17\times {10}^{-20}$ J
• D. $7.16\times {10}^{-20}$ kJ

Answer 1

Answer: (B)

$6.17\times {10}^{-21}$ J

The expression for the average K.E. of an ideal gas per molecule is
$K.E.=\frac{3}{2}kT=\frac{3}{2}\frac{RT}{N}$
Here, $k$ is Boltzmann constant, $T$ is absolute temperature, $R$ is ideal gas constant and $N$ is Avogadro's number.
Substitute values in the above expression

$K.E.=\frac{3}{2}\times \frac{8.314J{K}^{-1}mo{l}^{-1}\times 298K}{6.023\times {10}^{23}mo{l}^{-1}}$

$K.E.=6.17\times {10}^{-21}J$