Question

The average kinetic energy (in joules) of molecules in $8.0\text{g}$ of methane at ${27}^{\circ }\text{C}$ is:

• A. $6.21\times {10}^{-20}\text{J/molecule}$
• B. $6.21\times {10}^{-21}\text{J/molecule}$
• C. $6.21\times {10}^{-22}\text{J/molecule}$
• D. $3.1\times {10}^{-22}\text{J/molecule}$

Answer 1

The correct option is B $6.21\times {10}^{-21}\text{J/molecule}$

K.E. $=\frac{3}{2}nRT$
As we know $n=\frac{N}{N_A}$
where N= No of molecules,
$N_A$=Avogadro's number
$\Rightarrow \frac{\text{K.E}}{\text{Molecule}}=\frac{3}{2}\times \frac{300\times 8.314}{N_A}$
$=6.21\times {10}^{-21}\text{J/molecule}$