Question

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

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

Answer 1

Answer: (B)

$6.21\times {10}^{-21}J/molecules$

$K{E}_{total}=\frac{3}{2}n{N}_{A}kT$

$K{E}_{Average}=\frac{3}{2}kT$

$KEtotal=\frac{3}{2}\times \frac{8}{16}\times 8.314\times 300=1870.65J$

$K{E}_{Average}=\frac{K{E}_{Total}}{n{N}_A}=\frac{1870.65}{0.5\times 6.023\times {10}^{23}}=6.21\times {10}^{-21}J/molecule$