Question

Calculate the kinetic energy in joule of $8.00$ of methane at ${27}^{\circ }C$.

Answer 1

Number of gram moles of methane $=\frac{8}{16}=\frac{1}{2}$
$T=(27+273)=300K$
$R=8.32J{K}^{-1}mo{l}^{-1}$
Kinetic energy of one gram mole $=\frac{3}{2}RT$
$=\frac{3}{2}\times 8.32\times 300J$
Thus, kinetic energy for $\frac{1}{2}$ gram mole $=\frac{1}{2}\times \frac{3}{2}\times 8.32\times 300$
$=1872J$