Question

The kinetic energy of $1g$ molecule of a gas, at normal temperature and pressure is: $(R=8.321J/mol-K)$

• A. $1.2\times 10^{2}J$
• B. $3.4\times 10^{3}J$
• C. $1.66\times 10^{4}J$
• D. $2.97\times 10^{4}J$

Answer 1

Answer: (B)

Average kinetic energy per molecule is equal to product of mass. of $1g$ molecule and square of mean square velocity.
The kinetic energy of $1g-mol$ is
$E=\frac{1}{2}M{\bar{v}}^2=\frac{1}{2}M\left(\frac{3RT}{M}\right)$
$\left[\because \bar{v}=\sqrt{\frac{3RT}{M}}\right]$
$E=\frac{3}{2}RT$ where $R$ is gas constant.
Putting the numerical values, we have
$E=\frac{3}{2}\times 8.31\times 273=3.4\times 10^{3}J$