Question

The kinetic energy of $1 \,g$ molecule of a gas, at normal temperature and pressure is $( R = 8.31 \,J \, mol^{-1} \, K^{-1} )$

• A. $ 3.4 \times 10^3 \, J$
• B. $ 2.97 \times 10^3 \, J$
• C. $ 1.2 \times 10^2 \, J$
• D. $ 0.66 \times 10^4 \, J$

Answer 1

The Correct Option is A

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