Q. The kinetic energy of one gram molecule of a gas at normal temperature and pressure is $(R = 8.31\ J/mol - K)$
A) $0.56 \times 10^{4}\ J$
B) $1.3 \times 10^{2}\ J$
C) $2.7 \times 10^{2}\ J$
D) $3.4 \times 10^{3}J$
Explanation
(d) Kinetic energy per $g\ mole\ E = \frac{f}{2}RT$
If nothing is said about gas then we should calculate the translational kinetic energy
$i.e.,\ E_{Trans} = \frac{3}{2}RT = \frac{3}{2} \times 8.31 \times (273 + 0) = 3.4 \times 10^{3}J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.