Question

The kinetic energy of one gram molecule of a gas at normal temperature and pressure is$(R = 8.31J/mole - K)$

• A. $0.56 \times 10^{4}J$
• B. $1.3 \times 10^{2}J$
• C. $2.7 \times 10^{2}J$
• D. $3.4 \times 10^{3}J$

Answer 1

Answer: (D)

$3.4 \times 10^{3}J$

Answer 2

Kinetic energy per gm mole $E = \frac{f}{2}RT$

If nothing is said about gas then we should calculate the translational kinetic energy

i.e., $E_{\text{Trans}} = \frac{3}{2}RT = \frac{3}{2} \times 8.31 \times (273 + 0) = 3.4 \times 10^{3}J$