Question

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

$E = \frac{3}{2}RT = \frac{3}{2} \times 8.31 \times 273 = 3.4 \times 10^{3}$Joule