The kinetic energy (E) of 'one gram molecule' (which equals 1 mole) of an ideal gas is given by the formula $E = \frac{3}{2}RT$. Given: $R = 8.31 \text{ J mol}^{-1} \text{ K}^{-1}$ . Standard Temperature (STP) $T = 273 \text{ K}$ (\approx $0^\circ\text{C}$). Substituting the values: $E = \frac{3}{2} \times 8.31 \times 273$. $E = 1.5 \times 2268.63 = 3402.9 \text{ J}$. In scientific notation, this is approximately $3.4 \times 10^3 \text{ J}$.