At what temperature would one mole of molecular nitrogen (N₂)have 7.0 × 10³ J of translational kinetic energy? Note: the atomic mass of N is 14 u.

Question

Quizplus · Accepted Answer

The translational kinetic energy (KE) of one mole of an ideal gas is given by the equation KE = (3/2)RT, where R is the ideal gas constant (8.314 J/mol·K), and T is the temperature in Kelvin. Rearranging the formula to solve for T gives T = (2KE)/(3R). Substituting the given kinetic energy (7.0 × 10³ J) into the equation yields T = (2 × 7.0 × 10³ J) / (3 × 8.314 J/mol·K) ≈ 562 K. To convert Kelvin to Celsius, subtract 273 from the Kelvin temperature, resulting in approximately 289 °C, which is closest to option B, 290 °C.

At What Temperature Would One Mole of Molecular Nitrogen (N₂)have