The kinetic energy for 14 grams of nitrogen gas at 127ºC is nearly (mol. mass of nitrogen = 28 and g — States of Matter and Gaseous State Chemistry Question

💡 Solution & Explanation

# Solution: Kinetic Energy of Nitrogen Gas **Given:** - Mass of $N_2$ = 14 g - Temperature = 127°C = 400 K - Molar mass of $N_2$ = 28 g/mol - Gas constant $R$ = 8.31 J K⁻¹ mol⁻¹ **Step 1: Calculate number of moles** $$n = \frac{\text{mass}}{\text{molar mass}} = \frac{14}{28} = 0.5 \text{ mol}$$ **Step 2: Apply kinetic energy formula** For an ideal gas, the average kinetic energy is given by: $$KE = \frac{3}{2}nRT$$ where the factor 3/2 accounts for three translational degrees of freedom at any temperature. **Step 3: Substitute values** $$KE = \frac{3}{2} \times 0.5 \times 8.31 \times 400$$ $$KE = 1.5 \times 0.5 \times 8.31 \times 400$$ $$KE = 0.75 \times 3324 = 2493 \text{ J}$$ $$KE \approx 2.5 \text{ kJ} \text{ or } 2500 \text{ J}$$ **Answer: C ≈ 2500 J (or 2.5 kJ)** The kinetic energy depends on temperature and the number of moles through the equipartition theorem—higher temperature and more molecules both increase total kinetic energy.