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

# 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: Find 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, average kinetic energy is: $$KE = \frac{3}{2}nRT$$ The factor 3/2 comes from translational kinetic energy in three dimensions. **Step 3: Calculate** $$KE = \frac{3}{2} \times 0.5 \times 8.31 \times 400$$ $$KE = \frac{3}{2} \times 0.5 \times 3324$$ $$KE = 1.5 \times 1662 = 2493 \text{ J}$$ $$KE \approx 2.5 \text{ kJ or } 2500 \text{ J}$$ **Answer: The kinetic energy is approximately 2.5 kJ** (Option C assumes this value or equivalent)