At what temperature most probable speed of molecules have the same value of root mean square speed o — States of Matter and Gaseous State Chemistry Question

💡 Solution & Explanation

# Solution **Given Information:** - Find temperature where most probable speed of $O_2$ equals RMS speed of $O_2$ at 300 K **Step 1: Write the formulas** Most probable speed: $$v_p = \sqrt{\frac{2RT}{M}}$$ RMS speed: $$v_{rms} = \sqrt{\frac{3RT}{M}}$$ **Step 2: Calculate RMS speed at 300 K** $$v_{rms}(300K) = \sqrt{\frac{3R \times 300}{M}}$$ **Step 3: Set most probable speed at unknown temperature T equal to RMS speed at 300 K** $$v_p(T) = v_{rms}(300K)$$ $$\sqrt{\frac{2RT}{M}} = \sqrt{\frac{3R \times 300}{M}}$$ **Step 4: Square both sides and solve for T** $$\frac{2RT}{M} = \frac{3R \times 300}{M}$$ $$2T = 900$$ $$T = 450 \text{ K}$$ **Answer: 450 K (Option D)** The most probable speed increases with temperature. At 450 K, the slower-moving molecules (represented by $v_p$) have the same speed as the average of all molecular speeds (represented by $v_{rms}$) at 300 K.