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

I need to find the temperature at which the most probable speed of $O_2$ equals the RMS speed of $O_2$ at 300 K. **Step 1: Recall speed formulas** Most probable speed: $v_p = \sqrt{\frac{2RT}{M}}$ RMS speed: $v_{rms} = \sqrt{\frac{3RT}{M}}$ **Step 2: Find RMS speed at 300 K** $$v_{rms}(300K) = \sqrt{\frac{3R(300)}{M}}$$ **Step 3: Set up the equation** We need most probable speed at temperature $T$ to equal RMS speed at 300 K: $$v_p(T) = v_{rms}(300K)$$ $$\sqrt{\frac{2RT}{M}} = \sqrt{\frac{3R(300)}{M}}$$ **Step 4: Solve for T** Squaring both sides: $$\frac{2RT}{M} = \frac{3R(300)}{M}$$ $$2T = 3(300)$$ $$2T = 900$$ $$T = 450 \text{ K}$$ **Answer: 450 K** The most probable speed at 450 K equals the RMS speed at 300 K because the ratio of temperatures is 3:2, which matches the ratio of the coefficients in their respective speed formulas (3 for RMS, 2 for most probable).