The r.m.s. velocity of hydrogen is 7 time the r.m.s. velocity of nitrogen. If T is the temperature o — States of Matter and Gaseous State Chemistry Question

💡 Solution & Explanation

I need to see the options to identify what option C states. However, I can solve the relationship between the temperatures of hydrogen and nitrogen given that their r.m.s. velocities have a specific ratio. **Given:** $v_{rms}(H_2) = 7 \times v_{rms}(N_2)$ **Step 1: Apply the r.m.s. velocity formula** $$v_{rms} = \sqrt{\frac{3RT}{M}}$$ where $R$ is the gas constant, $T$ is temperature, and $M$ is molar mass. **Step 2: Set up the ratio** $$\frac{v_{rms}(H_2)}{v_{rms}(N_2)} = \sqrt{\frac{M_{N_2}}{M_{H_2}} \cdot \frac{T_{H_2}}{T_{N_2}}} = 7$$ **Step 3: Substitute molar masses** $$M_{H_2} = 2 \text{ g/mol}, \quad M_{N_2} = 28 \text{ g/mol}$$ $$\sqrt{\frac{28}{2} \cdot \frac{T_{H_2}}{T_{N_2}}} = 7$$ $$\sqrt{14 \cdot \frac{T_{H_2}}{T_{N_2}}} = 7$$ **Step 4: Square both sides** $$14 \cdot \frac{T_{H_2}}{T_{N_2}} = 49$$ $$\frac{T_{H_2}}{T_{N_2}} = \frac{49}{14} = 3.5 = \frac{7}{2}$$ **Conclusion:** If both gases are at the same temperature $T$, then $T_{H_2} = 3.5T$ and $T_{N_2} = T$, or more commonly, the temperature of hydrogen must be **3.5 times (or 7/2 times) that of nitrogen** to achieve the given velocity