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 don't see the answer options (A, B, C, D) in your question. However, I can provide the solution framework for this r.m.s. velocity problem: **Step-by-step solution:** The r.m.s. velocity is given by: $$v_{rms} = \sqrt{\frac{3RT}{M}}$$ where $R$ is the gas constant, $T$ is temperature, and $M$ is molar mass. For hydrogen ($H_2$, $M = 2$ g/mol) and nitrogen ($N_2$, $M = 28$ g/mol): $$\frac{v_{rms}(H_2)}{v_{rms}(N_2)} = \sqrt{\frac{M_{N_2}}{M_{H_2}}} = \sqrt{\frac{28}{2}} = \sqrt{14} \approx 3.74$$ **The given ratio is 7**, not 3.74. This means the two gases are **at different temperatures**. Let $T_H$ and $T_N$ be the temperatures of hydrogen and nitrogen respectively: $$7 = \sqrt{\frac{M_{N_2}}{M_{H_2}} \times \frac{T_H}{T_N}} = \sqrt{\frac{28}{2} \times \frac{T_H}{T_N}}$$ $$49 = 14 \times \frac{T_H}{T_N}$$ $$\frac{T_H}{T_N} = \frac{49}{14} = 3.5$$ **Without seeing option C, the answer likely states:** The temperature of hydrogen is **3.5 times** the temperature of nitrogen, or $T_H = 3.5T_N$ (or similar relationship). Please share the options so I can confirm which is correct.