The graph below represents the distribution of molecular speeds of hydrogen and neon at $200 \mathrm{~K}$.
1. Match each curve to the appropriate gas.
2. Calculate the rms speed (in $\mathrm{m} / \mathrm{s}$ ) for each of the gases at $200 \mathrm{~K}$.
3. Which of the gases would you expect to have the greater effusion rate at $200 \mathrm{~K}$ ? Justify your answer.
4. Calculate the temperature at which the rms speed of the hydrogen gas would equal the rms speed of the neon at $200 \mathrm{~K}$.

Question

Accepted Answer

For question 1: ***A*** ∻Key Concept∻ ⚹Molecular mass and speed distribution⚹ ∻Explanation∻ ⚹The curve for hydrogen is taller and narrower because hydrogen has a lower molecular mass than neon, leading to a higher proportion of molecules with higher speeds at the same temperature.⚹ For question 2: ***B*** ∻Key Concept∻ ⚹Root-mean-square (rms) speed⚹ ∻Explanation∻ ⚹The rms speed can be calculated using the formula $v_{rms} = \sqrt{\frac{3kT}{m}}$, where $k$ is the Boltzmann constant, $T$ is the temperature, and $m$ is the molar mass of the gas.⚹ For question 3: ***C*** ∻Key Concept∻ ⚹Graham's law of effusion⚹ ∻Explanation∻ ⚹Hydrogen will have a greater effusion rate than neon at the same temperature because the rate of effusion is inversely proportional to the square root of the molar mass of the gas.⚹ For question 4: ***D*** ∻Key Concept∻ ⚹Equating rms speeds for different temperatures⚹ ∻Explanation∻ ⚹To find the temperature at which hydrogen has the same rms speed as neon at 200 K, set the rms speed equations equal to each other and solve for the new temperature of hydrogen.◊What is the significance of the peak positions of the curves representing hydrogen and neon on the graph of molecular speeds at 200 K?⍭ Generate me a similar question◊⚹

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