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15240672380's Question
Posted 2 months ago

The graph below represents the distribution of molecular speeds of hydrogen and neon at 200 K200 \mathrm{~K}.
1. Match each curve to the appropriate gas.
2. Calculate the rms speed (in m/s\mathrm{m} / \mathrm{s} ) for each of the gases at 200 K200 \mathrm{~K}.
3. Which of the gases would you expect to have the greater effusion rate at 200 K200 \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 K200 \mathrm{~K}.
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Answer from Sia
Posted 2 months ago
For question 1:

Key Concept
Molecular mass and speed distribution
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:

Key Concept
Root-mean-square (rms) speed
The rms speed can be calculated using the formula vrms=3kTmv_{rms} = \sqrt{\frac{3kT}{m}}, where kk is the Boltzmann constant, TT is the temperature, and mm is the molar mass of the gas.

For question 3:

Key Concept
Graham's law of effusion
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:

Key Concept
Equating rms speeds for different temperatures
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.

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