Maxwell Boltzmann Distribution Pogil Answer Key Extension Questions
$$v_p = \sqrt\frac2kTm$$
This question tests whether students confuse the distribution of energy with the required threshold .
As temperature increases, what happens to the peak of the Maxwell-Boltzmann distribution curve? Explain why. The curve is not symmetric
The curve is not symmetric. It starts at the origin , rises to a peak, and then tails off slowly to the right.
) will have a lower average speed and a narrower, higher peak. Higher peak, faster average speed. Heavier Gas ( O2cap O sub 2 ): Lower peak, slower average speed. Higher peak, faster average speed
a)
), the kinetic energy of particles is theoretically zero. The distribution curve would not be a curve at all; it would be a single vertical line (a Dirac delta function) at a speed of slower average speed. a) )
As temperature increases, the average kinetic energy of the particles increases. This causes two major changes to the graph: Shift to the Right: The peak ( vmpv sub m p end-sub ) shifts to the right, indicating higher average speeds.
When self-checking your extension answers or preparing for an exam based on this guided inquiry, keep these core principles in mind:
"The M-B curves for isotopes are nearly identical because mass difference is small relative to absolute mass. However, the effusion rate depends on the inverse square root of mass. Over many stages, this tiny difference in the distribution's average velocity accumulates into measurable separation."