. Ball 1.72: Calculate the average grade using equation 1.32, 3Ball 1.32: We saw
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Question
. Ball 1.72: Calculate the average grade using equation 1.32,
3Ball 1.32: We saw in Lecture 3 that the relative change in volume 2: We saw in Lecture 3 that the relative change in volume when we vary tem perature and pressure is given by dV At a pressure of 1.08 atm and 350 K for one mole of ideal gas, what is the predicted 0.10 atm) absolute change in volume if the pressure changes by 0.10 atm (that is, dp and the temperature change is 10.0 K? 3 Van der Waals Gases Ball 1.52: Under what conditions of volume does a van der Waals gas behave like an ideal gas? Use the van der Waals equation of state to justify your answer. Ball 1.44: The van der Waals constant b can be used to estimate molecular sizes, assum- ing the molecules are shaped like spheres: 1. Convert b to units of m3/mol, using the fact that 1 m3 = 1000 L. 2. Divide by Avogadro's number to get the individual molecular contribution to b. 3. Use V = (4/3/Tr3 to estimate the radius of the molecule. Using these steps, estimate the sizes of (a) He (b) H2O (c) C2H6 "Ball 1.37: Liquid nitrogen comes in large cylinders that require special tank carts and hold 120 L of liquid at 77 K. Given the density of liquid nitrogen of 0.840 g/cm3, use the van der Waals equation to estimate the volume of nitrogen gas after it evaporates at 77 K and atmospheric pressure. (Hint: Solve the cubic form of the van der Waals equation we obtained in the Math Diagnostic. If your calculator can't solve equations, feel free to use Wolfram Alpha, http://www.wolframalpha.com.,) 4 Basic Statistical Mechanics 1. Ball 1.72: Calculate the average grade using equation 1.32, if the scores on a 10-point quiz are 5, 5, 5, 5, 10, 10, and 10. 2. Ball 1.74: What is the ratio of the populations of two energy states whose energies differ by 1000 J at (a) 200 K (b) 500 K (c) 1000 K? What trend do you see in your answers? 3. Ball 1.76: Determine the expected(E) for translational and rotational motions, in RT units, for these gases. (a) cyanogen, (CN)2 (b) H20 (c) Kr (d) CgH6Explanation / Answer
Ball 1.72:
Equation 1.32 is given in the question, so
Out of seven grades four are 5, so (4/7 x 5). Here, Pi = 4/7 and I = 5.
Out of seven grades three are 10, so (3/7 x 10). Here, Pi = 3/7 and I = 10.
Just substitute the arrived values in the given equation.
Average, <I> = [(4/7 x 5) + (3/7 x 10)] / [4/7 + 3/7]
= [20/7 + 30/7] / [7/7]
= [50/7] / 1
Average = 7.143
One can also verify the answer by using conventional method,
Average = [5+5+5+5+10+10+10] / 7 = 50/7 = 7.143
Go through my answer carefully, I used different brackets since Latex is not working right now.
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