Show complete solutions and enclose all final answers in a box. Round off final answers to 4 decimal places and floating values for intermediate answers. Use 2 decimal places for molecular weights. 8) Compare the values obtained for the pressure(atm) of 2 mols 0z at 298.15K held in a8.25 dm' bulb usinga) Redlick-Kwong equation: For 02, Tc = 154.6 K and Pc = 50.43 bar%3Db) Beattie – Bridgeman equationFor O2, the Beattie-Bridgeman constants:Ao = 0.14911 Pa-m6-mol2Bo = 46.24 x 10-6 m²-mol-a= 25.62 x 10 -6 m³-mol-1b= 4.208 x 106 m³-mol-1C = 4.80 x 10-1 m3-K3-mol1Ра-т3Use R = 8.314mole-K

Given: moles of O2 (n) = 2 moles Temperature (T) = 298.15 K Volume of bulb = 8.25 dm3 = 0.00825 m3…

8) Compare the values obtained for the pressure(atm) of 2 mols Oz at 298.15K held in a 8.25 dm bulb using a) Redlick-Kwong equation: For O2, Tc = 154.6 K and Pc = 50.43 bar b) Beattie – Bridgeman equation

8) Compare the values obtained for the pressure(atm) of 2 mols Oz at 298.15K held in a 8.25 dm bulb using a) Redlick-Kwong equation: For O2, Tc = 154.6 K and Pc = 50.43 bar b) Beattie – Bridgeman equation

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

Show complete solutions and enclose all final answers in a box. Round off final answers to 4 decimal places and floating values for intermediate answers. Use 2 decimal places for molecular weights.

8) Compare the values obtained for the pressure(atm) of 2 mols 0z at 298.15K held in a
8.25 dm' bulb using
a) Redlick-Kwong equation: For 02, Tc = 154.6 K and Pc = 50.43 bar
%3D
b) Beattie – Bridgeman equation
For O2, the Beattie-Bridgeman constants:
Ao = 0.14911 Pa-m6-mol2
Bo = 46.24 x 10-6 m²-mol-
a= 25.62 x 10 -6 m³-mol-1
b= 4.208 x 106 m³-mol-1
C = 4.80 x 10-1 m3-K3-mol1
Ра-т3
Use R = 8.314
mole-K

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