Bonus. Given n, R, T, and V, through an EoS p is uniquely determined, however the same isn't true of other variables like . EoS will in general have multiple roots. Algebraically the van der Waals equation can be written as the following cubic equation RT b+ P a + ab = 0 P Given a = 5.5088 dm . · atm · mol -2 and b = 0.065144 dm³ · mol! for ethane. Compute the molar volume of ethane at 300 K and 200 atm. You may use Mathematica or any software to facilitate calculation. Report your answer in L. mol- to three decimal places. Note. 1atm = 1.01325bar and 1dm3 = 1L.
Bonus. Given n, R, T, and V, through an EoS p is uniquely determined, however the same isn't true of other variables like . EoS will in general have multiple roots. Algebraically the van der Waals equation can be written as the following cubic equation RT b+ P a + ab = 0 P Given a = 5.5088 dm . · atm · mol -2 and b = 0.065144 dm³ · mol! for ethane. Compute the molar volume of ethane at 300 K and 200 atm. You may use Mathematica or any software to facilitate calculation. Report your answer in L. mol- to three decimal places. Note. 1atm = 1.01325bar and 1dm3 = 1L.
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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Transcribed Image Text:Bonus. Given n, R, T, and V, through an EoS p is uniquely determined, however the same isn't
true of other variables like V. EoS will in general have multiple roots. Algebraically the van der
Waals equation can be written as the following cubic equation
『-(+)ア+マ--0
ab
=D0
P
RT
a
V
P
Given a =
5.5088 dm6 · atm · mol-2 and b
0.065144 dm . mol- for ethane. Compute the molar
volume of ethane at 300 K and 200 atm. You may use Mathematica or any software to facilitate
calculation. Report your answer in L mol- to three decimal places. Note. 1atm
1dm = 1L.
: 1.01325bar and
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