Give an example of Comparison of Z factors for various gases. 1.11.0T= 2.000.9TR=1.500.8ххT=1.300.7리호0.6TR= 1.20ofso9040.5TR= 1.100.4Legend:x Methaneo EthyleneA Ethaneo PropaneIso-pentaneо т-НеptaneA Nitrogen• Carbon dioxideTR=1.000.30.2O n-ButaneWaterAverage curve based on data onhydrocarbons0.10.51.01.52.02.53.03.54.04.55.05.56.06.57.0Reduced pressure PR=z

Answered: 1.1 1.0 T= 2.00 0.9 TR=1.50 0.8 х х…

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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Give an example of Comparison of Z factors for various gases.

1.1
1.0
T= 2.00
0.9
TR=1.50
0.8
х
х
T=1.30
0.7
리호
0.6
TR= 1.20
of
so
904
0.5
TR= 1.10
0.4
Legend:
x Methane
o Ethylene
A Ethane
o Propane
Iso-pentane
о т-Неptane
A Nitrogen
• Carbon dioxide
TR=1.00
0.3
0.2
O n-Butane
Water
Average curve based on data on
hydrocarbons
0.1
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
Reduced pressure PR
=z

Expert Solution

Step 1

The gaseous state is the state of matter with least intermolecular force of attraction between molecules and highest kinetic energy. The standard properties of gases like Pressure, volume, and temperature can be calculated with the help of ideal gas equation.

Step 2

The ideal gas equation can be written as:

$P \times V = n \times R \times T$

P = Pressure

V = volume

n = moles

R = gas constant

T = temperature

Step 3

Under normal T and P, gases show deviation from ideal behavior and it can show by Vander waal equation. The compressibility factor for real gases can be written as:

$Z = \frac{P V}{R T}$

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