Virial Equations of State express the compressibility (z) of a gas as a series e
ID: 516615 • Letter: V
Question
Virial Equations of State express the compressibility (z) of a gas as a series expansion in volume or pressure. Oxidation of metals is frequently carried out under high pressure. Using Virial EoS up to the third virial coefficients:
a) Calculate the volumetric flow of oxygen fed to a reactor at 50 atm, 298 K, with a molar flow rate of 10 mol/s. The second (B) and third (C) virial coefficients at this temperature are -16.1 x 10 ^-6 m^3/mol and 1.2 x 10 ^ -9 m^6/mol^2.
b) What is the percentage error if the ideal gas law is used and the virial equation truncated after the second term is used?
Problem 1 sibility (z) of a gas as a series expansion in volume or pressure. oxidation of metals is frequently carried out under high pressure. Using Virial EOS up to the third virial coefficients: 298 K, with a a) Calculate the volumetric flow rate of oxygen fed to a reactor at 50 atm, molar rate of m The second (B) and third (C) virial coefficients at this flow 10 temperature are -16.1 x10% m /mol and 1.2 x 10 m /mol b) hat is the percentage error if the ideal gas law is used and the virial equation truncated after the second term is used?Explanation / Answer
Virial EOS Can be written as PV/nRT= 1 +B*n/V + C*n2/v2
Given P= 50 atm, n= 10 mol/s, R= 0.0821 L.atm/mole.se B= -16.1*10-6 m3/mol= -16.1*10-6*1000 L/mol=16.1*10-3 L/mol, C= 1.2*10-9 m6/mol2= 1.2*10-9*106 L2/mol2=1.2*10-3 L2/mole2
Hence Virial equation of state becomes
50*V/(10*0.0821*298)= 1-16.1*10-3*10/V +1.2*10-3*100/V2,
0.204V = 1- 0.161/V +0.12/V2
When the equation is solved using excel, x= 4.762 L/s
Based on ideal gas equation, V= nRT/P= 10*0.0821*298/50=4.89316 L/s
Percentage error = 100*{(4.762-4.89316)/4.762}=-2.7543%
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