A dry gas reservoir is initially at an average pressure of 4000 psia and temperature of 160°F. The gas has an apparent molecular weight of 18.85 Ib'mol. Assume the volume occupied by the gas in the reservoir remains constant. If the rescrvoir originally contained IMM Cuft of reservoir gas, how much gas has been produced at a final rescrvoir pressure of 500 psia? Given:Initial pressure Pi =4000 psiaFinal pressure Pf =500 psiatemperature Ti =160 oF =619.67ORInitial volume Vi =final volume Vf =1 MM ft3apparent molecular weight = 18.85 lb/molR=10.7315 psia ft3lb. mol. ORVolume of reservoir is constantPV = NRTPV = wgMgRTwg = PVMGRT=4000x1012x18.8510.7315x619.67= 11.34x1012 lbPV = nRT PV = wgMgRTwg = PVMGRT wg = 500x1012x18.8510.7315x619.67 wg = 1.42x1012 lbTotal gas has been produced at final reservoir pressure of 500psia is== 11.34x1012 - 1.42x1012=9.92x1012 lb

The gas law is applied to find out the amount of gas if pressure, volume and temperature is known.…

Answered: A dry gas reservoir is initially at an…

Introduction to Chemical Engineering Thermodynamics

8th Edition

ISBN:9781259696527

Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Chapter1: Introduction

Section: Chapter Questions

Problem 1.1P

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A dry gas reservoir is initially at an average pressure of 4000 psia and temperature of 160°F. The gas has an apparent molecular weight of 18.85 Ib'mol. Assume the volume occupied by the gas in the reservoir remains constant. If the rescrvoir originally contained IMM Cuft of reservoir gas, how much gas has been produced at a final rescrvoir pressure of 500 psia?

Given:Initial pressure Pi =4000 psiaFinal pressure Pf =500 psiatemperature Ti =160 oF =619.67
ORInitial volume Vi =final volume Vf =1 MM ft3apparent molecular weight = 18.85 lb/mol
R=10.7315 psia ft3lb. mol. OR
Volume of reservoir is constant
PV = NRTPV = wgMgRTwg = PVMGRT=4000x1012x18.8510.7315x619.67= 11.34x1012 lb
PV = nRT PV = wgMgRTwg = PVMGRT wg = 500x1012x18.8510.7315x619.67 wg = 1.42x1012 lb
Total gas has been produced at final reservoir pressure of 500psia is
== 11.34x1012 - 1.42x1012=9.92x1012 lb

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