One mole of a monatomic ideal gas 1. One mole of a monatomic ideal gas is held initially at a pressure of 6 atm and 3 L. The gasundergoes isothermal expansion to 5 L followed by adiabatic expansion to 7 L. The gas is thenisothermally compressed to back to 4 L and adiabatically compressed back to 2 L.Complete the table belowа.P (atm)V (L)т (к)State334

Ideal gas is one which obeys Boyle's law, Charles's law, Gay lussac's law and Avogadro's law, i.e. which obeys PV=nRT equation.[where, P=pressure, V=volume, n=number of moles, R=universal gas constant, T =temperature Isothermal process is one where the temperature is constant during the process. Adiabatic process is one where the heat change during the process is zero. The given problem is represented as P=6 atm,V=3Ln=1 mole ideal gasT=?→Isothermal ExpansionP=?, V=5L, n=1mole→Adiabatic ExpansionV=7L, P=?, T=?→Isothermal CompressionV=4L, P=?, T=?→Adiabatic CompressionV=2L, P=?, T=? At initial stage, Pressure(P)=6 atm, Volume(V)=3L, Number of moles(n)=1 mole, Temperature(T)=? From the ideal gas equationPV=nRT, ...........(1)By repalacing the value of P, V, n, R6×3=1×0.083×T, R=constant=0.083L/atm.mol.Kor, T=6×31×0.083=216.87 K For isothermal expansion Given, initial pressure (P1)=6 atm, initial volume (V1)=3L, Initial temperature (T1)=216.87K Final volume (V2)=5L, We have to calculate, Final pressure(P2)=? and final temperature (T2)=? Since the process is isothermal process, therefore temperature remains constant during the process So final temperature (T2)=216.87K Now from Boyle's lawPV=constant ............(1)or, P1V1= P2V2By replacing the value of P1,V1and V2, we get6×3=P2×5or, 18=P2×5or, P2=185=3.6 atm For adiabatic expansion Given, initial pressure (P1)=3.6 atm, initial volume (V1)=5L, Initial temperature (T1)=216.87K Final volume (V2)=7L, We have to calculate, Final pressure(P2)=? and final temperature (T2)=? Since the process is adiabatic process, therefore there is no heat change during the process For adiabatic process, PVγ=constant.......(3)Where, γ=CPCV, CP= heat capacity at constant pressure, CV=heat capacity at constant volumeFor monoatomic gases, CP=52R, CV=32R, therefore, γ=(5/2)R(3/2)R=53=1.67Therefore from equation 3P1V1γ=P2V2γBy replacing the value of P1,V1, V2 and γ, we get3.6×(5)1.67=P2(7)1.67 or, P2=3.6×(57)1.67=2.05 atmAgaing in adabatic process, from the relation between T and VTVγ-1=constant........(4)or, T1V1γ-1=T2V2γ-1By replacing the value of T1, V1, V2 and γ216.87×(5)1.67-1=T2(7)1.67-1 or, T2=216.87×(57)1.67-1=173.09 K For isothermal compression Given, initial pressure (P1)=2.05 atm, initial volume (V1)=7L, Initial temperature (T1)=173.09k Final volume (V2)=4L, We have to calculate, Final pressure(P2)=? and final temperature (T2)=? Since the process is isothermal process, therefore temperature remains constant during the process So final temperature (T2)=173.09 K Now from Boyle's lawPV=constant or, P1V1= P2V2By replacing the value of P1,V1and V2, we get2.05×7=P2×4or, 14.35=P2×4or, P2=14.354=3.59 atm For adiabatic compression Given, initial pressure (P1)=3.59 atm, initial volume (V1)=4L, Initial temperature (T1)=173.09K Final volume (V2)=2L, We have to calculate, Final pressure(P2)=? and final temperature (T2)=? Since the process is adiabatic process, therefore there is no heat change during the process For adiabatic process, PVγ=constantWhere, γ=CPCV, CP= heat capacity at constant pressure, CV=heat capacity at constant volumeFor monoatomic gases, CP=52R, CV=32R, therefore, γ=(5/2)R(3/2)R=53=1.67Therefore P1V1γ=P2V2γBy replacing the value of P1,V1, V2 and γ, we get3.59×(4)1.67=P2(2)1.67 or, P2=3.59×(42)1.67=11.42 atmAgaing in adabatic process, from the relation between T and VTVγ-1=constantor, T1V1γ-1=T2V2γ-1By replacing the value of T1, V1, V2 and γ173.09×(4)1.67-1=T2(2)1.67-1 or, T2=173.09×(42)1.67-1=275.4 KBy putting the value of P,T and V calculated in each step in the given table we get final answer. Answer: The given table completed as, State P (atm) V (L) T (K) 1 6 3 216.87 2 3.6 5 216.87 3 2.05 7 173.09 4 3.59 4 173.09 5 11.42 2 275.4

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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

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