INTRO.TO CHEM.ENGR.THERMO.-EBOOK>I<
INTRO.TO CHEM.ENGR.THERMO.-EBOOK>I<
8th Edition
ISBN: 9781260940961
Author: SMITH
Publisher: INTER MCG
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Chapter 6, Problem 6.47P
Interpretation Introduction

Interpretation:

Determine the molar volume of the propane in the final state and enthalpy and entropy changes if propane gas is compressed from initial state to final.

Concept Introduction:

From Lee and Kesler generalized correlations, the molar volume of the propane in the final state, enthalpy and entropy changes are calculated as:

  V=ZRTP

Where, Z=Z0+ωZ1

                      ....(1)

And enthalpy and entropy changes are calculated as:

  ΔH=T1T2CPigdT+H2RH1R

Where, HR=(HR)0+ω(HR)1

                      ....(2)

And

  ΔS=T1T2CPigdTTRlnP2P1+S2RS1R

Where, SR=(SR)0+ω(SR)1

                      ....(3)

Expert Solution & Answer
Check Mark

Answer to Problem 6.47P

The molar volume is:

  V=184.2cm3mol

And

  ΔH=6723.42Jmol

  ΔS=15.905Jmol K

Explanation of Solution

Given information:

It is given that in the initial state, propane is assumed as an ideal gas. And at initial state

And final state pressure and temperature are given as, P1=1bar, T1=308.15K, P2=135bar,  T2=468.15K

For pure species propane, the properties can be written down using Appendix B, Table B.1

ω=0.152, Tc=369.8K, Pc=42.48bar, M=44.1gmmol .

Now For final state

  Tr=T2TcTr=468.15K369.8K=1.266

And

  Pr=PPcPr=135bar42.48bar=3.178

So, at above values of Tr and Pr, The values of Z0 and Z1 can be written from Appendix D

Tr=1.266 lies between reduced temperatures Tr=1.2 and Tr=1.3 and Pr=3.178 lies in between reduced pressures Pr=3.00 and Pr=5.00 .

At Tr=1.2 and Pr=3.00

Z0=0.5425, ( H R)RTC0=2.801, ( S R)R0=1.727

At Tr=1.3 and Pr=3.00

Z0=0.6344, ( H R)RTC0=2.274, ( S R)R0=1.299

At Tr=1.2 and Pr=5.00

Z0=0.7069, ( H R)RTC0=3.166, ( S R)R0=1.827

At Tr=1.3 and Pr=5.00

Z0=0.7358, ( H R)RTC0=2.825, ( S R)R0=1.554

And

At Tr=1.2 and Pr=3.00

Z1=0.1095, ( H R)RTC1=0.934, ( S R)R1=0.991

At Tr=1.3 and Pr=3.00

Z1=0.2079, ( H R)RTC1=0.300, ( S R)R1=0.481

At Tr=1.2 and Pr=5.00

Z1=0.0141, ( H R)RTC1=1.840, ( S R)R1=1.767

At Tr=1.3 and Pr=5.00

Z1=0.0875, ( H R)RTC1=1.066, ( S R)R1=1.147

Applying linear interpolation of two independent variables, From linear double interpolation, if M is the function of two independent variable X and Y then the value of quantity M at two independent variables X and Y adjacent to the given values are represented as follows:

  X1XX2Y1M1,1M1,2YM=?Y2M2,1M2,2

So,

  M=[( X 2 X X 2 X 1 )M1,1+( X X 1 X 2 X 1 )M1,2]Y2YY2Y1+[( X 2 X X 2 X 1 )M2,1+( X X 1 X 2 X 1 )M2,2]YY1Y2Y1Z0=[( 53.178 53)×0.5425+( 3.1783 53)×0.7069]×1.31.2661.31.2+[( 53.178 53)×0.6344+( 3.1783 53)×0.7358]×1.266-1.21.31.2Z0=0.614

  M=[( X 2 X X 2 X 1 )M1,1+( X X 1 X 2 X 1 )M1,2]Y2YY2Y1+[( X 2 X X 2 X 1 )M2,1+( X X 1 X 2 X 1 )M2,2]YY1Y2Y1( H R )RTC0=[( 53.178 53)×2.801+( 3.1783 53)×3.166]×1.31.2661.31.2+[( 53.178 53)×2.274+( 3.1783 53)×2.825]×1.266-1.21.31.2( H R )RTC0=2.497

  M=[( X 2 X X 2 X 1 )M1,1+( X X 1 X 2 X 1 )M1,2]Y2YY2Y1+[( X 2 X X 2 X 1 )M2,1+( X X 1 X 2 X 1 )M2,2]YY1Y2Y1( S R )R0=[( 53.178 53)×1.727+( 3.1783 53)×1.827]×1.31.2661.31.2+[( 53.178 53)×1.299+( 3.1783 53)×1.554]×1.266-1.21.31.2( S R )R0=1.462

And

  M=[( X 2 X X 2 X 1 )M1,1+( X X 1 X 2 X 1 )M1,2]Y2YY2Y1+[( X 2 X X 2 X 1 )M2,1+( X X 1 X 2 X 1 )M2,2]YY1Y2Y1Z1=[( 53.178 53)×0.1095+( 3.1783 53)×0.0141]×1.31.2661.31.2+[( 53.178 53)×0.2079+( 3.1783 53)×0.0875]×1.266-1.21.31.2Z1=0.164

  M=[( X 2 X X 2 X 1 )M1,1+( X X 1 X 2 X 1 )M1,2]Y2YY2Y1+[( X 2 X X 2 X 1 )M2,1+( X X 1 X 2 X 1 )M2,2]YY1Y2Y1( H R )RTC1=[( 53.178 53)×0.934+( 3.1783 53)×1.84]×1.31.2661.31.2+[( 53.178 53)×0.300+( 3.1783 53)×1.066]×1.266-1.21.31.2( H R )RTC1=0.588

  M=[( X 2 X X 2 X 1 )M1,1+( X X 1 X 2 X 1 )M1,2]Y2YY2Y1+[( X 2 X X 2 X 1 )M2,1+( X X 1 X 2 X 1 )M2,2]YY1Y2Y1( S R )R1=[( 53.178 53)×0.991+( 3.1783 53)×1.767]×1.31.2661.31.2+[( 53.178 53)×0.481+( 3.1783 53)×1.147]×1.266-1.21.31.2( S R )R1=0.717

Now from equation (1) at final state

  Z=Z0+ωZ1Z=0.614+0.152×0.164Z=0.639

And

  V=ZRTPV=0.639×83.14 cm 3barmol K×468.15K135barV=184.2cm3mol

From equation (2) at final state,

  HR=(HR)0+ω(HR)1

  ( HR )0RTC=2.497(HR)0=2.497×8.314Jmol K×369.8K(HR)0=7677.069Jmol

And

  ( H R )RTC1=0.588(HR)1=0.588×8.314Jmol K×369.8K(HR)1=1807.816Jmol

So,

  HR=(HR)0+ω(HR)1HR=7677.069Jmol+0.152×1807.816JmolHR=7951.86Jmol

  ΔH=T1T2CPigdT+H2RH1R

At final state H1R=0

  ΔH=T1T2CPigdT+H2R

  T0TΔCPRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)

Where τ=TT0

  τ=TT0=468.15308.15=1.519

Values of above constants for propane in above equation are given in appendix C table C.1 and noted down below:

  iA103B106C105Dpropane1.21328.7858.8240

  T0TΔ C PRdT=AT0(τ1)+B2T02(τ21)+C3T03(τ31)+DT0(τ1τ)T0TΔCPRdT=1.213×308.15×(1.5191)+28.785×1032308.152(1.51921)+8.824×1063×308.153(1.51931)+0×105308.15(1.51911.519)T0TΔCPRdT=1765.1286K

  ΔH=RT1T2 C P igRdT+H2RΔH=8.314JKmol×1765.1286K7951.86JmolΔH=6723.42Jmol

  ( SR )0R=1.462(SR)0=1.462×8.314Jmol K(SR)0=12.15Jmol K

And

  ( S R )R1=0.717(SR)1=8.314Jmol K×0.717(SR)1=5.96Jmol K

So,

  SR=(SR)0+ω(SR)1SR=12.15Jmol K+0.152×5.96Jmol KSR=13.06Jmol K

  ΔS=T1T2CPigdTTRlnP2P1+S2RS1R

At final state S1R=0

  ΔS=RT1T2CPigRdTTRlnP2P1+S2R

  T1T2CPigRdTT=[A+{BT0+(CT02+Dτ2T02)(τ+12)}(τ1lnτ)]×lnτ

  τ=TT0=468.15308.15=1.519

Values of above constants for propane in above equation are given in appendix C table C.1 and noted down below:

  iA103B106C105Dpropane1.21328.7858.8240

  T1T2 C P igRdTT=[A+{BT0+(CT02+D τ 2 T 0 2 )( τ+12)}(τ1lnτ)]×lnτT1T2CP igRdTT=[1.213+{28.785×103×308.15+(8.824×106×308.152+0× 10 5 1.519 2× 308.15 2)(1.519+12)}(1.5191ln1.519)]×ln1.519T1T2CPigRdTT=4.563

Hence,

  ΔS=RT1T2 C P igRdTTRln P 2 P 1+S2RΔS=8.314Jmol K×4.5638.314Jmol K×ln135bar1bar+(13.06Jmol K)ΔS=15.905Jmol K

Conclusion

The molar volume is:

  V=184.2cm3mol

And

  ΔH=6723.42Jmol

  ΔS=15.905Jmol K

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Chapter 6 Solutions

INTRO.TO CHEM.ENGR.THERMO.-EBOOK>I<

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