CENGEL'S 9TH EDITION OF THERMODYNAMICS:
CENGEL'S 9TH EDITION OF THERMODYNAMICS:
9th Edition
ISBN: 9781260917055
Author: CENGEL
Publisher: MCG
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Chapter 13.3, Problem 44P

(a)

To determine

The volume of the mixture using the ideal gas mixture.

(a)

Expert Solution
Check Mark

Answer to Problem 44P

The volume of the mixture using the ideal gas mixture is 0.8m3_.

Explanation of Solution

Write the equation to calculate the mole number of oxygen and nitrogen gas using an ideal gas equation of state.

NO2=(PVRuT)O2 (I)

NN2=(PVRuT)N2 (II)

Here, pressure and temperature of oxygen and nitrogen gas is PO2,PN2, TO2,TN2, volume of oxygen and nitrogen gas is VO2,VN2, universal gas constant of oxygen and nitrogen gas is Ru,O2andRu,N2 respectively.

Calculate the mole number of the mixture.

Nm=NO2+NN2 (III)

Calculate the volume of the mixture.

Vm=NmRuTmPm (IV)

Here, pressure and temperature of the mixture is Pm and Tm.

Conclusion:

Substitute 8.314kPam3kmolK for Ru, 200 K for TO2, 0.3m3 for VO2, and 8 MPa for PO2 in Equation (I).

NO2=(8MPa)(0.3m3)(8.314kPam3kmolK)200K=(8MPa×1000kPa1MPa)(0.3m3)(8.314kPam3kmolK)200K=1.443kmol

Substitute 8.314kPam3kmolK for Ru, 200 K for TN2, 0.5m3 for VN2, and 8 MPa for PN2 in Equation (II).

NN2=(8MPa)(0.5m3)(8.314kPam3kmolK)200K=(8MPa×1000kPa1MPa)(0.5m3)(8.314kPam3kmolK)200K=2.406kmol

Substitute 1.443 kmol for NO2 and 2.406 kmol for NN2 in Equation (III).

Nm=1.443kmol+2.406kmol=3.849kmol

Substitute 3.849 kmol for Nm, 8.314kPam3kmolK for Ru, 200 K for Tm, and 8 MPa for Pm in Equation (IV).

Vm=(3.849kmol)(8.314kPam3kmolK)(200K)8MPa=(3.849kmol)(8.314kPam3kmolK)(200K)8MPa×1000kPa1MPa=0.8m3

Thus, the volume of the mixture using the ideal gas mixture is 0.8m3_.

(b)

To determine

The volume of the mixture using Kay’s rule.

(b)

Expert Solution
Check Mark

Answer to Problem 44P

The volume of the mixture using Kay’s rule is 0.79m3_.

Explanation of Solution

Write the equation of pseudo-reduced temperatures and pressures of O2,andN2.

TR,O2=TmTcr,O2 (V)

PR,O2=PmPcr,O2 (VI)

TR,N2=TmTcr,N2 (VII)

PR,N2=PmPcr,N2 (VIII)

Here, the critical temperature of O2 is Tcr,O2, mixture temperature is Tm, mixture pressure is Pm, critical temperature of N2 is Tcr,N2, critical pressure of N2 is Pcr,N2, and critical pressure of O2 is Pcr,O2.

Write the equation to calculate the mole number of oxygen and nitrogen gas using the kay’s rule.

NO2=(PVZRuT)O2 (IX)

NN2=(PVZRuT)N2 (X)

Here, compressibility factor of O2andN2 are ZO2 and ZN2.

Calculate the mole fraction of O2andN2.

yO2=NO2Nm (XI)

yN2=NN2Nm (XII)

Write the critical temperature of a gas mixture.

Tcr,m=yiTcr,i=yO2Tcr,O2+yN2Tcr,N2 (XIII)

Write the critical pressure of a gas mixture.

Pcr,m=yiPcr,i=yO2Pcr,O2+yN2Pcr,N2 (XIV)

Write the equation of reduced temperature and pressure.

TR=TmTcr,m (XV)

PR=PmPcr,m (XVI)

Calculate the volume of the mixture.

Vm=ZmNmRuTmPm (XVII)

Conclusion:

Refer to Table A-1, obtain the critical temperatures and pressures of O2,andN2.

Tcr,O2=154.8KPcr,O2=5.08MPaTcr,N2=126.2KPcr,N2=3.39MPa

Substitute 200 K for Tm, 8 MPa for Pm and obtained values from Table A-1 in Equations (V) to (VIII).

TR,O2=1.292PR,O2=1.575TR,N2=1.585PR,N2=2.360

Refer to Figure A-15, obtain the compressibility factor for O2,andN2 by reading the above values of reduced temperatures and reduced pressure.

ZO2=0.77ZN2=0.863

Substitute 0.77 for ZO2, 8.314kPam3kmolK for Ru, 200 K for TO2, 0.3m3 for VO2, and 8 MPa for PO2 in Equation (IX).

NO2=(8MPa)(0.3m3)0.77(8.314kPam3kmolK)200K=(8MPa×1000kPa1MPa)(0.3m3)0.77(8.314kPam3kmolK)200K=1.874kmol

Substitute 0.863 for ZN2, 8.314kPam3kmolK for Ru, 200 K for TN2, 0.5m3 for VN2, and 8 MPa for PN2 in Equation (X).

NN2=(8MPa)(0.5m3)0.863(8.314kPam3kmolK)200K=(8MPa×1000kPa1MPa)(0.5m3)0.863(8.314kPam3kmolK)200K=2.787kmol

Substitute 1.874 kmol for NO2 and 2.787 kmol for NN2 in Equation (III).

Nm=1.874kmol+2.787kmol=4.661kmol

Substitute 1.874 kmol for NO2, 4.661 kmol for Nm, and 2.787 kmol for NN2 in Equations (XI) and (XII).

yO2=0.402yN2=0.598

Substitute 0.402 for yO2, 154.8 K for Tcr,O2, 0.598 for yN2, and 126.2 K for Tcr,N2 in Equation (XIII).

Tcr,m=(0.402)(154.8K)+(0.598)(126.2K)=137.7K

Substitute 0.402 for yN2, 5.08 MPa for Pcr,O2, 0.598 for yN2, and 3.39 MPa for Pcr,N2 in Equation (XIV).

Pcr,m=(0.402)(5.08MPa)+(0.598)(3.39MPa)=4.07MPa

Substitute 200 K for Tm, 8000 kPa for Pm, 137.7 K for Tcr,m, and 4.07 MPa for Pcr,m in Equations (XV) and (XVI).

TR=1.452PR=1.966

Refer to Figure A-15, obtain the compressibility factor of the mixture by reading the pseudo reduced temperature and pseudo reduced pressure of 1.452 and 1.966.

Zm=0.82

Substitute 0.82 for Zm, 4.661 kmol for Nm, 8.314kPam3kmolK for Ru, 200 K for Tm, and 8 MPa for P in Equation (XVII).

Vm=(0.82)(4.661kmol)(8.314kPam3kmolK)(200K)(8MPa)=(0.82)(4.661kmol)(8.314kPam3kmolK)(200K)(8MPa×1000kPa1MPa)=0.79m3

Thus, the volume of the mixture using Kay’s rule is 0.79m3_.

(c)

To determine

The volume of the mixture using the compressibility chart and Amagat’s law.

(c)

Expert Solution
Check Mark

Answer to Problem 44P

The volume of the mixture using the compressibility chart and Amagat’s law is 0.80m3_.

Explanation of Solution

Write the equation of compressibility factor of the mixture.

Zm=yiZi=yO2ZO2+yN2ZN2 (XVIII)

Here, the mole fraction of O2andN2 are yO2andyN2 respectively.

Calculate the volume of the mixture.

Vm=ZmNmRuTmPm (XIX)

Conclusion:

Substitute 0.402 for yO2, 0.77 for ZO2, 0.598 for yN2, and 0.863 for ZN2, in Equation (XVIII).

Zm=(0.402)(0.77)+(0.598)(0.863)=0.83

Substitute 0.863 for Zm, 4.661 kmol for Nm, 8.314kPam3kmolK for Ru, 200 K for Tm, and 8 MPa for P in Equation (XIX).

Vm=(0.863)(4.661kmol)(8.314kPam3kmolK)(200K)(8MPa)=(0.863)(4.661kmol)(8.314kPam3kmolK)(200K)(8MPa×1000kPa1MPa)=0.80m3

Thus, the volume of the mixture using the compressibility chart and Amagat’s law is 0.80m3_.

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Chapter 13 Solutions

CENGEL'S 9TH EDITION OF THERMODYNAMICS:

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