Physics for Scientists and Engineers with Modern Physics, Technology Update
Physics for Scientists and Engineers with Modern Physics, Technology Update
9th Edition
ISBN: 9781305401969
Author: SERWAY, Raymond A.; Jewett, John W.
Publisher: Cengage Learning
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Chapter 20, Problem 34P
To determine

The difference in internal energy Eint,BEint,A .

Expert Solution & Answer
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Answer to Problem 34P

The difference in internal energy Eint,BEint,A is 4.29×104 J .

Explanation of Solution

Since the gas goes through a cyclic process, the net change in internal energy of the gas must be zero.

Write the equation for the net change in internal energy of the gas.

  ΔEint=0                                                                                                                   (I)

Here, ΔEint is the net change in internal energy of the gas after one complete cycle.

The net change in internal energy must be equal to the sum of the change in internal energy of the gas during the different processes in the cycle.

Write the expression for ΔEint .

  ΔEint=ΔEint,AB+ΔEint,BC+ΔEint,CD+ΔEint,DA

Here, ΔEint,AB is the change in internal energy of the gas during the process AB, ΔEint,BC is the change in internal energy of the gas during the process BC, ΔEint,CD is the change in internal energy of the gas during the process CD and ΔEint,DA is the change in internal energy of the gas during the process DA.

Put the above equation in equation (I) and rewrite it for ΔEint,AB .

  ΔEint,AB+ΔEint,BC+ΔEint,CD+ΔEint,DA=0ΔEint,AB=ΔEint,BCΔEint,CDΔEint,DA                    (II)

Write the expression for ΔEint,AB .

  ΔEint,AB=Eint,BEint,A

Here, Eint,B is the internal energy of the gas at the point B and Eint,A is the internal energy of the gas at the point A.

Put the above equation in equation (II).

  Eint,BEint,A=ΔEint,BCΔEint,CDΔEint,DA                                                        (III)

The process BC is an isochoric process. The work done in an isochoric process is zero. This implies W is 0 for the process BC.

Write the first law of thermodynamics.

  ΔEint=Q+W                                                                                                          (IV)

Here, ΔEint is the change in internal energy of the gas, Q is the heat added to the gas and W is the work done on the gas.

Use equation (IV) to write the expression for ΔEint,BC .

  ΔEint,BC=QBC+WBC                                                                                                (V)

Here, QBC is the heat added to the gas during the process BC and WBC is the work done on the gas during the process BC.

The process BC is isobaric.

Write the expression for WBC .

  WBC=PBΔVBC

Here, PB is the pressure of the gas at the point B and ΔVBC is the change in volume of the gas during the process BC.

Put the above equation in equation (V).

  ΔEint,BC=QBC+(PBΔVBC)=QBCPBΔVBC                                                                                    (VI)

The process CD is isothermal. The change in internal energy for an isothermal process is zero.

Write the expression for ΔEint,CD .

    ΔEint,CD=0                                                                                                           (VII)

Use equation (IV) to write the expression for ΔEint,DA .

  ΔEint,DA=QDA+WDA                                                                                           (VIII)

Here, QDA is the heat added to the gas during the process DA and WDA is the work done on the gas during the process DA.

The process DA is isobaric.

Write the expression for WDA .

  WDA=PDΔVDA

Here, PD is the pressure of the gas at the point D and ΔVDA is the change in volume of the gas during the process DA.

Put the above equation in equation (VIII).

  ΔEint,DA=QDA+(PDΔVDA)=QDAPDΔVDA                                                                                    (IX)

Put equations (VI), (VII) and (IX) in equation (III).

  Eint,BEint,A=(QBCPBΔVBC)0(QDAPDΔVDA)=QBC+PBΔVBCQDA+PDΔVDA=(QBC+QDA)+PBΔVBC+PDΔVDA                                         (X)

Write the equation for ΔVBC .

  ΔVBC=VCVB                                                                                                    (XI)

Here, VC is the volume of the gas at the point C and VB is the volume of the gas at the point B.

Write the equation for ΔVDA .

  ΔVDA=VAVD                                                                                                       (XII)

Here, VA is the volume of the gas at the point A and VD is the volume of the gas at the point D.

Conclusion:

It is given that the heat entering the system from B to C is 345 kJ and the heat leaving the system from D to A is 371 kJ . From figure P20.34, the volume of the gas at A is 0.2 m3 , volume at B is 0.09 m3 , the volume at C is 0.4 m3 and the volume at D is 1.2 m3 , the pressure at the point B is 3 atm and the pressure at the point D is 1 atm .

Substitute 0.4 m3 for VC and 0.09 m3 for VB in equation (XI) to find ΔVBC .

  ΔVBC=0.4 m30.09 m3=0.31 m3

Substitute 0.2 m3 for VA and 1.2 m3 for VD in equation (XII) to find ΔVDA .

  ΔVDA=0.2 m31.2 m3=1.00 m3

Substitute 345 kJ for QBC , 371 kJ for QDA , 3 atm for PB , 0.31 m3 for ΔVBC , 1 atm for PD and 1.00 m3 for ΔVDA in equation (X) to find Eint,BEint,A .

Eint,BEint,A=(345 kJ1000 J1 kJ+(371 kJ1000 J1 kJ))+(3 atm1.013×105 Pa1 atm)(0.31 m3)+(1 atm1.013×105 Pa1 atm)(1.00 m3)=4.29×104 J

Therefore, the difference in internal energy Eint,BEint,A is 4.29×104 J .

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