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Chapter 18, Problem 61P

(a)

To determine

Pressure and volume at given points.

(a)

Expert Solution
Check Mark

Answer to Problem 61P

Volume at A, B, C, D are 1.97×103m3_, 11.9×103m3_, 32.8×103m3_ and 5.44×103m3_ respectively

Pressure at B and D are 4.14atm_ and 6.03atm_ respectively.

Explanation of Solution

Write the ideal gas equation and rearrange for VA and VC to determine the volume at point A and C.

    PAVA=nRTAVA=nRTAPAPAVC=nRTCVC=nRTCPC        (I)

Here PA, and PC is the pressure at A and C respectively, R is the real gas constant, n is the number of moles, VA and VC is the volume at A and C respectively, TA and TC is the temperature at A and C respectively.

Bundle: Principles of Physics: A Calculus-Based Text, 5th + WebAssign Printed Access Card for Serway/Jewett's Principles of Physics: A Calculus-Based Text, 5th Edition, Multi-Term, Chapter 18, Problem 61P

From figure, AB is isothermal.

Therefore.

    PAVA=PBVB        (II)

Here PB is the pressure at B and VB is the volume at B.

BC is adiabatic.

Therefore,

    PCVCγ=PBVBγ        (III)

Here γ is the specific heat ratio.

Rearrange (II) in terms of PB and substitute in (III)

    (PAVAVB)VBγ=PCVCγ        (IV)

Rearrange (IV) in terms of VB

    VB=(PCPA)(VCγVA)1γ1        (V)

Similarly,

CD is isothermal.

    PCVC=PDVD        (VI)

Here PD is the pressure at D and VD is the volume at D.

AD is adiabatic.

    PAVAγ=PDVDγ        (VII)

Substitute for PD from (VI) in (VII)

    PAVAγ=(PCVCVD)VDγ        (VIII)

Rearrange (VIII) in terms of VD

    VD=(PAPC)(VAγVC)1γ1        (IX)

Rearrange (II) and (VI) in terms of PB and PD

    PB=PAVAVBPD=PCVCVD        (X)

Conclusion:

Substitute 1mol for n, 8.314J/molK for R, 600K for TA, 400K for TC, 25atm for PA and 1atm for PC in (I)

    VA=(1mol)(8.314J/molK)(600K)25atm(1.013×105Pa)=1.97×103m

    VC=(1mol)(8.314J/molK)(400K)1atm(1.013×105Pa)=32.8×103m

Substitute 1atm for PC, 25atm for PA, 32.8×103m for VC, 1.97×103m for VA and 1.4 for γ in (V)

    VB=(1atm25atm)((32.8×103m)1.41.97×103m)11.41=11.9×103m3

Substitute 1atm for PC, 25atm for PA, 32.8×103m for VC, 1.97×103m for VA and 1.4 for γ in (IX)

    VD=(25atm1atm)((1.97×103m)1.432.8×103m)11.41=5.44×103m3

Substitute 1.97×103m for VA, 11.9×103m3 for VB, 32.8×103m for VC, 5.44×103m3 for VD, 1atm for PC and 25atm for PA in (X)

    PB=(1.97×103m)(25atm)(11.9×103m3)=4.14atmPD=(32.8×103m)(1atm)(5.44×103m3)=6.03atm

Volume at A, B, C, D are 1.97×103m3_, 11.9×103m3_, 32.8×103m3_ and 5.44×103m3_ respectively

Pressure at B and D are 4.14atm_ and 6.03atm_ respectively.

(b)

To determine

Net work done per cycle.

(b)

Expert Solution
Check Mark

Answer to Problem 61P

Net work done is 2.99kJ_.

Explanation of Solution

Write the first law of thermodynamics

    ΔE=Qh+WAB        (XI)

Here ΔE is the change in internal energy, Qh is the amount of heat transfer and WAB is the work done.

As the change in internal energy is zero, (XI) becomes,

    Qh=WAB

Write the equation for work done

    |Qh|=nRThln(VBVA)        (XII)

Here Th is the temperature of the hot reservoir.

Write the equation for efficiency

    e=Weng|Qh|        (XIII)

Here e is the efficiency of the engine and Weng is the work done by the engine.

Rewrite (XIIII) in terms of Weng and substitute for Qh from (XII)

    Weng=enRThln(VBVA)        (XIV)

Conclusion:

Substitute 1.97×103m for VA, 11.9×103m3 for VB, 1mol for n, 8.314J/molK for R, 600K for Th and 0.33 for e in (XIV)

    Weng=(0.33)(1mol)(8.314J/molK)(600K)ln(11.9×103m31.97×103m)=2.99kJ

Net work done is 2.99kJ_.

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Chapter 18 Solutions

Bundle: Principles of Physics: A Calculus-Based Text, 5th + WebAssign Printed Access Card for Serway/Jewett's Principles of Physics: A Calculus-Based Text, 5th Edition, Multi-Term

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