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Chapter 22, Problem 42PQ

(a)

To determine

The change in entropy of the cold reservoir.

(a)

Expert Solution
Check Mark

Answer to Problem 42PQ

The change in entropy of the cold reservoir is 64.2J/K.

Explanation of Solution

Write the expression for the change in entropy.

    ΔS=ΔQT

Here, ΔS is the change in entropy, ΔQ is the energy transferred, and T is the temperature.

Conclusion:

Substitute 2.35atm for P, 0.340m3 for Vf , and 0.525m3 for Vi to find W.

    W=(2.35atm)(0.340m30.525m3)=(2.35atm)(1.01×105Pa1atm)(0.340m30.525m3)=4.40×104J

Therefore, the work done on the gas is 4.40×104J.

(b)

To determine

The heat transferred when a monoatomic carbon gas undergoes a constant pressure process.

(b)

Expert Solution
Check Mark

Answer to Problem 42PQ

The heat transferred when a monoatomic carbon gas undergoes a constant pressure process

is 1.10×105J

Explanation of Solution

Write the expression initial temperature from ideal gas equation.

    Ti=PVinR                                                                                                                      (I)

Here, Ti is the initial temperature, n is the number of moles, and R is the ideal gas constant.

Write the expression final temperature from ideal gas equation.

    Tf=PVfnR                                                                                                                   (II)

Here, Tf is the final temperature.

Write the expression for the heat transfer.

    QP=nCP(TfTi)                                                                                                   (III)

Here, QP is the amount of heat transfer and CP is the specific heat of the monoatomic gas at constant pressure.

For monoatomic gas,

    CP=52R                                                                                                                (IV)

Use (I), (II),(IV) to rewrite (III).

    QP=n(52R)(PVfnRPVinR)=52P(VfVi)                                                                                   (V)

Conclusion:

Substitute 2.35atm for P, 0.340m3 for Vf , and 0.525m3 for Vi to find W.

    W=(52)(2.35atm)(0.340m30.525m3)=(52)(2.35atm)(1.01×105Pa1atm)(0.340m30.525m3)=1.10×105J

Therefore, the heat transferred when a monoatomic carbon gas undergoes a constant pressure process is 1.10×105J.

(c)

To determine

The change in entropy.

(c)

Expert Solution
Check Mark

Answer to Problem 42PQ

The change in entropy is 271J/K

Explanation of Solution

Write the expression for change in entropy.

    ΔS=ifn(52R)dTT=n(52R)ifdTT=n(52R)[ln(T)]if=n(52R)ln(Tf)ln(Ti)=n(52R)ln(TfTi)                                                                                  (VI)

Here, ΔS is the change in entropy.

Use (I) and (II) to rewrite (VI).

    ΔS=n(52R)ln(PVfnRPVinR)=n(52R)ln(VfVi)                                                                                        (VII)

Conclusion:

Substitute 30.0mol for n , 8.314JmolK for R , 0.340m3 for Vf , and 0.525m3 for Vi  in (VII) to find QP.

    QP=(30.0mol)(52)(8.314JmolK)ln(0.340m30.525m3)=270.9J/K271J/K

Therefore, the coefficient of performance of an ideal heat pump is 271J/K

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Chapter 22 Solutions

Bundle: Physics For Scientists And Engineers: Foundations And Connections, Volume 1, Loose-leaf Version + Webassign Printed Access Card For Katz's ... And Connections, Single-term Courses

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