EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
16th Edition
ISBN: 8220100546716
Author: Katz
Publisher: CENGAGE L
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Chapter 22, Problem 41PQ

(a)

To determine

The work done on the gas.

(a)

Expert Solution
Check Mark

Answer to Problem 41PQ

The work done on the gas is 4.40×104J.

Explanation of Solution

Write the expression for the work done for an isobaric process.

    W=P(VfVi)

Here, W is the work done on the gas, P is the pressure, Vf is the final volume, and Vi is the intial volume.

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 41PQ

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 41PQ

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

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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