Physics for Scientists and Engineers: Foundations and Connections
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN: 9781133939146
Author: Katz, Debora M.
Publisher: Cengage Learning
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Chapter 22, Problem 73PQ

Figure P22.73 illustrates the cycle ABCA for a 2.00-mol sample of an ideal diatomic gas, where the process CA is a reversible isothermal expansion. What is a. the net work done by the gas during one cycle? b. How much energy is added to the gas by heat during one cycle? c. How much energy is exhausted from the gas by heat during one cycle? d. What is the efficiency of the cycle? e. What would be the efficiency of a Carnot engine operated between the temperatures at points A and B during each cycle?

Chapter 22, Problem 73PQ, Figure P22.73 illustrates the cycle ABCA for a 2.00-mol sample of an ideal diatomic gas, where the

(a)

Expert Solution
Check Mark
To determine

The net work done by the gas during one cycle.

Answer to Problem 73PQ

The net work done by the gas during one cycle is 411J.

Explanation of Solution

The work done by the gas along AB is WAB, has magnitude zero since the process is isochoric process.

Write the expression to calculate the work done by the gas along BC.

  WBC=P(VfVi)                                                                                                      (I)

Here, WBC is the work done along BC, P is the pressure of the gas, Vf is the final volume and Vi is the initial volume.

Write the expression to calculate the work done by the gas along CA.

  WCA=PViln(VfVi)                                                                                                   (II)

Here, WCA is the work done by the gas along CA, Vi is the initial volume and Vf is the final volume.

Write the expression to calculate the total work done by the gas.

  W=WAB+WBC+WCA                                                                                              (III)

Here, W is the total work done by the gas.

Conclusion:

Substitute 4.50atm for P, 3.00L for Vi and 1.00L for Vf in the above equation (I) to calculate WBC.

  WBC=(4.50atm(1.013×105Pa1atm))(3.00L(103m31L)1.00L(103m31L))=912J

Substitute 4.50atm for P, 1.00L for Vi and 3.00L for Vf in the above equation (II) to calculate WBC.

  WCA=(4.50atm(1.013×105Pa1atm))1.00L(103m31L)iln(3.00L1L)=501J

Substitute 0 for WAB, 912J for WBC and 501J for WCA in the above equation (IV) to calculate W.

  W=0+(912J)+501J=411J

Therefore, the net work done by the gas during one cycle is 411J.

(b)

Expert Solution
Check Mark
To determine

The heat added in one cycle.

Answer to Problem 73PQ

The heat added in one cycle is 2780J.

Explanation of Solution

For the diatomic gases, the specific heat capacity at constant volume is 52R and specific heat capacity at constant pressure is 72R. The heat absorbed during the isothermal process along CA is same as the work done the gas along the same path.

The temperature at A and C of the gas is same.

Write the expression to calculate the temperature of the gas at A and C.

  T=PVnR                                                                                                                     (V)

Here, T is the temperature of the gas at A and C, P is the pressure at A, V is the pressure at A n is the number of moles and R is the universal gas constant.

Write the expression to calculate the temperature at B.

  T=PVnR                                                                                                                (VI)

Here, T is he temperature at B, P is the pressure at B and V is the volume at B.

Write the expression to calculate the energy added to the system.

  Q=ncv(TT)+WCA

Here, Q is the energy added to the system, cv is the specific heat capacity at constant volume and cp is the specific heat capacity at constant pressure.

Substitute 52R for cv in the above equation to rewrite.

  Q=n(52R)(TT)+WCA                                                                     (VII)

Conclusion:

Substitute 1.50atm for P, 3.00L for V, 2.00mol for n and 8.314J/molK for in the equation (V) to calculate T.

    T=(1.50atm(1.013×105Pa1atm))3.00L(103m31L)(2.00mol)8.314J/molK=27.4K

Substitute 4.50atm for P, 3.00L for V, 2.00mol for n and 8.314J/molK for R in the equation (VI) to calculate T.

    T=(4.50atm(1.013×105Pa1atm))3.00L(103m31L)(2.00mol)8.314J/molK=82.2K

Substitute 2.00mol for n and 8.314J/molK for R, 27.4K for T, 82.2K for T and 501J for WCA in the above equation (VII) to calculate Q.

    Q=(2.00mol)(52(8.314J/molK))(82.2K27.4K)+501J2780J

Therefore, the heat added in one cycle is 2780J.

(c)

Expert Solution
Check Mark
To determine

The energy exhausted from the gas.

Answer to Problem 73PQ

The energy exhausted from the gas is 3.19×103J.

Explanation of Solution

For the diatomic gases, the specific heat capacity at constant pressure is 72R.

Write the expression to calculate the heat exhausted from the gas.

  |Q|=|ncp(TT)|

Here, Q is the heat exhausted from the gas.

Substitute 72R for cp in the above equation to rewrite.

  |Q|=|n(72R)(TT)|

Conclusion:

Substitute 2.00mol for n and 8.314J/molK for R, 27.4K for T and 82.2K for T in the above equation to calculate Q.

    |Q|=|(2.00mol)(72(8.314J/molK))(27.4K82.2K)|=3.19×103J

Therefore, the energy exhausted from the gas is 3.19×103J.

(d)

Expert Solution
Check Mark
To determine

The efficiency of the cycle.

Answer to Problem 73PQ

The efficiency of the cycle is 12.9%.

Explanation of Solution

Write the expression to calculate the efficiency of one cycle.

  e=(|W||Q|)×100%

Here, e is the efficiency.

Conclusion:

Substitute 3.19×103J for Q and 411J for W in the above equation to calculate e.

  e=(|411J||3.19×103J|)×100%=12.9%

Therefore, the efficiency of the cycle is 12.9%.

(e)

Expert Solution
Check Mark
To determine

The efficiency of the Carnot engine.

Answer to Problem 73PQ

The efficiency of the Carnot engine is 66.7%.

Explanation of Solution

Write the expression to calculate the efficiency of the Carnot engine.

  e=(1TT)×100%

Here, e is the efficiency of the Carnot engine.

Conclusion:

Substitute 27.4K for T and 82.2K for T in the above equation to calculate e.

  e=(127.4K82.2K)×100%=66.7%

Therefore, the efficiency of the Carnot engine is 66.7%.

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Chapter 22 Solutions

Physics for Scientists and Engineers: Foundations and Connections

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