EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
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Chapter 18, Problem 77P

(a)

To determine

The P-V diagram.

(a)

Expert Solution
Check Mark

Answer to Problem 77P

The P-V diagram is shown in figure 1.

Explanation of Solution

Calculation:

Consider the given data, the pressure versus volume graph for the process is shown below.

  EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 18, Problem 77P

Figure 1

Conclusion:

Therefore, the P-V diagram is shown in figure 1.

(b)

To determine

The proof that Qh=nRThln(V2V1) .

(b)

Expert Solution
Check Mark

Answer to Problem 77P

It is proved that Qh=nRThln(V2V1) .

Explanation of Solution

Formula used:

The expression for heat absorbed is given by,

  Qh=ΔEint,12+W12

The expression for work done is given by,

  W12=nRThln(V2V1)

Calculation:

The heat absorbed is calculated as,

  Qh=ΔEint,12+W12=0+nRThln( V 2 V 1 )=nRThln( V 2 V 1 )

Conclusion:

Therefore, it is proved that Qh=nRThln(V2V1) .

(c)

To determine

The proof that Qc=nRTcln(V3V4) .

(c)

Expert Solution
Check Mark

Answer to Problem 77P

It is proved that Qc=nRTcln(V3V4) .

Explanation of Solution

Formula used:

The expression for heat absorbed is given by,

  Qc=ΔEint,34W34

The expression for work done is given by,

  W34=nRTcln(V3V4)

Calculation:

The heat absorbed is calculated as,

  Qc=ΔEint,34W34=0nRTcln( V 3 V 4 )=nRTcln( V 3 V 4 )

Conclusion:

Therefore, it is proved that Qc=nRTcln(V3V4) .

(d)

To determine

The proof that V2V1=V3V4 .

(d)

Expert Solution
Check Mark

Answer to Problem 77P

It is proved that V2V1=V3V4 .

Explanation of Solution

Formula used:

The expression for quasistatic adiabatic process at point 4 is given by,

  TcV4γ1=ThV1γ1

The expression for quasistatic adiabatic process at point 2 and 3 is given by,

  ThV2γ1=TcV3γ1

Calculation:

The expression for quasistatic adiabatic process at point 4 is given by,

  TcV4γ1=ThV1γ1V1V4=( T c T h )1 γ1

The expression for quasistatic adiabatic process at point 2 and 3 is given by,

  ThV2γ1=TcV3γ1V2V3=( T c T h )1 γ1=V1V4

Conclusion:

Therefore, it is proved that V2V1=V3V4 .

(e)

To determine

The proof that efficiency is 1QcQh .

(e)

Expert Solution
Check Mark

Answer to Problem 77P

It is proved that efficiency is 1QcQh .

Explanation of Solution

Formula used:

The expression for efficiency is given by,

  ε=WonQh

The expression for work done is given by,

  Won=(QhQc)

Calculation:

The efficiency is calculated as,

  ε=( ( Q h Q c ))Qh=1QcQh

Conclusion:

Therefore, it is proved that efficiency is 1QcQh .

(f)

To determine

The proof that QcQh=TcTh .

(f)

Expert Solution
Check Mark

Answer to Problem 77P

It is proved that QcQh=TcTh .

Explanation of Solution

Formula used:

The expression for heat is given by,

  Qh=nRThlnV2V1

Calculation:

The ratio of Qc and Qh is calculated as,

  QcQh=nRTcln V 3 V 4 nRThln V 2 V 1 =TcTh( V 3 V 4 = V 2 V 1 )

Conclusion:

Therefore, it is proved that QcQh=TcTh .

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Chapter 18 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

Ch. 18 - Prob. 11PCh. 18 - Prob. 12PCh. 18 - Prob. 13PCh. 18 - Prob. 14PCh. 18 - Prob. 15PCh. 18 - Prob. 16PCh. 18 - Prob. 17PCh. 18 - Prob. 18PCh. 18 - Prob. 19PCh. 18 - Prob. 20PCh. 18 - Prob. 21PCh. 18 - Prob. 22PCh. 18 - Prob. 23PCh. 18 - Prob. 24PCh. 18 - Prob. 25PCh. 18 - Prob. 26PCh. 18 - Prob. 27PCh. 18 - Prob. 28PCh. 18 - Prob. 29PCh. 18 - Prob. 30PCh. 18 - Prob. 31PCh. 18 - Prob. 32PCh. 18 - Prob. 33PCh. 18 - Prob. 34PCh. 18 - Prob. 35PCh. 18 - Prob. 36PCh. 18 - Prob. 37PCh. 18 - Prob. 38PCh. 18 - Prob. 39PCh. 18 - Prob. 40PCh. 18 - Prob. 41PCh. 18 - Prob. 42PCh. 18 - Prob. 43PCh. 18 - Prob. 44PCh. 18 - Prob. 45PCh. 18 - Prob. 46PCh. 18 - Prob. 47PCh. 18 - Prob. 48PCh. 18 - Prob. 49PCh. 18 - Prob. 50PCh. 18 - Prob. 51PCh. 18 - Prob. 52PCh. 18 - Prob. 53PCh. 18 - Prob. 54PCh. 18 - Prob. 55PCh. 18 - Prob. 56PCh. 18 - Prob. 57PCh. 18 - Prob. 58PCh. 18 - Prob. 59PCh. 18 - Prob. 60PCh. 18 - Prob. 61PCh. 18 - Prob. 62PCh. 18 - Prob. 63PCh. 18 - Prob. 64PCh. 18 - Prob. 65PCh. 18 - Prob. 66PCh. 18 - Prob. 67PCh. 18 - Prob. 68PCh. 18 - Prob. 69PCh. 18 - Prob. 70PCh. 18 - Prob. 71PCh. 18 - Prob. 72PCh. 18 - Prob. 73PCh. 18 - Prob. 74PCh. 18 - Prob. 75PCh. 18 - Prob. 76PCh. 18 - Prob. 77PCh. 18 - Prob. 78PCh. 18 - Prob. 79PCh. 18 - Prob. 80PCh. 18 - Prob. 81PCh. 18 - Prob. 82PCh. 18 - Prob. 83PCh. 18 - Prob. 84PCh. 18 - Prob. 85PCh. 18 - Prob. 86PCh. 18 - Prob. 87PCh. 18 - Prob. 88PCh. 18 - Prob. 89PCh. 18 - Prob. 90PCh. 18 - Prob. 91PCh. 18 - Prob. 92PCh. 18 - Prob. 93PCh. 18 - Prob. 94PCh. 18 - Prob. 95PCh. 18 - Prob. 96PCh. 18 - Prob. 97PCh. 18 - Prob. 98P
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