The work done by engine during the 4 steps and the net work done.

Question

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Answer 1

Question

Chapter 15, Problem 86P

(a)

To determine

The work done by engine during the 4 steps and the net work done.

(a)

Expert Solution

Answer to Problem 86P

The work done by engine during the 4 steps are $692 J$ , $0 J$ , $−506 J$ , $0 J$ and the net work done is $186 J$ .

Explanation of Solution

Write the expression for work done during process A.

$WA=nRTln(VfVi)$

Here, T is the temperature, $WA$ is the work done, $Vi$ is the volume, $Vf$ is the final volume and R is the gas constant.

B and D are isochoric processes. Therefore, work done is zero.

Write the expression for work done during process C.

$WC=nRTln(VfVi)$

Here, T is the temperature, $WC$ is the work done, $Vi$ is the volume, $Vf$ is the final volume and R is the gas constant.

Conclusion:

For the process A,

Substitute 373 K for T, $0.020 m3$ for $Vi$ , $0.025 m3$ for $Vf$ , 1.00 mol for n and $8.314 Jmol−1K−1$ for R to get $WA$ .

$WA=(1.00 mol)(8.314 Jmol−1K−1)(373 K)ln(0.025 m30.020 m3)=692 J$

For the process C,

Substitute 273 K for T, $0.025 m3$ for $Vi$ , $0.020 m3$ for $Vf$ , 1.00 mol for n and $8.314 Jmol−1K−1$ for R to get $WC$ .

$WC=(1.00 mol)(8.314 Jmol−1K−1)(273 K)ln(0.020 m30.025 m3)=−506 J$

The net work done is,

$Wnet=(692 J)+(0 J)+(−506 J)+(0 J)=186 J$

Therefore, work done by engine during the 4 steps are $692 J$ , $0 J$ , $−506 J$ , $0 J$ and the net work done is $186 J$ .

(b)

To determine

The efficiency of the engine.

(b)

Expert Solution

Answer to Problem 86P

The efficiency of the engine is $0.0670$ .

Explanation of Solution

Write the expression for efficiency of the engine.

$e=WnetQin$

Here, e is the efficiency and $Qin$ is input heat.

Write the expression for input heat.

$Qin=52nRΔT+WA$

Here, $ΔU$ is the change in internal energy and $ΔT$ is the change intemperature.

Therefore, the input heat is,

$Qin=52(1.00 mol)(8.314 Jmol−1K−1)(373 K−273 K)+(692 J)=2792 J$

Conclusion:

Substitute $186 J$ for $Wnet$ and $2792 J$ for $Qin$ to get e.

$e=186 J2792 J=0.0670$

Therefore, efficiency of the engine is $0.0670$ .

(c)

To determine

The ratio of efficiency of reversible engine and the given engine.

(c)

Expert Solution

Answer to Problem 86P

The ratio of efficiency of reversible engine and the given engine is $4.00$ .

Explanation of Solution

Write the expression for the efficiency of reversible engine.

$er=1−TCTH$

Here, $TC$ is the temperature of cold reservoir and $TH$ is the temperature of the hot reservoir.

Conclusion:

Substitute 273 K for $TC$ and 373 K for $TH$ to get $er$ .

$er=1−273 K373 K=0.268$

The ratio is given by,

$ere=0.2680.0670=4.00$

The ratio of efficiency of reversible engine and the given engine is $4.00$ .

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Answer 2

Question

Chapter 15, Problem 86P

(a)

To determine

The work done by engine during the 4 steps and the net work done.

(a)

Expert Solution

Answer to Problem 86P

The work done by engine during the 4 steps are $692 J$ , $0 J$ , $−506 J$ , $0 J$ and the net work done is $186 J$ .

Explanation of Solution

Write the expression for work done during process A.

$WA=nRTln(VfVi)$

Here, T is the temperature, $WA$ is the work done, $Vi$ is the volume, $Vf$ is the final volume and R is the gas constant.

B and D are isochoric processes. Therefore, work done is zero.

Write the expression for work done during process C.

$WC=nRTln(VfVi)$

Here, T is the temperature, $WC$ is the work done, $Vi$ is the volume, $Vf$ is the final volume and R is the gas constant.

Conclusion:

For the process A,

Substitute 373 K for T, $0.020 m3$ for $Vi$ , $0.025 m3$ for $Vf$ , 1.00 mol for n and $8.314 Jmol−1K−1$ for R to get $WA$ .

$WA=(1.00 mol)(8.314 Jmol−1K−1)(373 K)ln(0.025 m30.020 m3)=692 J$

For the process C,

Substitute 273 K for T, $0.025 m3$ for $Vi$ , $0.020 m3$ for $Vf$ , 1.00 mol for n and $8.314 Jmol−1K−1$ for R to get $WC$ .

$WC=(1.00 mol)(8.314 Jmol−1K−1)(273 K)ln(0.020 m30.025 m3)=−506 J$

The net work done is,

$Wnet=(692 J)+(0 J)+(−506 J)+(0 J)=186 J$

Therefore, work done by engine during the 4 steps are $692 J$ , $0 J$ , $−506 J$ , $0 J$ and the net work done is $186 J$ .

(b)

To determine

The efficiency of the engine.

(b)

Expert Solution

Answer to Problem 86P

The efficiency of the engine is $0.0670$ .

Explanation of Solution

Write the expression for efficiency of the engine.

$e=WnetQin$

Here, e is the efficiency and $Qin$ is input heat.

Write the expression for input heat.

$Qin=52nRΔT+WA$

Here, $ΔU$ is the change in internal energy and $ΔT$ is the change intemperature.

Therefore, the input heat is,

$Qin=52(1.00 mol)(8.314 Jmol−1K−1)(373 K−273 K)+(692 J)=2792 J$

Conclusion:

Substitute $186 J$ for $Wnet$ and $2792 J$ for $Qin$ to get e.

$e=186 J2792 J=0.0670$

Therefore, efficiency of the engine is $0.0670$ .

(c)

To determine

The ratio of efficiency of reversible engine and the given engine.

(c)

Expert Solution

Answer to Problem 86P

The ratio of efficiency of reversible engine and the given engine is $4.00$ .

Explanation of Solution

Write the expression for the efficiency of reversible engine.

$er=1−TCTH$

Here, $TC$ is the temperature of cold reservoir and $TH$ is the temperature of the hot reservoir.

Conclusion:

Substitute 273 K for $TC$ and 373 K for $TH$ to get $er$ .

$er=1−273 K373 K=0.268$

The ratio is given by,

$ere=0.2680.0670=4.00$

The ratio of efficiency of reversible engine and the given engine is $4.00$ .

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Answer 3

Question

Chapter 15, Problem 86P

(a)

To determine

The work done by engine during the 4 steps and the net work done.

(a)

Expert Solution

Answer to Problem 86P

The work done by engine during the 4 steps are $692 J$ , $0 J$ , $−506 J$ , $0 J$ and the net work done is $186 J$ .

Explanation of Solution

Write the expression for work done during process A.

$WA=nRTln(VfVi)$

Here, T is the temperature, $WA$ is the work done, $Vi$ is the volume, $Vf$ is the final volume and R is the gas constant.

B and D are isochoric processes. Therefore, work done is zero.

Write the expression for work done during process C.

$WC=nRTln(VfVi)$

Here, T is the temperature, $WC$ is the work done, $Vi$ is the volume, $Vf$ is the final volume and R is the gas constant.

Conclusion:

For the process A,

Substitute 373 K for T, $0.020 m3$ for $Vi$ , $0.025 m3$ for $Vf$ , 1.00 mol for n and $8.314 Jmol−1K−1$ for R to get $WA$ .

$WA=(1.00 mol)(8.314 Jmol−1K−1)(373 K)ln(0.025 m30.020 m3)=692 J$

For the process C,

Substitute 273 K for T, $0.025 m3$ for $Vi$ , $0.020 m3$ for $Vf$ , 1.00 mol for n and $8.314 Jmol−1K−1$ for R to get $WC$ .

$WC=(1.00 mol)(8.314 Jmol−1K−1)(273 K)ln(0.020 m30.025 m3)=−506 J$

The net work done is,

$Wnet=(692 J)+(0 J)+(−506 J)+(0 J)=186 J$

Therefore, work done by engine during the 4 steps are $692 J$ , $0 J$ , $−506 J$ , $0 J$ and the net work done is $186 J$ .

(b)

To determine

The efficiency of the engine.

(b)

Expert Solution

Answer to Problem 86P

The efficiency of the engine is $0.0670$ .

Explanation of Solution

Write the expression for efficiency of the engine.

$e=WnetQin$

Here, e is the efficiency and $Qin$ is input heat.

Write the expression for input heat.

$Qin=52nRΔT+WA$

Here, $ΔU$ is the change in internal energy and $ΔT$ is the change intemperature.

Therefore, the input heat is,

$Qin=52(1.00 mol)(8.314 Jmol−1K−1)(373 K−273 K)+(692 J)=2792 J$

Conclusion:

Substitute $186 J$ for $Wnet$ and $2792 J$ for $Qin$ to get e.

$e=186 J2792 J=0.0670$

Therefore, efficiency of the engine is $0.0670$ .

(c)

To determine

The ratio of efficiency of reversible engine and the given engine.

(c)

Expert Solution

Answer to Problem 86P

The ratio of efficiency of reversible engine and the given engine is $4.00$ .

Explanation of Solution

Write the expression for the efficiency of reversible engine.

$er=1−TCTH$

Here, $TC$ is the temperature of cold reservoir and $TH$ is the temperature of the hot reservoir.

Conclusion:

Substitute 273 K for $TC$ and 373 K for $TH$ to get $er$ .

$er=1−273 K373 K=0.268$

The ratio is given by,

$ere=0.2680.0670=4.00$

The ratio of efficiency of reversible engine and the given engine is $4.00$ .

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Answer 4

Question

Chapter 15, Problem 86P

(a)

To determine

The work done by engine during the 4 steps and the net work done.

(a)

Expert Solution

Answer to Problem 86P

The work done by engine during the 4 steps are $692 J$ , $0 J$ , $−506 J$ , $0 J$ and the net work done is $186 J$ .

Explanation of Solution

Write the expression for work done during process A.

$WA=nRTln(VfVi)$

Here, T is the temperature, $WA$ is the work done, $Vi$ is the volume, $Vf$ is the final volume and R is the gas constant.

B and D are isochoric processes. Therefore, work done is zero.

Write the expression for work done during process C.

$WC=nRTln(VfVi)$

Here, T is the temperature, $WC$ is the work done, $Vi$ is the volume, $Vf$ is the final volume and R is the gas constant.

Conclusion:

For the process A,

Substitute 373 K for T, $0.020 m3$ for $Vi$ , $0.025 m3$ for $Vf$ , 1.00 mol for n and $8.314 Jmol−1K−1$ for R to get $WA$ .

$WA=(1.00 mol)(8.314 Jmol−1K−1)(373 K)ln(0.025 m30.020 m3)=692 J$

For the process C,

Substitute 273 K for T, $0.025 m3$ for $Vi$ , $0.020 m3$ for $Vf$ , 1.00 mol for n and $8.314 Jmol−1K−1$ for R to get $WC$ .

$WC=(1.00 mol)(8.314 Jmol−1K−1)(273 K)ln(0.020 m30.025 m3)=−506 J$

The net work done is,

$Wnet=(692 J)+(0 J)+(−506 J)+(0 J)=186 J$

Therefore, work done by engine during the 4 steps are $692 J$ , $0 J$ , $−506 J$ , $0 J$ and the net work done is $186 J$ .

(b)

To determine

The efficiency of the engine.

(b)

Expert Solution

Answer to Problem 86P

The efficiency of the engine is $0.0670$ .

Explanation of Solution

Write the expression for efficiency of the engine.

$e=WnetQin$

Here, e is the efficiency and $Qin$ is input heat.

Write the expression for input heat.

$Qin=52nRΔT+WA$

Here, $ΔU$ is the change in internal energy and $ΔT$ is the change intemperature.

Therefore, the input heat is,

$Qin=52(1.00 mol)(8.314 Jmol−1K−1)(373 K−273 K)+(692 J)=2792 J$

Conclusion:

Substitute $186 J$ for $Wnet$ and $2792 J$ for $Qin$ to get e.

$e=186 J2792 J=0.0670$

Therefore, efficiency of the engine is $0.0670$ .

(c)

To determine

The ratio of efficiency of reversible engine and the given engine.

(c)

Expert Solution

Answer to Problem 86P

The ratio of efficiency of reversible engine and the given engine is $4.00$ .

Explanation of Solution

Write the expression for the efficiency of reversible engine.

$er=1−TCTH$

Here, $TC$ is the temperature of cold reservoir and $TH$ is the temperature of the hot reservoir.

Conclusion:

Substitute 273 K for $TC$ and 373 K for $TH$ to get $er$ .

$er=1−273 K373 K=0.268$

The ratio is given by,

$ere=0.2680.0670=4.00$

The ratio of efficiency of reversible engine and the given engine is $4.00$ .

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 15, Problem 86P

(a)

To determine

The work done by engine during the 4 steps and the net work done.

(a)

Expert Solution

Answer to Problem 86P

The work done by engine during the 4 steps are $692 J$ , $0 J$ , $−506 J$ , $0 J$ and the net work done is $186 J$ .

Explanation of Solution

Write the expression for work done during process A.

$WA=nRTln(VfVi)$

Here, T is the temperature, $WA$ is the work done, $Vi$ is the volume, $Vf$ is the final volume and R is the gas constant.

B and D are isochoric processes. Therefore, work done is zero.

Write the expression for work done during process C.

$WC=nRTln(VfVi)$

Here, T is the temperature, $WC$ is the work done, $Vi$ is the volume, $Vf$ is the final volume and R is the gas constant.

Conclusion:

For the process A,

Substitute 373 K for T, $0.020 m3$ for $Vi$ , $0.025 m3$ for $Vf$ , 1.00 mol for n and $8.314 Jmol−1K−1$ for R to get $WA$ .

$WA=(1.00 mol)(8.314 Jmol−1K−1)(373 K)ln(0.025 m30.020 m3)=692 J$

For the process C,

Substitute 273 K for T, $0.025 m3$ for $Vi$ , $0.020 m3$ for $Vf$ , 1.00 mol for n and $8.314 Jmol−1K−1$ for R to get $WC$ .

$WC=(1.00 mol)(8.314 Jmol−1K−1)(273 K)ln(0.020 m30.025 m3)=−506 J$

The net work done is,

$Wnet=(692 J)+(0 J)+(−506 J)+(0 J)=186 J$

Therefore, work done by engine during the 4 steps are $692 J$ , $0 J$ , $−506 J$ , $0 J$ and the net work done is $186 J$ .

(b)

To determine

The efficiency of the engine.

(b)

Expert Solution

Answer to Problem 86P

The efficiency of the engine is $0.0670$ .

Explanation of Solution

Write the expression for efficiency of the engine.

$e=WnetQin$

Here, e is the efficiency and $Qin$ is input heat.

Write the expression for input heat.

$Qin=52nRΔT+WA$

Here, $ΔU$ is the change in internal energy and $ΔT$ is the change intemperature.

Therefore, the input heat is,

$Qin=52(1.00 mol)(8.314 Jmol−1K−1)(373 K−273 K)+(692 J)=2792 J$

Conclusion:

Substitute $186 J$ for $Wnet$ and $2792 J$ for $Qin$ to get e.

$e=186 J2792 J=0.0670$

Therefore, efficiency of the engine is $0.0670$ .

(c)

To determine

The ratio of efficiency of reversible engine and the given engine.

(c)

Expert Solution

Answer to Problem 86P

The ratio of efficiency of reversible engine and the given engine is $4.00$ .

Explanation of Solution

Write the expression for the efficiency of reversible engine.

$er=1−TCTH$

Here, $TC$ is the temperature of cold reservoir and $TH$ is the temperature of the hot reservoir.

Conclusion:

Substitute 273 K for $TC$ and 373 K for $TH$ to get $er$ .

$er=1−273 K373 K=0.268$

The ratio is given by,

$ere=0.2680.0670=4.00$

The ratio of efficiency of reversible engine and the given engine is $4.00$ .

Answer 6

Chapter 15, Problem 86P

(a)

To determine

The work done by engine during the 4 steps and the net work done.

(a)

Expert Solution

Answer to Problem 86P

The work done by engine during the 4 steps are $692 J$ , $0 J$ , $−506 J$ , $0 J$ and the net work done is $186 J$ .

Explanation of Solution

Write the expression for work done during process A.

$WA=nRTln(VfVi)$

Here, T is the temperature, $WA$ is the work done, $Vi$ is the volume, $Vf$ is the final volume and R is the gas constant.

B and D are isochoric processes. Therefore, work done is zero.

Write the expression for work done during process C.

$WC=nRTln(VfVi)$

Here, T is the temperature, $WC$ is the work done, $Vi$ is the volume, $Vf$ is the final volume and R is the gas constant.

Conclusion:

For the process A,

Substitute 373 K for T, $0.020 m3$ for $Vi$ , $0.025 m3$ for $Vf$ , 1.00 mol for n and $8.314 Jmol−1K−1$ for R to get $WA$ .

$WA=(1.00 mol)(8.314 Jmol−1K−1)(373 K)ln(0.025 m30.020 m3)=692 J$

For the process C,

Substitute 273 K for T, $0.025 m3$ for $Vi$ , $0.020 m3$ for $Vf$ , 1.00 mol for n and $8.314 Jmol−1K−1$ for R to get $WC$ .

$WC=(1.00 mol)(8.314 Jmol−1K−1)(273 K)ln(0.020 m30.025 m3)=−506 J$

The net work done is,

$Wnet=(692 J)+(0 J)+(−506 J)+(0 J)=186 J$

Therefore, work done by engine during the 4 steps are $692 J$ , $0 J$ , $−506 J$ , $0 J$ and the net work done is $186 J$ .

Answer 7

Chapter 15, Problem 86P

(a)

To determine

The work done by engine during the 4 steps and the net work done.

Answer 8

(a)

Expert Solution

Answer to Problem 86P

The work done by engine during the 4 steps are $692 J$ , $0 J$ , $−506 J$ , $0 J$ and the net work done is $186 J$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Write the expression for work done during process A.

$WA=nRTln(VfVi)$

Here, T is the temperature, $WA$ is the work done, $Vi$ is the volume, $Vf$ is the final volume and R is the gas constant.

B and D are isochoric processes. Therefore, work done is zero.

Write the expression for work done during process C.

$WC=nRTln(VfVi)$

Here, T is the temperature, $WC$ is the work done, $Vi$ is the volume, $Vf$ is the final volume and R is the gas constant.

Conclusion:

For the process A,

Substitute 373 K for T, $0.020 m3$ for $Vi$ , $0.025 m3$ for $Vf$ , 1.00 mol for n and $8.314 Jmol−1K−1$ for R to get $WA$ .

$WA=(1.00 mol)(8.314 Jmol−1K−1)(373 K)ln(0.025 m30.020 m3)=692 J$

For the process C,

Substitute 273 K for T, $0.025 m3$ for $Vi$ , $0.020 m3$ for $Vf$ , 1.00 mol for n and $8.314 Jmol−1K−1$ for R to get $WC$ .

$WC=(1.00 mol)(8.314 Jmol−1K−1)(273 K)ln(0.020 m30.025 m3)=−506 J$

The net work done is,

$Wnet=(692 J)+(0 J)+(−506 J)+(0 J)=186 J$

Therefore, work done by engine during the 4 steps are $692 J$ , $0 J$ , $−506 J$ , $0 J$ and the net work done is $186 J$ .

Answer 13

(b)

To determine

The efficiency of the engine.

(b)

Expert Solution

Answer to Problem 86P

The efficiency of the engine is $0.0670$ .

Explanation of Solution

Write the expression for efficiency of the engine.

$e=WnetQin$

Here, e is the efficiency and $Qin$ is input heat.

Write the expression for input heat.

$Qin=52nRΔT+WA$

Here, $ΔU$ is the change in internal energy and $ΔT$ is the change intemperature.

Therefore, the input heat is,

$Qin=52(1.00 mol)(8.314 Jmol−1K−1)(373 K−273 K)+(692 J)=2792 J$

Conclusion:

Substitute $186 J$ for $Wnet$ and $2792 J$ for $Qin$ to get e.

$e=186 J2792 J=0.0670$

Therefore, efficiency of the engine is $0.0670$ .

Answer 14

(b)

To determine

The efficiency of the engine.

Answer 15

(b)

Expert Solution

Answer to Problem 86P

The efficiency of the engine is $0.0670$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Write the expression for efficiency of the engine.

$e=WnetQin$

Here, e is the efficiency and $Qin$ is input heat.

Write the expression for input heat.

$Qin=52nRΔT+WA$

Here, $ΔU$ is the change in internal energy and $ΔT$ is the change intemperature.

Therefore, the input heat is,

$Qin=52(1.00 mol)(8.314 Jmol−1K−1)(373 K−273 K)+(692 J)=2792 J$

Conclusion:

Substitute $186 J$ for $Wnet$ and $2792 J$ for $Qin$ to get e.

$e=186 J2792 J=0.0670$

Therefore, efficiency of the engine is $0.0670$ .

Answer 20

(c)

To determine

The ratio of efficiency of reversible engine and the given engine.

(c)

Expert Solution

Answer to Problem 86P

The ratio of efficiency of reversible engine and the given engine is $4.00$ .

Explanation of Solution

Write the expression for the efficiency of reversible engine.

$er=1−TCTH$

Here, $TC$ is the temperature of cold reservoir and $TH$ is the temperature of the hot reservoir.

Conclusion:

Substitute 273 K for $TC$ and 373 K for $TH$ to get $er$ .

$er=1−273 K373 K=0.268$

The ratio is given by,

$ere=0.2680.0670=4.00$

The ratio of efficiency of reversible engine and the given engine is $4.00$ .

Answer 21

(c)

To determine

The ratio of efficiency of reversible engine and the given engine.

Answer 22

(c)

Expert Solution

Answer to Problem 86P

The ratio of efficiency of reversible engine and the given engine is $4.00$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Write the expression for the efficiency of reversible engine.

$er=1−TCTH$

Here, $TC$ is the temperature of cold reservoir and $TH$ is the temperature of the hot reservoir.

Conclusion:

Substitute 273 K for $TC$ and 373 K for $TH$ to get $er$ .

$er=1−273 K373 K=0.268$

The ratio is given by,

$ere=0.2680.0670=4.00$

The ratio of efficiency of reversible engine and the given engine is $4.00$ .

Answer 27

Ch. 15.1 - Prob. 15.1PP Ch. 15.2 - Prob. 15.2CP Ch. 15.3 - Prob. 15.2PP Ch. 15.3 - Prob. 15.3PP Ch. 15.4 - Prob. 15.4PP Ch. 15.4 - Prob. 15.4CP Ch. 15.5 - Prob. 15.5PP Ch. 15.6 - Prob. 15.6PP Ch. 15.7 - Prob. 15.7PP Ch. 15.7 - Prob. 15.8PP

The work done by engine during the 4 steps and the net work done.

Videos

The work done by engine during the 4 steps and the net work done.

Answer to Problem 86P

Explanation of Solution

The efficiency of the engine.

Answer to Problem 86P

Explanation of Solution

The ratio of efficiency of reversible engine and the given engine.

Answer to Problem 86P

Explanation of Solution

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