The total work done by the gas and heat absorbed by gas in each portion of cycle.

Question

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Answer 1

Question

Chapter 18, Problem 75P

To determine

The total work done by the gas and heat absorbed by gas in each portion of cycle.

Expert Solution & Answer

Answer to Problem 75P

The total work done is $6.62 kJ$ . The heat absorbed at point A, B, C and D are $13.2 kJ$ , $−8.98 kJ$ , $−6.58 kJ$ and $8.98 kJ$ respectively.

Explanation of Solution

Given:

The initial pressure is $2.00 atm$ .

The temperature is $360 K$

Formula used:

The expression for volume at D is given by,

$VD=nRTDPD$

The expression for pressure at point C is given by,

$PC=nRTCVC$

The expression for temperature at point A and B is given by,

$TA=PAVAnR$

The expression for heat absorbed at D is given by,

$QD=32nRΔTD$

The expression for heat absorbed at A is given by,

$QA=nRTAlnVBVA$

The expression for heat absorbed at B is given by,

$QB=−32nR(TC−TB)$

The expression for heat absorbed at C is given by,

$QC=nRTClnVDVC$

The expression for total work done is given by,

$Wby tot=WA+WB+WC+WD$

Calculation:

The volume at point D is calculated as,

$VD=nRTDPD=( 2 mol)( 8.314 J/ mol ⋅K)( 360 K)( ( 2.0 atm )( 101.3 kPa 1 atm ))=29.5 L$

The pressure at point C is calculated as,

$PC=nRTCVC=( 2 mol)( 8.314 J/ mol ⋅K)( 360 K)( 3 V D )=( 2 mol)( 8.314 J/ mol ⋅K)( 360 K)( 3( ( 29.5 L )( 101.3 atm 1 L ) ))=0.667 atm$

The temperature at point A and B is calculated as,

$TA=PAVAnR=( 2 P C )( 88.6 L)( 2 mol)( 8.206× 10 −2 L⋅atm/ mol⋅K )=( 2( 0.667 atm ))( 88.6 L)( 2 mol)( 8.206× 10 −2 L⋅atm/ mol⋅K )=720 K$

The heat absorbed at point D is calculated as,

$QD=32nRΔTD=32(2 mol)(8.314 J/mol⋅K)(720 K−360 K)=(( 8.98× 10 3 J)( 10 −3 kJ 1 kJ ))=8.98 kJ$

The heat absorbed at point A is calculated as,

$QA=nRTAlnVBVA=(2 mol)(8.314 J/mol⋅K)(720 K)ln( 88.6 L 29.5 L)=(( 13.2× 10 3 J)( 10 −3 kJ 1 kJ ))=13.2 kJ$

The heat absorbed at point B is calculated as,

$QB=−32nRΔTC=−32(2 mol)(8.314 J/mol⋅K)(720 K−360 K)=−(( 8.98× 10 3 J)( 10 −3 kJ 1 kJ ))=−8.98 kJ$

The heat absorbed at point C is calculated as,

$QC=nRTClnVDVC=(2 mol)(8.314 J/mol⋅K)(360 K)ln( 29.5 L 88.6 L)=(( −6.58× 10 3 J)( 10 −3 kJ 1 kJ ))=−6.58 kJ$

The total work done is calculated as,

$Wby tot=WA+WB+WC+WD=0+13.2 kJ+0−6.58 kJ=6.62 kJ$

Conclusion:

Therefore, the total work done is $6.62 kJ$ . The heat absorbed at point A, B, C and D are $13.2 kJ$ , $−8.98 kJ$ , $−6.58 kJ$ and $8.98 kJ$ respectively.

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Answer 2

Question

Chapter 18, Problem 75P

To determine

The total work done by the gas and heat absorbed by gas in each portion of cycle.

Expert Solution & Answer

Answer to Problem 75P

The total work done is $6.62 kJ$ . The heat absorbed at point A, B, C and D are $13.2 kJ$ , $−8.98 kJ$ , $−6.58 kJ$ and $8.98 kJ$ respectively.

Explanation of Solution

Given:

The initial pressure is $2.00 atm$ .

The temperature is $360 K$

Formula used:

The expression for volume at D is given by,

$VD=nRTDPD$

The expression for pressure at point C is given by,

$PC=nRTCVC$

The expression for temperature at point A and B is given by,

$TA=PAVAnR$

The expression for heat absorbed at D is given by,

$QD=32nRΔTD$

The expression for heat absorbed at A is given by,

$QA=nRTAlnVBVA$

The expression for heat absorbed at B is given by,

$QB=−32nR(TC−TB)$

The expression for heat absorbed at C is given by,

$QC=nRTClnVDVC$

The expression for total work done is given by,

$Wby tot=WA+WB+WC+WD$

Calculation:

The volume at point D is calculated as,

$VD=nRTDPD=( 2 mol)( 8.314 J/ mol ⋅K)( 360 K)( ( 2.0 atm )( 101.3 kPa 1 atm ))=29.5 L$

The pressure at point C is calculated as,

$PC=nRTCVC=( 2 mol)( 8.314 J/ mol ⋅K)( 360 K)( 3 V D )=( 2 mol)( 8.314 J/ mol ⋅K)( 360 K)( 3( ( 29.5 L )( 101.3 atm 1 L ) ))=0.667 atm$

The temperature at point A and B is calculated as,

$TA=PAVAnR=( 2 P C )( 88.6 L)( 2 mol)( 8.206× 10 −2 L⋅atm/ mol⋅K )=( 2( 0.667 atm ))( 88.6 L)( 2 mol)( 8.206× 10 −2 L⋅atm/ mol⋅K )=720 K$

The heat absorbed at point D is calculated as,

$QD=32nRΔTD=32(2 mol)(8.314 J/mol⋅K)(720 K−360 K)=(( 8.98× 10 3 J)( 10 −3 kJ 1 kJ ))=8.98 kJ$

The heat absorbed at point A is calculated as,

$QA=nRTAlnVBVA=(2 mol)(8.314 J/mol⋅K)(720 K)ln( 88.6 L 29.5 L)=(( 13.2× 10 3 J)( 10 −3 kJ 1 kJ ))=13.2 kJ$

The heat absorbed at point B is calculated as,

$QB=−32nRΔTC=−32(2 mol)(8.314 J/mol⋅K)(720 K−360 K)=−(( 8.98× 10 3 J)( 10 −3 kJ 1 kJ ))=−8.98 kJ$

The heat absorbed at point C is calculated as,

$QC=nRTClnVDVC=(2 mol)(8.314 J/mol⋅K)(360 K)ln( 29.5 L 88.6 L)=(( −6.58× 10 3 J)( 10 −3 kJ 1 kJ ))=−6.58 kJ$

The total work done is calculated as,

$Wby tot=WA+WB+WC+WD=0+13.2 kJ+0−6.58 kJ=6.62 kJ$

Conclusion:

Therefore, the total work done is $6.62 kJ$ . The heat absorbed at point A, B, C and D are $13.2 kJ$ , $−8.98 kJ$ , $−6.58 kJ$ and $8.98 kJ$ respectively.

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Answer 3

Question

Chapter 18, Problem 75P

To determine

The total work done by the gas and heat absorbed by gas in each portion of cycle.

Expert Solution & Answer

Answer to Problem 75P

The total work done is $6.62 kJ$ . The heat absorbed at point A, B, C and D are $13.2 kJ$ , $−8.98 kJ$ , $−6.58 kJ$ and $8.98 kJ$ respectively.

Explanation of Solution

Given:

The initial pressure is $2.00 atm$ .

The temperature is $360 K$

Formula used:

The expression for volume at D is given by,

$VD=nRTDPD$

The expression for pressure at point C is given by,

$PC=nRTCVC$

The expression for temperature at point A and B is given by,

$TA=PAVAnR$

The expression for heat absorbed at D is given by,

$QD=32nRΔTD$

The expression for heat absorbed at A is given by,

$QA=nRTAlnVBVA$

The expression for heat absorbed at B is given by,

$QB=−32nR(TC−TB)$

The expression for heat absorbed at C is given by,

$QC=nRTClnVDVC$

The expression for total work done is given by,

$Wby tot=WA+WB+WC+WD$

Calculation:

The volume at point D is calculated as,

$VD=nRTDPD=( 2 mol)( 8.314 J/ mol ⋅K)( 360 K)( ( 2.0 atm )( 101.3 kPa 1 atm ))=29.5 L$

The pressure at point C is calculated as,

$PC=nRTCVC=( 2 mol)( 8.314 J/ mol ⋅K)( 360 K)( 3 V D )=( 2 mol)( 8.314 J/ mol ⋅K)( 360 K)( 3( ( 29.5 L )( 101.3 atm 1 L ) ))=0.667 atm$

The temperature at point A and B is calculated as,

$TA=PAVAnR=( 2 P C )( 88.6 L)( 2 mol)( 8.206× 10 −2 L⋅atm/ mol⋅K )=( 2( 0.667 atm ))( 88.6 L)( 2 mol)( 8.206× 10 −2 L⋅atm/ mol⋅K )=720 K$

The heat absorbed at point D is calculated as,

$QD=32nRΔTD=32(2 mol)(8.314 J/mol⋅K)(720 K−360 K)=(( 8.98× 10 3 J)( 10 −3 kJ 1 kJ ))=8.98 kJ$

The heat absorbed at point A is calculated as,

$QA=nRTAlnVBVA=(2 mol)(8.314 J/mol⋅K)(720 K)ln( 88.6 L 29.5 L)=(( 13.2× 10 3 J)( 10 −3 kJ 1 kJ ))=13.2 kJ$

The heat absorbed at point B is calculated as,

$QB=−32nRΔTC=−32(2 mol)(8.314 J/mol⋅K)(720 K−360 K)=−(( 8.98× 10 3 J)( 10 −3 kJ 1 kJ ))=−8.98 kJ$

The heat absorbed at point C is calculated as,

$QC=nRTClnVDVC=(2 mol)(8.314 J/mol⋅K)(360 K)ln( 29.5 L 88.6 L)=(( −6.58× 10 3 J)( 10 −3 kJ 1 kJ ))=−6.58 kJ$

The total work done is calculated as,

$Wby tot=WA+WB+WC+WD=0+13.2 kJ+0−6.58 kJ=6.62 kJ$

Conclusion:

Therefore, the total work done is $6.62 kJ$ . The heat absorbed at point A, B, C and D are $13.2 kJ$ , $−8.98 kJ$ , $−6.58 kJ$ and $8.98 kJ$ respectively.

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Answer 4

Question

Chapter 18, Problem 75P

To determine

The total work done by the gas and heat absorbed by gas in each portion of cycle.

Expert Solution & Answer

Answer to Problem 75P

The total work done is $6.62 kJ$ . The heat absorbed at point A, B, C and D are $13.2 kJ$ , $−8.98 kJ$ , $−6.58 kJ$ and $8.98 kJ$ respectively.

Explanation of Solution

Given:

The initial pressure is $2.00 atm$ .

The temperature is $360 K$

Formula used:

The expression for volume at D is given by,

$VD=nRTDPD$

The expression for pressure at point C is given by,

$PC=nRTCVC$

The expression for temperature at point A and B is given by,

$TA=PAVAnR$

The expression for heat absorbed at D is given by,

$QD=32nRΔTD$

The expression for heat absorbed at A is given by,

$QA=nRTAlnVBVA$

The expression for heat absorbed at B is given by,

$QB=−32nR(TC−TB)$

The expression for heat absorbed at C is given by,

$QC=nRTClnVDVC$

The expression for total work done is given by,

$Wby tot=WA+WB+WC+WD$

Calculation:

The volume at point D is calculated as,

$VD=nRTDPD=( 2 mol)( 8.314 J/ mol ⋅K)( 360 K)( ( 2.0 atm )( 101.3 kPa 1 atm ))=29.5 L$

The pressure at point C is calculated as,

$PC=nRTCVC=( 2 mol)( 8.314 J/ mol ⋅K)( 360 K)( 3 V D )=( 2 mol)( 8.314 J/ mol ⋅K)( 360 K)( 3( ( 29.5 L )( 101.3 atm 1 L ) ))=0.667 atm$

The temperature at point A and B is calculated as,

$TA=PAVAnR=( 2 P C )( 88.6 L)( 2 mol)( 8.206× 10 −2 L⋅atm/ mol⋅K )=( 2( 0.667 atm ))( 88.6 L)( 2 mol)( 8.206× 10 −2 L⋅atm/ mol⋅K )=720 K$

The heat absorbed at point D is calculated as,

$QD=32nRΔTD=32(2 mol)(8.314 J/mol⋅K)(720 K−360 K)=(( 8.98× 10 3 J)( 10 −3 kJ 1 kJ ))=8.98 kJ$

The heat absorbed at point A is calculated as,

$QA=nRTAlnVBVA=(2 mol)(8.314 J/mol⋅K)(720 K)ln( 88.6 L 29.5 L)=(( 13.2× 10 3 J)( 10 −3 kJ 1 kJ ))=13.2 kJ$

The heat absorbed at point B is calculated as,

$QB=−32nRΔTC=−32(2 mol)(8.314 J/mol⋅K)(720 K−360 K)=−(( 8.98× 10 3 J)( 10 −3 kJ 1 kJ ))=−8.98 kJ$

The heat absorbed at point C is calculated as,

$QC=nRTClnVDVC=(2 mol)(8.314 J/mol⋅K)(360 K)ln( 29.5 L 88.6 L)=(( −6.58× 10 3 J)( 10 −3 kJ 1 kJ ))=−6.58 kJ$

The total work done is calculated as,

$Wby tot=WA+WB+WC+WD=0+13.2 kJ+0−6.58 kJ=6.62 kJ$

Conclusion:

Therefore, the total work done is $6.62 kJ$ . The heat absorbed at point A, B, C and D are $13.2 kJ$ , $−8.98 kJ$ , $−6.58 kJ$ and $8.98 kJ$ respectively.

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Answer 5

Chapter 18, Problem 75P

To determine

The total work done by the gas and heat absorbed by gas in each portion of cycle.

Expert Solution & Answer

Answer to Problem 75P

The total work done is $6.62 kJ$ . The heat absorbed at point A, B, C and D are $13.2 kJ$ , $−8.98 kJ$ , $−6.58 kJ$ and $8.98 kJ$ respectively.

Explanation of Solution

Given:

The initial pressure is $2.00 atm$ .

The temperature is $360 K$

Formula used:

The expression for volume at D is given by,

$VD=nRTDPD$

The expression for pressure at point C is given by,

$PC=nRTCVC$

The expression for temperature at point A and B is given by,

$TA=PAVAnR$

The expression for heat absorbed at D is given by,

$QD=32nRΔTD$

The expression for heat absorbed at A is given by,

$QA=nRTAlnVBVA$

The expression for heat absorbed at B is given by,

$QB=−32nR(TC−TB)$

The expression for heat absorbed at C is given by,

$QC=nRTClnVDVC$

The expression for total work done is given by,

$Wby tot=WA+WB+WC+WD$

Calculation:

The volume at point D is calculated as,

$VD=nRTDPD=( 2 mol)( 8.314 J/ mol ⋅K)( 360 K)( ( 2.0 atm )( 101.3 kPa 1 atm ))=29.5 L$

The pressure at point C is calculated as,

$PC=nRTCVC=( 2 mol)( 8.314 J/ mol ⋅K)( 360 K)( 3 V D )=( 2 mol)( 8.314 J/ mol ⋅K)( 360 K)( 3( ( 29.5 L )( 101.3 atm 1 L ) ))=0.667 atm$

The temperature at point A and B is calculated as,

$TA=PAVAnR=( 2 P C )( 88.6 L)( 2 mol)( 8.206× 10 −2 L⋅atm/ mol⋅K )=( 2( 0.667 atm ))( 88.6 L)( 2 mol)( 8.206× 10 −2 L⋅atm/ mol⋅K )=720 K$

The heat absorbed at point D is calculated as,

$QD=32nRΔTD=32(2 mol)(8.314 J/mol⋅K)(720 K−360 K)=(( 8.98× 10 3 J)( 10 −3 kJ 1 kJ ))=8.98 kJ$

The heat absorbed at point A is calculated as,

$QA=nRTAlnVBVA=(2 mol)(8.314 J/mol⋅K)(720 K)ln( 88.6 L 29.5 L)=(( 13.2× 10 3 J)( 10 −3 kJ 1 kJ ))=13.2 kJ$

The heat absorbed at point B is calculated as,

$QB=−32nRΔTC=−32(2 mol)(8.314 J/mol⋅K)(720 K−360 K)=−(( 8.98× 10 3 J)( 10 −3 kJ 1 kJ ))=−8.98 kJ$

The heat absorbed at point C is calculated as,

$QC=nRTClnVDVC=(2 mol)(8.314 J/mol⋅K)(360 K)ln( 29.5 L 88.6 L)=(( −6.58× 10 3 J)( 10 −3 kJ 1 kJ ))=−6.58 kJ$

The total work done is calculated as,

$Wby tot=WA+WB+WC+WD=0+13.2 kJ+0−6.58 kJ=6.62 kJ$

Conclusion:

Therefore, the total work done is $6.62 kJ$ . The heat absorbed at point A, B, C and D are $13.2 kJ$ , $−8.98 kJ$ , $−6.58 kJ$ and $8.98 kJ$ respectively.

Answer 6

Chapter 18, Problem 75P

To determine

The total work done by the gas and heat absorbed by gas in each portion of cycle.

Expert Solution & Answer

Answer to Problem 75P

The total work done is $6.62 kJ$ . The heat absorbed at point A, B, C and D are $13.2 kJ$ , $−8.98 kJ$ , $−6.58 kJ$ and $8.98 kJ$ respectively.

Explanation of Solution

Given:

The initial pressure is $2.00 atm$ .

The temperature is $360 K$

Formula used:

The expression for volume at D is given by,

$VD=nRTDPD$

The expression for pressure at point C is given by,

$PC=nRTCVC$

The expression for temperature at point A and B is given by,

$TA=PAVAnR$

The expression for heat absorbed at D is given by,

$QD=32nRΔTD$

The expression for heat absorbed at A is given by,

$QA=nRTAlnVBVA$

The expression for heat absorbed at B is given by,

$QB=−32nR(TC−TB)$

The expression for heat absorbed at C is given by,

$QC=nRTClnVDVC$

The expression for total work done is given by,

$Wby tot=WA+WB+WC+WD$

Calculation:

The volume at point D is calculated as,

$VD=nRTDPD=( 2 mol)( 8.314 J/ mol ⋅K)( 360 K)( ( 2.0 atm )( 101.3 kPa 1 atm ))=29.5 L$

The pressure at point C is calculated as,

$PC=nRTCVC=( 2 mol)( 8.314 J/ mol ⋅K)( 360 K)( 3 V D )=( 2 mol)( 8.314 J/ mol ⋅K)( 360 K)( 3( ( 29.5 L )( 101.3 atm 1 L ) ))=0.667 atm$

The temperature at point A and B is calculated as,

$TA=PAVAnR=( 2 P C )( 88.6 L)( 2 mol)( 8.206× 10 −2 L⋅atm/ mol⋅K )=( 2( 0.667 atm ))( 88.6 L)( 2 mol)( 8.206× 10 −2 L⋅atm/ mol⋅K )=720 K$

The heat absorbed at point D is calculated as,

$QD=32nRΔTD=32(2 mol)(8.314 J/mol⋅K)(720 K−360 K)=(( 8.98× 10 3 J)( 10 −3 kJ 1 kJ ))=8.98 kJ$

The heat absorbed at point A is calculated as,

$QA=nRTAlnVBVA=(2 mol)(8.314 J/mol⋅K)(720 K)ln( 88.6 L 29.5 L)=(( 13.2× 10 3 J)( 10 −3 kJ 1 kJ ))=13.2 kJ$

The heat absorbed at point B is calculated as,

$QB=−32nRΔTC=−32(2 mol)(8.314 J/mol⋅K)(720 K−360 K)=−(( 8.98× 10 3 J)( 10 −3 kJ 1 kJ ))=−8.98 kJ$

The heat absorbed at point C is calculated as,

$QC=nRTClnVDVC=(2 mol)(8.314 J/mol⋅K)(360 K)ln( 29.5 L 88.6 L)=(( −6.58× 10 3 J)( 10 −3 kJ 1 kJ ))=−6.58 kJ$

The total work done is calculated as,

$Wby tot=WA+WB+WC+WD=0+13.2 kJ+0−6.58 kJ=6.62 kJ$

Conclusion:

Therefore, the total work done is $6.62 kJ$ . The heat absorbed at point A, B, C and D are $13.2 kJ$ , $−8.98 kJ$ , $−6.58 kJ$ and $8.98 kJ$ respectively.

Answer 7

Chapter 18, Problem 75P

To determine

The total work done by the gas and heat absorbed by gas in each portion of cycle.

Answer 8

Expert Solution & Answer

Answer to Problem 75P

The total work done is $6.62 kJ$ . The heat absorbed at point A, B, C and D are $13.2 kJ$ , $−8.98 kJ$ , $−6.58 kJ$ and $8.98 kJ$ respectively.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given:

The initial pressure is $2.00 atm$ .

The temperature is $360 K$

Formula used:

The expression for volume at D is given by,

$VD=nRTDPD$

The expression for pressure at point C is given by,

$PC=nRTCVC$

The expression for temperature at point A and B is given by,

$TA=PAVAnR$

The expression for heat absorbed at D is given by,

$QD=32nRΔTD$

The expression for heat absorbed at A is given by,

$QA=nRTAlnVBVA$

The expression for heat absorbed at B is given by,

$QB=−32nR(TC−TB)$

The expression for heat absorbed at C is given by,

$QC=nRTClnVDVC$

The expression for total work done is given by,

$Wby tot=WA+WB+WC+WD$

Calculation:

The volume at point D is calculated as,

$VD=nRTDPD=( 2 mol)( 8.314 J/ mol ⋅K)( 360 K)( ( 2.0 atm )( 101.3 kPa 1 atm ))=29.5 L$

The pressure at point C is calculated as,

$PC=nRTCVC=( 2 mol)( 8.314 J/ mol ⋅K)( 360 K)( 3 V D )=( 2 mol)( 8.314 J/ mol ⋅K)( 360 K)( 3( ( 29.5 L )( 101.3 atm 1 L ) ))=0.667 atm$

The temperature at point A and B is calculated as,

$TA=PAVAnR=( 2 P C )( 88.6 L)( 2 mol)( 8.206× 10 −2 L⋅atm/ mol⋅K )=( 2( 0.667 atm ))( 88.6 L)( 2 mol)( 8.206× 10 −2 L⋅atm/ mol⋅K )=720 K$

The heat absorbed at point D is calculated as,

$QD=32nRΔTD=32(2 mol)(8.314 J/mol⋅K)(720 K−360 K)=(( 8.98× 10 3 J)( 10 −3 kJ 1 kJ ))=8.98 kJ$

The heat absorbed at point A is calculated as,

$QA=nRTAlnVBVA=(2 mol)(8.314 J/mol⋅K)(720 K)ln( 88.6 L 29.5 L)=(( 13.2× 10 3 J)( 10 −3 kJ 1 kJ ))=13.2 kJ$

The heat absorbed at point B is calculated as,

$QB=−32nRΔTC=−32(2 mol)(8.314 J/mol⋅K)(720 K−360 K)=−(( 8.98× 10 3 J)( 10 −3 kJ 1 kJ ))=−8.98 kJ$

The heat absorbed at point C is calculated as,

$QC=nRTClnVDVC=(2 mol)(8.314 J/mol⋅K)(360 K)ln( 29.5 L 88.6 L)=(( −6.58× 10 3 J)( 10 −3 kJ 1 kJ ))=−6.58 kJ$

The total work done is calculated as,

$Wby tot=WA+WB+WC+WD=0+13.2 kJ+0−6.58 kJ=6.62 kJ$

Conclusion:

Therefore, the total work done is $6.62 kJ$ . The heat absorbed at point A, B, C and D are $13.2 kJ$ , $−8.98 kJ$ , $−6.58 kJ$ and $8.98 kJ$ respectively.

Answer 12

Ch. 18 - Prob. 1P Ch. 18 - Prob. 2P Ch. 18 - Prob. 3P Ch. 18 - Prob. 4P Ch. 18 - Prob. 5P Ch. 18 - Prob. 6P Ch. 18 - Prob. 7P Ch. 18 - Prob. 8P Ch. 18 - Prob. 9P Ch. 18 - Prob. 10P

The total work done by the gas and heat absorbed by gas in each portion of cycle.

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The total work done by the gas and heat absorbed by gas in each portion of cycle.

Answer to Problem 75P

Explanation of Solution

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