The P-V diagram.

Question

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Answer 1

Question

Chapter 18, Problem 74P

(a)

To determine

The P-V diagram.

(a)

Expert Solution

Answer to Problem 74P

The P-V diagram is shown in figure 1.

Explanation of Solution

Given:

The initial pressure is 2.00 atm .

The initial volume is 2.00 L

The final volume is 4.00 L

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 74P

Figure 1

Conclusion:

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

(b)

To determine

The temperature at end of each part of cycle.

(b)

Expert Solution

Answer to Problem 74P

The temperature at the end of cycle 1, 2 and 3 are 24.4 K , 24.4 K and 48.7 K respectively.

Explanation of Solution

Formula used:

The expression for temperature for portion 1 is given by,

T1=P1V1nR

The expression for temperature for portion 2 is given by,

T2=T1

The expression for temperature for portion 3 is given by,

T3=P3V3nR

Calculation:

The temperature for portion1 is calculated as,

T1=P1V1nR=( 2.00 atm)( 2 L)( ( 2.00 mole )( 0.08206 L⋅atm/ mol⋅K ))=24.4 K

The temperature for portion2 is calculated as,

T2=T1=24.4 K

The temperature for portion3 is calculated as,

T3=P3V3nR=( 2.00 atm)( 4 L)( ( 2.00 mole )( 0.08206 L⋅atm/ mol⋅K ))=48.7 K

Conclusion:

Therefore, the temperature at the end of cycle 1, 2 and 3 are 24.4 K , 24.4 K and 48.7 K respectively.

(c)

To determine

The head absorbed and work done during each cycle.

(c)

Expert Solution

Answer to Problem 74P

The heat absorbed for process 1, 2 and 3 are 281 J , 606 J and 1010.15 J respectively. The work done for process 1, 2 and 3 are 281 J , 0 and 405 J respectively.

Explanation of Solution

Formula used:

The expression for heat absorbed for portion 1 is given by,

Qin1=nRTlnV2V1

The expression for heat absorbed for portion 2 is given by,

Qin2=34nR(T3−T2)

The expression for heat absorbed for portion 3 is given by,

Qin3=52nR(T1−T3)

The expression for work for portion 1 is given by,

W1=Qin1

The expression for work done for portion 3 is given by,

W3=−P1,3(V1−V3)

Calculation:

The heat absorbed for portion1 is calculated as,

Qin1=nRTlnV2V1=(2.00 mol)(8.314 J/mol⋅K)(24.4 K)ln( 4 L 2 L)=281 J

The work done for portion1 is calculated as,

W1=Qin1=281 J

The heat absorbed for portion2 is calculated as,

Qin2=34nR(T3−T2)=32(2.00 mol)(8.314 J/mol⋅K)(48.7 K−24.4 K)=606 J

The heat absorbed for portion3 is calculated as,

Qin3=52nR(T1−T3)=52(2.00 mol)(8.314 J/mol⋅K)(48.7 K−24.4 K)=1010.15 J

The work done for portion2 is zero because it is process with constant volume.

The work done for portion3 is calculated as,

W3=−P1,3(V1−V3)=−(2.00 atm)(2 L−4 L)=−(( −4 atm⋅L)( 101.3 J 1 atm⋅L ))=405 J

Conclusion:

Therefore, the heat absorbed for process 1, 2 and 3 are 281 J , 606 J and 1010.15 J respectively. The work done for process 1, 2 and 3 are 281 J , 0 and 405 J respectively.

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Answer 2

Question

Chapter 18, Problem 74P

(a)

To determine

The P-V diagram.

(a)

Expert Solution

Answer to Problem 74P

The P-V diagram is shown in figure 1.

Explanation of Solution

Given:

The initial pressure is 2.00 atm .

The initial volume is 2.00 L

The final volume is 4.00 L

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 74P

Figure 1

Conclusion:

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

(b)

To determine

The temperature at end of each part of cycle.

(b)

Expert Solution

Answer to Problem 74P

The temperature at the end of cycle 1, 2 and 3 are 24.4 K , 24.4 K and 48.7 K respectively.

Explanation of Solution

Formula used:

The expression for temperature for portion 1 is given by,

T1=P1V1nR

The expression for temperature for portion 2 is given by,

T2=T1

The expression for temperature for portion 3 is given by,

T3=P3V3nR

Calculation:

The temperature for portion1 is calculated as,

T1=P1V1nR=( 2.00 atm)( 2 L)( ( 2.00 mole )( 0.08206 L⋅atm/ mol⋅K ))=24.4 K

The temperature for portion2 is calculated as,

T2=T1=24.4 K

The temperature for portion3 is calculated as,

T3=P3V3nR=( 2.00 atm)( 4 L)( ( 2.00 mole )( 0.08206 L⋅atm/ mol⋅K ))=48.7 K

Conclusion:

Therefore, the temperature at the end of cycle 1, 2 and 3 are 24.4 K , 24.4 K and 48.7 K respectively.

(c)

To determine

The head absorbed and work done during each cycle.

(c)

Expert Solution

Answer to Problem 74P

The heat absorbed for process 1, 2 and 3 are 281 J , 606 J and 1010.15 J respectively. The work done for process 1, 2 and 3 are 281 J , 0 and 405 J respectively.

Explanation of Solution

Formula used:

The expression for heat absorbed for portion 1 is given by,

Qin1=nRTlnV2V1

The expression for heat absorbed for portion 2 is given by,

Qin2=34nR(T3−T2)

The expression for heat absorbed for portion 3 is given by,

Qin3=52nR(T1−T3)

The expression for work for portion 1 is given by,

W1=Qin1

The expression for work done for portion 3 is given by,

W3=−P1,3(V1−V3)

Calculation:

The heat absorbed for portion1 is calculated as,

Qin1=nRTlnV2V1=(2.00 mol)(8.314 J/mol⋅K)(24.4 K)ln( 4 L 2 L)=281 J

The work done for portion1 is calculated as,

W1=Qin1=281 J

The heat absorbed for portion2 is calculated as,

Qin2=34nR(T3−T2)=32(2.00 mol)(8.314 J/mol⋅K)(48.7 K−24.4 K)=606 J

The heat absorbed for portion3 is calculated as,

Qin3=52nR(T1−T3)=52(2.00 mol)(8.314 J/mol⋅K)(48.7 K−24.4 K)=1010.15 J

The work done for portion2 is zero because it is process with constant volume.

The work done for portion3 is calculated as,

W3=−P1,3(V1−V3)=−(2.00 atm)(2 L−4 L)=−(( −4 atm⋅L)( 101.3 J 1 atm⋅L ))=405 J

Conclusion:

Therefore, the heat absorbed for process 1, 2 and 3 are 281 J , 606 J and 1010.15 J respectively. The work done for process 1, 2 and 3 are 281 J , 0 and 405 J respectively.

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Answer 3

Question

Chapter 18, Problem 74P

(a)

To determine

The P-V diagram.

(a)

Expert Solution

Answer to Problem 74P

The P-V diagram is shown in figure 1.

Explanation of Solution

Given:

The initial pressure is 2.00 atm .

The initial volume is 2.00 L

The final volume is 4.00 L

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 74P

Figure 1

Conclusion:

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

(b)

To determine

The temperature at end of each part of cycle.

(b)

Expert Solution

Answer to Problem 74P

The temperature at the end of cycle 1, 2 and 3 are 24.4 K , 24.4 K and 48.7 K respectively.

Explanation of Solution

Formula used:

The expression for temperature for portion 1 is given by,

T1=P1V1nR

The expression for temperature for portion 2 is given by,

T2=T1

The expression for temperature for portion 3 is given by,

T3=P3V3nR

Calculation:

The temperature for portion1 is calculated as,

T1=P1V1nR=( 2.00 atm)( 2 L)( ( 2.00 mole )( 0.08206 L⋅atm/ mol⋅K ))=24.4 K

The temperature for portion2 is calculated as,

T2=T1=24.4 K

The temperature for portion3 is calculated as,

T3=P3V3nR=( 2.00 atm)( 4 L)( ( 2.00 mole )( 0.08206 L⋅atm/ mol⋅K ))=48.7 K

Conclusion:

Therefore, the temperature at the end of cycle 1, 2 and 3 are 24.4 K , 24.4 K and 48.7 K respectively.

(c)

To determine

The head absorbed and work done during each cycle.

(c)

Expert Solution

Answer to Problem 74P

The heat absorbed for process 1, 2 and 3 are 281 J , 606 J and 1010.15 J respectively. The work done for process 1, 2 and 3 are 281 J , 0 and 405 J respectively.

Explanation of Solution

Formula used:

The expression for heat absorbed for portion 1 is given by,

Qin1=nRTlnV2V1

The expression for heat absorbed for portion 2 is given by,

Qin2=34nR(T3−T2)

The expression for heat absorbed for portion 3 is given by,

Qin3=52nR(T1−T3)

The expression for work for portion 1 is given by,

W1=Qin1

The expression for work done for portion 3 is given by,

W3=−P1,3(V1−V3)

Calculation:

The heat absorbed for portion1 is calculated as,

Qin1=nRTlnV2V1=(2.00 mol)(8.314 J/mol⋅K)(24.4 K)ln( 4 L 2 L)=281 J

The work done for portion1 is calculated as,

W1=Qin1=281 J

The heat absorbed for portion2 is calculated as,

Qin2=34nR(T3−T2)=32(2.00 mol)(8.314 J/mol⋅K)(48.7 K−24.4 K)=606 J

The heat absorbed for portion3 is calculated as,

Qin3=52nR(T1−T3)=52(2.00 mol)(8.314 J/mol⋅K)(48.7 K−24.4 K)=1010.15 J

The work done for portion2 is zero because it is process with constant volume.

The work done for portion3 is calculated as,

W3=−P1,3(V1−V3)=−(2.00 atm)(2 L−4 L)=−(( −4 atm⋅L)( 101.3 J 1 atm⋅L ))=405 J

Conclusion:

Therefore, the heat absorbed for process 1, 2 and 3 are 281 J , 606 J and 1010.15 J respectively. The work done for process 1, 2 and 3 are 281 J , 0 and 405 J respectively.

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Answer 4

Question

Chapter 18, Problem 74P

(a)

To determine

The P-V diagram.

(a)

Expert Solution

Answer to Problem 74P

The P-V diagram is shown in figure 1.

Explanation of Solution

Given:

The initial pressure is 2.00 atm .

The initial volume is 2.00 L

The final volume is 4.00 L

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 74P

Figure 1

Conclusion:

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

(b)

To determine

The temperature at end of each part of cycle.

(b)

Expert Solution

Answer to Problem 74P

The temperature at the end of cycle 1, 2 and 3 are 24.4 K , 24.4 K and 48.7 K respectively.

Explanation of Solution

Formula used:

The expression for temperature for portion 1 is given by,

T1=P1V1nR

The expression for temperature for portion 2 is given by,

T2=T1

The expression for temperature for portion 3 is given by,

T3=P3V3nR

Calculation:

The temperature for portion1 is calculated as,

T1=P1V1nR=( 2.00 atm)( 2 L)( ( 2.00 mole )( 0.08206 L⋅atm/ mol⋅K ))=24.4 K

The temperature for portion2 is calculated as,

T2=T1=24.4 K

The temperature for portion3 is calculated as,

T3=P3V3nR=( 2.00 atm)( 4 L)( ( 2.00 mole )( 0.08206 L⋅atm/ mol⋅K ))=48.7 K

Conclusion:

Therefore, the temperature at the end of cycle 1, 2 and 3 are 24.4 K , 24.4 K and 48.7 K respectively.

(c)

To determine

The head absorbed and work done during each cycle.

(c)

Expert Solution

Answer to Problem 74P

The heat absorbed for process 1, 2 and 3 are 281 J , 606 J and 1010.15 J respectively. The work done for process 1, 2 and 3 are 281 J , 0 and 405 J respectively.

Explanation of Solution

Formula used:

The expression for heat absorbed for portion 1 is given by,

Qin1=nRTlnV2V1

The expression for heat absorbed for portion 2 is given by,

Qin2=34nR(T3−T2)

The expression for heat absorbed for portion 3 is given by,

Qin3=52nR(T1−T3)

The expression for work for portion 1 is given by,

W1=Qin1

The expression for work done for portion 3 is given by,

W3=−P1,3(V1−V3)

Calculation:

The heat absorbed for portion1 is calculated as,

Qin1=nRTlnV2V1=(2.00 mol)(8.314 J/mol⋅K)(24.4 K)ln( 4 L 2 L)=281 J

The work done for portion1 is calculated as,

W1=Qin1=281 J

The heat absorbed for portion2 is calculated as,

Qin2=34nR(T3−T2)=32(2.00 mol)(8.314 J/mol⋅K)(48.7 K−24.4 K)=606 J

The heat absorbed for portion3 is calculated as,

Qin3=52nR(T1−T3)=52(2.00 mol)(8.314 J/mol⋅K)(48.7 K−24.4 K)=1010.15 J

The work done for portion2 is zero because it is process with constant volume.

The work done for portion3 is calculated as,

W3=−P1,3(V1−V3)=−(2.00 atm)(2 L−4 L)=−(( −4 atm⋅L)( 101.3 J 1 atm⋅L ))=405 J

Conclusion:

Therefore, the heat absorbed for process 1, 2 and 3 are 281 J , 606 J and 1010.15 J respectively. The work done for process 1, 2 and 3 are 281 J , 0 and 405 J respectively.

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Answer 5

Chapter 18, Problem 74P

(a)

To determine

The P-V diagram.

(a)

Expert Solution

Answer to Problem 74P

The P-V diagram is shown in figure 1.

Explanation of Solution

Given:

The initial pressure is 2.00 atm .

The initial volume is 2.00 L

The final volume is 4.00 L

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 74P

Figure 1

Conclusion:

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

(b)

To determine

The temperature at end of each part of cycle.

(b)

Expert Solution

Answer to Problem 74P

The temperature at the end of cycle 1, 2 and 3 are 24.4 K , 24.4 K and 48.7 K respectively.

Explanation of Solution

Formula used:

The expression for temperature for portion 1 is given by,

T1=P1V1nR

The expression for temperature for portion 2 is given by,

T2=T1

The expression for temperature for portion 3 is given by,

T3=P3V3nR

Calculation:

The temperature for portion1 is calculated as,

T1=P1V1nR=( 2.00 atm)( 2 L)( ( 2.00 mole )( 0.08206 L⋅atm/ mol⋅K ))=24.4 K

The temperature for portion2 is calculated as,

T2=T1=24.4 K

The temperature for portion3 is calculated as,

T3=P3V3nR=( 2.00 atm)( 4 L)( ( 2.00 mole )( 0.08206 L⋅atm/ mol⋅K ))=48.7 K

Conclusion:

Therefore, the temperature at the end of cycle 1, 2 and 3 are 24.4 K , 24.4 K and 48.7 K respectively.

(c)

To determine

The head absorbed and work done during each cycle.

(c)

Expert Solution

Answer to Problem 74P

The heat absorbed for process 1, 2 and 3 are 281 J , 606 J and 1010.15 J respectively. The work done for process 1, 2 and 3 are 281 J , 0 and 405 J respectively.

Explanation of Solution

Formula used:

The expression for heat absorbed for portion 1 is given by,

Qin1=nRTlnV2V1

The expression for heat absorbed for portion 2 is given by,

Qin2=34nR(T3−T2)

The expression for heat absorbed for portion 3 is given by,

Qin3=52nR(T1−T3)

The expression for work for portion 1 is given by,

W1=Qin1

The expression for work done for portion 3 is given by,

W3=−P1,3(V1−V3)

Calculation:

The heat absorbed for portion1 is calculated as,

Qin1=nRTlnV2V1=(2.00 mol)(8.314 J/mol⋅K)(24.4 K)ln( 4 L 2 L)=281 J

The work done for portion1 is calculated as,

W1=Qin1=281 J

The heat absorbed for portion2 is calculated as,

Qin2=34nR(T3−T2)=32(2.00 mol)(8.314 J/mol⋅K)(48.7 K−24.4 K)=606 J

The heat absorbed for portion3 is calculated as,

Qin3=52nR(T1−T3)=52(2.00 mol)(8.314 J/mol⋅K)(48.7 K−24.4 K)=1010.15 J

The work done for portion2 is zero because it is process with constant volume.

The work done for portion3 is calculated as,

W3=−P1,3(V1−V3)=−(2.00 atm)(2 L−4 L)=−(( −4 atm⋅L)( 101.3 J 1 atm⋅L ))=405 J

Conclusion:

Therefore, the heat absorbed for process 1, 2 and 3 are 281 J , 606 J and 1010.15 J respectively. The work done for process 1, 2 and 3 are 281 J , 0 and 405 J respectively.

Answer 6

Chapter 18, Problem 74P

(a)

To determine

The P-V diagram.

(a)

Expert Solution

Answer to Problem 74P

The P-V diagram is shown in figure 1.

Explanation of Solution

Given:

The initial pressure is 2.00 atm .

The initial volume is 2.00 L

The final volume is 4.00 L

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 74P

Figure 1

Conclusion:

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

Answer 7

Chapter 18, Problem 74P

(a)

To determine

The P-V diagram.

Answer 8

(a)

Expert Solution

Answer to Problem 74P

The P-V diagram is shown in figure 1.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

The initial pressure is 2.00 atm .

The initial volume is 2.00 L

The final volume is 4.00 L

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 74P

Figure 1

Conclusion:

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

Answer 13

(b)

To determine

The temperature at end of each part of cycle.

(b)

Expert Solution

Answer to Problem 74P

The temperature at the end of cycle 1, 2 and 3 are 24.4 K , 24.4 K and 48.7 K respectively.

Explanation of Solution

Formula used:

The expression for temperature for portion 1 is given by,

T1=P1V1nR

The expression for temperature for portion 2 is given by,

T2=T1

The expression for temperature for portion 3 is given by,

T3=P3V3nR

Calculation:

The temperature for portion1 is calculated as,

T1=P1V1nR=( 2.00 atm)( 2 L)( ( 2.00 mole )( 0.08206 L⋅atm/ mol⋅K ))=24.4 K

The temperature for portion2 is calculated as,

T2=T1=24.4 K

The temperature for portion3 is calculated as,

T3=P3V3nR=( 2.00 atm)( 4 L)( ( 2.00 mole )( 0.08206 L⋅atm/ mol⋅K ))=48.7 K

Conclusion:

Therefore, the temperature at the end of cycle 1, 2 and 3 are 24.4 K , 24.4 K and 48.7 K respectively.

Answer 14

(b)

To determine

The temperature at end of each part of cycle.

Answer 15

(b)

Expert Solution

Answer to Problem 74P

The temperature at the end of cycle 1, 2 and 3 are 24.4 K , 24.4 K and 48.7 K respectively.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Formula used:

The expression for temperature for portion 1 is given by,

T1=P1V1nR

The expression for temperature for portion 2 is given by,

T2=T1

The expression for temperature for portion 3 is given by,

T3=P3V3nR

Calculation:

The temperature for portion1 is calculated as,

T1=P1V1nR=( 2.00 atm)( 2 L)( ( 2.00 mole )( 0.08206 L⋅atm/ mol⋅K ))=24.4 K

The temperature for portion2 is calculated as,

T2=T1=24.4 K

The temperature for portion3 is calculated as,

T3=P3V3nR=( 2.00 atm)( 4 L)( ( 2.00 mole )( 0.08206 L⋅atm/ mol⋅K ))=48.7 K

Conclusion:

Therefore, the temperature at the end of cycle 1, 2 and 3 are 24.4 K , 24.4 K and 48.7 K respectively.

Answer 20

(c)

To determine

The head absorbed and work done during each cycle.

(c)

Expert Solution

Answer to Problem 74P

The heat absorbed for process 1, 2 and 3 are 281 J , 606 J and 1010.15 J respectively. The work done for process 1, 2 and 3 are 281 J , 0 and 405 J respectively.

Explanation of Solution

Formula used:

The expression for heat absorbed for portion 1 is given by,

Qin1=nRTlnV2V1

The expression for heat absorbed for portion 2 is given by,

Qin2=34nR(T3−T2)

The expression for heat absorbed for portion 3 is given by,

Qin3=52nR(T1−T3)

The expression for work for portion 1 is given by,

W1=Qin1

The expression for work done for portion 3 is given by,

W3=−P1,3(V1−V3)

Calculation:

The heat absorbed for portion1 is calculated as,

Qin1=nRTlnV2V1=(2.00 mol)(8.314 J/mol⋅K)(24.4 K)ln( 4 L 2 L)=281 J

The work done for portion1 is calculated as,

W1=Qin1=281 J

The heat absorbed for portion2 is calculated as,

Qin2=34nR(T3−T2)=32(2.00 mol)(8.314 J/mol⋅K)(48.7 K−24.4 K)=606 J

The heat absorbed for portion3 is calculated as,

Qin3=52nR(T1−T3)=52(2.00 mol)(8.314 J/mol⋅K)(48.7 K−24.4 K)=1010.15 J

The work done for portion2 is zero because it is process with constant volume.

The work done for portion3 is calculated as,

W3=−P1,3(V1−V3)=−(2.00 atm)(2 L−4 L)=−(( −4 atm⋅L)( 101.3 J 1 atm⋅L ))=405 J

Conclusion:

Therefore, the heat absorbed for process 1, 2 and 3 are 281 J , 606 J and 1010.15 J respectively. The work done for process 1, 2 and 3 are 281 J , 0 and 405 J respectively.

Answer 21

(c)

To determine

The head absorbed and work done during each cycle.

Answer 22

(c)

Expert Solution

Answer to Problem 74P

The heat absorbed for process 1, 2 and 3 are 281 J , 606 J and 1010.15 J respectively. The work done for process 1, 2 and 3 are 281 J , 0 and 405 J respectively.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Formula used:

The expression for heat absorbed for portion 1 is given by,

Qin1=nRTlnV2V1

The expression for heat absorbed for portion 2 is given by,

Qin2=34nR(T3−T2)

The expression for heat absorbed for portion 3 is given by,

Qin3=52nR(T1−T3)

The expression for work for portion 1 is given by,

W1=Qin1

The expression for work done for portion 3 is given by,

W3=−P1,3(V1−V3)

Calculation:

The heat absorbed for portion1 is calculated as,

Qin1=nRTlnV2V1=(2.00 mol)(8.314 J/mol⋅K)(24.4 K)ln( 4 L 2 L)=281 J

The work done for portion1 is calculated as,

W1=Qin1=281 J

The heat absorbed for portion2 is calculated as,

Qin2=34nR(T3−T2)=32(2.00 mol)(8.314 J/mol⋅K)(48.7 K−24.4 K)=606 J

The heat absorbed for portion3 is calculated as,

Qin3=52nR(T1−T3)=52(2.00 mol)(8.314 J/mol⋅K)(48.7 K−24.4 K)=1010.15 J

The work done for portion2 is zero because it is process with constant volume.

The work done for portion3 is calculated as,

W3=−P1,3(V1−V3)=−(2.00 atm)(2 L−4 L)=−(( −4 atm⋅L)( 101.3 J 1 atm⋅L ))=405 J

Conclusion:

Therefore, the heat absorbed for process 1, 2 and 3 are 281 J , 606 J and 1010.15 J respectively. The work done for process 1, 2 and 3 are 281 J , 0 and 405 J respectively.