The PV diagram for each process in the cycle.

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Answer 1

Question

Chapter 18, Problem 72P

(a)

To determine

The PV diagram for each process in the cycle.

(a)

Expert Solution

Answer to Problem 72P

The required PV diagram is shown in Figure 1.

Explanation of Solution

Given data:

The amount n of sample of N2 gas is 1 mol .

The temperature T1 of the gas is 20° C .

The pressure P1 of the gas is 5 atm .

The pressure P2 of the gas after adiabatic expansion is 1 atm .

The temperature T2 of the gas after adiabatic process is t1° C .

The pressure P3 of the gas after isobaric expansion is 1 atm .

The temperature T3 of the gas after isobaric expansion process is 20° C .

The pressure P4 of the gas after isometric expansion is 5 atm .

The temperature T4 of the gas after isometric process is t2° C .

The pressure P5 of the gas after isobaric compression expansion is 5 atm .

The temperature T5 of the gas after isobaric compression process is 20° C .

Formula:

The expression to determine the initial volume of the gas is given by,

V1=nRT1P1

The volume of the gas after the adiabatic process is given by,

V2=( P 1 P 2 )1γV1

The expression for the temperature of the gas after adiabatic process is given by,

T2=( V 1 V 2 )γ−1T1

The volume of the gas after the isobaric process is given by,

V3=(T3T2)V2

The expression to determine the temperature of the gas after the isometric expansion is given by,

T4=P4P3T3

The expression for the volume of the gas after the isobaric process is given by,

V5=(T5T4)V4

Calculation:

The initial volume of the gas is calculated as,

V1=nRT1P1=(1)( 0.08206)( 20°C)5 atm=(1)( 0.08206)( 20 K+273 K)5 atm=4.81 L

The volume of the gas after the adiabatic process is calculated as,

V2=( P 1 P 2 )1γV1=( 5 atm 1 atm)1 1.44.81 L=15.18 L

The temperature of the gas after adiabatic process is calculated as,

T2=( V 1 V 2 )γ−1T1=( 4.81 L 15.18 L)γ−1(293 K)=185.02 K=(185.02 K−273°C)

Solve further as,

T2=−88.14°C

The volume of the gas after the isobaric process is calculated as,

V3=( T 3 T 2 )V2=( 293 K 185.02)(15.18 L)=24.04 L

The expression to determine the temperature of the gas after the isometric expansion is given by,

T4=P4P3T3=51(293 K)=(1465 K−273 K)=1192° C

The expression for the volume of the gas after the isobaric process is given by,

V5=( T 5 T 4 )V4=( 293 K 1465 K)(24.04° C)=4.81 L

From the above calculations the PV diagram for the different process is shown below.

The required diagram is shown in Figure 1

Physics for Scientists and Engineers, Vol. 1, Chapter 18, Problem 72P

Figure 1

Conclusion:

Therefore, the required PV diagram is shown in Figure 1.

(b)

To determine

The work done by the gas during the complete cycle.

(b)

Expert Solution

Answer to Problem 72P

The work done in the complete cycle is 6763 J .

Explanation of Solution

Formula Used:

The expression to determine the work done by the gas from A to B is given by,

WAB=P2V2−P1V1γ−1

The expression to determine the work done by the gas from B to C is given by,

WBC=−P2(V3−V2)

The expression to determine the work done by the gas from D to A is given by,

WDA=−P1(V1−V4)

The expression for the total work during the complete cycle is given by,

W=WAB+WBC+WDA=P2V2−P1V1γ−1−P2(V3−V2)−P1(V1−V4)

Calculation:

The total work during the complete cycle is calculated as,

W=P2V2−P1V1γ−1−P2(V3−V2)−P1(V1−V4)=( 1 atm)( 15.82 L)−( 1.5 atm)( 4.81 L)1.4−1−(1)(24.04 L−15.18 L)−(5 atm)(4.81 L−24.04 L)=−20.575 L⋅atm−8.86 L⋅atm+96.15 L⋅atm=(−20.575 L⋅atm−8.86 L⋅atm+96.15 L⋅atm)101.3 J

Solve further as,

W=6763 J

Conclusion:

Therefore, the work done in the complete cycle is 6763 J .

(c)

To determine

The heat absorbed by the gas during the complete cycle.

(c)

Expert Solution

Answer to Problem 72P

The heat absorbed by the gas is 6763 J .

Explanation of Solution

Formula:

The expression for the first law of thermodynamics is given by,

dQin=−dW

Calculation:

The heat absorbed by the gas is calculated as,

dQin=−dW=−6763 J

Conclusion:

Therefore, the heat absorbed by the gas is 6763 J .

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