The heat transfer during the process of piston-cylinder device.

Question

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Answer 1

Question

Chapter 4.5, Problem 118RP

To determine

The heat transfer during the process of piston-cylinder device.

Expert Solution & Answer

Answer to Problem 118RP

The heat transfer during the process of piston-cylinder device is −6.50 kJ_.

Explanation of Solution

Write the expression for mass of helium.

m=P1V1RT1 (I)

Here, the initial pressure of helium is P1, the initial volume of helium is V1, the universal gas constant is R, and the initial temperature is T1.

Write the expression of ideal gas relation for volume after compression.

P1V1RT1=P2V2RT2V2=T2P1T1P2V1 (II)

Here, the final pressure of helium is P1, the final volume of helium is V1, and the final temperature is T1.

Write the expression of polytropic index by using ideal gas relation.

P2V2n=P1V1n(P2P1)=(V1V2)n (III)

Here, the polytropic index exponent.

Write the expression of boundary work for the polytropic process.

Wb,in=−∫12P dV=−P2V2−P1V11−n=−mR(T2−T1)1−n (IV)

Here, the final temperature of helium is T2.

Write the expression for the energy balance equation.

Ein−Eout=ΔEsystem (V)

Here, the total energy entering the system is Ein, the total energy leaving the system is Eout, and the change in the total energy of the system is ΔEsystem.

Simplify Equation (V) and write energy balance relation of piston-cylinder device.

Qin+Wb,in=ΔUQin=m(u1−u2)−Wb,inQin=mcV(T1−T2)−Wb,in (VI)

Here, the heat to be transfer into the system is Qin, the constant-volume specific heat of helium is cV, the work to be done by the system is Wout, and the change in the internal energy of a system is ΔU.

Conclusion:

From the Table A-1 “Molar mass, gas constant, and critical-point properties”, obtain the value of gas constant of helium for piston-cylinder device as 2.0769 kPa⋅m3/kg ⋅K.

From the Table A-2 “Ideal-gas specific heats of various common gases”, obtain the value of constant-volume specific heat of helium for piston-cylinder device as 3.1156 kJ/kg ⋅K.

Substitute 100 kPa for P1, 0.2 m3 for V1, 2.0769 kPa⋅m3/kg ⋅K for R, and 283 K for T1 in Equation (I).

m=(100 kPa)(0.2 m3)(2.0769 kPa⋅m3/kg ⋅K)(283 K)=20 kPa⋅m3587.7627 kPa⋅m3/kg =0.03403 kg

Substitute 563 K for T2, 283 K for T1, 100 kPa for P1, 700 kPa for P2, and 0.2 m3 for V1 in Equation (II).

V2=(563 K)(100 kPa)(283 K)(700 kPa)×(0.2 m3)=56300 kPa⋅K198100 kPa⋅K×(0.2 m3)=(0.2842)×(0.2 m3)=0.05684 m3

Substitute 100 kPa for P1, 700 kPa for P2, 0.2 m3 for V1, and 0.05684 m3 for V2 in Equation (III).

700 kPa100 kPa=(0.2 m30.05684 m3)n7=(3.518649)nn=1.547

Substitute 0.03403 kg for m, 2.0769 kPa⋅m3/kg ⋅K for R, 1.547 for n, 563 K for T2, and 283 K for T1 in Equation (IV).

Wb,in=−(0.03403 kg)(2.0769 kPa⋅m3/kg ⋅K)(563 K−283 K)1−1.547=−(0.03403 kg)(2.0769 kJ/kg ⋅K)(280 K)(−0.547)=−19.7895 kJ−0.547=36.178 kJ

……≅36.18 kJ

Substitute 0.03403 kg for m, 3.1156 kJ/kg⋅K for cV, 563 K for T2, 283 K for T1, and 36.18 kJ for Wb,in in Equation (VI).

Qin=(0.03403 kg)(3.1156 kJ/kg⋅K)(563 K−283 K)−(36.18 kJ)=(0.03403 kg)(3.1156 kJ/kg⋅K)(280 K)−(36.18 kJ)=−6.493 kJ≅−6.50 kJ

Thus, the heat transfer during the process of piston-cylinder device is −6.50 kJ_.

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Answer 2

Question

Chapter 4.5, Problem 118RP

To determine

The heat transfer during the process of piston-cylinder device.

Expert Solution & Answer

Answer to Problem 118RP

The heat transfer during the process of piston-cylinder device is −6.50 kJ_.

Explanation of Solution

Write the expression for mass of helium.

m=P1V1RT1 (I)

Here, the initial pressure of helium is P1, the initial volume of helium is V1, the universal gas constant is R, and the initial temperature is T1.

Write the expression of ideal gas relation for volume after compression.

P1V1RT1=P2V2RT2V2=T2P1T1P2V1 (II)

Here, the final pressure of helium is P1, the final volume of helium is V1, and the final temperature is T1.

Write the expression of polytropic index by using ideal gas relation.

P2V2n=P1V1n(P2P1)=(V1V2)n (III)

Here, the polytropic index exponent.

Write the expression of boundary work for the polytropic process.

Wb,in=−∫12P dV=−P2V2−P1V11−n=−mR(T2−T1)1−n (IV)

Here, the final temperature of helium is T2.

Write the expression for the energy balance equation.

Ein−Eout=ΔEsystem (V)

Here, the total energy entering the system is Ein, the total energy leaving the system is Eout, and the change in the total energy of the system is ΔEsystem.

Simplify Equation (V) and write energy balance relation of piston-cylinder device.

Qin+Wb,in=ΔUQin=m(u1−u2)−Wb,inQin=mcV(T1−T2)−Wb,in (VI)

Here, the heat to be transfer into the system is Qin, the constant-volume specific heat of helium is cV, the work to be done by the system is Wout, and the change in the internal energy of a system is ΔU.

Conclusion:

From the Table A-1 “Molar mass, gas constant, and critical-point properties”, obtain the value of gas constant of helium for piston-cylinder device as 2.0769 kPa⋅m3/kg ⋅K.

From the Table A-2 “Ideal-gas specific heats of various common gases”, obtain the value of constant-volume specific heat of helium for piston-cylinder device as 3.1156 kJ/kg ⋅K.

Substitute 100 kPa for P1, 0.2 m3 for V1, 2.0769 kPa⋅m3/kg ⋅K for R, and 283 K for T1 in Equation (I).

m=(100 kPa)(0.2 m3)(2.0769 kPa⋅m3/kg ⋅K)(283 K)=20 kPa⋅m3587.7627 kPa⋅m3/kg =0.03403 kg

Substitute 563 K for T2, 283 K for T1, 100 kPa for P1, 700 kPa for P2, and 0.2 m3 for V1 in Equation (II).

V2=(563 K)(100 kPa)(283 K)(700 kPa)×(0.2 m3)=56300 kPa⋅K198100 kPa⋅K×(0.2 m3)=(0.2842)×(0.2 m3)=0.05684 m3

Substitute 100 kPa for P1, 700 kPa for P2, 0.2 m3 for V1, and 0.05684 m3 for V2 in Equation (III).

700 kPa100 kPa=(0.2 m30.05684 m3)n7=(3.518649)nn=1.547

Substitute 0.03403 kg for m, 2.0769 kPa⋅m3/kg ⋅K for R, 1.547 for n, 563 K for T2, and 283 K for T1 in Equation (IV).

Wb,in=−(0.03403 kg)(2.0769 kPa⋅m3/kg ⋅K)(563 K−283 K)1−1.547=−(0.03403 kg)(2.0769 kJ/kg ⋅K)(280 K)(−0.547)=−19.7895 kJ−0.547=36.178 kJ

……≅36.18 kJ

Substitute 0.03403 kg for m, 3.1156 kJ/kg⋅K for cV, 563 K for T2, 283 K for T1, and 36.18 kJ for Wb,in in Equation (VI).

Qin=(0.03403 kg)(3.1156 kJ/kg⋅K)(563 K−283 K)−(36.18 kJ)=(0.03403 kg)(3.1156 kJ/kg⋅K)(280 K)−(36.18 kJ)=−6.493 kJ≅−6.50 kJ

Thus, the heat transfer during the process of piston-cylinder device is −6.50 kJ_.

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Answer 3

Question

Chapter 4.5, Problem 118RP

To determine

The heat transfer during the process of piston-cylinder device.

Expert Solution & Answer

Answer to Problem 118RP

The heat transfer during the process of piston-cylinder device is −6.50 kJ_.

Explanation of Solution

Write the expression for mass of helium.

m=P1V1RT1 (I)

Here, the initial pressure of helium is P1, the initial volume of helium is V1, the universal gas constant is R, and the initial temperature is T1.

Write the expression of ideal gas relation for volume after compression.

P1V1RT1=P2V2RT2V2=T2P1T1P2V1 (II)

Here, the final pressure of helium is P1, the final volume of helium is V1, and the final temperature is T1.

Write the expression of polytropic index by using ideal gas relation.

P2V2n=P1V1n(P2P1)=(V1V2)n (III)

Here, the polytropic index exponent.

Write the expression of boundary work for the polytropic process.

Wb,in=−∫12P dV=−P2V2−P1V11−n=−mR(T2−T1)1−n (IV)

Here, the final temperature of helium is T2.

Write the expression for the energy balance equation.

Ein−Eout=ΔEsystem (V)

Here, the total energy entering the system is Ein, the total energy leaving the system is Eout, and the change in the total energy of the system is ΔEsystem.

Simplify Equation (V) and write energy balance relation of piston-cylinder device.

Qin+Wb,in=ΔUQin=m(u1−u2)−Wb,inQin=mcV(T1−T2)−Wb,in (VI)

Here, the heat to be transfer into the system is Qin, the constant-volume specific heat of helium is cV, the work to be done by the system is Wout, and the change in the internal energy of a system is ΔU.

Conclusion:

From the Table A-1 “Molar mass, gas constant, and critical-point properties”, obtain the value of gas constant of helium for piston-cylinder device as 2.0769 kPa⋅m3/kg ⋅K.

From the Table A-2 “Ideal-gas specific heats of various common gases”, obtain the value of constant-volume specific heat of helium for piston-cylinder device as 3.1156 kJ/kg ⋅K.

Substitute 100 kPa for P1, 0.2 m3 for V1, 2.0769 kPa⋅m3/kg ⋅K for R, and 283 K for T1 in Equation (I).

m=(100 kPa)(0.2 m3)(2.0769 kPa⋅m3/kg ⋅K)(283 K)=20 kPa⋅m3587.7627 kPa⋅m3/kg =0.03403 kg

Substitute 563 K for T2, 283 K for T1, 100 kPa for P1, 700 kPa for P2, and 0.2 m3 for V1 in Equation (II).

V2=(563 K)(100 kPa)(283 K)(700 kPa)×(0.2 m3)=56300 kPa⋅K198100 kPa⋅K×(0.2 m3)=(0.2842)×(0.2 m3)=0.05684 m3

Substitute 100 kPa for P1, 700 kPa for P2, 0.2 m3 for V1, and 0.05684 m3 for V2 in Equation (III).

700 kPa100 kPa=(0.2 m30.05684 m3)n7=(3.518649)nn=1.547

Substitute 0.03403 kg for m, 2.0769 kPa⋅m3/kg ⋅K for R, 1.547 for n, 563 K for T2, and 283 K for T1 in Equation (IV).

Wb,in=−(0.03403 kg)(2.0769 kPa⋅m3/kg ⋅K)(563 K−283 K)1−1.547=−(0.03403 kg)(2.0769 kJ/kg ⋅K)(280 K)(−0.547)=−19.7895 kJ−0.547=36.178 kJ

……≅36.18 kJ

Substitute 0.03403 kg for m, 3.1156 kJ/kg⋅K for cV, 563 K for T2, 283 K for T1, and 36.18 kJ for Wb,in in Equation (VI).

Qin=(0.03403 kg)(3.1156 kJ/kg⋅K)(563 K−283 K)−(36.18 kJ)=(0.03403 kg)(3.1156 kJ/kg⋅K)(280 K)−(36.18 kJ)=−6.493 kJ≅−6.50 kJ

Thus, the heat transfer during the process of piston-cylinder device is −6.50 kJ_.

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Answer 4

Question

Chapter 4.5, Problem 118RP

To determine

The heat transfer during the process of piston-cylinder device.

Expert Solution & Answer

Answer to Problem 118RP

The heat transfer during the process of piston-cylinder device is −6.50 kJ_.

Explanation of Solution

Write the expression for mass of helium.

m=P1V1RT1 (I)

Here, the initial pressure of helium is P1, the initial volume of helium is V1, the universal gas constant is R, and the initial temperature is T1.

Write the expression of ideal gas relation for volume after compression.

P1V1RT1=P2V2RT2V2=T2P1T1P2V1 (II)

Here, the final pressure of helium is P1, the final volume of helium is V1, and the final temperature is T1.

Write the expression of polytropic index by using ideal gas relation.

P2V2n=P1V1n(P2P1)=(V1V2)n (III)

Here, the polytropic index exponent.

Write the expression of boundary work for the polytropic process.

Wb,in=−∫12P dV=−P2V2−P1V11−n=−mR(T2−T1)1−n (IV)

Here, the final temperature of helium is T2.

Write the expression for the energy balance equation.

Ein−Eout=ΔEsystem (V)

Here, the total energy entering the system is Ein, the total energy leaving the system is Eout, and the change in the total energy of the system is ΔEsystem.

Simplify Equation (V) and write energy balance relation of piston-cylinder device.

Qin+Wb,in=ΔUQin=m(u1−u2)−Wb,inQin=mcV(T1−T2)−Wb,in (VI)

Here, the heat to be transfer into the system is Qin, the constant-volume specific heat of helium is cV, the work to be done by the system is Wout, and the change in the internal energy of a system is ΔU.

Conclusion:

From the Table A-1 “Molar mass, gas constant, and critical-point properties”, obtain the value of gas constant of helium for piston-cylinder device as 2.0769 kPa⋅m3/kg ⋅K.

From the Table A-2 “Ideal-gas specific heats of various common gases”, obtain the value of constant-volume specific heat of helium for piston-cylinder device as 3.1156 kJ/kg ⋅K.

Substitute 100 kPa for P1, 0.2 m3 for V1, 2.0769 kPa⋅m3/kg ⋅K for R, and 283 K for T1 in Equation (I).

m=(100 kPa)(0.2 m3)(2.0769 kPa⋅m3/kg ⋅K)(283 K)=20 kPa⋅m3587.7627 kPa⋅m3/kg =0.03403 kg

Substitute 563 K for T2, 283 K for T1, 100 kPa for P1, 700 kPa for P2, and 0.2 m3 for V1 in Equation (II).

V2=(563 K)(100 kPa)(283 K)(700 kPa)×(0.2 m3)=56300 kPa⋅K198100 kPa⋅K×(0.2 m3)=(0.2842)×(0.2 m3)=0.05684 m3

Substitute 100 kPa for P1, 700 kPa for P2, 0.2 m3 for V1, and 0.05684 m3 for V2 in Equation (III).

700 kPa100 kPa=(0.2 m30.05684 m3)n7=(3.518649)nn=1.547

Substitute 0.03403 kg for m, 2.0769 kPa⋅m3/kg ⋅K for R, 1.547 for n, 563 K for T2, and 283 K for T1 in Equation (IV).

Wb,in=−(0.03403 kg)(2.0769 kPa⋅m3/kg ⋅K)(563 K−283 K)1−1.547=−(0.03403 kg)(2.0769 kJ/kg ⋅K)(280 K)(−0.547)=−19.7895 kJ−0.547=36.178 kJ

……≅36.18 kJ

Substitute 0.03403 kg for m, 3.1156 kJ/kg⋅K for cV, 563 K for T2, 283 K for T1, and 36.18 kJ for Wb,in in Equation (VI).

Qin=(0.03403 kg)(3.1156 kJ/kg⋅K)(563 K−283 K)−(36.18 kJ)=(0.03403 kg)(3.1156 kJ/kg⋅K)(280 K)−(36.18 kJ)=−6.493 kJ≅−6.50 kJ

Thus, the heat transfer during the process of piston-cylinder device is −6.50 kJ_.

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Answer 5

Chapter 4.5, Problem 118RP

To determine

The heat transfer during the process of piston-cylinder device.

Expert Solution & Answer

Answer to Problem 118RP

The heat transfer during the process of piston-cylinder device is −6.50 kJ_.

Explanation of Solution

Write the expression for mass of helium.

m=P1V1RT1 (I)

Here, the initial pressure of helium is P1, the initial volume of helium is V1, the universal gas constant is R, and the initial temperature is T1.

Write the expression of ideal gas relation for volume after compression.

P1V1RT1=P2V2RT2V2=T2P1T1P2V1 (II)

Here, the final pressure of helium is P1, the final volume of helium is V1, and the final temperature is T1.

Write the expression of polytropic index by using ideal gas relation.

P2V2n=P1V1n(P2P1)=(V1V2)n (III)

Here, the polytropic index exponent.

Write the expression of boundary work for the polytropic process.

Wb,in=−∫12P dV=−P2V2−P1V11−n=−mR(T2−T1)1−n (IV)

Here, the final temperature of helium is T2.

Write the expression for the energy balance equation.

Ein−Eout=ΔEsystem (V)

Here, the total energy entering the system is Ein, the total energy leaving the system is Eout, and the change in the total energy of the system is ΔEsystem.

Simplify Equation (V) and write energy balance relation of piston-cylinder device.

Qin+Wb,in=ΔUQin=m(u1−u2)−Wb,inQin=mcV(T1−T2)−Wb,in (VI)

Here, the heat to be transfer into the system is Qin, the constant-volume specific heat of helium is cV, the work to be done by the system is Wout, and the change in the internal energy of a system is ΔU.

Conclusion:

From the Table A-1 “Molar mass, gas constant, and critical-point properties”, obtain the value of gas constant of helium for piston-cylinder device as 2.0769 kPa⋅m3/kg ⋅K.

From the Table A-2 “Ideal-gas specific heats of various common gases”, obtain the value of constant-volume specific heat of helium for piston-cylinder device as 3.1156 kJ/kg ⋅K.

Substitute 100 kPa for P1, 0.2 m3 for V1, 2.0769 kPa⋅m3/kg ⋅K for R, and 283 K for T1 in Equation (I).

m=(100 kPa)(0.2 m3)(2.0769 kPa⋅m3/kg ⋅K)(283 K)=20 kPa⋅m3587.7627 kPa⋅m3/kg =0.03403 kg

Substitute 563 K for T2, 283 K for T1, 100 kPa for P1, 700 kPa for P2, and 0.2 m3 for V1 in Equation (II).

V2=(563 K)(100 kPa)(283 K)(700 kPa)×(0.2 m3)=56300 kPa⋅K198100 kPa⋅K×(0.2 m3)=(0.2842)×(0.2 m3)=0.05684 m3

Substitute 100 kPa for P1, 700 kPa for P2, 0.2 m3 for V1, and 0.05684 m3 for V2 in Equation (III).

700 kPa100 kPa=(0.2 m30.05684 m3)n7=(3.518649)nn=1.547

Substitute 0.03403 kg for m, 2.0769 kPa⋅m3/kg ⋅K for R, 1.547 for n, 563 K for T2, and 283 K for T1 in Equation (IV).

Wb,in=−(0.03403 kg)(2.0769 kPa⋅m3/kg ⋅K)(563 K−283 K)1−1.547=−(0.03403 kg)(2.0769 kJ/kg ⋅K)(280 K)(−0.547)=−19.7895 kJ−0.547=36.178 kJ

……≅36.18 kJ

Substitute 0.03403 kg for m, 3.1156 kJ/kg⋅K for cV, 563 K for T2, 283 K for T1, and 36.18 kJ for Wb,in in Equation (VI).

Qin=(0.03403 kg)(3.1156 kJ/kg⋅K)(563 K−283 K)−(36.18 kJ)=(0.03403 kg)(3.1156 kJ/kg⋅K)(280 K)−(36.18 kJ)=−6.493 kJ≅−6.50 kJ

Thus, the heat transfer during the process of piston-cylinder device is −6.50 kJ_.

Answer 6

Chapter 4.5, Problem 118RP

To determine

The heat transfer during the process of piston-cylinder device.

Expert Solution & Answer

Answer to Problem 118RP

The heat transfer during the process of piston-cylinder device is −6.50 kJ_.

Explanation of Solution

Write the expression for mass of helium.

m=P1V1RT1 (I)

Here, the initial pressure of helium is P1, the initial volume of helium is V1, the universal gas constant is R, and the initial temperature is T1.

Write the expression of ideal gas relation for volume after compression.

P1V1RT1=P2V2RT2V2=T2P1T1P2V1 (II)

Here, the final pressure of helium is P1, the final volume of helium is V1, and the final temperature is T1.

Write the expression of polytropic index by using ideal gas relation.

P2V2n=P1V1n(P2P1)=(V1V2)n (III)

Here, the polytropic index exponent.

Write the expression of boundary work for the polytropic process.

Wb,in=−∫12P dV=−P2V2−P1V11−n=−mR(T2−T1)1−n (IV)

Here, the final temperature of helium is T2.

Write the expression for the energy balance equation.

Ein−Eout=ΔEsystem (V)

Here, the total energy entering the system is Ein, the total energy leaving the system is Eout, and the change in the total energy of the system is ΔEsystem.

Simplify Equation (V) and write energy balance relation of piston-cylinder device.

Qin+Wb,in=ΔUQin=m(u1−u2)−Wb,inQin=mcV(T1−T2)−Wb,in (VI)

Here, the heat to be transfer into the system is Qin, the constant-volume specific heat of helium is cV, the work to be done by the system is Wout, and the change in the internal energy of a system is ΔU.

Conclusion:

From the Table A-1 “Molar mass, gas constant, and critical-point properties”, obtain the value of gas constant of helium for piston-cylinder device as 2.0769 kPa⋅m3/kg ⋅K.

From the Table A-2 “Ideal-gas specific heats of various common gases”, obtain the value of constant-volume specific heat of helium for piston-cylinder device as 3.1156 kJ/kg ⋅K.

Substitute 100 kPa for P1, 0.2 m3 for V1, 2.0769 kPa⋅m3/kg ⋅K for R, and 283 K for T1 in Equation (I).

m=(100 kPa)(0.2 m3)(2.0769 kPa⋅m3/kg ⋅K)(283 K)=20 kPa⋅m3587.7627 kPa⋅m3/kg =0.03403 kg

Substitute 563 K for T2, 283 K for T1, 100 kPa for P1, 700 kPa for P2, and 0.2 m3 for V1 in Equation (II).

V2=(563 K)(100 kPa)(283 K)(700 kPa)×(0.2 m3)=56300 kPa⋅K198100 kPa⋅K×(0.2 m3)=(0.2842)×(0.2 m3)=0.05684 m3

Substitute 100 kPa for P1, 700 kPa for P2, 0.2 m3 for V1, and 0.05684 m3 for V2 in Equation (III).

700 kPa100 kPa=(0.2 m30.05684 m3)n7=(3.518649)nn=1.547

Substitute 0.03403 kg for m, 2.0769 kPa⋅m3/kg ⋅K for R, 1.547 for n, 563 K for T2, and 283 K for T1 in Equation (IV).

Wb,in=−(0.03403 kg)(2.0769 kPa⋅m3/kg ⋅K)(563 K−283 K)1−1.547=−(0.03403 kg)(2.0769 kJ/kg ⋅K)(280 K)(−0.547)=−19.7895 kJ−0.547=36.178 kJ

……≅36.18 kJ

Substitute 0.03403 kg for m, 3.1156 kJ/kg⋅K for cV, 563 K for T2, 283 K for T1, and 36.18 kJ for Wb,in in Equation (VI).

Qin=(0.03403 kg)(3.1156 kJ/kg⋅K)(563 K−283 K)−(36.18 kJ)=(0.03403 kg)(3.1156 kJ/kg⋅K)(280 K)−(36.18 kJ)=−6.493 kJ≅−6.50 kJ

Thus, the heat transfer during the process of piston-cylinder device is −6.50 kJ_.

Answer 7

Chapter 4.5, Problem 118RP

To determine

The heat transfer during the process of piston-cylinder device.

Answer 8

Expert Solution & Answer

Answer to Problem 118RP

The heat transfer during the process of piston-cylinder device is −6.50 kJ_.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Write the expression for mass of helium.

m=P1V1RT1 (I)

Here, the initial pressure of helium is P1, the initial volume of helium is V1, the universal gas constant is R, and the initial temperature is T1.

Write the expression of ideal gas relation for volume after compression.

P1V1RT1=P2V2RT2V2=T2P1T1P2V1 (II)

Here, the final pressure of helium is P1, the final volume of helium is V1, and the final temperature is T1.

Write the expression of polytropic index by using ideal gas relation.

P2V2n=P1V1n(P2P1)=(V1V2)n (III)

Here, the polytropic index exponent.

Write the expression of boundary work for the polytropic process.

Wb,in=−∫12P dV=−P2V2−P1V11−n=−mR(T2−T1)1−n (IV)

Here, the final temperature of helium is T2.

Write the expression for the energy balance equation.

Ein−Eout=ΔEsystem (V)

Here, the total energy entering the system is Ein, the total energy leaving the system is Eout, and the change in the total energy of the system is ΔEsystem.

Simplify Equation (V) and write energy balance relation of piston-cylinder device.

Qin+Wb,in=ΔUQin=m(u1−u2)−Wb,inQin=mcV(T1−T2)−Wb,in (VI)

Here, the heat to be transfer into the system is Qin, the constant-volume specific heat of helium is cV, the work to be done by the system is Wout, and the change in the internal energy of a system is ΔU.

Conclusion:

From the Table A-1 “Molar mass, gas constant, and critical-point properties”, obtain the value of gas constant of helium for piston-cylinder device as 2.0769 kPa⋅m3/kg ⋅K.

From the Table A-2 “Ideal-gas specific heats of various common gases”, obtain the value of constant-volume specific heat of helium for piston-cylinder device as 3.1156 kJ/kg ⋅K.

Substitute 100 kPa for P1, 0.2 m3 for V1, 2.0769 kPa⋅m3/kg ⋅K for R, and 283 K for T1 in Equation (I).

m=(100 kPa)(0.2 m3)(2.0769 kPa⋅m3/kg ⋅K)(283 K)=20 kPa⋅m3587.7627 kPa⋅m3/kg =0.03403 kg

Substitute 563 K for T2, 283 K for T1, 100 kPa for P1, 700 kPa for P2, and 0.2 m3 for V1 in Equation (II).

V2=(563 K)(100 kPa)(283 K)(700 kPa)×(0.2 m3)=56300 kPa⋅K198100 kPa⋅K×(0.2 m3)=(0.2842)×(0.2 m3)=0.05684 m3

Substitute 100 kPa for P1, 700 kPa for P2, 0.2 m3 for V1, and 0.05684 m3 for V2 in Equation (III).

700 kPa100 kPa=(0.2 m30.05684 m3)n7=(3.518649)nn=1.547

Substitute 0.03403 kg for m, 2.0769 kPa⋅m3/kg ⋅K for R, 1.547 for n, 563 K for T2, and 283 K for T1 in Equation (IV).

Wb,in=−(0.03403 kg)(2.0769 kPa⋅m3/kg ⋅K)(563 K−283 K)1−1.547=−(0.03403 kg)(2.0769 kJ/kg ⋅K)(280 K)(−0.547)=−19.7895 kJ−0.547=36.178 kJ

……≅36.18 kJ

Substitute 0.03403 kg for m, 3.1156 kJ/kg⋅K for cV, 563 K for T2, 283 K for T1, and 36.18 kJ for Wb,in in Equation (VI).

Qin=(0.03403 kg)(3.1156 kJ/kg⋅K)(563 K−283 K)−(36.18 kJ)=(0.03403 kg)(3.1156 kJ/kg⋅K)(280 K)−(36.18 kJ)=−6.493 kJ≅−6.50 kJ

Thus, the heat transfer during the process of piston-cylinder device is −6.50 kJ_.

Answer 12

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The heat transfer during the process of piston-cylinder device.

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