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An ideal gas is initially at a pressure of 0.1 MPa and a total volume of 2m3. It is first compressed
to 1MPa by a reversible adiabatic process and then cooled at constant pressure to a final volume
of 0.2m3. The total work done (in kJ) on the gas for the entire process (up to one decimal place) is
____________
Data: R = 8.314 J/molK; heat capacity at constant pressure (CP) = 2.5R
to 1MPa by a reversible adiabatic process and then cooled at constant pressure to a final volume
of 0.2m3. The total work done (in kJ) on the gas for the entire process (up to one decimal place) is
____________
Data: R = 8.314 J/molK; heat capacity at constant pressure (CP) = 2.5R
Imp : you should answer only the numeric value
Correct answer is '750'. Can you explain this answer?
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Verified Answer
An ideal gas is initially at a pressure of 0.1 MPa and a total volume ...
1→2 : Re versible adiabatic process
2→3: Re versible isobaric process
2→3: Re versible isobaric process
Most Upvoted Answer
An ideal gas is initially at a pressure of 0.1 MPa and a total volume ...
Given data:
Initial pressure (P1) = 0.1 MPa = 0.1 * 10^6 Pa
Initial volume (V1) = 2 m^3
Final pressure (P2) = 1 MPa = 1 * 10^6 Pa
Final volume (V2) = 0.2 m^3
R = 8.314 J/molK
Heat capacity at constant pressure (Cp) = 2.5R
Step 1: Calculating the work done during compression (W1)
Using the equation for reversible adiabatic process:
P1 * V1^γ = P2 * V2^γ
Where γ is the heat capacity ratio, given by:
γ = Cp / (Cp - R)
Calculating the value of γ:
γ = 2.5R / (2.5R - R)
γ = 2.5
Using the equation for work done during adiabatic compression:
W1 = (P2 * V2 - P1 * V1) / (γ - 1)
Substituting the given values:
W1 = (1 * 10^6 * 0.2 - 0.1 * 10^6 * 2) / (2.5 - 1)
W1 = (200000 - 200000) / 1.5
W1 = 0
Step 2: Calculating the work done during cooling at constant pressure (W2)
Using the equation for work done at constant pressure:
W2 = P2 * (V2 - V1)
Substituting the given values:
W2 = 1 * 10^6 * (0.2 - 2)
W2 = 1 * 10^6 * (-1.8)
W2 = -1.8 * 10^6
Step 3: Calculating the total work done (W_total)
W_total = W1 + W2
W_total = 0 + (-1.8 * 10^6)
W_total = -1.8 * 10^6 J
Converting the work done to kJ:
W_total = -1.8 * 10^3 kJ
The negative sign indicates that work is done on the gas.
Final Answer:
The total work done on the gas for the entire process is -1800 kJ.
Initial pressure (P1) = 0.1 MPa = 0.1 * 10^6 Pa
Initial volume (V1) = 2 m^3
Final pressure (P2) = 1 MPa = 1 * 10^6 Pa
Final volume (V2) = 0.2 m^3
R = 8.314 J/molK
Heat capacity at constant pressure (Cp) = 2.5R
Step 1: Calculating the work done during compression (W1)
Using the equation for reversible adiabatic process:
P1 * V1^γ = P2 * V2^γ
Where γ is the heat capacity ratio, given by:
γ = Cp / (Cp - R)
Calculating the value of γ:
γ = 2.5R / (2.5R - R)
γ = 2.5
Using the equation for work done during adiabatic compression:
W1 = (P2 * V2 - P1 * V1) / (γ - 1)
Substituting the given values:
W1 = (1 * 10^6 * 0.2 - 0.1 * 10^6 * 2) / (2.5 - 1)
W1 = (200000 - 200000) / 1.5
W1 = 0
Step 2: Calculating the work done during cooling at constant pressure (W2)
Using the equation for work done at constant pressure:
W2 = P2 * (V2 - V1)
Substituting the given values:
W2 = 1 * 10^6 * (0.2 - 2)
W2 = 1 * 10^6 * (-1.8)
W2 = -1.8 * 10^6
Step 3: Calculating the total work done (W_total)
W_total = W1 + W2
W_total = 0 + (-1.8 * 10^6)
W_total = -1.8 * 10^6 J
Converting the work done to kJ:
W_total = -1.8 * 10^3 kJ
The negative sign indicates that work is done on the gas.
Final Answer:
The total work done on the gas for the entire process is -1800 kJ.
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An ideal gas is initially at a pressure of 0.1 MPa and a total volume of 2m3. It is first compressedto 1MPa by a reversible adiabatic process and then cooled at constant pressure to a final volumeof 0.2m3. The total work done (in kJ) on the gas for the entire process (up to one decimal place) is____________Data: R = 8.314 J/molK; heat capacity at constant pressure (CP) = 2.5RImp : you should answer only the numeric valueCorrect answer is '750'. Can you explain this answer?
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An ideal gas is initially at a pressure of 0.1 MPa and a total volume of 2m3. It is first compressedto 1MPa by a reversible adiabatic process and then cooled at constant pressure to a final volumeof 0.2m3. The total work done (in kJ) on the gas for the entire process (up to one decimal place) is____________Data: R = 8.314 J/molK; heat capacity at constant pressure (CP) = 2.5RImp : you should answer only the numeric valueCorrect answer is '750'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about An ideal gas is initially at a pressure of 0.1 MPa and a total volume of 2m3. It is first compressedto 1MPa by a reversible adiabatic process and then cooled at constant pressure to a final volumeof 0.2m3. The total work done (in kJ) on the gas for the entire process (up to one decimal place) is____________Data: R = 8.314 J/molK; heat capacity at constant pressure (CP) = 2.5RImp : you should answer only the numeric valueCorrect answer is '750'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An ideal gas is initially at a pressure of 0.1 MPa and a total volume of 2m3. It is first compressedto 1MPa by a reversible adiabatic process and then cooled at constant pressure to a final volumeof 0.2m3. The total work done (in kJ) on the gas for the entire process (up to one decimal place) is____________Data: R = 8.314 J/molK; heat capacity at constant pressure (CP) = 2.5RImp : you should answer only the numeric valueCorrect answer is '750'. Can you explain this answer?.
An ideal gas is initially at a pressure of 0.1 MPa and a total volume of 2m3. It is first compressedto 1MPa by a reversible adiabatic process and then cooled at constant pressure to a final volumeof 0.2m3. The total work done (in kJ) on the gas for the entire process (up to one decimal place) is____________Data: R = 8.314 J/molK; heat capacity at constant pressure (CP) = 2.5RImp : you should answer only the numeric valueCorrect answer is '750'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about An ideal gas is initially at a pressure of 0.1 MPa and a total volume of 2m3. It is first compressedto 1MPa by a reversible adiabatic process and then cooled at constant pressure to a final volumeof 0.2m3. The total work done (in kJ) on the gas for the entire process (up to one decimal place) is____________Data: R = 8.314 J/molK; heat capacity at constant pressure (CP) = 2.5RImp : you should answer only the numeric valueCorrect answer is '750'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An ideal gas is initially at a pressure of 0.1 MPa and a total volume of 2m3. It is first compressedto 1MPa by a reversible adiabatic process and then cooled at constant pressure to a final volumeof 0.2m3. The total work done (in kJ) on the gas for the entire process (up to one decimal place) is____________Data: R = 8.314 J/molK; heat capacity at constant pressure (CP) = 2.5RImp : you should answer only the numeric valueCorrect answer is '750'. Can you explain this answer?.
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