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An ideal gas is taken through a cyclic thermodynamics process through four steps.The amount of heat involved in the steps are Q1 = 5960 J,Q2 = - 5600 J,Q3= -3000 J,Q4 = -3600 J respectively. The corresponding quantities of Internal energy changes are ΔU1 = 3.760 J ,ΔU2 = - 4800 J,ΔU3 = -1800 J,ΔU4 = ?find the value ΔU4 & net work done *?
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An ideal gas is taken through a cyclic thermodynamics process through ...
Calculating ΔU4 and Net Work Done:
- Calculating ΔU4:
Given: Q4 = -3600 J
From the first law of thermodynamics, we know that ΔU = Q - W, where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system.
Since Q4 is the heat removed from the system in step 4, we can write: ΔU4 = Q4 - W4
Substituting the values, we get: ΔU4 = -3600 J - W4
We need to calculate W4 to find the value of ΔU4.
- Calculating Net Work Done:
Net work done in a cyclic process is the sum of work done in all the steps.
Net work done = W1 + W2 + W3 + W4
We already know the values of Q1, Q2, Q3, and Q4. We need to find the work done in each step to calculate the net work done.
Work done in each step can be calculated using the formula: W = Q - ΔU
- Conclusion:
By calculating the work done in each step using the given data and the formula mentioned above, we can find the value of ΔU4 and the net work done in the cyclic thermodynamics process.
- Calculating ΔU4:
Given: Q4 = -3600 J
From the first law of thermodynamics, we know that ΔU = Q - W, where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system.
Since Q4 is the heat removed from the system in step 4, we can write: ΔU4 = Q4 - W4
Substituting the values, we get: ΔU4 = -3600 J - W4
We need to calculate W4 to find the value of ΔU4.
- Calculating Net Work Done:
Net work done in a cyclic process is the sum of work done in all the steps.
Net work done = W1 + W2 + W3 + W4
We already know the values of Q1, Q2, Q3, and Q4. We need to find the work done in each step to calculate the net work done.
Work done in each step can be calculated using the formula: W = Q - ΔU
- Conclusion:
By calculating the work done in each step using the given data and the formula mentioned above, we can find the value of ΔU4 and the net work done in the cyclic thermodynamics process.
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An ideal gas is taken through a cyclic thermodynamics process through four steps.The amount of heat involved in the steps are Q1 = 5960 J,Q2 = - 5600 J,Q3= -3000 J,Q4 = -3600 J respectively. The corresponding quantities of Internal energy changes are ΔU1 = 3.760 J ,ΔU2 = - 4800 J,ΔU3 = -1800 J,ΔU4 = ?find the value ΔU4 & net work done *?
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An ideal gas is taken through a cyclic thermodynamics process through four steps.The amount of heat involved in the steps are Q1 = 5960 J,Q2 = - 5600 J,Q3= -3000 J,Q4 = -3600 J respectively. The corresponding quantities of Internal energy changes are ΔU1 = 3.760 J ,ΔU2 = - 4800 J,ΔU3 = -1800 J,ΔU4 = ?find the value ΔU4 & net work done *? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about An ideal gas is taken through a cyclic thermodynamics process through four steps.The amount of heat involved in the steps are Q1 = 5960 J,Q2 = - 5600 J,Q3= -3000 J,Q4 = -3600 J respectively. The corresponding quantities of Internal energy changes are ΔU1 = 3.760 J ,ΔU2 = - 4800 J,ΔU3 = -1800 J,ΔU4 = ?find the value ΔU4 & net work done *? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An ideal gas is taken through a cyclic thermodynamics process through four steps.The amount of heat involved in the steps are Q1 = 5960 J,Q2 = - 5600 J,Q3= -3000 J,Q4 = -3600 J respectively. The corresponding quantities of Internal energy changes are ΔU1 = 3.760 J ,ΔU2 = - 4800 J,ΔU3 = -1800 J,ΔU4 = ?find the value ΔU4 & net work done *?.
An ideal gas is taken through a cyclic thermodynamics process through four steps.The amount of heat involved in the steps are Q1 = 5960 J,Q2 = - 5600 J,Q3= -3000 J,Q4 = -3600 J respectively. The corresponding quantities of Internal energy changes are ΔU1 = 3.760 J ,ΔU2 = - 4800 J,ΔU3 = -1800 J,ΔU4 = ?find the value ΔU4 & net work done *? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about An ideal gas is taken through a cyclic thermodynamics process through four steps.The amount of heat involved in the steps are Q1 = 5960 J,Q2 = - 5600 J,Q3= -3000 J,Q4 = -3600 J respectively. The corresponding quantities of Internal energy changes are ΔU1 = 3.760 J ,ΔU2 = - 4800 J,ΔU3 = -1800 J,ΔU4 = ?find the value ΔU4 & net work done *? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An ideal gas is taken through a cyclic thermodynamics process through four steps.The amount of heat involved in the steps are Q1 = 5960 J,Q2 = - 5600 J,Q3= -3000 J,Q4 = -3600 J respectively. The corresponding quantities of Internal energy changes are ΔU1 = 3.760 J ,ΔU2 = - 4800 J,ΔU3 = -1800 J,ΔU4 = ?find the value ΔU4 & net work done *?.
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An ideal gas is taken through a cyclic thermodynamics process through four steps.The amount of heat involved in the steps are Q1 = 5960 J,Q2 = - 5600 J,Q3= -3000 J,Q4 = -3600 J respectively. The corresponding quantities of Internal energy changes are ΔU1 = 3.760 J ,ΔU2 = - 4800 J,ΔU3 = -1800 J,ΔU4 = ?find the value ΔU4 & net work done *?, a detailed solution for An ideal gas is taken through a cyclic thermodynamics process through four steps.The amount of heat involved in the steps are Q1 = 5960 J,Q2 = - 5600 J,Q3= -3000 J,Q4 = -3600 J respectively. The corresponding quantities of Internal energy changes are ΔU1 = 3.760 J ,ΔU2 = - 4800 J,ΔU3 = -1800 J,ΔU4 = ?find the value ΔU4 & net work done *? has been provided alongside types of An ideal gas is taken through a cyclic thermodynamics process through four steps.The amount of heat involved in the steps are Q1 = 5960 J,Q2 = - 5600 J,Q3= -3000 J,Q4 = -3600 J respectively. The corresponding quantities of Internal energy changes are ΔU1 = 3.760 J ,ΔU2 = - 4800 J,ΔU3 = -1800 J,ΔU4 = ?find the value ΔU4 & net work done *? theory, EduRev gives you an
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