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In a reversible isothermal expansion process, the fluid expands from 10 bar and 2 m3 to 2 bar and 10 m3. During the process the heat supplied is at the rate of 100 kW. What is the rate of work done during the process?
- a)20 kW
- b)35 kW
- c)80 kW
- d)100 kW
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
In a reversible isothermal expansion process, the fluid expands from 1...
For reversible isothermal expansion heat supplied is equal to work done during the process and equal to 
∵ Temperature constant so no change in internal energy dQ = dU + dW; dU = 0 Therefore dQ = dW.
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In a reversible isothermal expansion process, the fluid expands from 1...
Given information:
- Reversible isothermal expansion process
- Initial conditions: 10 bar and 2 m3
- Final conditions: 2 bar and 10 m3
- Heat supplied: 100 kW
To determine the rate of work done during the process, we can use the First Law of Thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat (Q) supplied to the system minus the work (W) done by the system:
ΔU = Q - W
Since the process is isothermal, the internal energy remains constant. Therefore, ΔU = 0, and the equation becomes:
0 = Q - W
We are given that the heat supplied (Q) is 100 kW. Thus, the equation becomes:
0 = 100 kW - W
To find the rate of work done (W), we need to determine the work done during the expansion process. The work done during an isothermal expansion can be calculated using the equation:
W = nRT ln(Vf/Vi)
where:
- W is the work done
- n is the number of moles of the fluid
- R is the universal gas constant
- T is the temperature
- Vi and Vf are the initial and final volumes, respectively
Since the process is isothermal, the temperature remains constant. Therefore, we can simplify the equation to:
W = nRT ln(Vf/Vi)
To calculate the number of moles of the fluid (n), we can use the ideal gas law equation:
PV = nRT
where:
- P is the pressure
- V is the volume
- R is the gas constant
Rearranging the equation to solve for n, we get:
n = PV/RT
Substituting the given values, we have:
n = (10 bar)(2 m3)/(8.314 J/(mol·K))(273 K)
n ≈ 0.956 mol
Substituting the values of n, R, T, Vi, and Vf into the equation for work done, we get:
W = (0.956 mol)(8.314 J/(mol·K))(273 K) ln(10 m3/2 m3)
W ≈ 100 kW
Therefore, the rate of work done during the process is 100 kW. Hence, the correct answer is option 'D'.
- Reversible isothermal expansion process
- Initial conditions: 10 bar and 2 m3
- Final conditions: 2 bar and 10 m3
- Heat supplied: 100 kW
To determine the rate of work done during the process, we can use the First Law of Thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat (Q) supplied to the system minus the work (W) done by the system:
ΔU = Q - W
Since the process is isothermal, the internal energy remains constant. Therefore, ΔU = 0, and the equation becomes:
0 = Q - W
We are given that the heat supplied (Q) is 100 kW. Thus, the equation becomes:
0 = 100 kW - W
To find the rate of work done (W), we need to determine the work done during the expansion process. The work done during an isothermal expansion can be calculated using the equation:
W = nRT ln(Vf/Vi)
where:
- W is the work done
- n is the number of moles of the fluid
- R is the universal gas constant
- T is the temperature
- Vi and Vf are the initial and final volumes, respectively
Since the process is isothermal, the temperature remains constant. Therefore, we can simplify the equation to:
W = nRT ln(Vf/Vi)
To calculate the number of moles of the fluid (n), we can use the ideal gas law equation:
PV = nRT
where:
- P is the pressure
- V is the volume
- R is the gas constant
Rearranging the equation to solve for n, we get:
n = PV/RT
Substituting the given values, we have:
n = (10 bar)(2 m3)/(8.314 J/(mol·K))(273 K)
n ≈ 0.956 mol
Substituting the values of n, R, T, Vi, and Vf into the equation for work done, we get:
W = (0.956 mol)(8.314 J/(mol·K))(273 K) ln(10 m3/2 m3)
W ≈ 100 kW
Therefore, the rate of work done during the process is 100 kW. Hence, the correct answer is option 'D'.
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In a reversible isothermal expansion process, the fluid expands from 1...
D
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In a reversible isothermal expansion process, the fluid expands from 10 bar and 2 m3 to 2 bar and 10 m3. During the process the heat supplied is at the rate of 100 kW. What is the rate of work done during the process?a)20 kWb)35 kWc)80 kWd)100 kWCorrect answer is option 'D'. Can you explain this answer?
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In a reversible isothermal expansion process, the fluid expands from 10 bar and 2 m3 to 2 bar and 10 m3. During the process the heat supplied is at the rate of 100 kW. What is the rate of work done during the process?a)20 kWb)35 kWc)80 kWd)100 kWCorrect answer is option 'D'. Can you explain this answer? 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 In a reversible isothermal expansion process, the fluid expands from 10 bar and 2 m3 to 2 bar and 10 m3. During the process the heat supplied is at the rate of 100 kW. What is the rate of work done during the process?a)20 kWb)35 kWc)80 kWd)100 kWCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a reversible isothermal expansion process, the fluid expands from 10 bar and 2 m3 to 2 bar and 10 m3. During the process the heat supplied is at the rate of 100 kW. What is the rate of work done during the process?a)20 kWb)35 kWc)80 kWd)100 kWCorrect answer is option 'D'. Can you explain this answer?.
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