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A stationary mass of gas is compressed without friction from an initial state of 0.3 m3 and 1 MPa to a final state of 0.15 m3 and 1 MPa, the pressure remaining constant during the process. There is a transfer of 40 kJ of heat from the gas during the process. What is the change in internal energy of the gas?
- a)-5 kJ
- b)+25 kJ
- c)-25 kJ
- d)+15 kJ
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
A stationary mass of gas is compressed without friction from an initia...
= 0.1 χ 106 (0.15-0.3)
= -15Kj
-25KjMost Upvoted Answer
A stationary mass of gas is compressed without friction from an initia...
Given parameters:
Initial state: V1 = 0.3 m3, P1 = 0.1 MPa
Final state: V2 = 0.15 m3, P2 = 0.1 MPa
Heat transfer: Q = -40 kJ (negative sign indicates heat transfer from the gas)
To find: Change in internal energy (ΔU)
Using the first law of thermodynamics, we can write:
ΔU = Q - W
where ΔU is the change in internal energy, Q is the heat transfer, and W is the work done by the gas.
Since the process is isobaric (constant pressure), the work done can be calculated as:
W = PΔV
where P is the constant pressure and ΔV is the change in volume.
ΔV = V2 - V1 = -0.15 m3 (since the gas is compressed)
Substituting the given values, we get:
W = PΔV = 0.1 MPa x (-0.15 m3) = -15 kJ
Note that the negative sign indicates work done on the gas.
Substituting the values of Q and W in the first law equation, we get:
ΔU = Q - W = (-40 kJ) - (-15 kJ) = -25 kJ
Therefore, the change in internal energy of the gas is -25 kJ, which corresponds to option (c).
Answer: c) -25 kJ
Initial state: V1 = 0.3 m3, P1 = 0.1 MPa
Final state: V2 = 0.15 m3, P2 = 0.1 MPa
Heat transfer: Q = -40 kJ (negative sign indicates heat transfer from the gas)
To find: Change in internal energy (ΔU)
Using the first law of thermodynamics, we can write:
ΔU = Q - W
where ΔU is the change in internal energy, Q is the heat transfer, and W is the work done by the gas.
Since the process is isobaric (constant pressure), the work done can be calculated as:
W = PΔV
where P is the constant pressure and ΔV is the change in volume.
ΔV = V2 - V1 = -0.15 m3 (since the gas is compressed)
Substituting the given values, we get:
W = PΔV = 0.1 MPa x (-0.15 m3) = -15 kJ
Note that the negative sign indicates work done on the gas.
Substituting the values of Q and W in the first law equation, we get:
ΔU = Q - W = (-40 kJ) - (-15 kJ) = -25 kJ
Therefore, the change in internal energy of the gas is -25 kJ, which corresponds to option (c).
Answer: c) -25 kJ
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Community Answer
A stationary mass of gas is compressed without friction from an initia...
Here, Work transfer is negative due to compression.
W = p*(V2 - V1)
then
Q = U + W
Heat is rejected by tha gas hence Q is also negative
Hence, U = -Q - (-W)
U = -Q + W
U will be negative quantity.
U = -25 KJ
W = p*(V2 - V1)
then
Q = U + W
Heat is rejected by tha gas hence Q is also negative
Hence, U = -Q - (-W)
U = -Q + W
U will be negative quantity.
U = -25 KJ
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A stationary mass of gas is compressed without friction from an initial state of 0.3 m3 and 1 MPato a final state of 0.15 m3 and 1 MPa, the pressureremaining constant during the process. There is a transfer of 40 kJ of heat from the gas during the process. What is the change in internal energy of the gas?a)-5 kJb)+25 kJc)-25 kJd)+15 kJCorrect answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A stationary mass of gas is compressed without friction from an initial state of 0.3 m3 and 1 MPato a final state of 0.15 m3 and 1 MPa, the pressureremaining constant during the process. There is a transfer of 40 kJ of heat from the gas during the process. What is the change in internal energy of the gas?a)-5 kJb)+25 kJc)-25 kJd)+15 kJCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A stationary mass of gas is compressed without friction from an initial state of 0.3 m3 and 1 MPato a final state of 0.15 m3 and 1 MPa, the pressureremaining constant during the process. There is a transfer of 40 kJ of heat from the gas during the process. What is the change in internal energy of the gas?a)-5 kJb)+25 kJc)-25 kJd)+15 kJCorrect answer is option 'C'. Can you explain this answer?.
A stationary mass of gas is compressed without friction from an initial state of 0.3 m3 and 1 MPato a final state of 0.15 m3 and 1 MPa, the pressureremaining constant during the process. There is a transfer of 40 kJ of heat from the gas during the process. What is the change in internal energy of the gas?a)-5 kJb)+25 kJc)-25 kJd)+15 kJCorrect answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A stationary mass of gas is compressed without friction from an initial state of 0.3 m3 and 1 MPato a final state of 0.15 m3 and 1 MPa, the pressureremaining constant during the process. There is a transfer of 40 kJ of heat from the gas during the process. What is the change in internal energy of the gas?a)-5 kJb)+25 kJc)-25 kJd)+15 kJCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A stationary mass of gas is compressed without friction from an initial state of 0.3 m3 and 1 MPato a final state of 0.15 m3 and 1 MPa, the pressureremaining constant during the process. There is a transfer of 40 kJ of heat from the gas during the process. What is the change in internal energy of the gas?a)-5 kJb)+25 kJc)-25 kJd)+15 kJCorrect answer is option 'C'. Can you explain this answer?.
Solutions for A stationary mass of gas is compressed without friction from an initial state of 0.3 m3 and 1 MPato a final state of 0.15 m3 and 1 MPa, the pressureremaining constant during the process. There is a transfer of 40 kJ of heat from the gas during the process. What is the change in internal energy of the gas?a)-5 kJb)+25 kJc)-25 kJd)+15 kJCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mechanical Engineering.
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