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A gas is allowed to expand in a well insulated container against a constant external of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L .The change in internal energy Of the gas in joules will be A)1136.25J B)-500J C)-505 J D) 505 J Ans is (C).Can anyone explain this for me plzzz frndz.?
Most Upvoted Answer
A gas is allowed to expand in a well insulated container against a con...
For insulated container dq=0,so du=dw =-pdv=2.5(4.5-2.5)=5litre atm and 1litre atm = 101.3joule so -5 *101-505 joule
Community Answer
A gas is allowed to expand in a well insulated container against a con...
Given:
- External pressure (P) = 2.5 atm
- Initial volume (V1) = 2.50 L
- Final volume (V2) = 4.50 L
To find:
Change in internal energy (ΔU) in joules
Formula:
The change in internal energy of a system can be calculated using the formula:
ΔU = q + w
where,
- ΔU = change in internal energy
- q = heat transferred to the system
- w = work done by the system
Explanation:
In this case, the container is well insulated, which means there is no heat transfer between the system (gas) and the surroundings. Therefore, q = 0.
The work done by the system can be calculated using the formula:
w = -PΔV
where,
- P = external pressure
- ΔV = change in volume
Step 1: Calculate the change in volume (ΔV)
ΔV = V2 - V1
= 4.50 L - 2.50 L
= 2.00 L
Step 2: Calculate the work done (w)
w = -PΔV
= -(2.5 atm)(2.00 L)
= -5.00 atm L
Step 3: Calculate the change in internal energy (ΔU)
ΔU = q + w
= 0 + (-5.00 atm L)
To convert atm L to joules, we can use the conversion factor:
1 atm L = 101.3 J
Therefore,
ΔU = (-5.00 atm L) x (101.3 J/atm L)
= -505 J
Conclusion:
The change in internal energy of the gas is -505 J. Therefore, the correct answer is option C) -505 J.
- External pressure (P) = 2.5 atm
- Initial volume (V1) = 2.50 L
- Final volume (V2) = 4.50 L
To find:
Change in internal energy (ΔU) in joules
Formula:
The change in internal energy of a system can be calculated using the formula:
ΔU = q + w
where,
- ΔU = change in internal energy
- q = heat transferred to the system
- w = work done by the system
Explanation:
In this case, the container is well insulated, which means there is no heat transfer between the system (gas) and the surroundings. Therefore, q = 0.
The work done by the system can be calculated using the formula:
w = -PΔV
where,
- P = external pressure
- ΔV = change in volume
Step 1: Calculate the change in volume (ΔV)
ΔV = V2 - V1
= 4.50 L - 2.50 L
= 2.00 L
Step 2: Calculate the work done (w)
w = -PΔV
= -(2.5 atm)(2.00 L)
= -5.00 atm L
Step 3: Calculate the change in internal energy (ΔU)
ΔU = q + w
= 0 + (-5.00 atm L)
To convert atm L to joules, we can use the conversion factor:
1 atm L = 101.3 J
Therefore,
ΔU = (-5.00 atm L) x (101.3 J/atm L)
= -505 J
Conclusion:
The change in internal energy of the gas is -505 J. Therefore, the correct answer is option C) -505 J.
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A gas is allowed to expand in a well insulated container against a constant external of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L .The change in internal energy Of the gas in joules will be A)1136.25J B)-500J C)-505 J D) 505 J Ans is (C).Can anyone explain this for me plzzz frndz.?
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A gas is allowed to expand in a well insulated container against a constant external of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L .The change in internal energy Of the gas in joules will be A)1136.25J B)-500J C)-505 J D) 505 J Ans is (C).Can anyone explain this for me plzzz frndz.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A gas is allowed to expand in a well insulated container against a constant external of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L .The change in internal energy Of the gas in joules will be A)1136.25J B)-500J C)-505 J D) 505 J Ans is (C).Can anyone explain this for me plzzz frndz.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A gas is allowed to expand in a well insulated container against a constant external of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L .The change in internal energy Of the gas in joules will be A)1136.25J B)-500J C)-505 J D) 505 J Ans is (C).Can anyone explain this for me plzzz frndz.?.
A gas is allowed to expand in a well insulated container against a constant external of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L .The change in internal energy Of the gas in joules will be A)1136.25J B)-500J C)-505 J D) 505 J Ans is (C).Can anyone explain this for me plzzz frndz.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A gas is allowed to expand in a well insulated container against a constant external of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L .The change in internal energy Of the gas in joules will be A)1136.25J B)-500J C)-505 J D) 505 J Ans is (C).Can anyone explain this for me plzzz frndz.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A gas is allowed to expand in a well insulated container against a constant external of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L .The change in internal energy Of the gas in joules will be A)1136.25J B)-500J C)-505 J D) 505 J Ans is (C).Can anyone explain this for me plzzz frndz.?.
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