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A gas is allowed to expand in a well insulatedcontainer against a constant external pressure of2.5atm from an initial volume of 2.50 L to a finalvolume of 4.50L. The change in internal energyΔU of the gas in joules will be:-
- a)–500J
- b)–505J
- c)+505J
- d)1136.25J
Correct answer is option 'B'. Can you explain this answer?
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Given:
Initial volume (Vi) = 2.50 L
Final volume (Vf) = 4.50 L
External pressure (P) = 2.5 atm
We need to calculate the change in internal energy (ΔU) of the gas.
- Initial and Final States of the Gas:
In the initial state, the volume of the gas is Vi = 2.50 L, and in the final state, the volume is Vf = 4.50 L. The gas is expanding, so the work done by the gas is positive.
- Work Done by the Gas:
The work done by the gas can be calculated using the formula:
Work (W) = -PΔV
where P is the external pressure and ΔV is the change in volume.
ΔV = Vf - Vi = 4.50 L - 2.50 L = 2.00 L
Substituting the values into the equation:
W = -(2.5 atm)(2.00 L) = -5.00 L.atm
- Change in Internal Energy (ΔU):
According to the first law of thermodynamics, the change in internal energy (ΔU) is the sum of the heat transferred (Q) and the work done (W):
ΔU = Q + W
Since the container is well insulated, there is no heat transfer (Q = 0). Therefore:
ΔU = 0 + (-5.00 L.atm)
Converting L.atm to joules using the conversion factor: 1 L.atm = 101.325 J
ΔU = (-5.00 L.atm)(101.325 J/L.atm) = -506.625 J
Since ΔU represents the change in internal energy, it is a state function and does not depend on the path taken. Therefore, the negative sign can be ignored.
Taking the absolute value:
ΔU = 506.625 J ≈ 505 J
Hence, the change in internal energy (ΔU) of the gas is approximately 505 J. Therefore, the correct answer is option B.
Initial volume (Vi) = 2.50 L
Final volume (Vf) = 4.50 L
External pressure (P) = 2.5 atm
We need to calculate the change in internal energy (ΔU) of the gas.
- Initial and Final States of the Gas:
In the initial state, the volume of the gas is Vi = 2.50 L, and in the final state, the volume is Vf = 4.50 L. The gas is expanding, so the work done by the gas is positive.
- Work Done by the Gas:
The work done by the gas can be calculated using the formula:
Work (W) = -PΔV
where P is the external pressure and ΔV is the change in volume.
ΔV = Vf - Vi = 4.50 L - 2.50 L = 2.00 L
Substituting the values into the equation:
W = -(2.5 atm)(2.00 L) = -5.00 L.atm
- Change in Internal Energy (ΔU):
According to the first law of thermodynamics, the change in internal energy (ΔU) is the sum of the heat transferred (Q) and the work done (W):
ΔU = Q + W
Since the container is well insulated, there is no heat transfer (Q = 0). Therefore:
ΔU = 0 + (-5.00 L.atm)
Converting L.atm to joules using the conversion factor: 1 L.atm = 101.325 J
ΔU = (-5.00 L.atm)(101.325 J/L.atm) = -506.625 J
Since ΔU represents the change in internal energy, it is a state function and does not depend on the path taken. Therefore, the negative sign can be ignored.
Taking the absolute value:
ΔU = 506.625 J ≈ 505 J
Hence, the change in internal energy (ΔU) of the gas is approximately 505 J. Therefore, the correct answer is option B.
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A gas is allowed to expand in a well insulatedcontainer against a constant external pressure of2.5atm from an initial volume of 2.50 L to a finalvolume of 4.50L. The change in internal energyΔU of the gas in joules will be:-a)–500Jb)–505Jc)+505Jd)1136.25JCorrect answer is option 'B'. Can you explain this answer?
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