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Answer:
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Explanation:
The Mathematical expression of the first law is given as
ΔU = Q - W
Note that depending on convention, the mathematical expression for the first law can be written as
ΔU = Q + W
a) Yes, the gas does work on its surroundings. Since it expands; indicating that there is a change in volume.
The work done is given as -PΔV if we are using the convention of (ΔU = Q + W). And this work is negative work done, since the system expands and does work on the surroundings.
But if we are using the convention of (ΔU = Q - W), the work done is given as PΔV and the workdone by the system on the environment is positive work during expansion.
b) Since this all takes place at constant temperature and against a constant atmospheric pressure, the change in internal energy for this system is 0. Change in internal energy depends on a change in temperature.
So, from the mathematical expression,
ΔU = Q + W or ΔU = Q - W
If ΔU = 0,
Q = - W or Q = W
But either ways,
Q = PΔV
(because, W = -PΔV for ΔU=Q+W and W = PΔV for ΔU=Q-W)
So, either ways, the heat transfer is the same and it is positive. This indicates heat is transferred from the surroundings to the system.
(c) What is ΔU for the gas for this process?
Since this all takes place at constant temperature and against a constant atmospheric pressure, the change in internal energy for this system is 0. Change in internal energy depends on a change in temperature.
Hope this Helps!!!
Based on the given question, we can state that Yes, the gas does work on its surroundings because it expands, this shows that there is a change in volume.
Based on the second question, we can state that because this takes place at constant temperature and against a constant atmospheric pressure, then the change in internal energy for this system is zero.
Based on the third question, we can see that the ΔU for the gas for this process is zero because of the constant temperature and atmospheric pressure.
The Mathematical expression of the first law
This is given as
ΔU = Q - W
Hence, from the mathematical expression,
- ΔU = Q + W or ΔU = Q - W
- If ΔU = 0,
- Q = - W or Q = W
Regardless,
Q = PΔV
This means that the heat transfer remains the same and is positive
Read more about heat transfer here:
https://brainly.com/question/12072129
