(2-8) Chemistry - Science (Dr. Bob222)

Which of the following statements is correct?
A) The internal energy of a system increases when more work is done by the system than the heat that flows into the system.
B) The system does work on the surroundings when an ideal gas expands against a constant external pressure.
C) The internal energy of a system decreases when work is done on the system and heat flows into the system.
D) The internal energy of the system cannot change if T does not change.
E) All the statements are false.

The correct answer is B, but I wonder why D is not considered correct? I guess I am confused with the wording

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  1. For B, think of the ideal gas as a compressed spring.
    When the compressed spring expands, it does work on the surroundings.

    The external pressure has to stay constant to ensure work is done. If the external pressure decreases, the system can expand without doing any work.

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  2. dE = q + w
    q changes with T.
    w, though, is pdV and you can change w willy nilly which will always change dE is not necessarily involved in changing volume.

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  3. Thank you for your answers.

    Dr.Bob, you said that "q changes with T", thus dE changes with T, then why the answer D in the problem below is not true.

    IT is just a bit confusing, thank you.

    In heating an ideal gas,
    A) It takes more heat to raise T by one degree K in constant P heating than in constant V heating.
    B) The gas does work W on the environment in heating at constant P.
    C) H does not change during heating.
    D) E does not change during heating
    E) Both A and B are true.

    E is correct answer.

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  4. In the first question, D says that dE cannot change if T doesn't change and I was answering that it certainly can because pdV changes work and that changes dE and no T is necessarily involved. The point in the first one is that changing T changes q and changing T CAN change volume (and thus work) but w change be changed without a change in T. The second question is different in that it specifically changes volume and constant P and that produces work.

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  5. Ahhhh, I got it now. I should have been more careful on my reading. Thank you

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  6. Also, could you tell me please what processes would result in work=0 (beside the constant volume)?

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  7. I would just look at it like this:

    Work done by the system: Work=- and q= N/A

    Work done onto the system: Work=+ and q=N/A

    Heat released from the system: Work=N/A and q=-

    Evaluate each answer choice with the chart.

    A) The internal energy of a system increases when more work is done by the system than the heat that flows into the system.

    ***This is false. This may be true for a certain process, but may be false for another; you can not apply this statement to all systems.

    B) The system does work on the surroundings when an ideal gas expands against a constant external pressure.

    ***True. When a system does work against its surrounding, its energy decreases and the energy of the surrounding increases.

    C) The internal energy of a system decreases when work is done on the system and heat flows into the system.

    ***False, the flow of heat and the work done on the system will increase energy not decrease it.

    D) The internal energy of the system cannot change if T does not change.

    ***False. Work done by or on the system can change the internal energy of the system.

    E) All the statements are false.

    ***One is true, so the rest are false.

    =================================

    In heating an ideal gas,

    A) It takes more heat to raise T by one degree K at constant P than heating at constant V.

    ***This answer choice makes no since. So, I do not know why it is true.

    B) The gas does work W on the environment in heating at constant P.

    ***True. Look at Dr. Bob222's equation and explanation to see why. And this is only true as long as the volume changes.

    C) H does not change during heating.

    ***False. If H is enthalpy, then enthalpy does change because heating changes temperature.

    D) E does not change during heating

    ***False. Look at the other answer choices to see why.

    E) Both A and B are true.

    ***I know that B is true. Reword or correct typos for answer choice A.

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  8. Also, could you tell me please what processes would result in work=0 (beside the constant volume)?

    I could be wrong, but I think constant volume is the only process resulting in W=0.

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  9. Copied from Wikipedia.
    "When no heat flows into or out of the gas because its container is at the same temperature, then there is no work done. Thus, work=0 which means external pressure on any moving surface is zero. This is called free expansion."

    So when the term pdV has dV = 0 or when p = 0, work = 0

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  10. Thank you for your explications.

    Could you tell me if the expansion of a gas into the vacuum would be considered zero work?

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  11. work = 0
    http://physics.bu.edu/~duffy/semester1/c27_process_expansion_sim.html

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  12. Okay, Dr. Bob222 is discussing isothermal processes, which he quoted from Wikipedia and is true. However, this also causes no change in internal energy, because no change in heat takes place. So, you rarely see one of theses calculations. But it is true.

    =================================

    I believe that answer to you question is yes.

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