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scott harvey

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Everything posted by scott harvey

  1. would that be an exchange of pressure for temperature ? if we expand a gas it would increase the temperature ?
  2. you haven't got to be like that mate a closed system can exchange heat or work with its surroundings isothermal is a process or change taking place at constant temperature
  3. i can only tell you what i have in front of me that's the question. forgetting that i will try and answer your question, please list them
  4. an isothermal gas expansion when the gas expansion follows the law pV1 = C
  5. what i wrote is what is stated on the material that was given to me to help with the course, i have asked my tutor to explain and all he has given me is this formula W = -m RT ln P1/ P2 if you don't wish to continue that's fine. i understand im coming across vague and doing thin completely different to what you ask. im new to this and this question has stomped me yes we can discuss in them terms, i somewhat understand.
  6. yeah i can do that, the first law states that energy can neither be created nor destroyed, thus power generation processes and energy sources actually involve conversion of energy from one form to another, rather than the creation of energy from nothing
  7. This is what i have come up with to solve the work done part
  8. I should of worded my question better, i have not attempted this question as i am searching for an example of it. i don't understand how to determine work done in general, i do not know where to begin with the question. sorry if this is a bit of a cheek
  9. Hi all i was wondering if any one could help with this question. i can do the calculations just need to see the formula and how it is set out the material i was sent to help me has nothing to do with this type of question. In a closed system 0.3kg of gas at 373 k is expanded isothermally and reversibly from 1MPA pressure to 200 KPA. given that Cv = 718 J/kg K and R = 287 J/kg K determine: the work done. the state of energy transfer within this isothermal process explaining your answer. many thanks scott
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