scott harvey Posted April 6, 2015 Share Posted April 6, 2015 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 Link to comment Share on other sites More sharing options...
swansont Posted April 6, 2015 Share Posted April 6, 2015 You have to show what you've done so far. What's changing and what's not changing in an isothermal process? Link to comment Share on other sites More sharing options...
studiot Posted April 6, 2015 Share Posted April 6, 2015 If you can do the 'calculations', why do you 'need the formulae' ? Which formulae? As I see it, this question is testing your understanding, not your arithmetic so explain what is happening. Link to comment Share on other sites More sharing options...
scott harvey Posted April 6, 2015 Author Share Posted April 6, 2015 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 Link to comment Share on other sites More sharing options...
studiot Posted April 6, 2015 Share Posted April 6, 2015 (edited) So start by following swansont's advice. Actually this is good advice for all physics questions(He was a physics teacher after all) List what you know. Then list a few things you don't know, but might like to. So What is a closed system, why did they specify one? If the question also specifies work can pass the boundary, what else can and can't? What formulae do you know connecting these things? I am going to cut the grass now, but I will look in for a cup of tea and to see how you are getting on, from time to time. Edited April 6, 2015 by studiot Link to comment Share on other sites More sharing options...
scott harvey Posted April 6, 2015 Author Share Posted April 6, 2015 This is what i have come up with to solve the work done part Link to comment Share on other sites More sharing options...
studiot Posted April 6, 2015 Share Posted April 6, 2015 (edited) Since when did they start measuring volume in kilograms? Why can you not simply answer the questions swansont and I asked? They are very simple and very basic and will lead you to the correct answers. Edited April 6, 2015 by studiot Link to comment Share on other sites More sharing options...
swansont Posted April 6, 2015 Share Posted April 6, 2015 You aren't given the volume. Do you have an equation that relates volume to other terms that you know? What is n in that equation? Link to comment Share on other sites More sharing options...
studiot Posted April 6, 2015 Share Posted April 6, 2015 Here is a hint. This question is about the first law of thermodynamics. Can you at least state what that is so we can take it further? Link to comment Share on other sites More sharing options...
scott harvey Posted April 6, 2015 Author Share Posted April 6, 2015 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 Link to comment Share on other sites More sharing options...
studiot Posted April 6, 2015 Share Posted April 6, 2015 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 Does it? I thought it stated that the change in internal energy = the sum of the work done and the heat exchanged dU = q + w or dU = q - w, depending upon your sign convention. Can we discuss the problem in these terms since that is what your lecturers are expecting? Link to comment Share on other sites More sharing options...
scott harvey Posted April 6, 2015 Author Share Posted April 6, 2015 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. Link to comment Share on other sites More sharing options...
swansont Posted April 6, 2015 Share Posted April 6, 2015 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/ [/size] P2[/size] 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 Do you know where that equation comes from? Also, do you see how your statement of the 1st law and studiot's say the same basic thing, except one is in much more detail? Link to comment Share on other sites More sharing options...
scott harvey Posted April 6, 2015 Author Share Posted April 6, 2015 an isothermal gas expansion when the gas expansion follows the law pV1 = C Link to comment Share on other sites More sharing options...
studiot Posted April 6, 2015 Share Posted April 6, 2015 (edited) The question also told you that Scott Harvey Cv = 718 J/kg K and R = 287 J/kg K determine: But you have made no attempt to use this information. So I'm quite sure that either your lecturers or your tutor gave you much more information than you seem willing to let on. Yes the equation you quote (from your lecturer) is appropriate, but only when you understand what you are doing and we have some way to go to reach that stage. So just please try to answer my simple questions and swansont's and don't offer any additional information. This trail will eventually lead to the answer you need. Edited April 6, 2015 by studiot Link to comment Share on other sites More sharing options...
scott harvey Posted April 6, 2015 Author Share Posted April 6, 2015 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 Link to comment Share on other sites More sharing options...
studiot Posted April 6, 2015 Share Posted April 6, 2015 (edited) Was it so long ago? I asked you what an open system was. Swansont asked you what isothermal meant. Both are mentioned in your set question. Edited April 6, 2015 by studiot Link to comment Share on other sites More sharing options...
scott harvey Posted April 6, 2015 Author Share Posted April 6, 2015 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 Link to comment Share on other sites More sharing options...
studiot Posted April 6, 2015 Share Posted April 6, 2015 scott harvey a closed system can exchange heat or work with its surroundings isothermal is a process or change taking place at constant temperature Both right. Just to add the system is closed because no matter (mass ) leaves or enters. So work and/or heat are exchanged with the surroundings and our goal is to evaluate each, bearing in mind that these exchanges take place at constant temperature. We are told that the gas expands So what changes and which way is any work 'exchanged' ie what does work on what? Once that is decided we can find a formula to calculate the value of that work, if any. If we expanded a gas what would normally happen to the temperature of that gas? Link to comment Share on other sites More sharing options...
scott harvey Posted April 7, 2015 Author Share Posted April 7, 2015 would that be an exchange of pressure for temperature ? if we expand a gas it would increase the temperature ? Link to comment Share on other sites More sharing options...
studiot Posted April 7, 2015 Share Posted April 7, 2015 We are assuming an ideal gas. Which brings us to the dreaded first law and internal energy. The internal energy of an ideal gas depends only upon the temperature. At the very beginning, swansont asked you what isothermal means. So what are the implications of this being an isothermal process? Link to comment Share on other sites More sharing options...
swansont Posted April 8, 2015 Share Posted April 8, 2015 We are assuming an ideal gas. Should we be assuming this? Cv for an ideal gas is 3/2 R, and for a gas with n degrees of freedom it's n/2 R. The information tells us we have a diatomic gas. (Though thermo was along time ago, so I don't recall offhand whether this matters for an isothermal process) edit: I can't think of a reason why we would care if it's monoatomic or not for this problem. We have PV = constant, and that should be enough to solve it. Link to comment Share on other sites More sharing options...
studiot Posted April 8, 2015 Share Posted April 8, 2015 (edited) The point is that if the internal energy depends only upon temperature and temperature does not change then the internal energy does not change. That is dU = 0. That is q = -w = RT ln(V2/V1) = RT ln(P1/P2) Which is an expanded version of the formula scott's tutor provided. Edited April 8, 2015 by studiot Link to comment Share on other sites More sharing options...
swansont Posted April 8, 2015 Share Posted April 8, 2015 The point is that if the internal energy depends only upon temperature and temperature does not change then the internal energy does not change. That is dU = 0. That is q = -w = RT ln(V2/V1) = RT ln(P1/P2) Which is an expanded version of the formula scott's tutor provided. Right. He should see where the formula comes from. Link to comment Share on other sites More sharing options...
studiot Posted April 8, 2015 Share Posted April 8, 2015 Furthermore , R, Cv and the quantity of gas are all given in engineer's units (kg- somethings) so n is not needed. Link to comment Share on other sites More sharing options...
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