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    Hey all,

    Thermo really confuses me these days...


    Forgot to label the corners, but the cycle is going clockwise btw.

    I've been given that P1 = 100 kPa, P2 = 200 kPa, V1 = 25L and V2 = 50L.

    It's also a monatomic gas with n = 1.

    a) Find the temperature of each state of the cycle
    I got this to be 300K at the bottom left corner (call it 1) and 600K at the top and right hand corners (call it 2 and 3, respectively) of the diagram. Is this okay?

    b)Find the heat flow for each part of the cycle
    From state 1 to 2, I found that the heat flow in, QH is the same thing as Cv.dT = (3/2)R.(600-300) = 3741J. This looked okay to me I guess!

    But I don't know how to find the heat flow QC out of the cycle... I've kind of guessed that heat flows out between states 3 and 1 as the temperature drops, but I'm not sure. Could I just use QC = CpdT + PdV? A detailed explanation of what's physically happening would be great. I keep getting lost in these questions!
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    ...bump :itsme:
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    (Original post by trm90)
    ...bump :itsme:
    I'm sure you've heard of the Carnot Cycle (http://en.wikipedia.org/wiki/Carnot_cycle). This is an example of a PV diagram (the ideal case, actually) - if you can understand the Carnot Cycle I'm sure you will be able to apply the same ideas to your problem.

    What is physically happening can be thought of in terms of a pistons and heat baths... Why don't you read the wiki on carnot cycles and see if that helps?
    Your other answers look fine though (not checked them, but working seems sensible).
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    (Original post by M_E_X)
    I'm sure you've heard of the Carnot Cycle (http://en.wikipedia.org/wiki/Carnot_cycle). This is an example of a PV diagram (the ideal case, actually) - if you can understand the Carnot Cycle I'm sure you will be able to apply the same ideas to your problem.

    What is physically happening can be thought of in terms of a pistons and heat baths... Why don't you read the wiki on carnot cycles and see if that helps?
    Your other answers look fine though (not checked them, but working seems sensible).
    Thanks! I do kind of understand how the carnot cycle works, but I'm sort of having trouble applying it to this question

    I was wondering if you could comment on my thought process though!

    So during an isothermal expansion, there is no increase in temperature (so dU = 0, is that right?) and the gas' volume is increasing, so W = PdV and hence Q = W = PdV (by the first law of thermodynamics).

    During the constant pressure process, there is decrease in volume AND temperature, so Q = -Cp dT - PdV, which is basically when heat is flowing out...

    But then during the constant volume process, the temperature is increasing hence dU = Q = Cv dT... so does that mean heat is flowing in in two processes in this diagram, and flowing out in the compression?

    I figured I could calculate Q out and Q in by using W = Qin - Qout, and finding work manually by finding the area of the triangle (which I did), but I'm still really confused as to where heat is flowing into and out of the system.

    Thanks again!
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    (Original post by trm90)
    I figured I could calculate Q out and Q in by using W = Qin - Qout, and finding work manually by finding the area of the triangle (which I did), but I'm still really confused as to where heat is flowing into and out of the system.
    That is exactly right, yeah Work done = area inside cycle.

    During isothermal expansion, heat flows in (but as volume expands temperature remains constant) - is that right?
    Opposite occurs during the isothermal contraction.

    I studied the carnot cycle for my exam in Jan (just gone), but sadly I've forgotten most of it now.
    Your method sounds right, though.
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    (Original post by M_E_X)
    That is exactly right, yeah Work done = area inside cycle.

    During isothermal expansion, heat flows in (but as volume expands temperature remains constant) - is that right?
    Opposite occurs during the isothermal contraction.

    I studied the carnot cycle for my exam in Jan (just gone), but sadly I've forgotten most of it now.
    Your method sounds right, though.
    Thanks!
 
 
 
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