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ΔU, Q, W of thermodynamic process

  1. Nov 3, 2016 #1
    1. The problem statement, all variables and given/known data
    A monatomic ideal gas undergoes the thermodynamic process shown in the PV diagram. Determine whether each of the values ΔU, Q, and W for the gas is positive, negative, or zero.
    upload_2016-11-3_7-41-47.png
    2. Relevant equations
    ΔU=Q-W
    W=integral under the curve
    PV=nRT T=PV/nR
    3. The attempt at a solution
    direction is negative then the integral negative then work is negative
    ΔU depends on change in temp, and here pressure increases and volume decreases so it was Ti=P0x2V0/nR and at final it becomes Tf=2P0xV0/nR Tf=Ti then it stays the same no change in temp then ΔU =0
    therefore 0=Q-W Q=W but W negative then Q also negative
    Is that correct? I need only to check my answer...
    and is there any other way to determine Q from Graph (without looking at W or ΔU) ???
     
  2. jcsd
  3. Nov 3, 2016 #2

    DrClaude

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    Staff: Mentor

    Yes.

    You could draw adiabats, one passing through point ##2V_0,P_0## and one through point ##V_0,2P_0##, and see if you are going closer to the origin (##Q<0##) or farther away from the origin (##Q>0##), but it is easier to calculate it as you did.
     
  4. Nov 3, 2016 #3
    how we draw them? do you mean to draw hyperbolas passing through them?
     
  5. Nov 3, 2016 #4

    DrClaude

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    Staff: Mentor

    I mean that you draw the curves ##P = V^{-\gamma}## that pass through the given points.
     
  6. Nov 3, 2016 #5

    rude man

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    Homework Helper
    Gold Member

    Careful with how W is defined. Physicists usually take W as the work done BY the gas, which is what you did & that's correct.

    But chemists often (or usually?) take W as the work done ON the gas in which case your W would be positive and the 1st law would read ΔU = Q + W.
     
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