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Thermodynamics - compressed gas P-V graph, how much work is done?

  1. Oct 5, 2008 #1
    1. The problem statement, all variables and given/known data

    A gas is compressed, allowed to expand linearly, and then cooled at constant volume as shown in the P-V graph. How much work is done by the gas in this cycle?

    in the graph: compression is a straight line at P(atm)=3 that goes from V(liters)=1 to V=5; cooling occurs at V=5 from P=4 to P=3; a diagonal line from (1,3) to (5,4) is the expansion; a graph of a right triangle should be drawn with this information


    2. Relevant equations

    PV = nRT or PV = NkT, P1/T1 = P2/T2 or V1/T1 = V2/T2 ?

    if it is adiabatic: f/2 dT/T = -dV/V where f is the number of degrees of freedom; further math will obtain PV^((f+2)/f) = constant

    3. The attempt at a solution

    Well, i'm not sure if it is adiabatic or isothemral... i think adiabatic since the cooling occurs all at once at the end of the graph?? from what i understand.. the volume goes from 1 to 5 liters during expansion while the pressure goes form 3 to 4 atm. then the cooling causes the pressure to go back to 3 while the volume is still 5?

    Vfinal Tfinal^(f/2) = Vinitial Tinitial^(f/2)

    5(Tfinal^(f/2))= 1(Tinitial^(f/2))

    ...

    okay, okay.. stumbled on something in notes:

    - integral P dV

    since Pinitian Vinitial^((f+2)/f) = Pfinal Vfinal^((f+2)/f) it is a constant

    so, P = (constant)V^-((f+2)/f)

    so,W = - integral (constant)V^-((f+2)/f) dV

    ends up to be

    {[-Pinitial Vinitial^((f+2)/f) ]/1 - ((f+2)/f) } (Vfinal^[1 - ((f+2)/f)] - Vinitial^[1 - ((f+2)/f)] )

    plugging values

    {[ -(3)(1)^((f+2)/f) ] / 1 - ((f+2)/f) } (5^[1 - ((f+2)/f)] - 1^[1 - ((f+2)/f)] )

    degrees of freedom: maybe the word "linearly" is a hint here?

    if the gas is monoatomic: ((f+2)/f) = 5/3

    if the gas is diatomic: 7/5,

    diatomic at high temperatures or triatomic: 8/5

    which degree of freedom should be used and why?

    for now i'm going to go with 5/3:

    {[ -(3)(1)^(5/3) ] / 1 - (5/3) } (5^[1 -(5/3)] - 1^[1 - (5/3)] )

    9/2 [5^(-2/3) - 1^(-2/3)]

    W ~ -3 J by the gas... so work is actually done on the gas?

    any explaination would be appreciated, thanks!
     
  2. jcsd
  3. Oct 5, 2008 #2

    Mute

    User Avatar
    Homework Helper

    "Expands linearly" describes the diagonal line in the PV graph since that line is, well... linear (i.e., not curved).

    The total work done is going to be the "sum" of the areas beneath each segment of the cycle. Note that the area can be negative depending on which way the cycle goes.

    Essentially what you're doing is calculating the area enclosed by the cycle. This should be easy in this case because you just have a right triangle; you just have to be careful to get the correct sign. For more general shapes enclosed by the curves, you have to calculate [itex]W_{segment} = -\int p dV[/itex] along each segment of the path, and sum all of the contributions up. On paths with constant volume, the work done is zero. On paths with constant pressure, the work done is [itex]-p\Delta V[/itex]. On paths where the pressure varies with volume, you do the integral. For example, in this problem for the linear section of the cycle you can write

    [tex]p = aV + b[/tex]

    where you'll have to determine the slope a and the intercept b. Then, on that part of the cycle, you would get [itex]W = -[a(V_f^2-V_i^2)/2 + b(V_f - V_i)][/itex].

    What you're doing right now isn't quite correct because your gas isn't expanding isothermally or adiabatically. Otherwise you'd be taking the correct approach (I think - it's hard to read non-latexed math).

    Does this help?
     
  4. Oct 5, 2008 #3
    yes... that is so awesome! I had been working on the problem just before I came here, and I had noticed that I was overcomplicating it. I did figure out that the work done would be the area enclosed, and I was able to show that graphically for the sloped line. Your formulas helped me to prove it mathematically. Thank you. :cool:
     
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