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Callen Thermodynamics Problem
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[QUOTE="Eldorin, post: 4480235, member: 485984"] [b]Still having a bit of trouble[/b] Alright, so I tried going through this and experienced some additional problems. So, I'd like to know how to solve it both graphically and not graphically. I linearized the equation, P=a*V+b, where P1=10^5, V1 = 10^-3, P2 = 10^5/32, V2 = 8e-3. From this I determined that a = -1.384, b=10^5 (Both rounded). I then set up dU = dW = -PdV, integrated and found that Ub-Ua = 0.5*(1.384*(Vb^2-Va^2))-(10^5(Vb-Va)) = -700J. From there, I graphically found (What I now assume to be Q, since it gets me the correct answer later in the progression. Can anyone explain why this is, but the Q answer I get doesn't match the answer?) by solving for the area under the triangle and got Q=339.06J. From there I set up Ub-Ua = Q - W, and solved for W and got -360.94J. At this point, I'm pretty thoroughly confused. Can anyone explain what I did wrong, how to find the solution without using the graph, and what solving it graphically actually gives, since it appears to not be Q or W. Thanks much! [/QUOTE]
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