The figure below shows how 3 moles of a monatomic ideal gas follows the three-pa

Physics
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Apr 22nd, 2015

A--->B: Q = W = integrate PdV = nRT/V dV = nRT ln (VB/VA) = 3RT ln(PA/PB)= 3RTln6 

PV = 3RT 

T = PV/3R = 100*1000*0.009/(3*8.31) =  36.1 K 

Q=W = 3RTln6 = 1612.5J 

T does not change so deltaU=0

B-->C 

Tc = Vc/VB TB = VA/VB TB = 1/6 TB = 6.02K

Q = 3*3/2R*(6.02-36.1) = -1121.9 J 

W = PdeltaV = 100*1000*(0.0015-0.009) =  -750J 

deltaU = Q-W =-1121.9+750 = 371.9J 

C-->A: 

Q = Q = 3*3/2R*(36.02-6.02) = 1121.9 J 

W= PdeltaV = 0 

deltaU = 1121.9 J

efficiency = (WAB+WBC)/QAB = (1612.5-750)/1612.5=53.5% 


May 2nd, 2015

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Apr 22nd, 2015
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Apr 22nd, 2015
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