Thermodynamics : Work in a Cyclic Process

  • Thread starter rizardon
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Homework Statement


Three moles of an ideal gas are taken around the cycle abc. For this gas, Cp= 29.1 J/mol K. Process ac is at constant pressure, process ba is at constant volume, and process cb is adiabatic. The temperatures of the gas in states a, c, and b are Ta= 300K, Tb= 490K, Tc= 600K. Calculate the total work W for the cycle.


Homework Equations


W=P(delta)V=nR(delta)T
W=-(delta)E=-nCv(delta)T

The Attempt at a Solution



Total work is the sum of the work done at each process

Total Work = Wac + Wcb + Wba

W at ac = isobaric process = nR(delta)T

W at cb = adiabatic process = -nCv(delta)T

W at ba = isochoric process, therefore W= 0

Cp = 29.1 = 3.5R so Cv = 2.5R = 20.8

Total Work = (3)(8.31)(190) - (3)(20.8)(110) + 0
= 4,736.7 - 6,864
= -2,127.3

I have to submit this for my homework, but I'm not sure if this is correct. Could you please check this for me?
 

Answers and Replies

  • #2
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Three moles of an ideal gas are taken around the cycle abc.
When a cycle is described as "ABC" I imagine they mean the process goes in the pattern A=>B=>C=>A, yet the way the rest of the problem unfolds is confusing to me.

Process ac is at constant pressure, process ba is at constant volume, and process cb is adiabatic.
I feel like this describes the cycle A=>C=>B=>A, contradicting what I previously thought.

If you could clarify the way the cycle occurs that would be helpful, and also make sure the contour that you follow around the cycle is the way they want, or else you might end up with the wrong sign on your answer. (Work calculated on a pV diagram going left to right will be positive, right to left will be negative.)

Hope that helps

-MG
 
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