(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Three moles of an ideal gas are taken around the cycle abc. For this gas, C_{p}= 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 T_{a}= 300K, T_{b}= 490K, T_{c}= 600K. Calculate the total work W for the cycle.

2. Relevant equations

W=P(delta)V=nR(delta)T

W=-(delta)E=-nC_{v}(delta)T

3. 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?

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# Homework Help: Thermodynamics : Work in a Cyclic Process

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