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