Thanks for helping me out guys! Thermodynamic cycle and efficiency of engine Con

Question

Thanks for helping me out guys!

Thermodynamic cycle and efficiency of engine Consider 1.0 mole of an ideal monatomic gas undergoing the cycle shown below The cycle consists of three segments : an isobaric path from c to a; an isochoric path from a to b; and an isothermal path from b to c. The temperature at a is 273 K. What is the ratio Vc, /Va in terms of the temperatures at band a? What is the work done along b to c? What is the work done along c to a? What is Qbc? What is Qab? What is the efficiency of the engine operating along the above cycle?

Explanation / Answer

a) Vc/Tc = Va/Ta

b) W = R*T*ln(Vb/Vc)

c) W = v(Vc-Va)

d) Qbc = W (since dU=0)

e) Qab = dU (since Wab=0)

f) efficiency, n = Qbc/Wac

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