A gas is taken through the cycle ABCA shown in the PV -diagram. In the diagram,

Question

A gas is taken through the cycle ABCA shown in the PV-diagram. In the diagram, P1 = 200000 Pa, P2 = 400000 Pa, V1 = 0.01 m3, and V2 = 0.02 m3.

(a) Determine the internal energy of the gas at each state in the cycle.

UA =  J
UB =  J
UC =  J

(b) Determine the work done by the gas for each process.  
WAB =  J
WBC =  J
WCA =  J

(c) Determine the heat added to the gas for each process. (Heat removed from the gas is negative.)

QAB =  J
QBC =  J
QCA =  J

(d) Determine the following for one full cycle based on your answers to (b) and (c):

(i) The net work done by the gas.

J

(ii) The heat added to the gas.

J

(iii) The efficiency of the engine (as a fraction).

Explanation / Answer

A) let the gas be monoatomic

UA =1.5 P1V1 = 1.5×200000×0.01 =3000J

UB=1.5P2V1 = 1.5×400000×0.01 =6000J

Uc=1.5P1V2 = 1.5×200000×0.02 = 6000J

b)Wab=0

Wbc=(400000+200000)×(0.02-0.01)/2 =3000 J

Wca=200000×(0.01-0.02) =-2000J

C)Qab=Ub-Ua + Wab = 3000J

Qbc = Uc-Ub + Wbc = 3000J

Qca=Ua-Uc + Wca = -3000 -2000= -5000J

D) net work done = 3000-2000 = 1000J

Total heat given = 1000J

Efficiency = 1000/1000= 1

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