The pV diagram in the figure to the right shows a heat engine operating on 0.850

Question

The pV diagram in the figure to the right shows a heat engine operating on 0.850 mol of H2. Segment ca is isothermal.

(a) Without doing any calculations, identify the segments during which heat enters the gas and those during which it leaves the gas. Explain your reasoning.

(b) Find the value of the temperature at all three verticies.

(c) Find the change in internal energy, work, and heat during each process of the cycle.

(d) Find the termal efficiency of this heat engine, treating the gas as ideal and monatomic.

(e) Calculate the coefficient of performace if the engine is run in reverse, as a refigerator.

Explanation / Answer

a)

From the given diagram, we can say that in the segment ab, the heat enters the gas because the volume expands, in the segment bc the beat leaves the gas znc in the segment ca is isobaric and it always leaves the heat.

b)

Here from the given PV diagram, the temperature at a is same as c because it isothermal process

Ta =Tc =PV/nR

Ta=Tc =PaVa/nR =0.700*1.013*105*0.030/0.85*8.314 =301K

Ta=Tc =301K

similalry at point b,

Tb =PbVb/nR =0.700*1.013*105*0.1/0.85*8.314 =1003.4K

The heat loss in the process of ca is given by

Qca =nRTcln(V2/V1) =0.85*8.314*301*ln(0.03/0.1)

=-2560J

Gain of heat in the process of ab is

Qab =nCpDeltaT

=(0.85)(28.78)(1003.4-301) =17180J

Loss of heat energy in the process of bc is

Qbc =nCvDeltaT =(0.85)(20.42)(-1003.4+301) =-12137J

d)

Therefore the work done is given by

W =Qca+Qab+Qbc =-2560J+17180J-12137J=2490J

Now the thermal efficiency is given by

n =work/Qab =2490J/17180J=0.145 =14.5%

e)

The coefficient of performace if the engine is run in reverse, as a refigerator.

n =1/K+1

Then K =(1/0.145)-1 =5.896

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