A heat engine is based on a cycle consisting of an isobaric, an isometric and an

Question

A heat engine is based on a cycle consisting of an isobaric, an isometric and an adiabatic process (Fig. 4). The working substance is 1.00 mol of ideal monatomic gas. The work done on the gas during the adiabatic process is 1023.3 J.

a) find the work done by the gas for the entire cycle

b) find the amount of heat for each process

c) find the efficiency of the heat engine

d) find the internal energy at points 1 and 2

Please include the necessary work, not just the answers

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Explanation / Answer

a)Work done by gas in adiabatic part= -1023.3 J

Work done in isobaric part= pV= 240*103*(18-12)*10-3=240*6=1440 J

Work done in isometric part=0

Total work done by gas=1440 J-1023.3 J=416.7J'

b) The process 3-1 is adiabatic, amount of heat=0.

For the cycle, the change in internal energy=0=U

Also, Q=U+W

Therefore, total heat supplied to cycle= W=416.7J

Q1-2= W1-2+U1-2=1440 J+ CvT=1440J+ 1.5R*PV/R=1440+1.5PV=1440J+2160J=3600J

Qtotal=Q1-2+Q2-3+Q3-1

Q3-1=0(adaibatic process)

Q2-3=416.7-3600J=-3183.3J

c)Efficiency=416.7/3183.3=13.1%

d)T at 1= PV/R=240*103*12*10-3/8.314=346.4 K

T at 2=519.6 K

Internal energy at 1= nCvT1=1*1.5*R*T=4320J

Internal energy at 2= nCvT2=6480J

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