A reversible engine contains 0.350 mol of ideal monatomic gas, initially at 586

Question

A reversible engine contains 0.350 mol of ideal monatomic gas, initially at 586 K and confined to a volume of 2.42 L . The gas undergoes the following cycle:
Isothermal expansion to 4.94 L
Constant-volume cooling to 282 K
Isothermal compression to 2.42 L
Constant-volume heating back to 586 K

Determine the engine's efficiency in percents, defined as the ratio of the work done to the heat absorbed during the cycle.

W/Qabsorbed= %

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Explanation / Answer

initial condition:

number of moles=n=0.35

temperature=T1=586 K

volume=V1=2.42 L

then pressure =P1=n*R*T1/V1=0.35*8.314*586/2.42 kPa=704.63 kPa

step 1:

isothermal expansion

as process is isothermal, tempeature remains constant at 586 K

so T2=T1=586 K

volume=V2=4.94 L

then pressure=P2=P1*V1/V2=704.63*2.42/4.94=345.18 kPa

as temperature remains constant, change in internal energy=0

as per first law of thermodynamics, heat supplied to the system=change in internal energy+work done by the system

work done by the system=W1=n*R*T*ln(V2/V1)=0.35*8.314*586*ln(4.94/2.42)=1216.8 J

then heat supplied to the system=Q1=1216.8 J

step 2:

constant volume cooling:

as volume is constant , work done=0

then heat supplied=change in internal energy

volume remain constant at 4.94 L

then V3=V2=4.94 L

temperature=T3=282 K

as P/T=constant

P2/T2=P3/T3

==>P3=P2*(T3/T2)=345.18*(282/586)=166.11 kPa

then heat supplied=Q2=change in internal energy=(3/2)*n*R*(T3-T2)

==>Q2=1.5*0.35*8.314*(282-586)=-1326.9 J

negative sign means heat is extracted from the system

work done=W2=0

step 3:

isothermal compression to 2.42 L

volume=V4=2.42 L

temperature remains constant.

T4=T3=282 K

then pressure=P4=n*R*T4/V4

=0.35*8.314*282/2.42=339.09 kPa

change in internal energy=0

work done=W3=n*R*T4*ln(V4/V3)

=0.35*8.314*282*ln(2.42/4.94)=-585.57 J

then heat supplied=Q3=W3=-585.57 J

negative sign means heat is extracted from the system

step 4:

constant volume heating back to 586 K

temperature=T5=586 K

volume =V5=V4=2.42 L

work done=W4=0

heat supplied=Q4=(3/2)*8.314*0.35*(586-282)=1326.9 J

so total heat supplied to the system=Q1+Q4=1216.8+1326.9=2543.7 J

total work done=W1+W2+W3+W4=1216.8+0-585.57+0=631.23 J

then efficiency=work done/total heat supplied=0.24815=24.815%

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