Question

Figure 1 of 1 p (atm) 6.0 5.0 4.0 3.0 2.0 1.0 V (L) O 1.0 3.0 5.0 7.0 2.0 4.0 6.0

(Figure 1) shows a pV diagram for a heat engine that uses 1.40 moles of an ideal gas. The internal energy of the gas changes by the following amounts:

ΔUa→b=+4040J, ΔUb→c=−4848J, ΔUc→d=−808J, and ΔUd→a=+1616J

Part A

How much heat goes into this gas per cycle?

Express your answer in joules to three significant figures.

Answer: ______ J

Part B

Where in the cycle does the heat go into the gas?

Select all that apply.

c→d

b→c

d→a

a→b

Part C

How much heat is ejected by the gas per cycle?

Express your answer in joules to three significant figures.

Answer: _______ J

Part D

Where in the cycle does the heat get ejected by the gas?

Select all that apply.

a→b

d→a

c→d

b→c

Part E

How much work does this engine do each cycle?

Express your answer in joules to three significant figures.

Answer: _______ J

Part F

What is the thermal efficiency of the engine?

Express your answer as a percentage to two significant figures.

Answer: ______ %

Answer 1

1 atm = 101325 pascals

1L = 10^-3 m³

Part A

W= area of squre = 4*101325*4*10^-3 = 1621.2 J answer

the heat goes into the gas

Q =W + ΔU

Qda = 0 +1616 = +1616 J > 0

Qab = Wab+ Uab > 0

Qbc = 0+Ubc < 0

Qcd =  −808 +Wcd <  

so heat enter into the system in d -> a and a -> b only  

and rejected in b->c and c- >d

heat entered = Qda + Qab

Qda = 1616 J becuase Wda = 0

Qab = Uab + Wab = 4040+ 5*101325*4*10^-3 = 6066.5

so entered heat = 1616 + 6066.5 = 7682.5 J

heat rejected = Qbc + Qcd

Qbc = - 4848 J

Qcd = −808 - 1*101325*4*10^-3 = -1213.3 J

so rejected heat = 4848+1213.3 = 6061.3 J

the thermal efficiency of the engine = Work/heat entered *100

=1621.2 /7682.5 *100 = 21.1025057 %