Engines

1. Plot the four given points. Connect them with appropriately shaped curves. Isochors are vertical lines. Adiabats are something like inverse curves, only less smooth. Add arrows to show the direction of each path. The overall cycle runs clockwise.

   

2. Compute the temperature at each point using the ideal gas law.

   T =	PV	⇐	PV = nRT
   	nR

   Do it four times. Note that the temperature at the top of each isochor is four times the temperature at the bottom. That's because each isochor corresponds to a quadrupling in pressure. The numbers were rigged to make the results easy to check.

   TA =	(100,000 Pa)(0.010 m3)	= 300 K
   	(0.40 mol)(8.31 J/mol K)
   TB =	(1,462,000 Pa)(0.002 m3)	= 880 K
   	(0.40 mol)(8.31 J/mol K)
   TC =	(5,850,000 Pa)(0.002 m3)	= 3520 K
   	(0.40 mol)(8.31 J/mol K)
   TD =	(400,000 Pa)(0.010 m3)	= 1200 K
   	(0.40 mol)(8.31 J/mol K)

3. Compute the change in internal energy on each segment using the function of state.

   ∆U = ³₂nR∆T

   Because it's a function of state, all that matters are the final and initial temperatures. The shape of the path does not. Do this four times, and watch the signs. They do matter.

   ∆UA→B =	32(0.40 mol)(8.31 J/mol K)(880 K − 300 K)
   ∆UA→B =	+2,886 J
   ∆UB→C =	32(0.40 mol)(8.31 J/mol K)(3520 K − 880 K)
   ∆UB→C =	+13,164 J
   ∆UC→D =	32(0.40 mol)(8.31 J/mol K)(1200 K − 3520 K)
   ∆UC→D =	−11,550 J
   ∆UD→A =	32(0.40 mol)(8.31 J/mol K)(300 K − 1200 K)
   ∆UD→A =	−4,500 J

   If you haven't already done so, now might be a good time to start filling in the table. I recommend that you include the sign on all the values associated with a path. Don't omit the positive signs. It shows that you know there is a sign associated with each part of the process. I also like to use red for negative values in a table, but that's just my personal style.

   A four step cycle (adiabatic & isochoric)

   state		A	B	C	D
   P	(Pa)	100,000	1,462,000	5,850,000	400,000
   V	(m3)	0.010	0.002	0.002	0.010
   T	(K)	300	880	3,520	1,200

   path		A→B	B→C	C→D	D→A	ABCDA
   descrip­tion		adia­batic	iso­choric	adia­batic	iso­choric	closed cycle
   ΔU	(J)	+2,886	+13,164	−11,550	−4,500
   Q	(J)
   W	(J)

   We are done with the heavy computations. The rest of the table can be filled in with simple numbers and simple arithmetic. Simple numbers first. No heat is exchanged on adiabatic processes (since they are too quick for heat to go anywhere), no work is done on isochoric processes (since there's no change in volume), and there is no change in internal energy on a closed cycle (since there is no change in temperature). The simple numbers for those values are all zero.

   A four step cycle (adiabatic & isochoric)

   state		A	B	C	D
   P	(Pa)	100,000	1,462,000	5,850,000	400,000
   V	(m3)	0.010	0.002	0.002	0.010
   T	(K)	300	880	3,520	1,200

   path		A→B	B→C	C→D	D→A	ABCDA
   descrip­tion		adia­batic	iso­choric	adia­batic	iso­choric	closed cycle
   ΔU	(J)	+2,886	+13,164	−11,550	−4,500	0
   Q	(J)	0		0
   W	(J)		0		0

   The remaining empty cells are filled in using the first law of thermodynamics…

   ∆U = Q + W

   or, in the case of the whole cycle ABCD, by just doing the sum across a row.

   A four step cycle (adiabatic & isochoric)

   state		A	B	C	D
   P	(Pa)	100,000	1,462,000	5,850,000	400,000
   V	(m3)	0.010	0.002	0.002	0.010
   T	(K)	300	880	3,520	1,200

   path		A→B	B→C	C→D	D→A	ABCDA
   descrip­tion		adia­batic	iso­choric	adia­batic	iso­choric	closed cycle
   ΔU	(J)	+2,886	+13,164	−11,550	−4,500	0
   Q	(J)	0	+13,164	0	−4,500	+8,664
   W	(J)	+2,886	0	−11,550	0	−8,664

You can add some of this information onto your PV graph if you'd like. It might make it easier for you to understand what's going on. It does for me.

4.  

   1. By definition, efficiency is the ratio of work output to energy input. In the case of a heat engine like this one, that's the same as the ratio of the net work to the heat gained from the hot reservoir. Note that values substituted into efficiency equations are always positive. Just as soon as we got used to worrying about signs, we no longer have to worry about signs.

      ηreal =	Wnet Qhot
      ηreal =	8,664 J
      	13,164 J
      ηreal = 66%

      After some algebra, you could also write efficiency as one minus the ratio of the heat exhausted to the cold reservoir to the heat gained from the hot reservoir.

      ηreal = 1 −	Qcold Qhot
      ηreal = 1 −	4,500 J
      	13,164 J
      ηreal = 66%

   2. The ideal efficiency is computed from an equation that uses the hottest and coldest temperatures of the cycle instead of the heats in and out of the system.

      ηideal = 1 −	Tcold Thot
      ηideal = 1 −	301 K
      	3520 K
      ηideal = 91%

   These efficiencies are unrealistically large, which happens when numbers are rigged to be easy to check. Problems with more realistic numbers will have to wait for another day.

Answer it.

Answer it.

Answer it.