Engines
Practice
practice problem 1
 Sketch the PV graph of this cycle.
 Determine the temperature at state A, B, C, and D.
 Calculate W, Q, and ΔU on the path A→B, B→C, C→D, D→A and for one complete cycle. (Include the algebraic sign with each value.)
 For an engine running this cycle, determine its…
 real efficiency
 ideal efficiency
state  A  B  C  D  

P  (Pa)  100, 
1,462, 
5,850, 
400, 

V  (m^{3})  0.010  0.002  0.002  0.010  
T  (K) 
path  A→B  B→C  C→D  D→A  ABCDA  

description  adiabatic  isochoric  adiabatic  isochoric  closed cycle  
ΔU  (J)  
Q  (J)  
W  (J) 
solution
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.
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.
T_{A} = (100,000 Pa)(0.010 m^{3}) = 300 K (0.40 mol)(8.31 J/mol K) T_{B} = (1,462,000 Pa)(0.002 m^{3}) = 880 K (0.40 mol)(8.31 J/mol K) T_{C} = (5,850,000 Pa)(0.002 m^{3}) = 3520 K (0.40 mol)(8.31 J/mol K) T_{D} = (400,000 Pa)(0.010 m^{3}) = 1200 K (0.40 mol)(8.31 J/mol K) Compute the change in internal energy on each segment using the function of state.
∆U = ^{3}_{2}nR∆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.
∆U_{A→B} = ^{3}_{2}(0.40 mol)(8.31 J/mol K) (880 K − 300 K) ∆U_{B→C} = ^{3}_{2}(0.40 mol)(8.31 J/mol K) (3520 K − 880 K) ∆U_{C→D} = ^{3}_{2}(0.40 mol)(8.31 J/mol K) (1200 K − 3520 K) ∆U_{D→A} = ^{3}_{2}(0.40 mol)(8.31 J/mol K) (300 K − 1200 K) 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 (m^{3}) 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 (m^{3}) 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 (m^{3}) 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.

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.
η_{real} = W_{net} Q_{hot} η_{real} = 8,664 J 13,164 J η_{real} = 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.
η_{real} = 1 − Q_{cold} Q_{hot} η_{real} = 1 − 4,500 J 13,164 J η_{real} = 66% 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.
η_{ideal} = 1 − T_{cold} T_{hot} η_{ideal} = 1 − 301 K 3520 K η_{ideal} = 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.
practice problem 2
solution
Answer it.
practice problem 3
solution
Answer it.
practice problem 4
solution
Answer it.