The Physics
Hypertextbook
Opus in profectus

# Engines

## Practice

### practice problem 1

0.40 moles of an ideal, monatomic gas runs through a four step cycle. All processes are either adiabatic or isochoric. The pressure and volume of the gas at the extreme points in the cycle are given in the table below.
1. Sketch the PV graph of this cycle.
2. Determine the temperature at state A, B, C, and D.
3. 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.)
4. For an engine running this cycle, determine its…
1. real efficiency
2. ideal efficiency
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)
path A→B B→C C→D D→A ABCDA
ΔU (J)
Q (J)
W (J)

#### solution

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 = 32nRT

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
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
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
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. 1.
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.

### practice problem 2

Write something completely different.

### practice problem 3

Write something different.