# 3.1: A first heat engine - the PV rectangle

Let’s solve a relatively simple problem first.

A monoatomic ideal gas begins with a volume of 2.00 m3 and under a pressure of 1.50 bar and at a temperature of 298 K. The gas first is compressed to a volume of 1.00 mat constant pressure, then (in the compressed state) pressurized to a pressure of 3.00 bar at constant volume, then decompressed back to 2.00 m3 at constant pressure, then restored to its original state in every way - the number of moles have not changed, and the temperature is again 298 K. Find the total work done by the gas at each step, and the total heat added during the cycle. I’ve provided a sketch of this cycle at the right. We can dub this a constant pressure / constant volume cycle, or simply a PV rectangle, because the constant pressure and constant volume steps make the cycle appear very simple. The cycle starts at the lower right corner of the rectangle, with the expanded gas at lowest pressure, and then proceeds clockwise around the rectangle.

Because of the integral definition of work (w = -∫ P dV), we could find the total work for the cycle by simply finding the area of the rectangle. However, at this point, it will be more instructive to work the exercise out, step by step.

In step 1, the compression step, we take a gas at 2.00 m3 and, while holding it at a constant volume of 1.50 bar (which equals 1.50 × 105 Pa), we halve its volume. This is the arrow at the bottom of the rectangle on the graph. It is very straightforward to find the work done on this gas (which is positive, because volume is decreasing):

$w 1 equals en-dash upper P 1 increment Subscript 1 Baseline upper V equals en-dash left-parenthesis 1.50 times 10 Superscript 5 Baseline Pa right-parenthesis left-parenthesis 1.00 m cubed en-dash 2.00 m cubed right-parenthesis equals 1.50 times 10 Superscript 5 Baseline upper J left-parenthesis compression right-parenthesis$

(There is new notation here that you are probably not accustomed to. The subscript is applied to the delta, writing Δ1V and not ΔV1, to emphasize that this is the first change in volume, and not a change in "volume 1". This first change in volume is the compression between our two given volumes; we'll distinguish this from the expansion to come.)

In the second step (represented by the arrow at the left of the graph), where the gas is pressurized, it is held at constant volume. If volume of a gas remains constant, no work is done. w2 = 0 for the second step.

In step 3, the expansion step, we now have a gas at a heightened pressure (3.00 bar, or 3.00 × 105 Pa), and we are letting the volume of the gas relax back to 2.00 m3. This work is also easily computed:

$w 3 equals en-dash upper P 3 increment Subscript 3 Baseline upper V equals en-dash left-parenthesis 3.00 times 10 Superscript 5 Baseline Pa right-parenthesis left-parenthesis 2.00 m cubed en-dash 1.00 m cubed right-parenthesis equals negative 3.00 times 10 Superscript 5 Baseline upper J left-parenthesis expansion right-parenthesis$

And again, in the fourth step, where the gas is depressurized, it is held at constant volume. If volume of a gas remains constant, no work is done. w4 = 0 for the fourth step.

Let’s summarize the calculations we’ve made:

$w 1 equals 1.50 times 10 Superscript 5 Baseline upper J left-parenthesis compression right-parenthesis$

$w 2 equals 0 left-parenthesis pressurization right-parenthesis$

$w 3 equals negative 3.00 times 10 Superscript 5 Baseline upper J left-parenthesis expansion right-parenthesis$

$w 4 equals 0 left-parenthesis depressurization right-parenthesis$

$w 1 plus w 2 plus w 3 plus w 4 equals 1.50 times 10 Superscript 5 Baseline upper J plus 0 minus 3.00 times 10 Superscript 5 Baseline upper J plus 0 equals negative bold 1.50 bold times bold 10 Superscript bold 5 Baseline bold upper J bold equals negative bold 150 bold kJ$

Note a couple of details in this work. We have completed a full cycle of the compression, pressurization, decompression, and depressurization. The total work done has a value; it is not zero. (Given the integration we talked about earlier, find the area of the rectangle and see if you understand where the total work done came from.)

What’s more, if the difference in the pressures is any greater, the value of this work increases in kind, without any kind of limit. This wouldn’t be offensive at all if it wasn’t for the entirely unrealistic constraint in the problem - no moles of gas lost, and the temperature returns to exactly the same value it started from.

This would require that, because the initial and final states are the same,

$normal upper Delta upper U equals three-halves n upper R normal upper Delta upper T equals three-halves n upper R left-parenthesis 0 right-parenthesis equals 0 left-parenthesis given normal upper Delta upper T equals 0 for full cycle right-parenthesis$

Because ΔU = 0, by the First Law:

$normal upper Delta upper U equals q plus w right-arrow if normal upper Delta upper U equals 0 comma q equals negative w$

Therefore, if the change in internal energy is zero, we don’t have to do any further computation to find the total heat transferred. If the work done by the system is 150 kJ, then the heat transferred into the system must be q = 150 kJ.

Our hope, therefore, is that when we compute the heat transfer for each step, we come back to the same result. To do this, we must now concern ourselves with the conditions at each step.

Initially we’re at 2.00 m3, 1.50 bar (which is equal to 1.50 × 105 Pa) and 298 K. We need a number of moles (note the value of R we use):

$n equals StartFraction upper P upper V Over upper R upper T EndFraction equals StartFraction left-parenthesis 1.50 times 10 Superscript 5 Baseline Pa right-parenthesis left-parenthesis 2.00 m cubed right-parenthesis Over left-parenthesis 8.3145 upper J mol Superscript negative 1 Baseline upper K Superscript negative 1 Baseline right-parenthesis left-parenthesis 298 upper K right-parenthesis EndFraction equals 12 ModifyingBelow 1 With bar .1 mol$

This allows us to compute the temperatures at each successive set of conditions, and prepare for the temperature to fluctuate more than a little bit:

$Compressed colon upper T equals StartFraction upper P upper V Over n upper R EndFraction equals StartFraction left-parenthesis 1.50 times 10 Superscript 5 Baseline Pa right-parenthesis left-parenthesis 1.00 m cubed right-parenthesis Over left-parenthesis 121.1 mol right-parenthesis left-parenthesis 8.3145 upper J mol Superscript negative 1 Baseline upper K Superscript negative 1 Baseline right-parenthesis EndFraction equals 149 upper K$

$Pressurized colon upper T equals StartFraction upper P upper V Over n upper R EndFraction equals StartFraction left-parenthesis 3.00 times 10 Superscript 5 Baseline Pa right-parenthesis left-parenthesis 1.00 m cubed right-parenthesis Over left-parenthesis 121.1 mol right-parenthesis left-parenthesis 8.3145 upper J mol Superscript negative 1 Baseline upper K Superscript negative 1 Baseline right-parenthesis EndFraction equals 298 upper K$

$Decompressed colon upper T equals StartFraction upper P upper V Over n upper R EndFraction equals StartFraction left-parenthesis 3.00 times 10 Superscript 5 Baseline Pa right-parenthesis left-parenthesis 2.00 m cubed right-parenthesis Over left-parenthesis 121.1 mol right-parenthesis left-parenthesis 8.3145 upper J mol Superscript negative 1 Baseline upper K Superscript negative 1 Baseline right-parenthesis EndFraction equals 596 upper K$

$Depressurized colon left-parenthesis you can demonstrate that the temperature returns to 298 upper K exclamation-mark right-parenthesis$

Knowing temperatures, we can compute heat transfers at each step. Let’s be very careful about how we separate the quantities we’re studying, because we’ve set ourselves up to solve a problem like this.

Two steps are steps of changing volume - the compression and the decompression. During these steps, the pressure of the container is held constant. And, because this is a monoatomic ideal gas, we know what the molar specific heat is - it’s 5/2R. The other two steps are the pressurization and depressurization - during those steps, it’s volume held constant, and the molar specific heat becomes 3/2R.

Let’s build out each of these computations:

$q 1 equals n upper C Subscript upper P Baseline overbar normal upper Delta upper T equals left-parenthesis 12 ModifyingBelow 1 With bar .1 mol right-parenthesis five-halves left-parenthesis 8.3145 upper J mol Superscript negative 1 Baseline upper K Superscript negative 1 Baseline right-parenthesis left-parenthesis 149 upper K negative 298 upper K right-parenthesis equals minus 37 ModifyingBelow 5 With bar .1 kJ left-parenthesis compression right-parenthesis$

$q 2 equals n upper C Subscript upper V Baseline overbar normal upper Delta upper T equals left-parenthesis 12 ModifyingBelow 1 With bar .1 mol right-parenthesis three-halves left-parenthesis 8.3145 upper J mol Superscript negative 1 Baseline upper K Superscript negative 1 Baseline right-parenthesis left-parenthesis 298 upper K negative 149 upper K right-parenthesis equals 22 ModifyingBelow 5 With bar .0 kJ left-parenthesis pressurization right-parenthesis$

$q 3 equals n upper C Subscript upper P Baseline overbar normal upper Delta upper T equals left-parenthesis 12 ModifyingBelow 1 With bar .1 mol right-parenthesis five-halves left-parenthesis 8.3145 upper J mol Superscript negative 1 Baseline upper K Superscript negative 1 Baseline right-parenthesis left-parenthesis 596 upper K negative 298 upper K right-parenthesis equals 75 ModifyingBelow 0 With bar .3 kJ left-parenthesis decompression right-parenthesis$

$q 4 equals n upper C Subscript upper V Baseline overbar normal upper Delta upper T equals left-parenthesis 12 ModifyingBelow 1 With bar .1 mol right-parenthesis three-halves left-parenthesis 8.3145 upper J mol Superscript negative 1 Baseline upper K Superscript negative 1 Baseline right-parenthesis left-parenthesis 298 upper K negative 596 upper K right-parenthesis equals minus 45 ModifyingBelow 0 With bar .0 kJ left-parenthesis depressurization right-parenthesis$

$q 1 plus q 2 plus q 3 plus q 4 equals minus 37 ModifyingBelow 5 With bar .1 kJ plus 22 ModifyingBelow 5 With bar .0 kJ plus 75 ModifyingBelow 0 With bar .3 kJ minus 45 ModifyingBelow 0 With bar .0 kJ equals bold 150 bold kJ$

It’s always reassuring when a long computation works correctly!

There’s one other detail hidden behind these heat computations. The entire point of a heat engine is to take heat in to do useful work. We have the work done by the engine: -150 kJ. The heat we took in, though, is much, much greater - it’s the sum of the two positive values of q, those associated with the pressurization and decompression steps, when the temperature of the engine was increasing. That’s 975 kJ.

Of those 975 kJ of energy, we only used 150 kJ of it. The rest of the energy - 825 kJ - was simply kicked out of the engine, as waste heat.

We can define a quantity for an engine, the efficiency, that describes the fraction of the the heat energy input that we use to do useful work. Because of the sign on work when the engine does work on the surroundings, we have to have an extra negative sign:

$e equals minus StartFraction w Over q Subscript upper A upper B upper S Baseline EndFraction equals minus StartFraction negative 150 kJ Over 975 kJ EndFraction equals 0.154 left-parenthesis equals 15.4 percent-sign right-parenthesis$

An engine that only uses 15.4% of absorbed heat to do useful work is pretty darn inefficient. We need to do better. We need to perform using cycles that go beyond simple constant-pressure and constant-volume steps.