# 3.1: A first heat engine - the PV rectangle

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- 329860

Let’s solve a relatively simple problem first.

*A monoatomic ideal gas begins with a volume of 2.00 **m*^{3} *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 **m*^{3 }*at** constant pressure, then (in the compressed state) pressurized to a pressure of 3.00 bar at constant volume, then decompressed back to 2.00 **m*^{3} *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 m^{3} and, while holding it at a constant volume of 1.50 bar (which equals 1.50 × 10^{5} 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):

$${w}_{1}=\u2013{P}_{1}{\u2206}_{1}V=\u2013(1.50\times {10}^{5}\text{Pa})(1.00{\text{m}}^{3}\u20132.00{\text{m}}^{3})=1.50\times {10}^{5}\text{J}\text{}\text{}\text{}\text{}\text{}\text{(compression)}$$

(There is new notation here that you are probably not accustomed to. The subscript is applied to the delta, writing Δ_{1}*V* and not Δ*V*_{1}, 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.* *w*_{2} = 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 × 10^{5} Pa), and we are letting the volume of the gas relax back to 2.00 m^{3}. This work is also easily computed:

$${w}_{3}=\u2013{P}_{3}{\u2206}_{3}V=\u2013(3.00\times {10}^{5}\text{Pa})(2.00{\text{m}}^{3}\u20131.00{\text{m}}^{3})=-3.00\times {10}^{5}\text{J}\text{}\text{}\text{}\text{}\text{}\text{(expansion)}$$

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.* *w*_{4} = 0 for the fourth step.

Let’s summarize the calculations we’ve made:

$${w}_{1}=1.50\times {10}^{5}\text{J}\text{}\text{}\text{}\text{}\text{}\text{(compression)}$$

$${w}_{2}=0\text{}\text{}\text{}\text{}\text{}\text{(pressurization)}$$

$${w}_{3}=-3.00\times {10}^{5}\text{J}\text{}\text{}\text{}\text{}\text{}\text{(expansion)}$$

$${w}_{4}=0\text{}\text{}\text{}\text{}\text{}\text{(depressurization)}$$

$${w}_{1}+{w}_{2}+{w}_{3}+{w}_{4}=1.50\times {10}^{5}\text{J}+0-3.00\times {10}^{5}\text{J}+0=\mathbf{-}\mathbf{1.50}\mathbf{\times}{\mathbf{10}}^{\mathbf{5}}\mathbf{\text{J}}\mathbf{=}\mathbf{-}\mathbf{150}\mathbf{\text{kJ}}$$

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,

$$\mathrm{\Delta}U=\frac{3}{2}nR\mathrm{\Delta}T=\frac{3}{2}nR(0)=0\text{}\text{}\text{}\text{}\text{}\text{(given}\mathrm{\Delta}T=0\text{for full cycle)}$$

Because Δ*U* = 0, by the First Law:

$$\mathrm{\Delta}U=q+w\text{}\to \text{}\text{if}\mathrm{\Delta}U=0,\text{}q=-w$$

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 m^{3}, 1.50 bar (which is equal to 1.50 × 10^{5} Pa) and 298 K. We need a number of moles (note the value of *R* we use):

$$n=\frac{PV}{RT}=\frac{(1.50\times {10}^{5}\text{Pa})(2.00{\text{m}}^{3})}{(8.3145{\text{J mol}}^{-1}{\text{K}}^{-1})(298\text{K})}=12\underset{\_}{1}.1\text{mol}$$

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:

$$\text{Compressed:}\text{}\text{}T=\frac{PV}{nR}=\frac{(1.50\times {10}^{5}\text{Pa})(1.00{\text{m}}^{3})}{(121.1\text{mol})(8.3145{\text{J mol}}^{-1}{\text{K}}^{-1})}=149\text{K}$$

$$\text{Pressurized:}\text{}\text{}T=\frac{PV}{nR}=\frac{(3.00\times {10}^{5}\text{Pa})(1.00{\text{m}}^{3})}{(121.1\text{mol})(8.3145{\text{J mol}}^{-1}{\text{K}}^{-1})}=298\text{K}$$

$$\text{Decompressed:}\text{}\text{}T=\frac{PV}{nR}=\frac{(3.00\times {10}^{5}\text{Pa})(2.00{\text{m}}^{3})}{(121.1\text{mol})(8.3145{\text{J mol}}^{-1}{\text{K}}^{-1})}=596\text{K}$$

$$\text{Depressurized: (you can demonstrate that the temperature returns to 298 K!)}$$

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}/_{2}*R*. The other two steps are the *pressurization *and *depressurization* - during those steps, it’s volume held constant, and the molar specific heat becomes ^{3}/_{2}*R*.

Let’s build out each of these computations:

$${q}_{1}=n\overline{{C}_{P}}\mathrm{\Delta}T=(12\underset{\_}{1}.1\text{mol})\frac{5}{2}(8.3145{\text{J mol}}^{-1}{\text{K}}^{-1})(149\text{K}-298\text{K})=-37\underset{\_}{5}.1\text{kJ}\text{}\text{}\text{}\text{}\text{}\text{(compression)}$$

$${q}_{2}=n\overline{{C}_{V}}\mathrm{\Delta}T=(12\underset{\_}{1}.1\text{mol})\frac{3}{2}(8.3145{\text{J mol}}^{-1}{\text{K}}^{-1})(298\text{K}-149\text{K})=22\underset{\_}{5}.0\text{kJ}\text{}\text{}\text{}\text{}\text{}\text{(pressurization)}$$

$${q}_{3}=n\overline{{C}_{P}}\mathrm{\Delta}T=(12\underset{\_}{1}.1\text{mol})\frac{5}{2}(8.3145{\text{J mol}}^{-1}{\text{K}}^{-1})(596\text{K}-298\text{K})=75\underset{\_}{0}.3\text{kJ}\text{}\text{}\text{}\text{}\text{}\text{(decompression)}$$

$${q}_{4}=n\overline{{C}_{V}}\mathrm{\Delta}T=(12\underset{\_}{1}.1\text{mol})\frac{3}{2}(8.3145{\text{J mol}}^{-1}{\text{K}}^{-1})(298\text{K}-596\text{K})=-45\underset{\_}{0}.0\text{kJ}\text{}\text{}\text{}\text{}\text{}\text{(depressurization)}$$

$${q}_{1}+{q}_{2}+{q}_{3}+{q}_{4}=-37\underset{\_}{5}.1\text{kJ}+22\underset{\_}{5}.0\text{kJ}+75\underset{\_}{0}.3\text{kJ}-45\underset{\_}{0}.0\text{kJ}=\mathbf{150}\mathbf{\text{kJ}}$$

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:

$$e=-\frac{w}{{q}_{ABS}}=-\frac{-150\text{kJ}}{975\text{kJ}}=0.154(=15.4\mathrm{\%})$$

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.