3.3: The Carnot engine

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Image from user Keta on Wikimedia Commons. CC BY-SA 3.0. Figure not to scale.

A single book was the entire scientific output of the military scientist Sadi Carnot; it was called Reflections on the Motive Power of Fire, and as the title implies, it describes the behavior of all kinds of heat engines. Carnot was driven to maximize the efficiency of the engine, and by so doing, he was attracted to the steps that both eliminated heat transfer and concentrated heat input into useful work.

The engine that Carnot devised that focused on these steps started in the upper-left corner of the new diagram at the top of this page, at the highest pressure and volume. The step where heat energy would be invested, in Carnot’s vision, would be an isothermal step that both increased the pressure on the gas and decreased the volume, it performed large quantities of work at high temperature. The step that followed would perform more work by further reducing pressure and volume without heat transfer, in a rapid adiabatic process.

Those steps would then need to be undone, but the adiabatic process would lower the temperature considerably, so the heat released by the engine would be less in the third step and the “wasted work” in the final two steps would be much lower.

Here’s an example problem laying out such a cycle, using the Carnot cycle figure:

A monatomic ideal gas begins with a volume of 1.00 m³ and under a pressure of 3.00 bar and at a temperature of 596 K. The gas first does work by decompressing and depressurizing isothermally at 596 K; the pressure at the end of the isothermal step is 2.00 bar). The gas then decompresses and depressurizes adiabatically to a temperature of 298 K. Work is then done to the gas to compress and pressurize it isothermally at 298 K, and then it is restored to its original state by an adiabatic step; the number of moles have not changed, and the temperature is again 596 K. Find the total work done by the gas at each step, and the total heat added during the cycle.

Note that this problem mirrors the problem given you in the PV rectangle, but for isothermal and adiabatic steps instead - the curves in lieu of the straight horizontal and vertical lines on the graph. We can't just use the ideal gas law all the time.

We can, however, use the ideal gas law to find the number of moles of gas present from the initial conditions of the gas:

$n = \frac{P_{1} V_{1}}{R T_{1}} = \frac{(3.00 \times 10^{5} Pa) (1.00 m^{3})}{(8.3145 {J mol}^{- 1} K^{- 1}) (596 K)} = 60. \underline{5} 4 mol$

Note from the problem that V₁ = 1.00 m³ and that the pressures given are P₁ = 3.00 × 10⁵ Pa and P₂ = 2.00 × 10⁵ Pa, respectively. From there, knowing that temperature is held constant, we can find the volume at the end of the cycle:

$V_{2} = \frac{n R T_{2}}{P_{2}} = \frac{(60.54 mol) (8.3145 {J mol}^{- 1} K^{- 1}) (596 K)}{2.00 \times 10^{5} Pa} = 1.5 \underline{0} m^{3}$

The other pressures and volumes we need come from the adiabatic relations. Because we are studying a monatomic ideal gas, γ = ⁵/₃:

$T_{2} V_{2}^{γ - 1} = T_{3} V_{3}^{γ - 1} \to V_{3} = {(\frac{T_{2} V_{2}^{γ - 1}}{T_{3}})}^{\frac{1}{γ - 1}} = {(\frac{(596 K) (1.50 m^{3})^{\frac{2}{3}}}{298 K})}^{\frac{3}{2}} = 4.2 \underline{4} 3 m^{3}$

$P_{2} V_{2}^{γ} = P_{3} V_{3}^{γ} \to P_{3} = P_{2} {(\frac{V_{2}}{V_{3}})}^{γ} = (2.00 \times 10^{5} Pa) {(\frac{1.50 m^{3}}{4.243 m^{3}})}^{\frac{5}{3}} = 3.5 \underline{3} 5 \times 10^{4} Pa$

The volume and pressure at the start of step 4 result from the adiabatic step taking us back to the initial condition:

$T_{1} V_{1}^{γ - 1} = T_{4} V_{4}^{γ - 1} \to V_{4} = {(\frac{T_{1} V_{1}^{γ - 1}}{T_{4}})}^{\frac{1}{γ - 1}} = {(\frac{(596 K) (1.00 m^{3})^{\frac{2}{3}}}{298 K})}^{\frac{3}{2}} = 2.8 \underline{2} 8 m^{3}$

$P_{1} V_{1}^{γ} = P_{4} V_{4}^{γ} \to P_{4} = P_{1} {(\frac{V_{1}}{V_{4}})}^{γ} = (3.00 \times 10^{5} Pa) {(\frac{1.00 m^{3}}{2.828 m^{3}})}^{\frac{5}{3}} = 5.3 \underline{0} 3 \times 10^{4} Pa$

This gives us all the data we need to work out the four steps of the cycle. Here, unlike the PV rectangle, we will compute work for all four steps, both decompressions and both compressions; on the other hand, we will only need to compute heat for two of the four:

$Isothermal dec.: w_{12} = n R T_{1} \ln (\frac{V_{1}}{V_{2}}) = (60.54 mol) (8.3145 {J mol}^{- 1} K^{- 1}) (596 K) \ln (\frac{1.00 m^{3}}{1.50 m^{3}}) = - 121.6 kJ$

$Adiabatic decompression: w_{23} = n \bar{C_{V}} Δ_{23} T = (60.54 mol) \frac{3}{2} (8.3145 {J mol}^{- 1} K^{- 1}) (298 K - 596 K) = - 225.0 kJ$

$Isothermal comp.: w_{34} = n R T_{3} \ln (\frac{V_{3}}{V_{4}}) = (60.54 mol) (8.3145 {J mol}^{- 1} K^{- 1}) (298 K) \ln (\frac{4.243 m^{3}}{2.828 m^{3}}) = 60.86 kJ$

$Adiabatic compression: w_{41} = n \bar{C_{V}} Δ_{41} T = (60.54 mol) \frac{3}{2} (8.3145 {J mol}^{- 1} K^{- 1}) (596 K - 298 K) = 225.0 kJ$

Note that, because they’re so closely equated to the state energy change ΔU, the work done in steps 23 and 41 undo one another. This will always be what we expect for an adiabatic step, because heat transferred is zero for those steps.

The total work done is therefore:

$w_{T O T} = w_{12} + w_{23} + w_{34} + w_{41} = - 12 \underline{1} .6 kJ - 22 \underline{5} .0 kJ + 60. \underline{8} 6 kJ + 22 \underline{5} .0 kJ = - 6 \underline{0} .8 kJ$

(It's just as likely, if you round, that you come up with 60.7 kJ instead of 60.8 kJ. That's OK. The significant figure rules dictate that the result of this addition and subtraction is only significant to the ones place, which will be what we pay attention when we get to the punch line of this calculation.)

The heat transferred in the isothermal steps will look incredibly similar, whereas the adiabatic steps simplify by definition:

$Isothermal decomp.: q_{12} = n R T_{1} \ln (\frac{V_{2}}{V_{1}}) = (60.54 mol) (8.3145 {J mol}^{- 1} K^{- 1}) (596 K) \ln (\frac{1.50 m^{3}}{1.00 m^{3}}) = 121.6 kJ$

$Adiabatic decompression: q_{23} = 0 by definition$

$Isothermal comp.: q_{34} = n R T_{3} \ln (\frac{V_{4}}{V_{3}}) = (60.54 mol) (8.3145 {J mol}^{- 1} K^{- 1}) (298 K) \ln (\frac{2.828 m^{3}}{4.243 m^{3}}) = - 60.86 kJ$

$Adiabatic compression: q_{41} = 0 by definition$

It should be a very straightforward exercise to demonstrate that the total heat input to the engine is the same magnitude as the total work - 60.8 kJ.

The efficiency of this engine is calculated in the same way as before, with the heat transferred from the isothermal decompression being the absorbed quantity:

$e = - \frac{w}{q_{A B S}} = - \frac{- 6 \underline{0} .8 kJ}{12 \underline{1} .6 kJ} = 0.5 \underline{0} 0 (= 50 %)$

The efficiency is 50%. It’s 50% when the low-temperature isothermal step is half the temperature of the high.

Huh.