4.3.1: Illustrations

Illustration 1: Carnot Engine

\(\color{red}{\text{There is a time delay-since the system must be in equilibrium-before the initial change of state occurs.}}\)

\(\color{red}{\text{You must go in order.}}\)

In this animation \(N = nR\) (i.e., \(k_{B} = 1\)). This, then, gives the ideal gas law as \(PV = NT\). Restart.

There are four steps to the Carnot cycle: a combination of isothermal and adiabatic expansions and contractions. Which of the steps are isothermal and which are adiabatic? Make sure to step through the animation in order. During step one the gas does a positive amount of work. Step two is adiabatic, with the gas doing a positive amount of work. During step three there is a negative amount of work done by the gas. Finally in step four, which is adiabatic, the work done by the gas is negative. Notice that the total work done (the remaining area) is positive because positive work is done at high temperatures and negative work is done at lower temperatures. During step one heat is absorbed (\(Q > 0\)) and during step three heat is released (\(Q < 0\)). More heat is absorbed than is released for the entire cycle. This is the basis of how engines work: Heat (from the hot reservoir) is transferred into mechanical work (piston moving).

Illustration authored by Anne J. Cox.

Illustration 2: Entropy and Reversible/Irreversible Processes

In this animation \(N = nR\) (i.e., \(k_{B} = 1\)). This, then, gives the ideal gas law as \(PV = NT\).

In Animation 1 you see an ensemble of particles that looks "natural" once the particles spread throughout the box. You would not expect to see the reverse of this process. Why? Consider Animation 2. This one looks the same running forward or backward. The first animation is an example of an irreversible process, while the second animation is an example of a reversible process. What differentiates the two processes? The concept of entropy. Restart.

Run both animations again. Look at the total energy (kinetic energy) in both cases. Does the energy change? No. The conservation of energy (stated in thermodynamics as the first law of thermodynamics) does not help us determine which of these animations is more likely (since energy is conserved in both cases).

To determine which animation is a more realistic picture of gas particles in a box, we must apply the second law of thermodynamics and the associated concept of entropy. Entropy is a measure of the disorder of a system. Which animation has greater entropy (disorder)? Why? Clearly Animation 2 is a much more ordered system. Animation 1 starts out ordered, yet ends up disordered.

As you watch the first animation, also notice the many different velocity distributions that are possible that still result in the same overall energy (temperature) and pressure. Statistically speaking, it is much, much, much, much more likely for the speeds of an ensemble of particles to be closer to a Maxwell-Boltzmann distribution than to all have the same speeds.

Entropy and the second law of thermodynamics describe what is more likely to happen. It is more likely for particles to go into states of greater disorder because there are more possible "disorderly states" than ordered ones (and the number of possible states is related to the entropy). For example, there are many more ways for a group of particles to follow a Maxwell-Boltzmann distribution than the distribution having identical speeds for each particle. The second law says that entropy either increases or stays the same; irreversible processes cause an increase in entropy. This is our way of knowing whether or not a movie is running forward or backward-as we move forward in time, entropy increases. If entropy decreases somewhere (when electrons are organized to light up a computer screen in a certain manner so you can read this, for example), energy is required, and the energy needed results in an increase in entropy elsewhere so that globally, entropy increases.

Illustration authored by Anne J. Cox.

Illustration 3: Entropy and Heat Exchange

Animation 1 shows two objects of the same size, same mass, and same specific heat (\(mc = 2\) for both), initially at different temperatures but in thermal contact with each other (temperature is given in kelvin and heat exchange is given in joules). The color-coded bars indicate the heat exchanged between the red and blue objects. Restart.

When two objects are in thermal contact with each other, we expect them to eventually reach the same temperature. However, there is nothing in the first law of thermodynamics that requires this. The only requirement from the first law is that energy be conserved, that the heat from one object goes into the other.

Try Animation 2. Is energy conserved? Does the heat from one go into the other? How does the heasxchange in this case compare with Animation 1? What you see in Animation 2, of course, does not happen, even though energy is conserved. The second law of thermodynamics, that entropy (in an isolated system) either increases or stays the same, governs this. The change in entropy, \(\Delta S\), is given by \(\Delta S =\Delta Q/T\) (for reversible processes at a constant temperature), and since \(Q = mc\Delta T\), with a bit of calculus, you get

\[\Delta S=mc\ln (T_{f}/T_{i})\nonumber\]

where \(c\) is the specific heat capacity of a material and \(m\) is the mass of the material. What is the change in entropy for each object in the first animation? What is the total change in entropy? What about for the second animation? Notice that the net change in entropy for Animation 1 is positive, and in Animation 2 the net change in entropy is less than zero. According to the second law, processes don't naturally decrease entropy (it requires the input of energy), so what you see in Animation 2 does not occur in an isolated system because that would violate the second law of thermodynamics.

Illustration authored by Anne J. Cox.

Illustration 4: Engines and Entropy

\(\color{red}{\text{There is a time delay-since the system must be in equilibrium-before the initial change of state occurs.}}\)

\(\color{red}{\text{You must go in order.}}\)

In this animation \(N = nR\) (i.e., \(k_{B} = 1\)). This, then, gives the ideal gas law as \(PV = NT\). Restart.

This engine cycle, the Carnot cycle,  is considered to be a reversible process because it can run forward or backward. For a reversible process, we define a change in entropy as \(dS = dQ/T\) or \(\Delta S =\Delta Q/T\). (Note that if \(T\) changes, some calculus is involved so that \(\Delta S = \int dQ/T\).)

For the Carnot cycle, we can calculate the change in entropy over a cycle by finding the change in entropy for each step. What is the heat input or output divided by the temperature in each step? Note that for the two adiabatic steps, although the temperature changes, the heat input is zero and the process is reversible, so you do not need to use calculus, and \(\Delta Q = 0\). Adding the two nonzero terms, you should find that the entropy change is zero for this cycle. The second law of thermodynamics says that to keep the entropy change at zero is the best you can do (the entropy of a cyclic process either increases or remains zero).

Entropy is important for calculations regarding engines because it tells you the best efficiency you can hope for. The efficiency of any engine is the \(\text{(work out)/(heat in) = }|W|/|Q_{H}|\) where \(Q_{H}\) is the heat input from the high temperature reservoir (from whatever heats the gas: burning gasoline, propane, boiling water, etc.). To calculate the efficiency of this engine, take the work done by the gas (\(698\) total for all steps) and divide by the heat absorbed by the engine (\(2079\) during step 1) and we get \(0.33\).

For an ideal engine (no frictional losses, reversible processes), \(|W| = |Q_{H}| -|Q_{L}|\), and the efficiency is \(|W|/|Q_{H}| = 1 - |Q_{H}|/|Q_{L}|\), where \(Q_{L}\) is the heat exhausted to the low temperature reservoir. Since the entropy change of this cycle is zero, this means that \(Q_{H}/T_{H} + Q_{L}/T_{L} = 0\). Thus, for an engine operating between two temperature reservoirs, the maximum efficiency is \(1 - |T_{H}|/|T_{L}|\).

Entropy, \(S\), is a state variable (does not depend on the process) just like pressure, volume, and temperature unlike work and heat, which do depend on the process. We can thus describe a thermodynamic process by including entropy on a graph. We often describe a process on a \(TS\) diagram, because the area under a curve in a \(TS\) diagram is the heat. Click here to change the graph from a PV diagram to a TS diagram. (Note that the initial entropy is arbitrary—we are simply interested in the change in entropy.)

Illustration authored by Anne J. Cox.

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