4.3.2: Explorations

Exploration 1: Engine Efficiency

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

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

In this animation \(N = nR\) (i.e., \(k_{B} = 1\)). This, then, gives the ideal gas law as \(PV = NT\). Assume an ideal monatomic gas. The efficiency of an engine is defined as \(\varepsilon =\text{ (work out)/(heat in) }= |W|/|Q_{H}|\). Restart.

1. Pick a temperature for the hot reservoir (between \(200\text{ K}\) and \(150\text{ K}\)) and a lower temperature (between \(150\text{ K}\) and \(100\text{ K}\)) for the cold reservoir. (Note that new reservoir temperatures only register at the beginning of the engine cycle and that you should run the engine steps in order for the animation to make sense.) Find the work done for each step and the heat absorbed or released (remember that \(\Delta U = (3/2)nR\Delta T = (3/2)N\Delta T\)).
2. Calculate the efficiency of the engine for these temperatures.
3. Pick a different pair of temperatures for the reservoirs. Is this engine more or less efficient?  (Calculate the efficiency for this new engine.)
4. Why is the engine in (c) more or less efficient?
5. What would make the engine even more efficient? Try it and explain.
6. Calculate the difference in temperatures of the reservoir divided by the temperature of the hot reservoir: \((T_{H} - T_{L})/T_{H} = 1 - T_{L}/T_{H}\). Compare this value with the efficiency number for each case above. For a Carnot engine, either calculation gives the efficiency because of the following:
7. For step 1, \(W = Q_{H} = nRT_{H}\ln(V_{1}/V_{0}) = NT_{H}\ln(V_{1}/V_{0})\), where \(V_{0}\) is the volume at the beginning of step 1 and \(V_{1}\) is the volume at the end of step 1. Explain why (do some algebra!) and verify with the animation.
8. Similarly, for step 3, \(W = Q_{L} = nRT_{L}\ln(V_{3}/V_{2}) = NT_{L}\ln(V_{3}/V_{2})\), so \(|Q_{L}| = NT_{L}\ln(V_{2}/V_{3})\), where \(V_{2}\) is the volume at the beginning of step 3 and \(V_{3}\) is the volume at the end of step 3. Explain why (do some algebra!) and verify with the animation.
9. Steps 2 & 4 are adiabatic, so \(Q = 0\). From steps 2 and 4, \(P_{1}V_{1}^{\gamma}= P_{2}V_{2}^{\gamma}\) and \(P_{3}V_{3}^{\gamma}= P_{0}V_{0}^{\gamma}\), where \(\gamma\) is the adiabatic coefficient (ratio of the specific heat at constant pressure to specific heat at constant volume). Using these relations and the ideal gas law, show (more algebra) that \((V_{1}/V_{0}) = (V_{2}/V_{3})\).
10. Therefore, show for a Carnot engine \(|W|/|Q_{H}| = 1 - |Q_{L}|/|Q_{H}| = 1 - T_{L}/T_{H}\).

The Carnot engine efficiency of \(1- T_{L}/T_{H}\) is the ideal efficiency for any engine operating between two reservoirs of \(T_{H}\) and \(T_{L}\), because the net change in entropy is zero for a Carnot cycle (see Illustration 21.4). Often, you will compare the efficiency of other engines, \(|W|/|Q_{H}|\), to the ideal Carnot engine efficiency. Note that you cannot have a \(100\%\) efficient engine because that would require \(T_{L} = 0\) (which is forbidden by the third law of thermodynamics). Another way to think about this is that to get an efficiency of \(100\%\) you would need to put the heat released into the low temperature reservoir back into an engine. But that engine would need to run between \(T_{L}\) and a lower temperature, and again you can't get to \(T_{L} = 0\).

Exploration authored by Anne J. Cox.

Exploration 2: Internal Combustion Engine

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

The Otto engine cycle is close to the cycle of an internal combustion engine (and closer to a real engine than the Carnot engine). This cycle consists of adiabatic and isochoric processes plus a cycle of exhausting smoke and taking in new gas. Identify which parts of the engine cycle correspond to which process. No net work is done in the complete process of exhausting smoke and taking in gas. Explain why. Notice that during this part of the cycle, the number of particles changes because the red valves at the top open and close to let gas in and out. Thus, in the release of high temperature particles and intake of low temperature particles, heat is exchanged (released to the environment).

1. For the adiabatic expansion, what are the initial pressure and volume? What are the final pressure and volume? (Remember you can click on the graph to read points from it.) From these values, find the adiabatic constant, \(\gamma\) (since \(PV^{\gamma} =\text{ constant}\) for an adiabatic expansion).
2. Is the gas monatomic (\(\gamma = 1.67\)), diatomic (\(\gamma = 1.4\)), or polyatomic (\(\gamma = 1.33\))?
3. What is the net work done during the cycle (the work out)?
4. Neglecting the gas exhaust and intake parts of the cycle, in which part of the cycle is heat absorbed? In which part of the cycle is heat released?
5. Calculate the heat absorbed. Remember that \(Q = \Delta U + W\) and that \(\Delta U = (f/2)N\Delta T\), where \(f = 3\) for monatomic gases, \(5\) for diatomic gases, and \(6\) for polyatomic gases.
6. What is the efficiency of this engine? The efficiency of an engine is \(\varepsilon =\text{ (work out)/(heat in) }= |W|/|Q_{H}|\).
7. Check that your answer is equal to \(1 - (V_{\text{min}}/V_{\text{max}})^{1-\gamma}\) and is therefore dependent on the ratio of the maximum and minimum volume (called the compression ratio).

Exploration authored by Anne J. Cox.
Script authored by Anne J. Cox and Wolfgang Christian.
Applet authored by CoLoS and modified by Wolfgang Christian.

Exploration 3: Entropy, Probability, and Microstates

In the animation, two containers are separated by a "membrane." Initially, no particles can cross the membrane. Note that the red and the blue particles are identical, they are colored so you can keep track of them. Once the particles are fairly evenly distributed in the left chamber, you are ready to let particles through. Try letting particles through the membrane. This animation allows about every other particle that hits the membrane to get through (equally in either direction). When there are about the same number of particles on both the left and right sides, pause the animation and count the number of red particles on each side and the number of blue particles on each side. Let the animation continue and stop it again a few seconds later when there are about the same number of particles on each side. Again, find the number of red and blue particles on each side. Restart.

1. Given that there are \(30\) blue particles and \(10\) red particles total, if you made many such measurements, what would you expect the average number of red and blue particles to be on each side (when there are a total of \(20\) particles on each side)?
2. Now, restart the animation. Once the particles are fairly evenly distributed in the left chamber, try letting particles through the membrane a different way. Again, this animation allows about every other particle through the membrane.
3. When there are about the same number of particles on each side, count the red and blue particles on each side. What is different about the way this membrane is set up?
4. Is it possible for the first membrane to have this outcome?
5. Is this outcome likely?

The reason that the second membrane does not appear "natural" is the second law of thermodynamics. One version of the second law is that the entropy of an isolated system always stays the same or increases (where entropy is defined as a measure of the disorder of the system). In other words, "natural" systems move in the direction of greater disorder. In the animations, the first membrane seems "natural" because it allows for the most disorder-a random distribution of reds and blues on both sides. This is compared to the second membrane that only allows blue particles through, and thus the right side will always have only blue particles in it.

Another way to interpret the second law is in terms of probability. It is possible with the first animation to get \(0\) red particles in the right chamber, but it is very unlikely (just like it is possible you will win the lottery, but it is very unlikely). It is also possible for the second animation to behave as it does, but again, it is very unlikely. Consider the animation above with only six particles: four blue and two red. To keep track of things, we've colored the blue ones different shades of blue and the red ones different shades of red. Run the animation and notice how often there are three blue ones on the right side when there are three particles on each side. What follows will allow you to calculate the probability of this happening and show that when there are three particles on each side there is a \(20\%\) chance that there will be three blue ones in the right chamber.

6. Considering the different arrangements of three particles on each side, note that there are four different ways to get three blues on the right and two reds and one blue on the left (list these and click here to show them)s. Similarly, there are the same four ways to get three blue particles in the left chamber.
7. There are six ways to get the light red on the left and the dark red on the right with two blues each (click here to show these). Again, there are the same six ways to have the dark red on the left and the light red on the right.
8. This gives a total of how many different arrangements (of three particles on each side)? Since all these states are equally likely, you have only a \(20\%\) chance of having three blues in the right chamber.

As we add more particles, it becomes less likely to get all of one color on one side. With \(40\) particles, \(30\) blue and \(10\) red, there is only around a \(0.02\%\) chance that when there are \(20\) particles on each side, there will be \(20\) blue ones on the left and \(10\) red and \(10\) blue on the right. This is not impossible, but not very likely (better odds than your local lottery, where your odds might be around one in a couple of million). A more ordered state (\(20\) blues on the right) is less likely, statistically, than a less ordered state (reds on both sides of the membrane, which is a more even mixing). Entropy is related to the number of available states that correspond to a given arrangement (mathematically, \(S = k_{B}\ln W\), where \(S\) is entropy, \(W\) is the number of equivalent arrangements or microstates, and \(k_{B}\) is the Boltzmann constant).

Going back to our six particle example, there were more states that corresponded to one red in each chamber than two reds in one chamber, so one red in each chamber is a more likely state for the system to be in. Most of the time, however, we are dealing with more than six particles (usually around Avogadro's number), so the very ordered state is even less likely to happen. Connecting entropy to probability, then, gives a version of the second law that does not forbid the system from being in a highly ordered state; it simply says that it is highly unlikely.

Exploration authored by Anne J. Cox.

Exploration 4: Entropy of Expanding Ideal Gas

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

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

In thermodynamic processes the entropy depends not on the path taken but on the end points. It is a "state function" (in contrast to heat and work, which depend on the process). Since \(Q = \Delta U + W\) and \(\Delta U = (3/2)nR\Delta T\) (for a monatomic gas) ,

\[\Delta S=\int dQ/T = \int (3/2)nRdT/T+\int PdV/T = nr[(3/2)\ln (T_{f}/T_{i})+\ln (V_{f}/V_{i})]\nonumber\]

Thus, \(\Delta S = (3/2)N \ln(T_{f}/T_{i}) + N \ln(V_{f}/V_{i})\) for an ideal monatomic gas (note that ln represents the natural log, base e).

In the animations, note that the area under the \(PV\) diagram is equal to the work.

1. What is the work done in each case?
2. What is the heat absorbed or released in each case?
3. What is the area under the associated TS diagram? (Note that the choice of the initial entropy is arbitrary.)
4. How does the change in entropy compare for the three processes?
5. Compare your measurements from the graphs to the calculated values found from using the equation above for an ideal monatomic gas.

Another way to measure the change in entropy is to use \(Q = mc\Delta T\) or, for a gas, \(Q = CN\Delta T\). In this case,

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

6. Show that, for the isobaric expansion where \(C = C_{P} = (5/2)\), you get this change in entropy.

Exploration authored by Anne J. Cox.

Physlets were developed at Davidson College and converted from Java to JavaScript using the SwingJS system developed at St. Olaf College.