In the Classroom
W
Thermodynamics for Visual Learners Todd M. Hamilton Department of Chemistry, Adrian College, Adrian, MI 49221;
[email protected] Example One: The Second Law of Thermodynamics The first example deals with the second law of thermodynamics. One way to state the second law is as follows: “It is impossible to build a cyclic machine that converts heat into work with 100% efficiency” (13). I use a diagram of the Carnot cycle (Figure 1) to illustrate the consequences of the second law. The values of work, w, and heat, q, for a representative cycle are listed in Table 1. Some students like to see the results in tabular form to make comparisons. In the first (isothermal, T = TH) step, q = ᎑w and the efficiency is 100%. In the second (adiabatic) step, even more work is produced, and it seems that the efficiency exceeds 100%. The “cost” comes in steps 3 and 4, and this is the point of my example. Because this is a cyclic process, the system must return to the initial starting point. Since the system performed work on the surroundings as it expanded in steps 1 and 2, the surroundings must do work on the system as it is compressed in steps 3 and 4. This is further illustrated by noting that the work is the integral of PdV. The work that the system performs corresponds to both gray areas in Figure 1. The work that the surroundings must perform on the system is the light gray area under curves 3 and 4. This “lost work” area is subtracted from the previous area, resulting in less than 100% efficiency, that is, the second law. The efficiency, e, is calculated using the following equation: e = ᎑w兾qH, where w is the net work done by the system and qH is the heat absorbed by the system from the “hot” reservoir. In this example (TC = 300 K and TH = 500 K), the efficiency is 40%. The net work done by the system on the surround-
1
P
Physical chemistry is a difficult subject for even our best students. Therefore, we must look for teaching strategies that address various learning styles. I am encouraged by the many articles in this Journal that describe the teaching of physical chemistry using folk culture and poetry (1), humor and music (2), paradoxes and puzzles (3), stories (4), writing (5), workshops and computers (6), and the Internet (7). Many physical chemistry students can learn an abstract concept more easily if they can visualize it. Visual examples have been developed for illustrating Raoult’s law (8), enthalpy (9, 10), the variational method (11), and entropy (12). I have developed several examples for visual learners in the area of thermodynamics and would like to share these with other professors.
TH
4 2
3
TC
V Figure 1. The Carnot cycle: Step 1 is an isothermal step occurring at the temperature of the “hot” reservoir, TH. Step 3 occurs at the temperature of the “cold” reservoir, TC. Steps 2 and 4 are adiabatic. The gray areas represent the work done by the system on the surroundings. The dark gray area represents the net work done by the system. The efficiency is the dark gray area divided by the area under curve 1.
ings is represented by the dark gray area in Figure 1. The efficiency is the dark gray area divided by the area under curve 1, which is equal to qH. Example Two: Isothermal Process (Ideal Gas) The second example centers around the fact that when an ideal gas undergoes an isothermal process, the internal energy, U, does not change; that is, (∂U/∂V )T = 0. Typically, I illustrate this point by starting with the expression dU = TdS + dw and the Gibbs equations (where A is the Helmholtz energy) dU = TdS − PdV and dA = SdT − PdV (along with the appropriate Maxwell relation) to prove that (∂U兾∂V )T = 0. Thus, for an ideal gas, temperature is a measure of the kinetic energy and a constant T implies a constant U. Additionally, I have found a visual example to illustrate the point (Figure 2). Imagine a system going from state A to state B. The change can occur in two steps, an irreversible isobaric step (step 1 in Figure 2) plus an irreversible isochoric step (step
Table 1. Values of Work and Heat for a Carnot Cycle (P/atm,V/cm3)initial → (P/atm,V/cm3)final
Work/J
Heat/J
1
Isothermal, T = 500 K
(8.21, 2000) → (6.84, 6000)
᎑4567
4567
2
Adiabatic
(6.84, 6000) → (1.91, 12900)
᎑2494
0
3
Isothermal, T = 300 K
(1.91, 12900) → (5.73, 4300)
2740
᎑2740
4
Adiabatic
(5.73, 4300) → (8.21, 2000)
2494
0
Step
Description
JChemEd.chem.wisc.edu • Vol. 80 No. 12 December 2003 • Journal of Chemical Education
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In the Classroom
1
A
1
/ atm
A
/ atm
2
2 3
Pext
Pext
2 3 1
B
10
B
20
V / (103 cm3)
V / (103 cm3) Figure 2. Diagram of an ideal gas illustrating an irreversible isobaric step (step 1), an irreversible isochoric step (step 2), and a reversible isothermal step (step 3).
Figure 3. Diagram of an ideal gas illustrating an irreversible isobaric step (step 1), an irreversible isochoric step (step 2), and an arbitrary P–V step (step 3).
2). In step 1, the thermal surroundings at Tf supply the heat needed for the expansion; in step 2 heat is lost to the thermal surroundings. The change can alternatively occur through a single, reversible, isothermal step (step 3). Note that the numerical values on the axes in Figure 2 are meaningful; in step 3, the pressure is halved and the volume is doubled, in accordance with Boyle’s Law. The change in internal energy ∆U for step 1 would be: nCV,m∆T (assuming that CV,m does not change significantly with temperature). Using a form of the ideal gas law, ∆T = Pext∆V兾nR, and the fact that CV,m = 1.5R for an ideal monatomic gas, we obtain:
gas P–V process (again, assuming that the heat capacities do not change significantly with temperature). I explain why using the following example. The P–V diagram in Figure 3 is similar to that in Figure 2, except step 3 is not an isothermal process; step 3 represents any arbitrary P–V change. The main point is that any arbitrary P–V change can be represented by a two-step process, an irreversible isobaric step (step 1 in Figure 3) and an irreversible isochoric step (step 2). Obviously, for the isochoric step, the above ∆U equation is valid. What about the isobaric step? For the isobaric step, q = ∆H = nCp,m∆T and the work, w = ᎑Pext∆V. Therefore, the internal energy change can be written:
∆U1 = 1.5Pext ∆V
(
)(
)
= 1.5 2 atm 10,000 cm 3 = 30,000 atm cm3 For step 2, ∆U2 = q = n C V,m ∆T = 1.5 V ∆Pext
(
)(
)
= 1.5 20,000 cm 3 −1 atm = −30,000 atm cm3
(
)
= 30,000 atm cm 3 + −30,000 atm cm 3 = 0
Since U is a state function, step 3 should give the same result. Therefore, ∆U = 0 for an ideal gas undergoing an isothermal process. Example Three: Ideal Gas Change of State In example two, I took for granted that ∆U = nCV,m∆T even for an isobaric process (step 1). My students are often puzzled by the fact that we can use this equation containing the constant-volume heat capacity in a constant-pressure process. In fact, we can use the above ∆U equation for any ideal 1426
Since Cp,m = 2.5R for an ideal, monatomic gas, and according to the ideal gas equation, P∆V = nR∆T, we have the following: ∆U = n 2.5R ∆T − n R ∆T = n1.5 R ∆T = n C V,m ∆T Hence, the ∆U equation holds also for isobaric processes. The essential change in both steps, and in any arbitrary P–V change, is the change in temperature, ∆T.
Adding steps 1 and 2, we obtain: ∆U total = ∆U1 + ∆U2
∆U = q + w = n C p,m ∆T − Pext ∆V
Example Four: Internal Energy, Temperature, and Heat Students may be quite successful in performing thermodynamics calculations and still have misconceptions about three fundamental quantities: internal energy, temperature, and heat. I use the following visual example to differentiate between them. Two ideal monatomic gas samples are represented in Figure 4. The gas in each container is the same and the amount is the same. The atoms of gas are represented by black circles and the velocities are represented by the arrows. The length of the arrow corresponds to the magnitude of the velocity, that is, the longer the arrow, the larger the magnitude of the velocity. The atoms in the gas sample A are mov-
Journal of Chemical Education • Vol. 80 No. 12 December 2003 • JChemEd.chem.wisc.edu
In the Classroom
A
B
75 °C
25 °C
strongly make the point that a gas sample does not contain heat; heat is energy in transfer. A sample contains internal energy (kinetic and potential) and temperature reflects the kinetic component of the internal energy. Developing Additional Examples
higher U higher T “hot”
|q|
lower U lower T “cold”
Figure 4. Representation of ideal gas samples illustrating the relationship of gas velocity (arrows), internal energy (U ), and heat (q) between (A) higher temperature and (B) lower temperature situations.
ing faster than the atoms in the sample B. If one covers the Boltzmann distribution before discussing thermodynamics, the point can be made that the arrows are not all the same length and the property of interest is the mean length. Since the potential energy is neglected in the ideal gas model, I will focus on the total kinetic energy of the gas sample. This total kinetic energy is directly related to the velocities of the atoms (again, the length of the arrows). The atoms in the gas on side A have larger velocities, and therefore a higher internal energy, than the atoms on side B. The temperature is a measure of the average kinetic energy of the atoms in the sample, so a connection can be made between the temperature reading (75 ⬚C) and the individual velocities (length of the arrows). If the students have been introduced to the kinetic theory of gases, the root-mean-square speed formula can be used to quantify this relationship. The atoms in sample A are moving faster resulting in a higher temperature reading than the atoms in sample B. Note that this example is a special case involving a monatomic gas in which I have ignored potential interactions and focused only on translational energy. The more fundamental relationship between temperature and internal energy is given by the equation, T = (∂U/∂S)V, where temperature reflects how a given change in energy is associated with an entropy change (i.e., how the added energy is distributed over molecular energy levels). Sample A would feel “hot”, while sample B would feel “cold”. This point provides a connection between the abstract concept of internal energy, the everyday notion of temperature, and something the student can actually feel (hot or cold). The illustration is completed by noting that if these two samples are placed in thermal contact, heat flows from the hot to the cold sample until both samples are at the same temperature (comparable velocities or arrow lengths). I
I find these visual examples to be useful in cementing difficult concepts. I still go through the rigorous derivations and traditional arguments; the visual examples help the students make connections. I have found the recent MathCad examples (14, 15) and the highly visual, interactive software package P-Chem (16) to be beneficial in developing and exploring additional visual examples in thermodynamics and other physical chemistry topics. Caveat I have presented simple visual examples to help illustrate thermodynamic concepts. I strongly recommend that professors present a full treatment of ideal gases, the second law, and the Carnot engine so that students understand the assumptions and special conditions under which these examples are valid. Acknowledgments I would like to thank the reviewers for helpful and valuable comments. W
Supplemental Material
The figures are available as electronic files in this issue of JCE Online. Literature Cited 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16.
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