I think we are spending too much time with PRS questions that are too easy. The reading is difficult, so I can tell I'm not getting enough from lecture. Perhaps either more difficult PRS questions or less time with PRS questions and more time lecturing on the material in greater detail. (1 student). There is no doubt that I am still trying to get the balance correct, but I would ask you to consider the following: Most of the people are still getting the wrong answer to most of the questions. So I am not sure that harder questions are the right answer (the homework helps to address this). It is also true that it is difficult to think up the questions, and the first time through, some will be good and others not. In my opinion, I thought this lecture had a little better balance with the blackborad vs. the active learning. Although, I would still like to move in the direction you suggest (a little more time covering material). Limiting the amount of time spent covering the mud helped. The short problems worked well, but the class seemed to get more engaged with the long problem at the end. It was nice to see, usually at the end of a lecture everyone is asleep.
Responses to 'Muddiest Part of the Lecture Cards'
1) For adiabatic why does Q=0? (1 student) That is the definition of the word "adiabatic"-- no heat transfer. There is frequently some misunderstanding surrounding the words "temperature" and "heat", and consequently confusion about the difference between isothermal and adiabatic. Temperature is a property of the system, it units are Kelvin. Heat is energy transfer by virtue of a temperature difference, its units are Joules. Temperature and heat are not the same thing. Note also that a body cannot contain "heat", it contains energy. Heat is what we call the energy when it crosses the system boundary. It is possible to have an isothermal process with heat addition (you just need to take energy out via work at the same rate), and it is possible to have an adiabatic process where the temperature changes (compress a thermally-insulated cylinder).
2) I am having trouble applying the concepts discussed in lecture to the actual problems (i.e. question 4 in class). (1 student) Practice makes perfect. There are a lot of good examples worked out in S, B, &VW, and you will get practice on the homework.
3) What are the types of work that a gas can do and have done on it? (1 student) Gas typically does work (or has work done on it) through exerting pressure forces on surfaces, like those we have been considering in the piston-cylinder problems. Gas can also exert shear forces on surfaces (think of trying to move your hand quickly through a bucket of honey. Air is less viscous than honey, but the forces are still there. In fact viscous forces make up an important part of aircraft drag--you will learn more about this in Unified Fluids.) One can also take a chunk of gas and raise or lower it in a potential field (push-pull work), or give the chunk of gas translational kinetic energy (also resulting from push- pull work). Of course due to gas's low density, changes in kinetic and potential energy are often small compared to changes in internal energy. This is why when working with gases we will often use DU=Q-W rather than DE=Q-W.
4) When are things a function of the state of the system and when are they not? (1 student) Whether or not something is a function of the state of a system or path dependent depends on its behavior in the world around us. The equations we are working with were developed over many years to model the world around us, and they do a pretty good job. So some parameters (internal energy, temperature, pressure, density, enthalpy) are a function of the thermodynamic state of the system and others are not (heat and work).
5) Does the gas moved by the fan gain kinetic energy? (1 student) and How do motors add work to a system, or how does power supplied to a motor turn into heat and pressure? (1 student). The energy from the motor eventually all ends up as internal energy. Through pressure and shear forces (see response #3 directly above) the spinning fan causes the gas in the box to move. After the fan is shut off, eventually viscous forces will cause the gas in the box to slow down and then stop moving. The kinetic energy is dissipated as heat in much the same way as it is when a block slides along a table, eventually coming to a stop. To be rigoursly correct, in the problem statement I should have said that after 50 seconds the motor is shut off and the system is allowed to come to thermodynamic equilibrium (then certainly all the kinetic energy of the gas would end up as internal energy).
6) In the battery problem why is the battery work negative? (1 student) The battery is having work done on it (energy is flowing into the system) when it is being charged. Our sign convention is that work done by the system is positive and work done on the system is negative.
7) How can two path dependent functions guarantee a path independent result? (1 student) They can't! That is one of the amazing things to me about the First Law of Thermodynamics (that the difference of two path dependent functions ends up in a property which does not depend on anything but the state of the system).
8) I don't understand the First Law equation for quasi-static processes. Why does putting pdv in make it applicable only to quasi-static? (7 students) One of the things I will do in recitation tomorrow is give you a bit of an equation roadmap. In short, we define work from the perspective of what the system does to the surroundings. Thus we distinguish between psys and pext, and we define work to be the integral of pext(dv). When the process is quasi-static psys is approximately equal to pext so in textbooks and in my notes the subscript is dropped and "p" is used, hence dw=pdv for a quasi-static process where "p" is the same as "psys". So when the First Law, de=dq-dw, is written as de=dq-pdv it means that w= the integral of pdv, or in other words, the system pressure is equal to the external pressure---quasi-static.
9) I was unclear on your explanation of the PRS question on the cyclic process. Why is answer 3 not possible in question 4? If two (independent) properties give you the thermodynamic state, doesn't returning to the same value for those two properties mean returning to the same state? (6 students) The point of the problem is to emphasize two useful corollaries of the First Law. Yes, specifying any two independent properties of the system (they need to be independent, e.g. density and specific volume will not work because they are simply reciprocals of each other) fully specifies the thermodynamic state of the system. So returning to those two values means that the system has returned to the same thermodynamic state. It does not mean however, that the surroundings have returned to the same thermodynamic state (although in an ideal world this is theoretically possible, in the world we live in it has never been observed--we will talk more about this when we get to the Second Law of Thermodynamics). Consider an automobile engine for example. The pistons go up and down over and over again, continually returning to the same state. But the surroundings do not return to the same state (fuel and oxygen are burned, waste heat is produced, etc.).
10) I don't understand how the difference between pext and psys has a bearing on the work done. Why does it always work with pext? (1 student) See response 8) above and T2 mud response #2. If it is still unclear, please contact me or one of the TA's or talk it through with one of your classmates.
11) Still unclear on the path-dependent, path-independent stuff. Is the change in energy of a system (DU and DE) always path-independent? (2 students) Yes. Energy is a function of the state of the system, not how you get there.
12) How many different ways are there to write the First Law? (1 student) Hundreds, perhaps thousands (no kidding). We will use about 20 different forms in this class (no kidding). This is why it is important to be clear on the nomenclature, and to recognize various terms as indicators of the assumptions that have been made. In recitation tomorrow, I will give you a bit of a First Law equation roadmap to help with this.
13) Don't understand why work is path independent if the process is adiabatic. (4 students) This idea is expressed in one of the corollaries of the First Law. Internal energy is a function of the thermodynamic state of the system. It does not depend on the process by which the system arrived at that state. If the process is adiabatic, then the First Law becomes DU=W. So work is equal to something that does not depend on path.
14) The differential forms of the First Law were not well explained. When would I use the differential form? (2 students) We will talk more about these in recitation tomorrow. The differential forms of the First Law are just a useful, convenient way to write them that sometimes make manipulating the equations easier. For example, sometimes it is easier to work with du=dq-pdv rather than Du=q- the integral of (pdv) even though both are identical expressions of the First Law for a quasi-static process for a system where the changes in kinetic and potential energy can be neglected.
15) When Q or W is added or taken away does that change U, KE, or PE, or all three? (1 student) Heat and work can change all three---it depends. For example, I can take a steel ball and throw it (i.e. do work on it) thus increasing its kinetic and possibly potential energy. I can take the same steel ball and squeeze it (i.e. do work on it) and thus increase its internal energy. I could also apply a voltage across the ball, allow a current to flow (i.e. do electrical work on it) and increase its internal energy.
16) You mentioned that DU is one of the properties that define the state of the system. Is DE the other or are there others as well? (1 student) Internal energy, U, is a thermodynamic property and is a function of the thermodynamic state of the system (like pressure, temperature, density, specific volume, enthalpy, entropy). Knowing any two independent properties (e.g. density and specific volume will not work because one is simply the inverse of the other) fully specifies the thermodynamic state of the system. You can think of kinetic energy and potential energy as functions of the dynamic state of the system.
17) Please work through the answer to the last PRS question. (19 students) I will work through this in detail in recitation tomorrow. Here is my answer. Note that the wording of the last part of the question is unclear. It can lead you either to say the work done by the gas is 598J or the net work done by the system is -1202J.
18) No mud (14 students). Uh oh, my percentage is dropping in this category.