Lecture T2: Changing the State of the System with Heat and Work

After a short review of the important concepts from lecture T1 (see T1 mud for more), I discussed changing the state of a thermodynamic system and how this is represented on various thermodynamic diagrams. On the blackboard I did an example of a piston-cylinder arrangement being heated at constant pressure. Then the class did an example for a small chunk of gas passing through a gas turbine engine. The results on this problem were very good. I then commented on the distinction between heat and work and noted that for a particular situation, determining whether it is heat or work sometimes depends on where you draw the system boundary. Remember that heat and work are transfers of energy to/from the system (heat by virtue of a temperature difference only, work is everything else), and don't confuse heat (energy transfer, joules) with temperature (thermodynamic property, Kelvin). I then briefly reviewed the formulation for work presented in the notes. Three of the more difficult things to understand about the material are 1) the signs to use for work (please review this), and 2) the distinction between thermodynamic equilibrium and quasi-equilibrium processes (or quasi-static processes) -- in particular when it is appropriate to assume a system is "close" to equilibrium, and 3) the fact that since we define work as what our system does to the surroundings we can sometimes calculate it even if the system isn't in quasi-equilibrium (e.g. by observing a weight being lifted over a know distance). There are some useful notes on the item (2) below and in the old mud responses. I finished the lecture by noting that work was a path dependent quantity. That is, it is a function of how you get from state 1 to state 2 (the area under a pext-V curve) not on the difference between state 1 and 2. To reinforce this we concluded the lecture with a turn-to-your-partner exercise where I asked you to move along two different thermodynamic paths using ice, a bunsen burner and weights. Note in the example that the amount of weight lifted (the work done) depended on which thermodynamic path you took even though the initial and final states for the two paths were identical.

Responses to 'Muddiest Part of the Lecture Cards'

(69 respondents out of 78 attendees)

1) Could you explain more about quasi-equilibrium and quasi-static? (14 students) Whether you can consider a process to be a quasi-static (same as quasi-equilibrium) depends on what you consider your system to be and in particular how big it is, and how quickly it comes to thermodynamic equilibrium. Only under situations where the time rate of change of the process is slow relative to the time it takes for the system to reach equilibrium can we assume that the process is quasi-static. Consider a very small chunk of gas (say 1E-6 m on a side) as it moves through an engine. We would like to say that one unique set of properties defines its thermodynamic state and that all the forces balance one another (i.e. everything is equal--thus equilibrium). Changes in the state of the system (the chunk of gas) must take place slowly relative to the time it takes for the chunk of gas to reach equilibrium. Now consider a change to the chunk of gas (say the pressure changes on one side because it bumps up against something). How long does it take for the information to travel from one side of the chunk of gas to the other and by what mechanism does it travel? Pressure information travels at the speed of sound through molecular collisions propagating as a pressure wave. For such a small chunk of gas, it is not hard to imagine that this time to reach equilibrium is very short indeed (go ahead and calculate it--the speed of sound is about 300m/s for air so it is about 3E-9s). So as long as the changes in state of the small chunk of gas are slow relative to this time we can consider the gas to be in thermodynamic equilibrium. One can make similar arguments for adding energy by heat transfer to a small chunk of gas, but the propagation speed is much slower as it occurs by molecular diffusion (you can look up the thermal diffusivity for a gas and figure out what the time is). Now how does this differ from the piston-cylinder example? In that case, we considered the entire volume of the cylinder as the system. If the cylinder was say 1m in length, it would take about 0.3s for a pressure wave to propagate the length of the piston (and even longer for temperature differences to be felt since they propagate by molecular diffusion). So our criterion for how slow the changes to the system have to be to consider it in equilibrium depends on the scale of the system or more appropriately, how quickly it achieves equilibrium. A nice discussion of this appears in the middle paragraph of p. 19 S, B, & VW. The residence time of a chunk of gas in a gas turbine engine is about 0.05s, and thus most of the processes that impact it occur much slower than the time it takes a small chunk of gas to reach equilibrium.

2) When can you use psys = pext and why? -- and related questions (9 students). In the case of the piston-cylinder arrangement the former is the pressure of the system and the latter is the pressure applied by the surroundings (weights, the atmosphere, etc.). When moving slowly relative to the time it takes the system to equilibrate, the two are approximately equal. But the important point to realize is that they are not always equal. For example, if you were to rapidly remove the weights, it takes some time for the gas in the cylinder to respond. Second, it is important to remember that pext is the relevant parameter for evaluating the work done by the system. And whether or not you assume them to be equal depends on what level of fidelity you require from your engineering model. In some situations, if they are within 10% it would be sufficient for you to get a rough estimate. In other situations, you might prefer the agreement to be within 0.01%. I am unclear on the derivation of the integral for work (2 students) Please read over the notes again. If it is still unclear, grab me after lecture and I would be happy to go over it with you.

3) Why do we have to remove weights and cool simultaneously (in the turn-to-your-partner exercise)? (2 students) First read over the explanation of the correct answer (link above). Consider the ideal gas law (pv=RT). We want v (or V) to be constant, while reducing the pressure. So T must be reduced in proportion to the reduction in p. Now you don't know enough thermodynamics yet to prove that you must remove energy so that T will be reduced in direct proportion with p, so you have to rely on intuition. If you were only to remove the weights (which is required to lower the pressure) without cooling, then you would expect the volume to expand. So some amount of cooling is necessary to hold the volume constant.

4) I didn't understand removing 1/2 the weights impulsively from the cylinder and still being able to determine the work done. (1 student) The system is not in quasi-equilibrium so you can not use the system properties to determine the work done by the system. However in this case, the work done by the system on the surroundings can be determined from observing what happens to the surroundings (a known weight being lifted over a know distance). What happens if you lift both weights? (1 student) If you lift both weights from the cylinder (and the piston is massless and there is no external pressure -- so it is expanding into a vacuum) then no work is done since no force is applied even though the massless piston moves. This is obviously an idealization because there are no frictionless, massless pistons.

5) Unclear about work and path dependence. (1 student). See discussion under general comments above.

6) How do we choose system boundaries to make the problem easier? (1 student) Experience helps. In some problems it will be obvious (in others less so). Often it depends on what information is provided and what question is being asked. For the simple example of the lightbulb being within or next to the system, if you were told the current and the voltage then it is easy to calculate the electrical work and represent it as such. On the other hand, if you were told that the lightbulb provides 1000J/s of heating to the system, then it would be more straightforward to represent it as heat transfer at the system boundary. Remember, you will get the same answer no matter what you do -- it just may be more direct or easier to solve one way or the other.

7) When you say there is no heat transfer between two items that are touching each other and they are at the same temperature, do you mean that there is no net energy transfer, but there is still energy transfer? (1 student) Certainly molecules from one block could cross into the other block and vice versa, but for the purposes of understanding the thermodynamic state of the two systems it doesn't matter.

8) Still unclear on the sign convention for work (4 students). Please read this over again.

9) Restate the three most confusing points (3 students). See above.

10) On the gas turbine problem, how do we determine the exact path. (2 students) Read over the answer--hopefully this will answer your question. Why doesn't combustion increase the pressure? (1 student) It can, it depends on how the system is designed. In a gas turbine engine the combustion occurs at roughly constant pressure. A candle sitting on a table burns at constant pressure. Combustion inside a closed box increases the pressure.

11) I understand that if the weight is the same then the pressure must be the same, but then how does the piston move. (1 student) Excellent question. The pressure on the inside of the system isn't exactly balanced by the weight -- but it is close -- quasi-static or quasi-equilibrium. See discussions above.

12) What is DE? (1 student) This is the change in total energy of the system (including potential, kinetic, internal, etc.).

13) For those of us who forgot 8.01, what is R in pv=RT? (1 student) This is the gas constant as described in the notes.

14) Why are some cylinders sideways, what is the significance, and how does the weight maintain pressure without g (thus no weight?)? (1 student). There is no particular significance, except that in one case a weight or an external pressure (or a mechanical pin) would be required to balance the system pressure. In the sideways case a weight would not suffice to balance the system pressure. And no g, no weight.