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welcome to our first section in our unit on the first law of thermodynamics in this section, we're going to look at how work is done by an ideal gas and use that to find what the first law of thermodynamics is now going back to our example of a piston with some gas inside. Looking at this, we say, Okay, well, if I have a piston with area A, that is, if you were to rotate it, you would see an area a on the cross section. And we have a piston here that gets pushed outwards by the gas on the inside. That means the gasses exert going to exert some force over a distance. Delta X. If we want to find the amount of work done by the gas than we can say, definition of work is work equals the integral of FDX. I'm going to substitute pressure times area here for force because we can calculate the pressure of the gas a little easier than the precise force putting that in. Then we have a times DX. Well, that's just the infinitesimal change in volume that this system experiences when the force is applied, which means we can rewrite it as this work is equal to the integral from some initial volume to some final volume of PDV. Now, this is going to be the work done by an ideal gas. Other books and other classes will define a work done on the ideal gas instead of the work done by the ideal gas. Because of this, they'll put a negative in front of their calculation. We're going to stick with the work done by an ideal gas as being a positive amount of work here. Now, um, remember that you have this relationship from the ideal gas law where we can actually write P in terms of V that will be very helpful when going to evaluate this integral. The other thing we can do is we can look at RPV diagram notice in R P V diagrams. What's gonna happen is the work will be the area underneath the curve. So, for example, when we have a constant volume, there is no UN area underneath this line. Therefore, the work done here is zero jewels. So a constant volume or is a coric process, does zero work. On the other hand, if we have a constant pressure process where we go from A to B. Then we take the area under the curve and that will be the amount of pressure which is constant multiplied by Delta V. So foran is a barrack or constant pressure process. This is going to be our work for a constant temperature of eso thermal process. It's a little more complicated. We have to take the integral underneath this oddly shaped line, though if we use the relationship we had before of p equals and R T over V, we will quickly come up with an answer of work is equal to N. R. T multiplied by the natural log of V two over V one. This works because we know that NRT is going to have to be constant because temperature is a constant value in an ice. A thermal process. Similarly, because we have NRT here, we can also plug in PV. You might ask well, which pressure in which volume? Well, anyone that sits on this line will do because the pressure multiplied by the volume at any point here will be equal to NRT, which is a constant now, looking at this and seeing what happens if we were to draw an odd looking PV diagram, for example, over here, if I had put a bubble in it instead, the area under the curve would change, which means that work is path dependent. Now. This is odd, because before we focused a lot on the path independence of work. But that's because we were dealing with conservative forces. Thes air, not conservative forces, thes air non conservative forces, which means that it is dependent on the path. So you don't need to be thinking about that path independence anymore, because it is not what we're looking at. So looking into conservation of energy, though, and what we did before with this path independence. We knew we had some amount of energy in our system and that if the energy in the system changed, that meant we had some non conservative work or something. Well, let's let's consider an odd situation where we see energy changing and nothing that looks like work. So if you have water vapor in a bottle, what can we do to change the energy of the molecules, especially increase it? Well, we saw before. If we squeeze it, that's going to do work on the particles, which means that we'll have a change in temperature. And so we added energy in because temperature is a measure of the energy of those particles, we could also just heat it, put it in a microwave or something hot and let it heat up just by transferring energy from the hotter object. So in this case, when we say heat what we mean as our definition and I say this very precisely because heat is used in a very different context. In other places, heat is the energy transferred due to a temperature difference. So if we have an object T A and an object, T B and A has a greater temperature than be, then a will pass heat to be until the to have the same temperature. So, uh, since we have these two different ways of heating are molecules. What we're going to say is that a change in thermal energy, which is related to temperature, as I've said before, Ah, change in thermal energy is going to be equal to the heat added to the system. Then we're going to say, minus the work done by the system, the reason, we say minus the work done by the system is because if the system does work, it's going to lose some of its heat energy. If we have thinking about that piston example, when the gas expands and pushes the piston out and does work on it, it's actually given energy to the external environment and lost it from its internal environment. Thus, Delta E thermal is equal to Q minus W, and this is the first law of thermodynamics again because some people to find work in the opposite way you will have you would have a equals Q plus W in this situation, Um, but based on the definition that I gave encircled, we're going to write it out this way. So thinking about this, then we have a change in the energy of our system is going to be equal to the change in our mechanical energy, plus the change in our thermal energy. So this is kind of our new conservation law for energy that any change to the system is going to be the some of these to notice. If there were no change in thermal energy and we only had conservative forces thes two numbers would both be zero, and the change in energy of the system would be zero jewels moving on we have. We're going to talk a little bit more about heat here, so we've got heat, which is given by a variable. Q is going to be obviously in units of jewels because it's added into jewels. We have a number of different units, though that are common to report for heat, because when people first started thinking about heat, they didn't realize that it was just the transfer of energy. Therefore, we have the calorie. This is the graham calorie where 1 g of water is heated up by one degree Celsius. So if we wanted to do that to 1 g of water, we would need 4.187 jewels to heat that graham of water up by one degree Celsius. We also have the British thermal unit, or BTU, which is common in some places, which is 1055 jewels. And this is really more by industry. Some industries in particular. The fossil fuel industry very often will measure things and be to use, and then we also have the K Cal which is the number of calories you see on the side of all of your nutrition labels. Eso When you see those calories, it's not the same as thes calories. This is a gram calorie. This is a kilocalories. It's the calories that you eat. So obviously there are 1000 of these little calories in the big food calories that you're eating. And then we also have the kilowatt hour, which can be may use two major energy that would be equal to 8.6 times 10 to the five of these gram calories over here. This isn't a very common relationship, because generally kilowatt hours has used in a different context than he is. But technically, they're both units of energy. Now, there are four important terms that we need to distinguish between here. Before we finished, the first one is temperature. Okay, so temperature is what we measure. It is T K or T C or T f. Even if we wanted. Okay, it is a state. Variable is something we use a macroscopic measurement we take to describe the system. Heat is the transfer of energy between objects with different temperature. So if we have a knob jek t a greater than TB. Then we're going to have some Q passed around until T A is equal to T B. It's what brings object into thermal equilibrium. Thermal energy is related to the temperature. It's also related directly to the kinetic energy of the individual. Particles in the system will look at more of that later in another unit. And then there's also a term internal energy, which might be the one that shows up in your book instead of thermal energy. Um, technically, internal energy also includes potential energy, but since we're in an ideal gas, that would be zero. That's one of our assumptions, so because we don't want any interactions between the molecules, so technically, internal energy and thermal energy, in this case for an ideal gas are the same thing

University of North Carolina at Chapel Hill