Internal Energy o a Gas Work Done by a Gas Special Processes The First Law o Thermodynamics p Diagrams The First Law o Thermodynamics is all about the energy o a gas: how much energy does the gas possess, how does it gain or lose energy, and how can we determine these rom the macroscopic parameters o the gas (i.e. pressure, volume and temperature.) As always, the important stu is in boxes. Internal Energy o a Gas At the end o Chapter 16 we ound in the topic o Kinetic Theory that the kinetic energy o gas molecules was directly related to the temperature o the gas. Speciically: Average translational kinetic energy o one gas molecule: K avg kt where k is Boltzmann s constant and T is the temperature o the gas in Kelvins (o course.) I there are N molecules o the gas, then the total kinetic energy o the gas is simply the average kinetic energy o one molecule times the number o molecules. So: Total translational kinetic energy o an ideal gas: K NkT (Note: this is only valid or an ideal gas because the IGL was used as part o the derivation or Kinetic Theory.) Recall that the ideal gas law could be written as p nrt or p NkT (see Chapter 16 notes) Which means we can now rewrite our expression in terms o n and R: Total translational kinetic energy o an ideal gas: K nrt or K p The only kind o energy the gas can possess is this kinetic energy, so we reer to this energy as the internal energy o the gas. Monatomic gases only have translational kinetic energy, but diatomic gases also have rotational kinetic energy. This means diatomic gases have more energy than monatomic gases or the same temperature. How much more? Instead o /, diatomic gases have 5/. Which means that: Page 1 o 1
Internal energy o a monatomic ideal gas: E nrt Internal energy o a diatomic ideal gas: 5 E nrt Since the energy is directly related to the temperature o the gas, a change in temperature relates directly to a change in internal energy o the gas. Or: Change in internal energy o a monatomic ideal gas: Change in internal energy o a diatomic ideal gas: E E nr T 5 nr T These expressions can be rewritten: Monatomic ideal gas: E ( nrt nrti ) ( p pi ) 5 Diatomic ideal gas: E ( nrt nrti ) ( p pi ) 5 where we used the act that nrt can be replaced with the corresponding p or an ideal gas. This means that to determine the internal energy o an ideal gas, we only need to know the pressure and volume o the gas. We will take advantage o this later when we create p diagrams. Work Done by a Gas Consider a gas in a sealed container. For simplicity, we can imagine the container is a rigid cylinder with a movable piston at the top. The volume o the cylinder can change as the piston moves up or down. The gas pushes against the piston with pressure p (i.e. the pressure o the gas) as shown in the picture below. Imagine the piston moves upward as the gas pushes upward on the piston. The gas applies a orce to the piston, and the piston moves... so the gas does work on the piston. Note that i the gas does work on Page o 1
the piston, by deinition the gas gives energy to the piston. Where does this energy come rom? It must come rom the gas, because the gas gives it to the piston. This idea will be important soon, when we write the First Law o Thermodynamics. How much work does the gas do on the piston? In the diagram above, we can see that i p is the pressure o the gas and A is the area o the piston, then the orce the gas applies to the piston is pa. I the piston moves upward a distance d, then the work done by the gas is Work orce displacement pa d Note that when the piston moves upward, the gas ills the extra volume created by the movement o the piston. We can deine this increase in volume or the gas as and simple geometry gives us Ad We can then use this in our expression or the work done by the gas to write: Work p This expression is important, because it deines the work done by the gas entirely in terms o the parameters o the gas, i.e. the pressure and volume o the gas itsel. So to determine the work done by the gas, we do not need to know anything about what the gas is doing the work on! We only need to know about the gas itsel. Page o 1
We can also make a simple observation about the sign, positive or negative, o the work done by the gas. Recall that work is positive i the displacement is in the same direction as the orce and negative i the displacement is opposite the direction o the orce. In the description above, the gas pushes up on the piston, the piston moves upward, so work is positive and is positive. On the other hand, i the piston was to move downward while the gas is pushing upward on the piston, then the work would be negative... and would be negative. Which means we can make a simple deinition: Work done by a gas: Work is positive i the volume o the gas increases, i.e. is positive Work is negative i the volume o the gas decreases, i.e. is negative There is only one small problem with the expression or the work: in general, the pressure o the gas will vary as the volume o the gas changes. Which means we cannot simply take the pressure times to get the work, because there is not simply one pressure to use. How do we deal with this? We have to imagine the pressure changing as the volume increases... and take the work in ininitesimal bits. That is, imagine a tiny change in volume such that the pressure is constant during that tiny volume change. Or: tiny volume change is d so a tiny bit o work is dw p d Then to get the total work or the entire volume change, we simply add all the tiny bits o work and: Work done by a gas: W i pd This is the technical way to express the work done by a gas as it changes its volume. Note that essentially this is the same idea as our original expression, i.e. the pressure times, but the integral simply acknowledges that the pressure could be changing. What are we going to do with this integral? Not much. The problem is that there are two variables, pressure and volume, and so in general there is nothing we can do with the integral. It s an idea. Special Processes What exactly do I mean by in general? We can now deine two things which we will reer to throughout our discussion o the First Law o Thermodynamics (which is coming very soon!) Page 4 o 1
A state o a gas is a unique combination o its pressure, volume and temperature. For the purpose o deining a state we assume that n is constant; that is, the gas is in a sealed container. So the three parameters o the gas are simply p and T. A speciic combination o these three parameters deines a state o the gas. A process or a gas is how the gas transers rom one state to another, i.e. rom some initial p and T to some inal p and T. We can deine a general process as one in which all three parameters change, i.e. the inal p and T are all dierent rom the initial values, and there is no special way in which they change. We can also deine a special process. A special process is one or which only two o the three parameters o the gas change; that is, one o the three parameters o the gas remains constant. So we can deine the three special processes by the one parameter that remains constant: constant volume or isochoric (i.e. pressure and temperature change) constant pressure or isobaric (i.e. volume and temperature change) constant temperature or isothermal (i.e. pressure and volume change) We will use these special processes in problems rom Chapter 17 and 18 that deal with the First Law o Thermodynamics. I stated above that the integral expression or the work done by a gas could not be used or a process in general, simply because o the math: you cannot do an integral with two variables. But we can use the integral expression or our three special processes. Which means we can write expressions or the work done by a gas or each o these three special processes: constant volume: since the volume o the gas does not change, work is zero. constant pressure: since pressure is constant, p can be actored out o the integral and then work p ( - i ) constant temperature: using the ideal gas law, we can replace p with nrt/ and then... W nrt d pd d nrt i i i nrt ln i Page 5 o 1
Important results: 1. The work done by a gas can be determined or a special process, i.e. i the gas changes through constant volume, constant pressure or constant temperature.. The work done by a gas is process dependent; that is, the expression we use to determine the work depends on the speciic process used by the gas. In summary: constant volume: W 0 constant pressure: W p( i ) constant temperature: W nrt ln i It is worth noting that the irst two expressions above are valid or all gases and come directly rom the deinition o work. The third expression, or the work done at constant temperature, is valid only or an ideal gas because the ideal gas law was used to derive the expression. The First Law o Thermodynamics We can now write the First Law o Thermodynamics, which is actually just a simple expression o conservation o energy. The First Law acknowledges that: The gas possesses energy, which we deined as internal energy above; The gas can give away energy when it does work; Energy in the orm o heat can be added to or removed rom the gas. Since the internal energy o a gas is directly related to its temperature, when the energy o a gas changes its temperature also changes. I its temperature changes, then its pressure and/or volume must also change. (To justiy this, think o the ideal gas law. I p nrt, when one side o the equation changes, so must the other side.) Which means that i the energy o the gas changes, the state o the gas changes... and vice versa! Which means the energy o a gas changes during a process, i.e. transition rom one state to another. Page 6 o 1
How much does the energy change? For any process we can write: change in energy o a gas heat added to the gas - work done by the gas (Note: we subtract the work done by the gas because when the gas does work it gives away energy.) We can write this in abbreviated orm as the First Law o Thermodynamics: A ew important observations: E Q - W 1. I the initial and inal temperatures o the gas, and whether the gas is monatomic or diatomic, are known, E can be calculated.. I the type o special process is known, W can be calculated.. I E and W are calculated, they can be used to calculate Q. There is nothing tricky about the First Law... it is just a way to calculate the energy entering or leaving a gas. I the pressures, volumes and temperatures (initial and inal) o a gas are known, these calculations are simple. It is useul to rearrange the First Law to read: because this is how we will use it. E + W Q p Diagrams In general, diagrams are a useul way to help us visualize and analyze abstract inormation. For example, in Chapter 4 we introduced ree body diagrams to analyze orces acting on an object. We can do something similar here to help us visualize the inormation associated with states and processes o a gas: pressure, volume and temperature. A p diagram is a simple way to graphically show the pressures and volumes associated with states o a gas, as well as how the gas changes rom one state to another... i.e. the processes. A p diagram looks like a simple xy plot, with pressure on the vertical axis and volume on the horizontal axis. Page 7 o 1
As such, each point on a p diagram represents a state o the gas, i.e. a speciic combination o pressure and volume. We assume that n or the gas is constant, that our p diagram represents a ixed amount o gas in a sealed container, so that i pressure and volume are known, then temperature can also be determined. For this reason, while temperature is not explicitly shown on a p diagram, it is implicit with the given pressure and volume. A line on a p diagram connects two points and represents the path taken by the gas to go rom one state (i.e. its initial state ) to another state (i.e. the inal state.) Thereore, a line on a p diagram represents a process or the gas. Creating a p diagram is airly simple: Draw the perpendicular axes Label p on the vertical axis, with units; label on the horizontal axis, with units Label the given points on the diagram, using given values o p and/or Draw appropriate lines (i.e. to represent the correct processes) to connect the points Consider this example: A gas starts with a pressure o 1.00 atm and volume 10.0 L (i.e. Point 1); its pressure is increased to 5.00 atm while its volume is held constant (i.e. now at Point ); it is then allowed to expand at constant temperature until its pressure is reduced to 1.00 atm (i.e. now at Point ); inally, it is cooled at constant pressure until it returns to the initial state. Note that we have: Point 1: p 1.00 atm 10.0 L Point : p 5.00 atm, 10.0 L Point : p 1.00 atm,?? We are not given the volume or Point, but since the gas expands rom Point to Point, we know that the volume at Point is greater than the volume at Point. We can diagram these three points as: Page 8 o 1
We can now connect the points by drawing lines that represent the three described processes: Point 1 to Point : constant volume Point to Point : constant temperature Point to Point 1: constant temperature a vertical line (because volume is the horizontal axis) a downward curving line (see below!) a horizontal line (because pressure is the vertical axis) It should be pretty easy to see why a constant volume and constant pressure process are represented by a vertical line and horizontal line, respectively. But why is a constant temperature process represented by a downward curving line? I we use the ideal gas law to represent the relationship between pressure, volume and temperature or our gas, we can write p nrt or p nrt / Note that we want to write the equation as p... because pressure is on our vertical axis. So our equation takes the orm o y..., which is how we express equations when y is on the vertical axis. Also note that nrt is constant i T is constant. And that is in the position where we traditionally ind x. Which means that the ideal gas law shows that p nrt / is like y constant / x or y 1 / x where I used just 1 or a constant. This means that on a p diagram, a line representing a constant pressure process must have the same o a y 1/x graph rom traditional algebra. This curve is high to the let, low to the right, and downwardly curving... just as shown on the diagram above. What can we do with a p diagram? The diagram is simply a way to display given inormation. The next step is to use this inormation to answer the question(s) posed in the problem. This will require that you organize the given inormation. This organization is best done in a table. Page 9 o 1
There are two tables necessary to organize the inormation. Each state o the gas is deined by its pressure, volume and temperature. So we create a table or these three parameters, as well as the product p, or each state. For the diagram above, this table would be: p (atm) (L) T (K) p (atm-l) 1 1.00 10.0 10.0 5.00 10.0 50.0 1.00 We can only calculate temperatures i we are given n or i we are given the temperature or at least one o the states. Otherwise, we can leave the temperature column blank. However, using the ideal gas law we can see that i initial and inal temperatures are equal, then initial and inal p values are also equal. Which means that two points connected by a constant temperature process must have the same p values. We can use this act to ill in the p or Point, and then we can ill in the volume or Point : p (atm) (L) T (K) p (atm-l) 1 1.00 10.0 10.0 5.00 10.0 50.0 1.00 50.0 50.0 We can now move on to the second table, which will have columns or E W and Q or each process: Page 10 o 1
1 to all in atm-l E + W Q to to 1 total We can use the inormation in the p column rom the irst table to complete the second table. First calculate E or each process. You must know whether the gas is monatomic or diatomic to calculate E. For this example we will take the gas to be diatomic. Then or each process: Or: So: E 5 ( ) p p i 1 to : E (5/) (50.0-10.0) atm-l 100 atm-l to : E (5/) (50.0-50.0) atm-l 0 to 1: E (5/) (10.0-50.0) atm-l -100 atm-l 1 to 100 all in atm-l E + W Q to 0 to 1-100 total 0 Next we can calculate W or each process. But work is process dependent, so we must use the correct expression or each process: Page 11 o 1
1 to : constant volume W 0 to : constant temperature W nrt ln( / ) (50.0 atm-l) ln(50.0/10.0) 80.5 atm-l So: to 1: E constant pressure W p 1 - p 10.0 atm-l - 50.0 atm-l -40.0 atm-l all in atm-l E + W Q 1 to 100 0.0 to 0 80.5 to 1-100 -40.0 total 0 40.5 Finally, we can complete the Q column by simply adding E and W or each process: all in atm-l E + W Q 1 to 100 0.0 100.0 to 0 80.5 80.5 to 1-100 -40.0-140.0 total 0 40.5 40.5 A ew important notes about the results o this table: For a complete cycle, the total o the E column must be zero. I the cycle is clockwise, the total o the work column must be positive. I the cycle is counterclockwise, the total o the work column must be negative. The total o the columns must also obey the First Law: E + W Q Page 1 o 1