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Internal Energy | 28 mins | 0 completed | Learn |

Introduction to Heat Capacity | 23 mins | 0 completed | Learn |

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Work & PV Diagrams | 14 mins | 0 completed | Learn |

Concept #1: Work and PV Diagrams

**Transcript**

Hey guys, in this video we're going to talk about work and PV diagrams. Alright let's get to it now remember guys PV diagrams are a diagram of gases pressure right the P versus volume pressure volume PV diagrams they're great specifically for analyzing thermal processes things that gases undergo that could change the temperature of the gas since the gases state is defined by the temperature and we know that the temperature is related to the pressure and the volume of a gas by the ideal gas law, so because P and V are related to the state of the gas the temperature PV diagrams are perfect for analyzing thermal processes of ideal gases now for a single process you can diagram it on a PV diagrams a change from some state A to some state B. Let me minimize myself so we can see this figure imagine this as a change from some hypothetical state A to some hypothetical state B the work done by this process is going to equal the area under the line representing the process on the PV diagram so right here there is some line. That diagrams the process from state A to state B and we can represent the work done as just the area which I shaded in blue the area here contained under the line now to find the area is a pretty straightforward process of simply approximating the shape under the line as something that you know the formula force some regular shape like a rectangle right or a triangle or possibly a circle if you can etc. Any regular shape for instance here we can approximate the upper part as a triangle and we can approximate the lower part as a rectangle and the total area is just going to be the sum of those two areas because we can easily find the area of a triangle it's one half base times height and we can easily find the area of a rectangle it's just the base times the height now the only thing that is really important here that you need to be aware of is the sign of the work because the sign depends on what direction the process is moving on a PV diagram if the process goes right to left meaning it starts on the right and is going to the left whether it goes up whether it goes down or whether it just goes straight from right to left the work is always positive. If the process goes from left to right regardless of whether it's going up or whether it's going down if it's going to the right the work is always negative. Those signs are really really important but those signs are linked entirely to how we define the first law of thermodynamics because we said delts U is Q plus W because we use a positive sign here we have chosen to define all positive works as work done on the gas, works to increase the gas's energy because we use a positive sign in the first law we've assigned all negative works to be worked done by the gas a removal of energy from the gas if you were to use a negative sign here which some books do and some professors do. Then you have to accordingly change the sign of the left to right and the right to left process but the best thing to do is to stick with the plus sign here because that makes the most conceptual sense just like with mechanical energy if work is positive you're increasing the energy of the system in this case the gas and if work is negative you're decreasing the energy of the system,so let's do a quick example. What is the work done for a gas to move from A to B so first from A to B is to the right to the right means that the work is negative now to find how much work is done the magnitude of the work we're going to approximate this as a triangle on top and as a rectangle below so exactly like I said in the figure above because this is the same figure I just added numbers, so let's find what those areas are this area is one half base times height the base is just 0.002 Cubic meters and the height is 1 times 10 to the 5 Pascal and when you plug this into a calculator you get 100 joules. Now the area for the lower rectangle is just going to be base times height now it's the same base it's still 0.002 but it's a different height now the number happens to be the same the number still 1 times 10 to the 5 but only because we're going from 0 to 1 time 10 to the 5 if the numbers on the vertical axis were different there's no guarantee that the height of the lower rectangle is the same as the height of the triangle it just happens to be the same in this case, if you plug this into a calculator we get 200 joules so now what's the work well the magnitude of the work is just going to be the sum of those two areas right which we know is 300 joules but the question the biggest question is going to be is this positive or is this negative remember because it's a right word process the work is negative so we had to put a negative sign out front so the work is -300 joules now is this work done on or work done by the gas in our sign convention in negative work is work done by the gas the gas loses energy in this process. Now for cyclical processes for processes that start and end at the same position the total work done is just the sum of the work done for each leg of that cycle, so imagine that we had some hypothetical cycle that goes to the right then goes up and then comes back down we can find the work due to the first cycle sorry the first leg of the cycle the work to the second leg of the cycle and the work to the third leg of the cycle independent one another exactly as I said before for single processes and then add them together that'll be the area of sorry that'll be the total work right find the area under each individual process and then sum those areas summing means with appropriate signs for instance step one is to the right so the work is going to be negative and step three is to the left even though it's down into the left its still to the left so the work is positive the area under step three is going to be larger than the area under step one so the work overall for the cycle is actually going to end up being positive now instead of finding the work due to each individual leg and summing them you can actually find the area enclosed by the cycle and that will be equivalent to the work done the work of the cycle will actually be equal to the area enclosed by the loop enclosed by that cycle now there also has to be a sine convention for area enclosed by a loop for area enclosed by cycle because in order to have a cycle you're always going to have to have at least one leg going one direction and one leg going back that way you can complete the cycle you can end where you started. Counterclockwise cycles always have positive works this example here that I drew is a counterclockwise cycle and it's positive all counterclockwise cycles have positive works clockwise cycles always have negative works. So not only do you have to memorize that process going to the left have Sorry going to the right have negative works and processes going to the left have positive works but you also have to memorize these that counterclockwise loops always have positive work the gas always gains energy and clockwise loops always have negative works the gas always loses energy what's the work done in this cycle A to C to B to A well A to C to B to A first of all is a counter-clockwise cycle, because it's counterclockwise that means that the work has to be positive right for all counterclockwise cycles the work is positive we can approximate this as a triangle and find the area enclosed by this loop and the area is going to be one half base times height right the base is 0.01 minus 0.003 which is 0.007 and the height is 7 minus 3 which is 4 right, times 10 to the 6. Plugging this into your calculator we get a positive 14000 well actually there's never going to be a negative here this answer can never be negative because we're just finding the area so 14000 joules which is equivalent to 14 kilojoules so the work is going to be the area which is 14 kilojoules the question is the work positive or is it negative remember because its counterclockwise loop the work is always going to be positive it's positive for all counterclockwise loops so this is a positive number a positive sign right there and lastly is the work done on or by the gas. Remember in our sign convention positive work means that the gas gains energy. This is work done on the gas. That wraps up our discussion guys on work and P.V. diagrams and more specifically how to find the work using the P.V. diagrams. Thanks for watching guys.

Example #1: Work and Heat Transfer in a Four-Step Cycle

**Transcript**

Hey guys, I hope you're able to solve this one on your own. If not here's a little bit of help. How much work is done in the four step cycle illustrated in the following PV diagram? Is it work done on or by the gas? How much heat is exchanged between the gas and the environment in this cycle? Is the heat entering or leaving the gas? For a cycle where we stop and start on the same point the area, the work is going to be equal to the area enclosed. Now this cycle is clockwise which means that the work is negative but we still need to find the area. The area is just going to be the base times the height right this is just a rectangle. The base is four fifths V not and the height here is just 3P not. From P not to 4P not. So this is 12 fifths P not V not. So the work done is twelve fifths P not V not but since it was clockwise the work has to be negative so we need that negative sign. And this is work done by the gas because it's a negative work. Now the third law of thermodynamics links the work done to the heat exchange, delta U is Q plus W. What's that heat sorry what's the change in internal energy delta U for a complete cycle which is what we're calculating here. It's always zero. So the heat transfer is always going to be negative work for a cycle. So that means that the heat transferred is actually positive 12 over 5P not V not. And now since it's a positive heat transfer we know that this is heat going into sorry little technical difficulty into the gas that's what it always means for a positive Q. This is what it always means for a negative work. Alright guys that wraps up this problem. Thanks for watching.

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Concept #1: Work and PV Diagrams

Example #1: Work and Heat Transfer in a Four-Step Cycle

A certain quantity of an ideal gas initially at temperature T0, pressure P0, and volume V0 is compressed to one-half its initial volume. As shown, the process may be adiabatic (process 1), isothermal (process 2), or isobaric (process 3).
Which of the following is true of the mechanical work done on the gas?
1. It is the same for process 1 and 2 and less for process 3.
2. It is greatest for process 1.
3. It is the same for all three processes.
4. It is the same for process 2 and 3 and less for process 1.
5. It is greatest for process 3.

An ideal gas is taken through the cycle shown in the diagram below. The process is carried out with 150 mg of Helium, which is a monoatomic gas with an atomic mass of 4 g/mol. First, the gas undergoes an adiabatic expansion (from point A to point C), followed by an increase in pressure at constant volume (from point C to point B). Finally, the gas is taken back to the initial state by an isothermal compression (from point B to point A). What is the total work done on the system? Note that the work during an isothermal process is W = nRT ln(V i / Vf).

Calculate the work W done by the gas during process 1 to 3 to 6.Express your answer in terms of p0 and V0.

Calculate the work W done by the gas during process 1→2. Express your answer in terms of p0 and V0.

How much work is done on the gas in the process shown in the figure?

Calculate the work W done by the gas during process 1→2→6→5→1.One important use for pV diagrams is in calculating work. The product pV has the units of Pa×m3=(N/m2)⋅m3=N⋅m=J; in fact, the absolute value of the work done by the gas (or on the gas) during any process equals the area under the graph corresponding to that process on the pV diagram. If the gas increases in volume, it does positive work; if the volume decreases, the gas does negative work (or, in other words, work is being done on the gas). If the volume does not change, the work done is zero.

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