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Heat Engines and the Second Law of Thermodynamics | 32 mins | 0 completed | Learn |

Heat Engines & PV Diagrams | 18 mins | 0 completed | Learn |

The Otto Cycle | 29 mins | 0 completed | Learn |

The Carnot Cycle | 21 mins | 0 completed | Learn |

Refrigerators | 23 mins | 0 completed | Learn |

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Concept #1: The Otto Cycle

**Transcript**

Hey guys, in this video we're going to talk about the auto cycle which is the theoretic cycle that the gas undergoes in a four stroke internal combustion engine alright let's get to it now remember guys that the common gasoline engine in cars is a four stroke internal combustion engine. Now those four strokes are the intake stroke where fuel air mixture is pulled into the cylinder the compression stroke where the piston compresses that fuel air mixture makes it very very dense very high pressure in between these compression and expansion strokes is ignition so ignition fits right up in here ignition is the injection of heat by a spark plug into that very very dense compressed fuel air mixture that ignites it causes it to change chemically into carbon dioxide and water and release a bunch of free energy that free energy causes the expansion stroke where the piston is pushed away from that gas allowing the exhaust to expand and release all that energy into the piston and finally there's the exhaust stroke where the piston pushes all of that exhaust all that burnt gasoline and air through the exhaust valve and out of the piston allowing the intake stroke to start again now the auto cycle as I said is the theoretic cycle it's a very idealized cycle that the gas is supposed to undergo in a four stroke internal combustion engine in reality it doesn't happen quite like the auto cycle but the auto cycle is a closed theoretic explanation of it. Now the auto cycle is given on a P.V diagram above me and it occurs in six steps the first step is the intake stroke where gas is pulled in at a constant pressure the second stroke is the compression stroke. Which is compressed very very rapidly the piston is moving very quickly in the cylinder during these strokes step three is the ignition stroke and this actually happens at a constant volume because ignitions intended theoretically to take place instantaneously instantly all of that air fuel is converted into exhaust and the pressure dramatically rises before the piston can move step four is that expansion stroke just like the compression stroke so both of these they both occur very very rapidly. Now step five is actually kind of like the second half of the expansion stroke the expansion stroke isn't technically finish until step five is done step five allows a depressurization of the depressurization of the exhaust by heat leaving the cylinder when the heat leaves the exhaust the exhaust drops in pressure and this also occurs at a constant volume like the ignition stroke and finally step six is the exhaust stroke which also occurs at a constant pressure so in the auto cycle these idealized theoretic steps are step one being at a constant pressure so it's an isobaric expansion step two remember the compression and the expansion strokes occur very very rapidly the piston is moving very quickly much too quick for heat to enter or leave the cylinder so this is an adiabatic

compression step three is isochoric it happens at this constant volume it's an isochoric pressurization the pressures increase at a constant volume, step four the expansion stroke just like I said about the compression stroke both of them occur very very rapidly so this is also adiabatic and step five which is sort of like that second half of the expanssion stroke allows heat to leave at a constant volume so this is also isochoric as the heat leaves the pressure drops so it's a depressurization and finally step six the exhaust stroke where exhaust is leaving against no resistance this is isobaric it occurs at just this initial pressure so these are the idealized steps isobaric, adiabatic, isochoric, adiabatic, isochoric, isobaric, Let's do an example. Estimate how much work is done by the gas in the auto cycle shown the falling figure is this work done on or by the gas estimate the work done by finding the area enclosed by the cycle. Now normally we would find the area enclosed by the cycle but the shape is weird for that the shape is kind of like this, where this height is larger than this height I don't know what the area of that shape is but we can break this down to two steps that contribute to work notice that this step the isochoric and this step the other isochoric don't do any work because they are at a constant volume and isochoric processes never do any work. Now these two isobaric processes do contribute to the work but they contribute the same amount of work right they're both horizontal lines right on top of each other in opposite directions they contribute the same magnitude of the work but since their opposite directions the signs of the work are opposite so they cancel so really the only thing that contributes to the work is this step and this step so we can find each of them independently and then add them together so the red step can be thought of like a triangle can be approximated as a triangle sitting on top of a rectangle so you can see it looks like a triangle sitting on top of a rectangle that is a shape that we can absolutely find the area for and it starts at 0.00005 and ends at 0.0005 and the triangle starts here at 7 and goes up to 170 and the base of the rectangles at 0 just the bottom alright and we can approximate the green process going to the left as that same shape.

A triangle sitting on top a rectangle, now the volume numbers are going to be the same but the pressure numbers are going to be different up here at the top of the triangle is 25 here at the bottom of the triangle is 1 and obviously it also starts at 0 so all we have to do is add the area of the triangle and the rectangle for both of these figures to find the two works done by each of these processes so the area is going to be that of a triangle plus that of a rectangle now they both have the same base but they're different heights some I'm going to put a little prime on the rectangle height just to indicate that it's a difference if you take the distance or sorry the difference here on a calculator you'll find that's 0.00045 so this becomes one half 0.00045 the height of the triangle is 163 right 170 minus 7 but don't forget that this is times 10 to the 5 pascals we need that the volume is just cubic meters so there's no times 10 to the anything here right plus 0.00045 the same base for the rectangle and the triangle but the height is 7 obviously times 10 to the 5 Pascals plugging this into a calculator we get 3982.5 joules for the area of the red look that red process let me minimize myself for this segment for the green process it's going to be the same exact equation right the area of the triangle plus the area of the rectangle the only thing that's going to change are the pressure numbers because the volume numbers are identical this is a change in pressure of 24 that's the height of the volume numbers is the same and this is a change clearly of 1.Plugging this into a calculator we get 585 joules so now the work is going to be the sum of each of these but with their appropriate sign that's very very important the sign is very important now lets look at the red process the red process is to the right so that work is negative so we get a negative sign in front of the first thing the work sorry the green processes to the left, so that work is positive so we get a positive sign here in between those works now plugging this into a calculater we get -3397.5 joules so that is how much work is done by sorry is done in this auto cycle now since it's negative this is by the gas right that's very important this is by the gas because it's negative and obviously the work done by the gas should be negative because this is an engine the gas should undergo a cycle that allows it to release work that allows the engine to release energy into the system. Alright guys that wraps up the idealized auto cycle as the cycle of the gas undergoes any four sorry four stroke piston engine. Thanks for watching guys.

Example #1: Finding the Compression Ratio

**Transcript**

Hey guys let's do an example three moles of an ideal diatomic gas which is treated with rigid molecular bonds undergoes the following Otto cycle. How much heat is input? How much heat is output? What's the work done by the engine and finally what's the efficiency of this engine? Now what's very important is do not estimate the work by finding the area enclosed by the cycle we saw a problem before where you can estimate the area of the cycle by approximating two triangles it turns out this is not a very accurate way to estimate the work done and when we are trying to calculate the efficiency you need a very precise calculation of the work done, so we're going to have to do this a different way so let's break this up into four steps. I'm going to call step one this step which is the compression stroke, step two this step which is ignition, step three which is that first part of the expansion stroke and step four which is like the second part of the expansion stroke we can ignore all contributions due to these steps because they don't do any work since its cyclic, since it starts and ends at the same position there is no change in internal energy which means by the first law of thermodynamics there is no heat exchange either so nothing is going on in those two steps they completely balance each other out. So lets look at step one, step one is adiabatic remember that's very important to keep in mind for the auto cycle which steps are adiabatic which steps are isochoric which steps are icothermal which steps are icobaric etc. Adiabatic by definition means that there is no heat transfer for this step so Q1 is 0. Now we know the first law of thermodynamics is delta U is Q plus W and since Q for step one is zero we can say that delta U for step one is just the work done in step one so if we can find the amount of work done sorry if we can find the change in internal energy for step one we can find the work done in step one I don't actually want to encircle those let me highlight them I want to circle the final answers, so let's find the change internal energy for the ideal gas during step one remember that the internal energy for any ideal gas is just F over 2 N, R, T. We were told that this was an ideal diatomic gas, diatomic gas is either have a degrees of freedom of 5 or degrees of freedom of 7 because we treat the moleculer bonds as rigid it's 5.

This means that the change in internal energy for step one is only going to be due to the change in temperature because the number of moles isn't changing gas doesn't enter or leave the cylinder except for the intake and the exhaust steps which we're ignoring right those are these steps which we're ignoring so during the cycle gas doesn't enter or leave the piston so the only thing that's changing is the temperature. So the question is what's the change in temperature for step one well if we want to relate pressure and volume like on a P.V. diagram to temperature we need to use the ideal gas law during step one both the pressure and the volume are changing and that's responsible for producing the change in temperature so the entirety of the left side is changing and that leads to a change in the temperature. So we can say that the change in the temperature is just the change of P times V during step one divided by N, R so how does P times V change in step one so delta T1 is going to be what's the final pressure times the volume, the final pressure is 25 right times 10 to the 5, don't forget this times 10 to the 5 the final volume is 0.00005 right during step one minus the initial pressure which was 1 times 10 to 5 and the initial volume which was 0.0005 only three zeros divided by the number of moles which is 3 and the ideal gas constant which is 8.314. Plugging this into your calculator this works out to be 3 Kelvin. So now that I know what the change in temperature is during step one I can plug that in to this equation which tells me how the internal energy changes in step one for a given temperature change.

The number of moles is still 3 right ideal gas constant still 8.314 and the temperature change is 3 Kelvin. Plugging this into your calculator we get 187 Joules as the change in internal energy for the first step but we don't care about how the internal energy cares I mean changes all we care about is the amount of heat input, the amount of heat output the work done and the efficiency. So this tells us remember guys that first law tells us that the work done during step one is the same as the change in internal energy in step one so this is 187 Joules and we're going to hold on to that for later now let's look at step two, this is step two. This is an isochoric process right. Isochoric processes mean that the change in volume for that step is 0 that also means that the work for that step is also 0 if the change in volume is 0 the work done is 0 now looking at the first law of thermodynamics, if the work done is 0 then the change in internal energy for that step is just the heat transferred for that step that's what the ideal sorry that's what the first law of thermodynamics tells us about this isochoric process and finally don't forget that any change in internal energy for this cycle since gas isn't coming in or leaving during the cycle it's always going to be five halves N R times delta T. This is because the only thing that leads to internal energy change is a temperature change in order to find how the temperature changes with the pressure and the volume that we have on the P.V diagram we have to look at the ideal gas law. Now for an isochoric process right the volume does not change only the pressure changes for an isochoric process right we're sitting here at this constant volume so we can say that the left side only changes with pressure, there's only a change of pressure here the volume is a constant and the right half changes with temperature so that means delta T2 is V2 delta P2 over N R, this volume that this isochoric process occurs at is just this right that minimum volume for the gas and the change in pressure is from 25 to 170 that's the increase in pressure that we're undergoing in this process and this is divided by the number of moles which is 3 times 8.314. Let me minimize myself here, which if you plug into your calculator is equivalent to 29.1 Kelvin. So now we know how the temperature changes in step two that leads us directly remember our equation for change in internal energy that leads us directly to how the internal energy changes in step two the number of moles is 3 the ideal gas constant is 8.314 the change in temperature is 29.1 right we just found that out so plugging this into a calculator tells us the amounts of sorry the amount of change of internal energy for the second step is 1815 Joules but we don't care about that all we care about is how much heat is transferred and don't forget our first law requirement for this step tells us that the heat transfer in this step is equivalent to the change internal energy which is 1815 Joules. So for step one and for step two we know how much heat is transferred and how much work is transferred all we have to do is match this for steps two in steps three sorry for steps three and steps four but remember step three is another adiabatic step we already know all the requirement and all the equations for the adiabatic process because we just used them in step one we know that this means that for step three the Heat transfer is 0 and we know this means for step three that the change in temperature can be given as the change in the pressure times the volume for step three over N R this is the same equation that we got from the ideal gas law in step one which is another adiabatic process all we have to do now is plug this in the final pressure after step three is 7 times 10 to the 5 the final volume after step three is 0.0005 the initial pressure of step three is 170 times 10 to the 5 Pascal and the initial volume is our very small volume 0.00005 the number of moles is always the same 3 and the ideal gas constants obviously always the same plugging this into our calculator we're going to get -20 Kelvin as the result of the temperature change, remember our first law requirement for an adiabatic process if you just check your notes it's that the amount of the sorry the change in internal energy equals the amount of work done so if we find the change in internal energy which we know is just five halves N R times the change in temperature for that step we'll find how much work is done the number of moles is just 3 ideal gas constanct 8.314 right we just found the temperature change was -20 plugging this into a calculater we get -1247 Joules which tells us that the work done in this process which is the same as a change in internal energy by the first law is -1247 Joules.

Finally we have step four the last step that we have to look at, step four is also an isochoric step just like step two was so we can use all the same equations in step two we know immediately that this means that the work done in step four is going to be 0, so we have that we know that the temperature change in step four is just going to be the volume in step four times the change in pressure of step four over N R this is the same equation that we used in step two which we got from the ideal gas law for an isochoric process in this case we're at the large volume 0.0005 our final pressure is 1 times 10 to the 5 right that minimum pressure we started on the ending point sorry the starting point which is the same as the ending point the initial pressure is 7 times 10 to the 5. The number of moles we know is 3 and the ideal gas constant is always 8.314, plugging this into a calculator we get -12 Kelvin as our temperature change now just like we did in every other step we can plug in this temperature change into the change internal energy right the change in internal energy for any gas process is going to be due to the change in temperature because the number of moles doesn't change in the cyclic process. The number of moles is 3 right 8.314 is ideal gas constant we just found the change in temperature is -12 so this is going to come out to be -748 Joules when you plug it into a calculator remember guys we just found this requirement by the first law for an isochoric processes that says that the heat transfer during that step is equivalent to the change in internal energy during that step so now we know the heat transfer during that step so now we've found how much work is done how much heat is transferred at every step we can answer the questions that we were given. How much heat is into the system? how much heat is out of the system? what's the work done by the system? And what is the efficiency ? those are our four questions well we only had two steps that heat was transferred step two and step three the two isochoric steps because the step one step three sorry step two and step four are the two isochoric steps step one and step three were both adiabatic so neither of them had a heat transfer step two had a positive heat that's the only instance where heat is entering the system so the heat input is just the heat for step two which is 1815 Joules. Now the only heat transfer that was out of the system that was leaving the system was the heat transfer in step four and this is a magnitude of 748 Joules that's how much heat left the system during this cycle the work done was done sorry there was work done in both adiabatic steps, step one and step three.Which is going to be 187 Joules minus 1247 Joules which works out to be when you plug it into your calculator -1060 Joules but remember what is this, this is actually the work done by the gas this isn't not the work done by the engine the work done by the engine is however much energy is released by the gas if you can see since this is -1060 Joules of work left the gas that's how much work was produced by the engine so the work output is positive 1060 Joules and finally the efficiency is just the work out divided by the heat in how much work do you get out for how much heat do you put in so this is 1060 over 1815 right which was the amount of work output and the amount of heat input and this works out to be 0.584 so the efficiency of the engine is 58.4 % this is a long process but this is how you solve P.V diagrams for these different properties of a heat engine. Alright guys that wraps up this problem. Thanks for watching.

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