Ch 23: The First Law of ThermodynamicsWorksheetSee all chapters
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Ch 01: Intro to Physics; Units
Ch 02: 1D Motion / Kinematics
Ch 03: Vectors
Ch 04: 2D Kinematics
Ch 05: Projectile Motion
Ch 06: Intro to Forces (Dynamics)
Ch 07: Friction, Inclines, Systems
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Ch 22: Kinetic Theory of Ideal Gasses
Ch 23: The First Law of Thermodynamics
Ch 24: The Second Law of Thermodynamics
Ch 23: Continuous Charge Distributions and Gauss' Law
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Ch 40: Nuclear Physics
Ch 41: Quantum Mechanics
Sections
Internal Energy
Introduction to Heat Capacity
First Law of Thermodynamics
Intro to Thermal Processes
Cyclic Thermal Processes
Work & PV Diagrams

Example #1: A Four-Step Cyclic Process, Part 1

Transcript

Hey guys, let's do an example. 3 moles of a monotonic ideal gas undergoes the following four step cyclic process. Part A, how much heat is transferred in steps one and steps two? B, how much work is done in steps one and two and C, what is the change in internal energy in steps one and two? So for this four step cyclic process all we have to deal with in this part is steps one and two. Step one, gas is pressurized isochorically from some initial pressure to some final pressure. Step two, the gas then undergoes an adiabatic expansion from some initial volume to some final volume. So we have an isochoric process and then an adiabatic process. So let's do this by step. Step one. First step is an isochoric process. What is the definition of an isochoric process? It's one in which there is no change in volume so immediately because there's no change in volume we can conclude that the work done during this step is zero. It's important to recognize that conclusion. Now something else that we can conclude is from the First Law of Thermodynamics. Since no work is being done in this step according to the First Law of Thermodynamics the heat transfer during this step should be equal to the change in internal energy of this step. So in order to figure out the heat transfer we need to figure out what's the change in internal energy. In order to do that we have to figure out the change in the temperature. Now the internal energy of an ideal gas is always going to be F over 2NRT. In this problem we're talking about a monoatomic ideal gas. The number of degrees of freedom F according to the equipartition theorem for a monoatomic ideal gas is just 3, the 3 translational directions. So this is 3 halves NRT. Now what will delta U be? The change in that internal energy. The number of moles isn't changing because the gas is just being compressed, being expanded, undergoing these processes but nothing in the problem indicates that the number of moles should change this is just one sample of gas undergoing a bunch of different processes. The only thing that changing is the temperature so the change in internal energy is just going to be equal to 3 halves times NR delta T. The delta just gets attached to the temperature because it's all that's changing so the question is what is the change in temperature going to be for this process this first step? Well the ideal gas law says that PV equals NRT so I can isolate the temperature as being the pressure times the volume over NR. In this problem what's changing in this first step? The number of moles isn't changing right like I said the volume isn't changing because it's an isochoric process the only thing that's changing is the pressure and since the pressure is in the numerator it's very easy to say delta T is just V delta P over NR. The delta just gets attached to the pressure. Now what is the volume that this isochoric process occurs at? It's not stated in the first step, it's actually stated in the second step. The second step says that the initial volume of step two is 0.004 cubic meters. That means that the final volume of step one is that same volume 0.004 cubic meters. Since the volume didn't change in step one that means both the initial and the final volume are 0.004. So we have 0.004 cubic meters now the pressure dropped sorry the pressure increased from 3 times 10 to the 5 to 5 time 10 to the 5. 5 times 10 to the 5 minus 3 times 10 to the 5 since delta P is positive, delta t is also going to be positive the temperature is going to increase, the number of moles is just 3 and the ideal gas constant is 8.314 and this turns out to be a temperature change of 32.1 Kelvin, this is delta T1. With that temperature change we can figure out what the change in internal energy is using this equation. So Delta U1 is just 3 halves NR delta T1 which is 3 halves the number, the number of moles is 3. Ideal gas constant we know 8.314 and delta T1 is 32.1 Kelvin multiplying this all together we get 1201 joules. That's how much the internal energy changed in step one. Knowing that we can instantly figure out what heat is transferred in step one because that's just going to be equal to delta U1 according to the first law so it'll also be 1201 joules. So those are our three questions. How much work is done? Check. How much the internal energy changes? Check. How much heat is transferred? Check. All those are done for step one. So let's move on to step two now step two, the problem says is adiabatic. What does adiabatic mean? By definition an adiabatic process is one where heat is not transferred. So we know that Q for step two is zero. Just like for step one there's also going to be a consequence to the First Law of Thermodynamics based on the fact that the process is adiabatic. Since it's adiabatic no heat is going to flow and that means according to the first law of thermodynamics that the work during step two is just going to be equal to the change in internal energy during step two. Now we already have an equation that gives us the change in internal energy. The change in internal energy is just 3 halves NR delta T. The question is what is delta T for our step two? That's what we need to know what delta T2 is and just like in step one we're going to analyze this by using the ideal gas law. The ideal gas law says PV equals, there we go a little bit computer glitching, NRT. What changes in this problem? We know that on the right hand side the only thing that changes is the temperature so the right hand side's just going to be NR delta T but what about the left hand side? Well the volume we're told explicitly is changing and we cannot assume that the pressure is constant so we have to assume that they both change so I put a delta for both of them. What delta PV just means is it's P2V2 minus P1V1. They're both the final minus the initial and that will tell me what the change in temperature is. The question is what is the final pressure? And what is the initial pressure? That's what we have to know. Let me put in spaces their minus. Now the volume we're told out right so that that's pretty easy, the volume we are told starts at 0.004 and ends at 0.006. So what we need to figure out is what's the initial pressure and what's the final pressure? N we know is 3, R we know is 8.314. Now let's look at the problem. The initial pressure of step two is going to be the final pressure of step one. If step one ends at 5 times 10 to the 5 Pascals, step two starts at that pressure. Likewise the final pressure of step two is going to be initial pressure of step three. So this is our initial and final pressures they are given to us in the problem we just have to go looking for them and do a little bit of deducting. So the final pressure is 2.5 times 10 to the 5 that's the initial pressure in step three. The initial pressure was the final pressure in step one which was 5 times 10 to the 5. So delta T2 the temperature change in step two is -20 Kelvin. So now that we know that temperature change we can use the equation for the change in internal energy to find our change in internal energy. So delta U2 is 3 halves NR delta T2 which is 3 halves, the number of moles is still 3 like it's been the whole problem, 8.314 and delta T2 -20 Kelvin. Plugging all this into a calculator we get -748 joules as our change in internal energy for step two. This means according to our first law of thermodynamics that the work during step two which is equal to the change in internal energy for step two is also -748 joules. So those are the three things we needed for step two. Now the problem is done you can write all the answers out explicitly for part A, part B, part C if you want. The only thing that we didn't address is whether or not the work was done on or by the gas and whether or not the heat was into or out of the gas. Remember guys work that's positives always work done on the gas, work that's negative is always work done by the gas. Heat that's positive is always heat flowing into the gas, a heat that's negative is always heat flowing out of the gas. For step one to the work was zero so we don't have to worry about that. For step one the Q was positive so that's heat going into the gas. For step two Q was zero so we don't have to worry about that. For step two as we can see right here the work is negative so this is work done by the gas. And given those answers that wraps up the entire problem. Alright guys be sure to watch part two and part three which finishes analyzing this four step process. Thanks for watching.

Example #2: A Four-Step Cyclic Process, Part 2

Transcript

Hey guys, I hope you're able to do this one on your own if not here's a little bit of help three moles of a monoatomic ideal gas undergoes the following four step cyclic process. A how much heat is transferred in steps three and four. B how much work is done in steps three and four and C what is the change internal energy in steps three and four so all we need to concern ourselves with in this problem is steps three and four. Step three the gases then de-pressurized isochorically from some initial pressure to some final pressure and four the gas then undergoes an adiabatic compression back to its original state and its original state is 0.004 cubic meters in volume and 3 times 10 to the 5 Pascal in pressure. Now step one sorry step three this is isochoric what does isochoric mean it means that the change in volume is 0 right off the bat if we know that the change in volume is 0 we know that the work done during this step is also 0. So we know the work done during step three that has a consequence on the first law of thermodynamics as we know as we know it's the change of internal energy is Q plus W, so W is 0 so we can say that the heat transfer during step three is equal to the change internal energy during step three so to find how much heat is transferred in step three we just have to find what the change in internal energy is for step three. We're going to use the ideal the internal energy for an ideal gas to solve this problem F over 2 N, R, T now remember this is a monoatomic ideal gas so the number of degrees of freedom in the eco partition theorem for monoatomic ideal gas is 3 just the 3 translational directions so this is three halves N, R, T now the number of moles isn't changing this gas is just undergoing various thermal processes for a particular sample but nothing about the problem implies that the number of moles is changing no gas is going in no gas is going out so the change in temperature sorry the change internal energy is just going to be three halves N R times the change in temperature the question is what's the change in temperature for step three that's what we have to answer first. The ideal gas law tells us that P.V. equals N R T Because this is an isochoric process the volume is a constant so only the pressure changes on the left and only the temperature changes on the right because the number of moles are not changing so this is N, R delta T. So Delta T for step three is just the volume for step three times delta P for step three over N, R now what volume is the gas at during step three for that we actually have to look at step two step two says the gas ends at 0.006 cubic meters that means that step three starts at that volume and since step three is isochoric the gas stays at that volume so the volume for step three is just 0.006 now what's the change in pressure well we're told that change in pressure explicitly right this is 1.5 times 10 to the 5 Pascal minus 2.5 times 10 to the 5 Pascals the number of moles N is just 3 R is just 8.314 in SI units and this gives us a change in temperature for step three of -24.1 Kalvin. So now that we know delta T we can use our equation and find delta U. So for step three delta U is three halves N R delta T for step three, three halves times three moles 8.314 is ideal gas constant and this is -24.1 Kelvin that's the change in temperature and this works out to -902 joules. So we know the change internal energy for step three and because of the first law of thermodynamics we can find the heat transfer sorry it's not a negative during step three which is just the change internal energy for step three which is -902 joules so we have an answer for part A right how much work it sorry how much heat is transferred in this part A during step three part B how much work was done during step three and Part C what is the change internal energy for step three so now we can move on to step four. Step four is an adiabatic expansion, what is an adiabatic process by definition it's one where heat is not exchanged so by definition the heat transfers for step four is 0 and and like for step three this has an implication about the first law when Q is 0 the first law just tells us that the work is equal to the change in internal energy so in order to find the work we have to find the change in internal energy just like in step three in order to find the heat transferred we have to find the change in internal energy. We use the same change in internal energy equation as we did in step three we just need to find the change in temperature for step four now the ideal gas law tells us P V equals N R T what changes on the left side what changes on the right side well on the right side we already know that T is what changes on the left side we cannot say for certain if one of them remains constant we know for a fact that the volume is changing the problem tells us what the change in volume is back to the original state we also can figure out based on what the problem is saying what the pressure changes to, so we need to say that there is a delta P times V both pressure and volume is changing and this just equals P2 V2 minus sorry V2 minus P1 V1 that's all the delta P times V means its the initial P times V sorry the final P times V minus the initial P times V. So this tells us that delta T is just delta P V over N R so what is delta T for step four well it's going to be the final pressure times the final volume minus the initial pressure times the initial volume, divided by the number of moles three times the ideal gas constant all we have to do is figure out what the final state is and what the initial state is and then we can plug in the numbers. Well the problem explicitly says that the final of step four is the initial state which was 0.004 cubic meters and 3 times 10 to the 5 Pascal. So those we can plug in 3 times 10 to the 5 Pascal 0.004 cubic meters that was the initial state so that is the final step four. Now what is the initial volume and the initial pressure of step four well the initial pressure of step four is whatever the final pressure of step three was and the initial volume of step four is whatever the volume of step three was remember guys that the volume we set for step three because step three was isochoric the volume was just 0.006 cubic meters which was the final volume in step two. So the this is 0.006 that was the initial volume for step four and the 1.5 times 10 to the 5 that was the initial pressure for step four which was the final pressure for step three if we plug all this into a calculator we get twelve Kelvins as the change in temperature for step four. So what does this tell us this tells us that the change internal energy for step four which is three halves N, R delta T for step four that's 3 halves times 3 write the number of moles 8.314 the gas constant times positive 12 Kelvin which is the change in temperature for step four and this is 449 joules that is our change in internal energy for step four and because of the first law of thermodynamics we can immediately find what the work was the work done was during step four. So we answered part A what was the heat transfer during step four part B what was the work done during step four and part C the only thing that we didn't answer in order to answer the question fully is whether the heats were into or out of the gas and whether the works were on or by the gas remember guys that all positive works are work done on the gas all negative work the work done by the gas all positive heats are heats into the gas all negative heats are heats out of the gas for step one we found that the work was 0 so we don't have to worry about whether it's on or by and Q was negative so that's heat leaving the gas first step four as we can see here Q was 0 so we didn't have to worry about whether it's into or out of the gas and the work was positive so this is work being done on the gas. Alright and that wraps up this problem. Thanks for watching guys.

Example #3: A Four-Step Cyclic Process, Part 3

Transcript

Hey guys, let's do an example. 3 moles of a monoatomic ideal gas undergoes the following four step cyclic process. A, how much heat is transferred in the cycle? B, how much work is done in the cycle? And C, what's the change in internal energy in the cycle? So the heat transfer during the cycle is simply the heat transfer during each step that makes up the cycle. The first two answers were found in part one of this problem, the second two answers were found in part two. Q1 was 1201 joules, Q2 was zero, Q3 was -902 joules, Q4 was zero. So this equals 299 joules. That is the total heat transferred during the full cycle. Part B, what's the work done during the cycle? Work one, work two, work three, work four. Works one and two were found in part one, works two and three were found in part two. Work one is zero, work two was -748 joules, work three is zero and work four was 449 joules and this ends up being -299 joules. The one thing I didn't address really quickly heat for the full cycle's positive so this is heat into the gas. The question didn't explicitly ask for that but I wanted to say it just so we cover everything, the work is negative so this is work done by the gas. Now part C, what is the change in internal energy for the cycle? Well there are three ways to answer this question. I want to do all three of them. First it's always zero this is always. Cyclic processes result in a change in internal energy that's zero. Now a second way to do it is to simply add up the change in internal energies for the cycle. The first two were found in part one of the problem, the second two were found in part two. So this is sorry positive 1201 joules, -748 joules, -902 joules and 449 joules and if you add those all up this does in fact equal zero. So that agrees with what we expect, now a third way that we can do this is we can just use the first law applied to the cycle. And we can see from the two answers given above that this is 299 plus -299 which is obviously zero. So you see no matter how you calculate the change in internal energy for the cycle it always comes out as zero and it should always come out a zero because that is a requirement for a cyclic process and we can see that it's absolutely satisfied in this problem. Alright guys that wraps up the third and final part to this problem. Thanks for watching.