1.Introduction and Definitions

1.1About us
Presenters
演讲人
Dr. Patrick O'Malley
帕特里克·奥马利博士
Prof Michael Anderson
迈克尔·安德森教授
Dr. Jonathan Agger
乔纳森·阿格博士
1.2Thermodynamics I Overview
热力学
Presented by Dr. Jonathan Agger
乔纳森·阿格博士主讲
Welcome everyone to our course in Basic Physical Chemistry.
欢迎大家来到我们的基础物理化学课程。
My name is Jonathan Agger and over the next two modules I am going to be teaching you thermodynamics.
我的名字是乔纳森·阿格，在接下来的两 个模块中将由我为你讲述有关热力学的学习。
In each module, I would like you to watch five video lectures, ranging in length from 5 to 15 minutes, and perform a formative 10 question multiple choice test to gain some feedback on your progress.
在每个模块中，我希望你观看五段时间长度在5分钟到10分钟的视频演讲，执行扮演运行形成的10个问题多选测试来获取反馈在你的进步中。
Your assessment for this section of the course will comprise a final 10 question multiple choice test and the virtual practical.
你这门课程这部分的的评估将会包含一个最终的10道单选题目测试和虚拟的实践课。
It just remains for me to wish you good luck with the course and I hope you enjoy it.
再有就是祝你们好运（在这门课程）以及希望你们享受它。
Additional material
附加材料
Important equations for thermodynamics. (pdf文件下载)
重要的热力学方程式
You will encounter these equations throughout the first two modules of the course and may find this document useful.
你会遭遇这些方程式在这个课程的头两个模块并且会发现这个文档非常有用。
Important unit conversion for Thermodynamics SI Unit Exercises.(pdf文件下载)
重要的热力学单元变换 SI单元练习。
SI Unit Exercises(pdf文件下载)
SI单元练习
These exercises are for your own practice only and will not count towards your final grade.
这些联系只是作为你的练习并不会出现在你的最终评分里。
Once you have completed all of the video lectures, you should be in a position to complete the first quiz of the course!
一旦你完成了所有的视频演讲，你应该在完成本课程的第一个小型测验中获得一席之地。

1.3Introduction
介绍
Welcome to Basic Physical Chemistry.
欢迎来到基础物理化学
My name is Jonathan Agar, and for the next two weeks, I'm going to be teaching you Thermodynamics.
我是乔纳森·阿格，在接下来的两周我将教授热力学
The science of thermodynamics was born of the Industrial Revolution in the late 1700s and early 1800s.
热力学这门科学诞生于十八世纪末期到十九世纪初期的工业革命
You have to imagine that up to that point, all heavy duty work was performed by horses.
从这一点你应该能想象到，所有的又脏又累的活都是通过马匹来进行的。
Indeed to this day, we still use horsepower as a measure of the rate at which work is performed.
直至今日，我们仍然使用马力作为衡量工作效率的标准。
The industrial revolution was all about the development and exploitation of a fantastic new invention, the heat engine.
工业革命是关于发展和开发一种非常棒的新发明，热发动机。
Heat engines use fuel to heat a working substance, usually a liquid or a gas, to high temperature.
热发动机使用燃料加热一个工作物质？，通常是液体或是气体，用来升高温度。
The working substance then generates mechanical work in the body of the engine, while transferring heat to a cold sink.
工作物质随后产生机械的工作在发动机内部，同时将热量转移到冷水池。
0:56
Thermodynamics was developed to gain the maximum efficiency from such heat engines, and the science was so successful, that in a very short space of time, steam engines completely transformed the world, from transportation, to manufacturing, to agriculture.
热力学被发展为获得最大量的功效通过热发动机，并且这门学科非常成功，在一个非常短的时间之内，蒸汽机改变了这个世界，从运输业（运输工具）到加工制造业和农业。
1:14
Out of the science of thermodynamics arose four fundamental laws that govern the interrelation and interconversion of heat energy and mechanical energy.
抛开热力学引起的四条治理热能和机械能相互关系和互换现象的基本定律。
1:24
We're going to study those four laws in detail. And in doing so, we're going to learn about concepts such as temperature, internal energy, enthalpy, entropy, and gives energy.
我们将要详细学习这四条定律。我们要学习关于温度，内能，焓，熵和 给予能量？。
1:37
Despite thermodynamics clearly being of immense巨大的浩瀚的 practical importance, it is often perceived by undergraduates as quite abstract, with concepts that are difficult to grasp.
尽管热力学很明显包含非常多的练习重点，它常常被未毕业生们认为(感觉)是相当抽象的，和非常难掌握的概念。
My aim in these lectures is to clearly explain such concepts, and hopefully give you a feel for the useful nature of this highly important science.
我在这门课程的目标是清楚地解释这些概念，抱有希望地给这门非常重要有用的自然科学渲染一种气氛。
1:59
Modern day heat engines, such as the one found in the world's fastest production car, the Bugatti Veyron, are testament to the progress we've made since the early days of the steam engine.
现代热力发动机，比如作为世界上发展最快的产品汽车，布加迪威龙跑车，作为自从蒸汽机时期以来我们所发展进步的证据
In 1829, Stevenson's rocket set the world land speed record at 35 miles an hour.
在1829年，著名的机车史蒂文森火箭号以35英里每小时刷新世界陆地速度记录
Today, the Veyron produces 883 kilowatts, a 1,184 brake horsepower, and has a top speed of 268 miles per hour.
在今天，布加迪威龙跑车产生883千瓦，一个1184刹车马力，并且最高时速228英里每小时
2:28
With regard to the structure of the costs, there are a series of video lectures to watch each week, ranging from around 5 to 15 minutes in length.
关于花费的结构，这里有每周可供观看的一系列视频演讲，长度范围在5到15分钟。
There is a formative, ten question, multiple choice test to complete each week, in order to gain feedback on your progress.
这里有一个构词要素？关键，10道问题，每周单选测试！为了获得你的发展进步反馈。
At the end of the course, there is an assessed, ten question multiple choice test.
在本课程结束，有一个10道单选题目测试的评定。
There is also a virtual practical（实践课） view to try, in which you will use a chemical reaction（化学反应） to calibrate(校准) a calorimeter（热量计/测热计）, and then use the calorimeter to determine（决定/判断） the heat released（释放） in the second chemical reaction.
还有一个虚拟练习……
The practical will also be assessed（评估）. If you have questions, then feel free to discuss them on the forum（论坛）.

Teaching assistants（助手/助理） will be there to help you, and I shall visit the forums regularly（定期地）.

It just reminds me to wish you good luck with the course, and I hope you enjoy it.
祝你好运，希望你享受这门课。
1.4Thermodynamics Definitions
热力学的解释（定义）
0:08
In this video, I shall present（介绍） you with a series of definitions and accepted conventions（会议/惯例/规矩） used in defining thermodynamic problems. This will equip （使具备）you with the language and basic concepts（观念/概念） that underpin（加强…的基础） this course.
0:22
Probably the most common term（术语） you'll encounter（遭遇/碰见/对抗） is thermodynamic system, or just system for short. A thermodynamic system is simply any part of the universe that we wish to study.
0:33
It could be something as simple as a beaker（烧杯） of hot water sitting on a laboratory（实验室） bench（工作台/长凳）. It could be a chemical reaction taking place（take place 发生） in a test tube（test tube试管）. It could be an oil refinery（精炼厂）（oil refinery炼油厂）, or it could even be something as complicated（结构复杂的） as an entire（整个的） galaxy. It purely （纯粹地）depends on exactly what we wish to understand the thermodynamic behavior of.
0:52
Once a system has been defined（解释/下定义）, then everything else in the entire universe is known as the thermodynamic surroundings（热力学环境？）, or just surroundings for short.
1:02
So, together, the system and the surroundings make up the entire universe.
1:10
One important property of the universe is that it is immensely（极大地/无限地）, mind-bogglingly large. The region visible from earth, known as the observable universe, is a sphere of radius 46.6 billion light years. One light year is the distance traveled by light in one year and is equal to 9.46 quadrillion meters. Thus, the observable universe is 880 septillion meters or 880 yotta meters across. This number is so massively large, that I had to actually look up how to quote it properly. And you may well never have even have heard of terms that I have just used. Just look at the number of zeros on the slide, as this gives the best visualization of the sense of vastness that I'm trying to communicate.
1:56
In order to understand why this is important from the point of view of thermodynamics, we need to consider a process that releases energy violently.
2:05
Imagine we'd perform the thermite reaction. This involves igniting a mixture of iron oxide and powdered aluminium to produce iron and aluminium oxide. The reaction is so intensely exothermic, i.e., gives out heat, the white hot molten iron is produced. Indeed it's used to this day to weld railway lines together in remote locations.
2:26
Even though a large amount of energy has been released, mainly in the form of heat and light, the impact on someone in the next town would be negligible. [BLANK_AUDIO]. Even if we consider the most violently energetic processes in the universe such as supernovae, i.e.,. when dying stars explode, the impact on distant galaxies is negligible. Think how many times in your life you've ever noticed the occurrence of a supernova in the sky above?
2:52
So, effectively, the universe is so vast that the energy released by any thermodynamic system has virtually no effect on the surroundings. We're thus able to make the assumption that the temperature and pressure of the surroundings of any thermodynamic system remain constant.
3:12
The boundary between the system and it's surroundings may be actual or notional. Where the system is a reaction occurring in a test tube, the walls of the test tube clearly form an actual physical boundary. Where the system is an entire galaxy, there is no physical boundary and we imagine a notional one.
3:30
The nature of the boundary will determine whether energy and matter can pass from the system to the surroundings and vice versa. And this leads to various different system definitions.
3:42
An open system, by definition, may exchange both energy and matter with its surroundings. Changes of composition are therefore possible. An example of an open system might be a reaction occurring in a test tube.
3:57
A closed system may exchange energy, but not matter with its surroundings. Pressure buildup is a distinct possibility. An example of a closed system might be a reaction occurring in a corked test tube.
4:11
An isolated system may exchange neither energy nor matter with its surroundings. Again, pressure buildup is a possibility. An example of an isolated system might be hot coffee in a thermos flask.
4:25
Simply considering heat transfer across the boundary, leads to two further definitions. A diathermic system allows heat flow into and out of the system.
4:37
And adiabatic system prevents heat flow into or out of the system. This slides show two attempts of different sophistication to create an adiabatic system.
4:51
There are three basic modes of heat transfer, conduction, convection and radiation. The thermos flask limits all three by means of vacuum sealed walls, a close fitting lid, and a silver mirrored internal surface. The plastic coated cardboard cup limits all three, by means of the insulating properties of cardboard, an optional plastic lid, not shown in this model and a shiny white internal surface. The cardboard cup will keep your coffee warm for the thick end of 20 minutes. The thermos flask will keep it warm for hours.
5:24
The state of a thermodynamic system is defined by a series of state functions. These will be discussed in detail on the next slide. Generally, a change of state of a system is accompanied by a change in it's state functions. It is however possible for a system to change state whilst on or more of it's states functions remains unchanged. This leads to three more definitions. And isothermal process implies constant temperature.
5:53
An Isobaric process implies constant pressure, and an isochoric process implies constant volume.
6:02
As I just mentioned, a state function is one which describes the state of a thermodynamic system.
6:08
There are a variety of state functions, some of which are listed here with their associated symbols shown. Functions such as pressure, volume, temperature, mass, and quantity, will probably already be familiar to you.
6:26
Functions such as internal energy, enthalpy, entropy, and Gibbs Energy, may or may not be familiar to you depending on your background in chemistry and of physics. If you don't recognize these terms at the moment, don't worry, as we're going to consider each one of them in detail as the course progresses.
6:47
So, one very important thing to bear in mind with state functions, is that whilst they describe the current state of a system, how the system actually came to be in that particular state, is of no consequence whatsoever to their value.
7:02
Different functions govern how a system changes from one state to another. They are aptly named path functions, and there are only two of them. Heat and Work. By convention, heat is given a symbol q and work is given a symbol w.
7:22
Heat and work are both forms of energy. So clearly, the state of a thermodynamic system is changed by the supply or removal of energy. The difference between the two is that heat energy transfer achieves or utilizes random molecular motion. Whereas work energy transfer achieves or utilizes uniform, concerted molecular motion.
7:48
One final convention in thermodynamics is energy supplied to a system is always considered to be positive. And conversely, the energy removed from a system is always considered to be negative.
8:04
To this end, heat and work are often defined as the heat supplied to the system, qin, and the work done on the system, won.
8:15
Okay, so you should now have an understanding of the various different commonly used definitions and conventions of thermodynamics. In the next lecture we will begin to cover the four fundamental laws of thermodynamics. Starting with the ironically named zeroth law given it was actually the last of the four to be discovered. Arising from a confidence crisis as to the true meaning of temperature. [BLANK_AUDIO]

2The Zeroth and First Laws

2.1The Zeroth Law of Thermodynamics and Temperature
0:00
[BLANK_AUDIO]. In this video I shall present the context for the creation of the 0th Law of Thermodynamics. And describe how it renewed confidence in our ability to measure temperature.
0:19
This context will involve a very brief description of the first three laws of thermodynamics, but we shall study each one in far greater detail later on.
0:29
The story begins around 1850, when a German scientist names Rudolf Clausius took the law of conservation of energy. Which states that energy can neither be created nor destroyed, merely changed from one form into another, and applied it to the idea of a thermodynamic system.
0:47
This resulted in the first law of thermodynamics, which defines change in the state function internal energy of a system.
0:56
Based on energy flow into or out of the system.
1:01
It subsequently then led to the definition of a further state function called enthalpy.
1:07
Following on from initial work by Sadi Carnot on heat engines in 1824, Clausius again around 1850 began to consider the direction in which processes occur spontaneously.
1:21
For example, hot objects always lose heat to their surroundings. A cold cup of coffee will never suck heat in from its surroundings to become hot again.
1:33
These considerations led to the definition of a further state function called entropy under second law of thermodynamics, which defines it and governs its behavior.
1:46
In 1912, another German chemist named Walter Nernst, was considering the behavior of matter, as temperature approaches absolute 0, or minus 273.15 degrees centigrade. The coldest temperature possible.
2:03
Consideration of the entropy of a perfect crystal led to formulation of the 3rd law of thermodynamics, which led to the definition of zero on the entropy scale and thus, it became possible to measure absolute entropy.
2:19
It wasn't until around 1935 that concerns about the nature of temperature measurement began to surface. Clearly, temperature and measurement of temperature underpinned every single facet of this new theory of thermodynamics. And yet, there was a fundamental issue with the way in which temperature was measured. Consider how a thermometer works. Imagine having your own temperature taken by a nurse in a hospital. The thermometer is placed under your tongue, heat flows from you through a thin wall of glass into a reservoir of liquid, which is usually either mercury or red dyed alcohol.
2:56
Thermal expansion causes the liquid to be pushed up a capillary tube with a graduated scale.
3:03
It is the temperature of the liquid, and its ensuing thermal expansion that is used to measure the temperature under your tongue. But at no point, does the liquid actually come into contact with your mouth. This Condray worked to the formal statement of a new law of thermodynamics.
3:22
When two objects are separately in thermodynamic equilibrium with a third object, then they are in thermodynamic equilibrium with each other.
3:34
Whenever two objects are in contact with one another energy will flow between them until they reach a state of thermodynamic equilibrium.
3:42
Upon reaching this state we say that the two objects are at the same temperature.
3:48
So in other words, this new law of thermodynamics states that when two objects are separately at the same temperature as a third object they are at the same temperature as each other. In terms of the thermometer, the two objects are your mouth and the mercury or alcohol, and the third object is the glass.
4:10
Put another way, once the mercury or alcohol level stops moving the liquid, the glass and your mouth are all in thermodynamic equilibrium with each other, and are all considered to be at the same temperature.
4:25
Now this might seem kind of obvious to you, and indeed this is probably why no one had really thought to state this law formally any earlier. However, because temperature's such an important concept, central to the entirety of thermodynamics, this new law was deemed so important that it required a name that would precede the first three.
4:47
Consequently, it was coined the zeroth law of thermodynamics.
4:53
Okay, so now we have a solid basis for the measurement of temperature. In the next lecture we shall consider the first law of thermodynamics, and the concepts of internal energy and enthalpy. [BLANK_AUDIO]

2.2The First Law of Thermodynamics and Enthalpy
0:00
[BLANK_AUDIO]. In this video, I shall present the first law of thermodynamics, and show how it governs the state function internal energy, and how considerations of the first law for isochoric and isobaric processes leads to the definition of the further state function called enthalpy. Around 1850 the German scientist, Rudolph Clausius, took the law of conservation of energy, which states that energy can neither be created nor destroyed, merely changed from one form into another, and applied it to the idea of a thermodynamic system.
0:43
All matter possesses internal energy, symbol U and thus, so do all thermodynamics systems.
0:50
We as human beings have a very intimate relationship with our own internal energy. As we live our everyday lives, we constantly release energy into our surroundings by a radiation of body heat, performing mechanical work going to movement, and even through sound waves when we speak.
1:07
As a result, our internal energy is constantly decreasing.
1:11
We boost our internal energy by keeping ourselves warm through heating systems and by consuming food. If our internal energy begins to fall to dangerously low values, our body responds by sending signals to the brain, causing for example, shivering, stomach ache, and headaches.
1:29
Measuring absolute internal energy is virtually impossible, owing to the complexity of matter. And so it's more normal for us to consider changes in internal energy, delta U.
1:41
What Clausius realized, was given energy can neither be created nor destroyed, any energy that enters or leaves a thermodynamic system must result in a change in the internal energy of that system.
1:56
This is expressed in mathematical form using the conventions laid out in the second lecture as equation one.
2:04
Delta U equals q in plus w on.
2:09
The heat supplied to, plus the work done, on a thermodynamic system, must result in a change in the internal energy of that system.
2:22
Q in, and w on, are positive for energy entering the system and result in an increase in internal energy. Whereas q in and w on are negative for energy leaving the system and result in a decrease in the internal energy.
2:41
As a side note, all the really important equations in this course will be boxed and highlighted in a similar fashion to equation one.
2:51
So thinking in terms of chemistry, it is usually obvious when a reaction releases heat energy. But how might a chemical reaction perform mechanical work?
3:02
The answer is when a gas is released.
3:05
The reaction between zinc metal and hydrochloric acid produces hydrogen gas, as shown in the equation on the slide.
3:13
The gas is released into the solution in the form of bubbles, which then rise to the surface and burst. Hydrogen molecules are thus released into the atmosphere whereby they collide with nitrogen and oxygen molecules, effectively pushing the atmosphere back.
3:31
This process is clearly invisible, however the emerging gas does perform mechanical work in pushing the atmosphere away.
3:40
To quantify the work done, imagine a gas stored in a cylinder with a frictionless, massless piston of cross sectional area A.
3:50
Now imagine the gas expands pushing the piston through a distance d, against an external pressure p.
3:58
The work done by the gas is the force exerted on the piston multiplied by the distance through which the piston moves.
4:07
The pressure on the piston is defined as the force exerted on the piston, divided by its cross-sectional area. And thus, the force exerted on the piston is equal to the pressure on the piston, multiplied by its cross-sectional area.
4:23
If we now substitute this expression for the force exerted on the piston into the expression for the work done by the gas, we find that the work done by the gas is equal to the pressure, multiplied by the cross section area of the piston, multiplied by the distance through which the piston moves.
4:42
But the cross-sectional area of the piston, multiplied by the distance through which the piston moves, is equal to the change in volume of the gas delta V.
4:53
So, the work done by the expanding gas is equal to the pressure multiplied by the change in volume of the gas.
5:01
Hopefully this intuitively makes sense. The greater the volume of gas released, the more work needs to be done to push the atmosphere away. And the greater the pressure, the more work needs to be done as it is harder to push the atmosphere away.
5:16
Finally when a gas is released by a chemical reaction, work is being done by the thermodynamic system. And so by convention, the work done on the system is negative. And thus w on equals minus p delta V. This is a very important result, and we shall denote it equation two.
5:40
If we now revisit the first law of thermodynamics, delta U equals q in plus w on, as we have just shown, w on is equal to minus p delta V. Thus, the first law becomes delta U equals q in minus p delta V.
5:59
Expansion work therefore depends on pressure and change in volume. There are thus two ways to make the expansion work zero.
6:09
The first way to make expansion more equal to zero is to perform the reaction under conditions of zero pressure.
6:16
When p equals zero, minus p delta V equals zero. No expansion work is done, because the gases expanding into a vacuum. And thus the first law of thermodynamics becomes simply delta u equals q in.
6:33
The change in the internal energy of a system is equal simply to the heat supplied to the system.
6:40
This is called free expansion.
6:43
However, given the number of human beings that have ever been into space, relatively very few experiments are carried out under such conditions.
6:53
The second way to make expansion work equal to zero is to perform the reaction at constant volume.
7:00
On the isochoric conditions, delta V must be zero and this minus p delta V equals zero.
7:09
No expansion work can be done, because any gas produced simply cannot expand.
7:14
Under these conditions, the first law of thermodynamics simply becomes delta U equals qv. The subscript v denotes the heat supplied q in at constant volume.
7:27
This equation is important and is given as equation three.
7:33
The major consequence of performing reactions at constant volume is the risk of serious pressure change. Consequently, constant volume reactors are invariably made of thick stainless steel and have lids secured with massive metal bolts.
7:48
Such engineering is extremely expensive. And consequently isochoric processes are most often the domain of the heavy chemical industry and chemical engineering.
8:01
Most chemists on the other hand perform chemical reactions in glassware under isobaric conditions. In this case, the first law of thermodynamics does not simplify.
8:14
It was therefore necessary to create a new state function called enthalpy, symbol H.
8:21
Enthalpy is defined as H equals U plus pV.
8:27
And therefore the change in enthalpy, delta h, equals delta u plus delta pv.
8:36
Since pV is a product, we must use the product rule to differentiate it.
8:43
Thus, delta H equals delta u plus p delta V, plus V delta p.
8:51
But, for an isobaric process, pressure remains constant, and therefore, delta p equals zero.
9:02
Thus, the V delta p term disappears to give delta H equals delta u, plus p delta V.
9:11
Now it should hopefully be apparent why enthalpy was defined in this specific way.
9:17
Because delta u plus p delta V is the left hand side of the second equation on the slide.
9:25
We can therefore substitute delta H for delta u plus p delta V. And the first law of thermodynamics at constant pressure becomes delta H equals qp. The subscript p denotes the heat supplied q in at constant pressure. This equation is very important and is given as equation four. So in summary, the first law of thermodynamics arises from the application of the law of conservation of energy to thermodynamic systems.
9:56
The expansion work done by a gas is p delta V, but this is the system performing work, and so in the context of work performed on the system, this becomes minus p delta V.
10:10
The heat supplied to a thermodynamic system at constant volume is equal to the internal energy change of the system, delta U. Well, so heat supplied to a thermodynamic system at constant pressure is equal to the enthalpy change of the system, delta H.
10:26
The former is very much the domain of heavy industry and chemical engineers, whilst the latter is the domain of chemists. In the next lecture, we shall examine the important concepts of reversible expansion work. [BLANK_AUDIO]

3Reversible Expansion and Heat Capacity

3.1Reversible Expansion
0:00
[BLANK_AUDIO]. In this video I shall present the concept of Reversible Isothermal Expansion of an Ideal Gas, and show how it leads to the maximum amount of work being performed by the gas.
0:22
We start with the ideal gas equation. pV equals nRT. Where p is the pressure of the gas, V is the volume of the gas, n is the quantity of gas measured in mols, R is the ideal gas constant equal to 8.314 joules per kelvin per mol and T is the absolute temperature measured in Kelvin.
0:50
We can rearrange the ideal gas equation to calculate the pressure of a gas. p equals nRT over V.
0:59
Now let's consider 1 mol of an ideal gas at 298 kelvin.
1:04
Substituting these values into our equation for pressure gives p equals 2477.572 joules divided by the volume.
1:17
Using this equation. Pause the video and calculate the pressure for one mol of an ideal gas at 298k at the following volumes. 1, 3, 5, 7, 9, and 11 meters cubed.
1:37
Make a list of the volumes and their associated pressures. Once you have done this, unpause the video. [BLANK_AUDIO]. Okay, so welcome back. You should have calculated 2,478, 826, 496, 354, 275, and 225 pascals respectively, for the 6 volumes mentioned.
2:10
If you didn't get these answers, don't worry overly much now. Just write them down. Keep this list handy as you're going to need it. [BLANK_AUDIO]. Now imagine this 1 mole of gas at 298 kelvin is stored in a cylinder with a frictionless, massless piston. Exactly the same as we used to determine expansion work in the last lecture.
2:36
If the external pressure is 2,478 pascals, then the gas will occupy a volume of 1 meter cubed, as per the calculations you have just performed.
2:47
What we now need to consider is what happens if the external pressure is suddenly dropped from 2,478 down to 826 pascals.
3:00
How will the gas react to the pressure drop?
3:04
The answer is that the gas will expand by two meters cubed to occupy a new volume of three meters cubed. Again, as shown by your calculations.
3:15
The graph shows the pressure of the gas against its volume.
3:19
As described, the gas occupies 1 meter cubed at 2,478 pascals of pressure, shown by the red dot.
3:29
When the pressure suddenly drops to 826 pascals, as shown, the gas will quickly expand to a volume of 3 meters cubed.
3:41
As the gas expands, it is working against the external pressure of 826 pascals, so the expansion work done, is p delta V, or the area of the orange rectangle, i.e, 826 pascals multiplied by 2 meters cube, which equals 1,652 joules.
4:06
Now suppose external pressure is further dropped to 496 pascals.
4:13
Your calculations should show that the gas will quickly expand by a further two meters cubed to occupy five meters cubed.
4:21
As the gas expands, it is now working against the external pressure of 496 pascals. So the expansion work done is the area of the second orange rectangle, i.e., 496 pascals multiplied by 2 meters cubed, which equals 982 Joules.
4:42
Further drops of the external pressure to 354, 275, and ultimately 225 pascals will result in 3 further 2 meter cubed expansions of the gas to 7, 9 and finally 11 meters cubed. The work performed during these expansions in each case is equal to the area of the orange rectangles swept out by the expansion, i.e., 354 pascals multiplied by 2 meters cubed, or 708 joules. 275 pascals multiplied by 2 meters cubed, or 550 joules,
5:28
and 225 pascals multiplied by 2 meters cubed, or 450 joules, respectively.
5:36
The total work done by the gas in it's expansion from 1 to 11 meters cubed is the sum of the 5 bursts of expansion work i.e., 4,342 joules.
5:51
All six points during the expansion are now shown on the graph.
5:55
You can see that they all fit the idea gas equation as they lie perfectly on the curve p equals nRT over V.
6:03
Given the gas's performing work as it expands, the system is performing work rather than work being done on the system. And so the work done on the system, w on is negative, and equal to minus 4,342 joules. Okay, so hopefully, the quantification of the work performed during this expansion now all makes sense.
6:31
But how could we change the expansion in order to make the gas perform more work?
6:37
The answer is to drop the pressure more gradually from 2,478 pascals down to 225 pascals.
6:46
If the pressure is dropped, so as to allow the gas to expand from 1 to 11 meters cubed in jumps of only 1 meter cubed each time, instead of 2 meters cubed, then in the Tan expansion bursts, the gas will perform a total of 5,004 joules of work. And the work done on the system, w on, is minus 5,004 joules.
7:12
If we persist with this idea and allow the gas to expand in 20 0.5 meter cubed bursts, then it will perform 5,428 joules of work. And w on equals minus 5,428 joules.
7:28
Notice how the total area of the 20 orange rectangles is becoming a better and better approximation to the area under the curve of p equals nRT over V. 40 expansions of 0.25 meters cubed leads to W on being minus 5,672 joules. 80 expansions of 0.125 meters cubed leads to w on being minus 5,803 joules and yes, I did work out the expansion work of each one individually.
8:05
If we allow the gas to expand by infinitesimally small amounts then this is called reversible expansion and w on is minus 5,941 joules. This is the maximum possible work that the gas can perform and the work done is represented by the area under the p equals nRT over V curve.
8:30
Obviously, a process of infinitesimally slow expansion would take an infinitely long period of time. And thus, reversible expansion work is only a theoretical, if extremely useful construct. W on for reversible expansion is calculated as follows.
8:53
W on is minus p delta V. But, pV equals nRT. Which can be rearranged to give p equals nRT over V. So, w on equals minus nRT over V dV.
9:13
If we know consider expansion from an initial volume, Vi, to a final volume, Vf, at constant temperature then w on equals minus nRT times the integral between Vi and Vf of 1 over V dV.
9:36
The integral of one over x dx is a standard integral quoted as log x plus c.
9:43
So, the integral of one over v dv is log v.
9:50
Substitution of the limits into the equation.
9:53
And grouping of the logarithmic terms using the logarithmic identity shown, yield equation five, for the work done on the system during the reversible isothermal expansion of an ideal gas. w on equals minus nRT log Vf over Vi.
10:17
Substitution of the values 1 mol, 8.314 joules per Kelvin per mol, 298 Kelvin and expansion from 1 to 11 meters cubed leads to the aforementioned value of minus 5,941 joules for w on.
10:38
In summary, reversible expansion is an infinitely slow theoretical construct.
10:46
But it is extremely useful as it enables determination of the maximum possible work that can be done by an expanding gas. You can imagine trying to optimize steam engines. This was an incredibly important property.
11:05
In the next lecture, we shall look at heat capacities of materials. [BLANK_AUDIO]

3.2Heat Capacity
0:00
[BLANK_AUDIO]. In this video, I shall present the concept of heat capacity. I shall review the various definitions of heat capacity, and consider the ramifications of carrying out processes at constant volume and constant pressure.
0:24
When a substance is heated, its temperature rises. The magnitude of the temperature change depends on the object's heat capacity.
0:33
When an object is heated, its temperature may rise rapidly, in which case it has a low heat capacity. Or it may rise slowly, in which case it has a high heat capacity.
0:44
Professional chefs invariably have a side towel, tucked into the waist bound of their apron, to enable them to pick up metal handled pans full of boiling water, as metals get hot very quickly.
0:56
Observe these two objects being heated at similar rates. [BLANK_AUDIO]. The object on the left heats up quickly, and therefore has low heat capacity. The object on the right heats up more slowly, and has high heat capacity.
1:18
Heat capacity is defined as the energy required to raise the temperature, of an object, by one degree centigrade or one kelvin, and it is measured in joules per kelvin.
1:32
Heat capacity, C, is equal to the heat energy supplied to an object, q, divided by it's corresponding change in temperature delta T.
1:43
This important equation is given as equation six. Heat capacity is an extensive property.
1:54
That is to say, its value is dependent on the amount of substance under consideration. For example, it will take significantly less heat to raise the temperature of one drop of water by one degree, than it will to raise the temperature of an entire swimming pool by one degree.
2:12
The swimming pool will therefore have a far higher heat capacity value, than the droplets of water, even though both are made of the same substance.
2:21
There are two ways to make heat capacity values intensive, ie independent of amount of substance.
2:30
Heat capacity divided by mass of substance is termed specific heat capacity, symbol Cs, units joules per kelvin per kilogram.
2:43
Alternatively, heat capacity divided by amount of substance, is termed molar heat capacity, symbol c m units joules per kelvin per mol.
2:58
Both specific heat capacity, and molar heat capacity, are intensive quantities.
3:08
The following slide will showcase a range of different materials and show their specific heat capacities in joules per kelvin per kilogram.
3:17
From lowest through to highest, the materials are gold, mercury, copper, bone, diamond, graphite and sulphur, sugar, paper, plastic, wood, and rubber.
3:45
Notice how the metallic substances have low heat capacity. This is in part due to their density, but also because metals absorb heat rapidly and heat up rapidly.
3:57
Materials that we would consider insulators, such as paper, plastic, wood, and rubber, have far higher heat capacities.
4:11
Isochoric heat capacity, Cv, is heat capacity recorded at constant volume, and is defined as the heat supplied at constant volume, qv, divided by the change in temperature, delta T. However, in lecture four, covering the first law of thermodynamics, we proved that qv is equal to the change in internal energy, delta U. And thus, the isochoric heat capacity is the change in internal energy divided by the change in temperature.
4:47
This result is quoted as equation seven.
4:52
Isobaric heat capacity, Cp, is heat capacity recorded at constant pressure, and is defined as the heat supplied at constant pressure, qp, divided by the change in temperature, delta T. However, again, in lecture four covering the first law of thermodynamics, we proved that qp is equal to the change in enthalpy, delta H. And thus the isobaric heat capacity is the change in enthalpy divided by the change in temperature. This result is quoted as equation eight.
5:29
Thus, the isochoric and isobaric heat capacities represent the slopes of the graphs of internal energy U and enthalpy H against temperature respectively.
5:46
Now let's consider the relationship between the molar isobaric and isochoric heat capacities for an ideal gas. By definition, H enthalpy is equal to U plus pV.
6:00
Thus, for one mol of substance, molar enthalpy is equal to molar internal energy, plus pressure multiplied by the molar volume.
6:12
For an ideal gas, pV equals nRT. And thus for one mol of gas, pV molar equals RT.
6:24
If we now replace pV molar with RT in the second equation on the slide, we get that the molar enthalpy is equal to the molar internal energy plus RT.
6:40
At 298 kelvin, RT is approximately 2.5 kilajoules per mol, and this is not negligible for a gas.
6:54
If we now assume a change of temperature, delta T, then the molar enthalpy, our molar internal energy, will change according to delta H molar equals delta U molar plus R delta T.
7:13
Dividing through by the change of temperature, delta T, yields delta H molar over delta T equals delta U molar over delta T plus R. But from equation seven and eight, this gives Cp molar equals Cv molar plus R, where r is 8.314 joules per kelvin per mol.
7:41
This equation is given as equation 9. Thus, the isobaric molar heat capacity of an ideal gas is greater than its isochoric molar heat capacity by 8.314 joules per kelvin per mole.
7:59
In practice, values may differ from this owing to the deviation from ideality of real world gases.
8:09
In the next lecture, we will begin to look at the second law of thermodynamics and introduce the concept of entropy. [BLANK_AUDIO]

4.2Thermodynamics I Recap
0:00
[BLANK_AUDIO]. Hi everyone. So we've now reached the end of week one. So far, we've covered the zeroth and first laws of thermodynamics, which has given us a solid interpretation of the concept of temperature, and develop the concepts of internal energy and enthalpy, which are vital for considering energy transfer processes on the isocoric and isobaric conditions.
0:29
We've also covered the idea of reversible expansion, and you should hopefully understand the importance of this in maximizing the mechanical work that can be achieved in a heat engine. Finally, we've looked to how matter responds to heat energy transfer, and define various forms of heat capacity. The ten question, multiple choice test hopefully gave you useful feedback to check understanding and help you to improve where necessary. Hopefully you're making good use of the forum, where there are teaching assistants, and often myself, available to help with any queries or discussion points you may have.