First Law of Thermodynamics: Classical Definition, Cyclic Integral, and Energy Analysis

we've already taken a pragmatic approach to coming up with a statement of the first law of thermodynamics

if you recall we came up with a statement for an isolated system and for a closed system that was in

terms of the works and heat transfer and the change in energy before we start doing examples using the first law of

thermodynamics i want to go back to the very classical definition of the first law

so the classical definition is very simply for a closed system the cyclic integral of heat transfer is

equal to the cyclic integral of work now a closed system we understand a closed system is

a system that permits the transfer of energy across the boundaries but not the passage of mass

so the other piece we have to understand is what is a cyclic integral so to demonstrate what a cyclic integral

is i've drawn this very simple diagram and i've drawn a point one and a path with a whole bunch of numbers here

a cyclic integral is an integral evaluated for a thermodynamic cycle so this is interesting so cyclic

integral we've got that linked now now we've introduced another term the thermodynamic cycle

so a thermodynamic cycle is one where the mechanical and thermodynamic properties return to their

starting point so this diagram demonstrates or illustrates what a thermodynamic cycle

is so we've got this point one on a pv diagram and i've indicated the path by the little arrow and it just

follows the path and comes back to the beginning okay something like a reciprocating

engine is a great example of a cycle that keeps returning to where it starts right it goes intake compression power

exhaust intake compression power exhaust so it keeps repeating itself so

this isn't really a foreign concept to anybody now how do we actually do the evaluation

well when i say um uh it's an integral evaluated across the cycle that means that it's got to be a path

integral and in calculus you you define that you're looking at a path integral by

putting this little path symbol on the integral so the path integral of the heat transfer

is simply a sum of all of the segments so the the integral not the path but the

integral from one to two plus the integral from two to three and on and on and on until we're

at the integral from n back to one which is the last one and the path integral of the work across

the same process is one to two of the work two to three of the work

and on and on all the way to n to one that gets us back to the starting point so what the first law of thermodynamics

tells us the classical statement is that these are the two integrals we're interested in

and the path integral of heat transfer is equal to the path integral of the work so the path

integrals are analogous to the cyclic integrals so this is the classical statement of the

first law so the next thing we're going to do is show how we get back to that other

statement we had by starting with this statement of the first law of thermodynamics

let's do this evaluation on a very simple system so i'm drawing another pv diagram between two states

one and two and i've got three different paths that can be taken to go from state one to

state two or vice versa so path a leads from one to two and then paths b

or c lead back to path one so either root a b or a ac is a thermodynamic cycle so as long as

we take the cyclic integrals we should have you know a definition of the first law so let's look at cycle a b

cycle a b is um the integral of heat a plus the integral of heat b is equal to

the integral of work a the integral of work b cyclic integral heat transfer cyclic

integral work it's just the segments of the complete path added up

now let's look at cycle ac same thing integral 1 to 2 on path a but this time integral 2 1 on path c for heat transfer

same thing for work one two on a two one on path c now if we subtract these two statements

the a's all disappear because q a q a w a w a both from one to two all gone so the only thing that's left

is integrals from two to one along paths b and c so the integral of heat transfer b

minus e transfer c because we're subtracting is equal to the work b minus the work c

we can rearrange that and put everything from path b on the left everything from path c on the right

everything is evaluated across the same uh well from the same two states two to one

and so because these integrals are equal it also means that the integrands must be equal

if the integration across two states of an integral is the same if they're equivalent then

it has to mean that there's equivalence between the integrands so that means that we can say that this

del q minus del w on path b is equal to del q minus del w on path c and this is a constant because we could

have done any number of other paths from two to one we would have got the same thing we have a whole list of these

and what the they equal is a constant and this constant is the difference in the energy level

from state one to state two or in this case two to one okay so this is a very important

statement we've used the classical definition of the first law we've conducted the cyclic integral

we've evaluated them for these three are these two different possible thermodynamic cycles

and we've come up with this statement that these quantities independent of the path are a constant

as long as you're working between the same two states and that's equal to d e

so just to clean up the connection between the classical statement of the first law

and the pragmatic approach that we use to to develop the first law uh we have the cyclic integral of work

equals the cyclic integral of heat transfer classical definition we've shown that

um if you conduct or evaluate the cyclic integrals on a thermodynamic cycle that there is equivalence

and the difference between you know the different types of paths is simply equal to the change in the

level of energy between the two states we've shown that now just a note and i've already

mentioned this in a previous lecture you've got to keep in mind

that the integral of del w is an amount of work it's not w2 minus w1

whereas if you did the integral from one to two of d e or du or dh or anything else

then it is a change in the level of energy or or whatever other quantity the

distinction though with heat and mass transfer is that amounts of heat and amounts of work

result in changes in the level of energy so we have to really keep in mind that i use this dell

to indicate that we're finding an amount of work not a change this is really meaningless

right here has no meaning whatsoever analogously for heat transfer so let's just you know close this little

segment by looking at different forms of the first line we've already been here about

eight lectures ago we looked at something that looked exactly like this so for a convention where

we've got you know any kind of closed system we've got our control volume drawn around it

and our convention for heat is positive heat goes in positive work goes out and we've got an

amount of mass and an energy level that changes depending on what we do with the

boundaries so using that convention q minus w is equal to changes in energy and

remember the energy is captured by internal kinetic and potential this is in joules

if we uh differentiate with respect to time then we see the rate at which you add

heat minus the rate at which you do work is equal to the rate of change of energy in the system

this is a very useful form of the first law it's the one that i like to use the most

and another form if you divide this statement by the mass or by the mass flux

you actually i put it in terms of lowercase letters so little q one two little w one two is

equal to lowercase e one two and this just puts it in terms of kilojoules per kilogram

okay so you solve it you solve the amounts of energy and work required to you know make some change to a kilogram

of the substance and then you find the total amount of heat and work by multiplying those values by the mass

of the system