Understanding Work in Thermodynamics: Force, Displacement & Energy

okay for today's lecture we're going to talk about the concept of work the concept of work is very important

and it's related to energy if i was standing in front of a class of students i would ask

you know why are we so consumed with characterizing the level of energy of certain systems

whether it's kinetic energy potential energy or other forms of energy all through your high school physics and

chemistry and so forth you've been you've been you've been taught how to characterize the levels of

energy now the reason that this is of interest to us is because

the fundamental measure of energy is work so if we understand the energy level of a system

at two different times or after something has happened if we really understand that system then

we can exploit the change in the energy level in the form of work and work is what's

important to us this is how we this is how we do things this is how we make things happen

so if we just look at the very basic definition of work and this is the one you've been studying all along

it's simply force times distance we're not going to change this this is the definition of work

so if you look at a very basic system we've got a block resting on the ground we're applying a

force in the positive x direction and we're moving the box from this position to

this position so we could write that as a small amount of work

and i'm using this delta right here to indicate that this is an amount of something

not a change in something this is an amount so in this case it's a small amount of

work that's being done when we apply a force to this box and it moves a distance dx in this case it's

infinitesimal if we do the integral of that from the starting position to the end

position we've got the integral of f dx and x2 minus x1 is d so the work that's done is f times d and

that's in units of newton meters or joules now this is very important what i want

to look at now is is the work for different processes okay and the first process we're going to

consider is the compression of the gas and this is a very important process in thermodynamics we're actually going to

we're gonna bring this up a lot as the course progresses so it's really important to understand this one

so let's just take a circular cylinder so i've got i'm showing it a circular this is the cylinder

we've got a piston inside of the cylinder and it encloses the gas okay so the gas can't leak out it can

either be compressed or it can expand so we're going to apply a force to this piston and we're going to do work

on the gas and the convention in this case is that work done on the gas is going to be

positive so work done in the direction of the force is positive if the piston is backed up to here then

the total length of the cylinder is l and the coordinate system shows x going in the compression direction so we

can characterize the work for this situation as the amount of work done to compress this gas is

this force which is clearly changing under compression multiplied by the distance

over which the force is applied now the way that a gas um applies a force on the piston so

we're going to look at the opposite to the force just to understand how this force grows as we move

in the positive direction so if we consider this side the way that a gas opposes compression

is by applying its pressure over an area in this case the area is the face of the piston looking this way

and the pressure is something that's growing as we compress but we're going to replace this force

by the pressure applied over an area okay and the reason why we don't have to

change the sign is because pressure is a positive number and a is obviously a positive number

because it's the area the other thing we need to know is how the volume is changing through this

process so we'll just write volume as equal to the maximum volume or the maximum

cylinder height subtract the position that the piston is in multiplied by the area so that's the

instantaneous volume of the system at some position x if we take the derivative of that then we have dv

is equal to minus a dx because l is a constant so when you take the derivative

it gets wiped out if we now replace this dx by this dx which is negative dv upon a then we get that this

small amount of work del w is equal to minus pdv and then if you make take the

integration of that you come up with the amount of work to go from

one volume to another volume is equal to the negative integral pdb and this is in joules

so this was derived starting from this very basic uh equation force times distance

integrating what we understand about the problem now one of the important things is that

we said that the positive convention for work is in the direction of positive x so when this is happening

dv is actually negative because the volume if work is being done on the system the

volume is getting smaller with increasing force applied to it if the volume is getting smaller then that

means the db is negative which makes the integral positive for compression work if the system

was allowed to expand against the piston then the volume would be growing which means negative work would be being

done and that would just mean that according to our convention the work is going in the other direction which is

also correct okay so this is kind of the final um the final expression

of what the work is to compress a gas inside of a cylinder and we're going to come back to this many times in

the course so we've spoken about the work required to compress a gas

let's look at a couple of other processes where work is also done we can quantify in a similar way

so the second example that your textbook covers is the work to extend a cable or a rod and in

this case let's just imagine something attached to a wall let's say it's a rod

and we're applying a force to stretch this rod the volume is conserved in this case

because we're not adding and removing volume so if you stretch the rod it's going to

appear a little narrower a little longer to keep the volume once again the amount of work that's going to be done

is this applied force done over a distance and analogous to the compression of a

gas when you're looking at extension or compression of a rod

the pressure is replaced by this normal stress component which you're going to talk about more on mechanics and

materials and that acts over an area so this is what replaces the force

and then once again by the distance in this case instead of the volume we introduce the strain rate the epsilon

which is the change in x divided by its length and then we can replace dx in this

expression by xd epsilon and that's multiplying the force which is normal stress times a

and if you combine a and x it'll always equal the volume because the volume is conserved so

the work in this case is uh sigma x v d epsilon across two different strains

and once again this is in joules and there's a couple of other examples that your textbook also talks about

the work from surface tension charged particles and there's a few others i think there's about six of them

but the whole idea is that the generalized form of the work always starts with this

expression where an amount of work is done when a force is applied over a distance and i'm just going to indicate

you've got a general displacement and you've got a general force and we can use this expression to derive

all the different forms of work all of which come out to be in joules of energy