Specific Heat, Ideal Gases & Thermodynamic Properties Explained

we've managed to get through the three regions on the phase diagram for water and

actually that's completely analogous to the phase diagram for other substances like

refrigeration 134a which we'll also be using what i'm going to do now is we're going

to introduce another property and this is another property that that really helps us to understand

how a substance responds to heating or pressurization so the property we're going to introduce

is called specific heat and by definition the specific heat is the limit as delta t

goes to zero of an amount of q denoted by this del divided by m delta t and the units for a specific

heat are kilojoules per kilogram kelvin or kilojoules per kilogram degree c because an increment of kelvin is the

same as an increment of degree c what this property means is it's the amount of energy required

to raise one kilogram of a substance by one degree c okay and this is what the unit actually

tells us it's the amount of energy required to raise a kilogram of the substance by a degree

now the value that this property takes depends on how the heat is added and so in the next couple of slides i'm going

to show you two different ways that we can add the heat and this provides us the definition

of two different types of specific heat specific heat at constant volume and specific heat at constant pressure

the first kind of process we're going to look at for specific heat is a constant volume process so in other

words we're going to keep the volume fixed so the experiment we're going to look at

is a piston cylinder device so cylinder piston we've got the control volume inside

containing some kind of a substance uh it's it's a closed system so it's going to

have a constant mass and it's got an energy level which is going to change depending on

on what we do with the boundaries so if we write our convention for this system as del q

an amount of q going in is positive and an amount of work leaving is positive now to keep the

volume fixed what we've done so we're keeping volume fixed

we've put a pin in the piston so this way the piston can't move which means that the volume can't change

so dv is zero which means that the amount of work being done

by the substance on the surroundings has to be zero because the integral of pdv will be zero

because dv is zero so i'm just indicating that here that del w is zero

and i'm going to indicate in our statement of the first law which states that

del q in minus del w according to our convention is equal to a small change in

energy and if the system is considered stationary which means that changes in the kinetic energy and the

potential energy are zero over this process so the system is sitting still

and it's sitting still at the beginning and the end of the process and it hasn't changed its position from the beginning

to the end of the process so that way kinetic and potential effects are zero which means the d e can

be approximated as d internal energy and we can write this in its intensive form as

m d small u so in this case the amount of q that we have to add is just

m times d u and if we insert that into our definition of the specific heat so this piece right here

we're replacing del q which was in the numerator with mdu that way the m s disappear

and if we take the limit as del t goes to zero we end up with d u d t so our first definition of

specific heat specific heat at a constant volume so c sub b is equal to d u dt at constant v

so we're just indicating constant b with that line second experiment we're going to do to

find a specific heat is we're going to take the the exact same arrangement where we've got the

piston cylinder device but in this case we're going to add heat at a constant pressure

so we've removed the pin and in this case we've kept the same convention positive q goes in positive work leaves

the system so the first law still says del q minus del w

is equal to d e and d e under the condition that the system is stationary is still d uppercase u which is md

now in this case we can't cross the workout because as you add heat if the piston is unpinned and the

pressure is staying constant if it's a compressible substance the piston will start to rise

okay so we have to keep both the work and the heat transfer so del q comes out to be mdu

we'll throw the work on the other side and we have dell w work done at a constant pressure well

work done at any pressure is pdv work done at a constant pressure is mp dv okay

if we now take this term right here substitute it for the work and add them together we get del q

is equal to m d u plus p dv and if you remember um u plus pv we introduced this new property

called enthalpy so du plus pdv is mdh so m times the change in enthalpy

so if we take this m times the change in enthalpy as our del q insert it into the definition of

specific heat let the limit of del t go to zero we end up with dh by dt

and so our second value of specific heat the specific heat at constant pressure

is equal to dh by dt and we just indicate that it's at a constant pressure

so let's make a couple of observations about specific heat we've got specific heat at constant

pressure specific heat at constant volume for liquids or sub-cooled liquids

compressed liquids however you want to look at it but for for a substance in its liquid

state we know that the amount of work because it's the pressurization

is a lot smaller than the amount of heat or the impact of the heating on the specific heat so in this case

cp and cv will equal just a single specific heat because pressurization doesn't lead to a change

in volume basically there's the the work is minuscule compared to the

amount of change you get in the energy level due to heating so the specific heats for a liquid are

basically the same thing and often it's only cp that's given in the tables for liquids

or simply denoted as c now for gases and vapors if i was if i was doing this

face-to-face and i may ask you this in subsequent um face-to-face interactions i would ask

you why is cp greater than c b for a gas okay now the answer to this is

that when you do the process under constant volume

all of the energy you're adding goes into raising the internal energy level of the substance

and that's what gives you the one degree temperature change you need to change you need to add

enough energy to make the internal energy change enough to give you a one degree

change in temperature when you're doing this process under constant pressure

some of the energy that you're putting into the system is going into warming the substance

but some of it is actually going into raising the piston or expanding the system

now that amount that's going into raising the piston is not changing the temperature of the substance

so what this means is that you need to add more energy in order to get that one degree

temperature change because some of the energy you're adding is actually going into expanding the

system so for gases and vapors cp is always bigger

than cv now the information for cp and cv is given in table a2 in your appendix

and this is for specific heats of many different substances so continuing with the use of specific

heat what i want to do is i want to revisit the compressed or sub-cooled liquid region

and i've just indicated again that that's region one that we have on the chart

it's to the left of the dome bounded by the critical point at the top and the reason why i want to go back to

the subcode liquid region is that the sub cooled liquid is an interesting phase of a substance because it's an

incompressible phase okay so this this actually has an impact on specific heat that we've already spoken

about now what we know about the sub cooled region

is that we've already taken specific volume and said that because the specific volume is so weakly

dependent on pressure that specific volume as a function of pressure and temperature can be

approximated on the saturated liquid line which is this

so on the saturated liquid line at the temperature t and we've already done an example showing that

it almost is exactly the same um we've done the same thing with internal energy we kind of argued that

because it takes a lot of pressurization to make any change in the volume

that basically by just pressurizing a sub cooled liquid you're not doing any work

which means you're not changing the energy level there's no temperature change when you compress

uh liquids even over huge amounts of pressure so we actually also argued that internal energy as a function of p and t

can be approximated as internal energy on the saturated liquid line at the

temperature and we've done an example of that as well to show that that is sound now what about other combinations like u

as a function of v and t well for this let's take the total derivative

so d u can be written as d u dv times dv plus du dt times dt and again because it's a compressed

liquid we know that there's no change in specific volume so that goes to zero

which just leaves the second term we've just evaluated that d u d t is c v or c because c p and c d are the

same thing so d u is c d d t or simply c d t now

the information this gives us is that we don't necessarily need to look up different

values of u in order to see what the energy difference is from one temperature to another if we do this

integration across two internal energies it's equivalent to

integrating this across two temperatures so cdt so if we're interested in finding a

change in internal energy of a substance that's a sub cooled liquid

then we just have to look at the change in temperature multiplied by its specific heat and as i said before

there's only going to be one specific heat for compressed liquids because cv and cp

would be exactly the same thing so this actually gives us it helps us because it simplifies the process

of finding delta u and this is quite often the case heat exchangers any kind of system where

you're warming a liquid this would be a lot easier way of understanding how much energy needs

to go in it with avoiding having to look up values in the charts the only other property we haven't

considered yet in the so-called liquid region other than simply looking it up is the

enthalpy so we've got a simplification of specific volume as being v

on the saturation line at t same thing for internal energy u on the saturation line

at t um for h we want to consider the same thing we did for

internal energy so h as a function of v and t is y well let's consider that h is a

combined property u plus pv and then considering the previous analysis

if we look at change in h it would be a change in u plus v times a change in p

now why is it v because v is basically a constant when you're looking at the subcode liquid right this is the

specific volume or the inverse of the density and that is doesn't change except over

huge changes in pressure so we consider that to be a constant so that changes in h then could be

written as the change in temperature times c plus the specific volume which can be found

at either temperature multiplied by the change in pressure this term is generally going to be

fairly small because the specific volume for a liquid is on the order of .001

so this change in pressure would have to be fairly substantial in order for this term to be the same

order as this term but we do have to include it because there are lots of situations where

the change in pressure is significant so finally this this is how you find changes

in h what if you're just looking up the value of h and we want to use the same approach we did for

values of specific volume values of internal energy well so if we want to use the saturation

values in a4 instead of using the compressed liquid tables then h as a function of pressure and

temperature should be written as follows h saturated liquid

as a function of t so on this line plus the specific volume at the temperature

multiplied by the difference between the pressure that we want and the saturation pressure at that

temperature and once again this term is only going to be significant for very large changes

in p but until you get some practice and you you're able to recognize how big of a

pressure difference you need you really should be including both terms so if you're asked for this on a

test or a quiz you should definitely use the full form of h

we've spent a lot of time looking at substances that have different phases so we've looked at

water we've alluded to refrigerants like refrigerant 134a and we've looked at you know drawing the

phase diagram looking at sub-cooled liquids mixture regions and superheated vapors

now i want to talk about ideal gases and when i say ideal gases i'm talking about things like

air argon nitrogen helium and so forth in this course we're not going to consider water vapor

to be an ideal gas although there is you know a region at fairly low pressures far away from the

dome where you can treat a vapor as a gas but like i said

and really pay attention to this in this course do not treat water vapor as an ideal gas

so let's look at what we have for ideal gases and again we're interested in ideal gases from the perspective that

we need to understand their properties in order to be able to use them in processes and

and understand the energy changes based on analysis using the first law so for an ideal gas this is something

you've done in high school and in first year chemistry we already have a law that relates p v

and t in an ideal gas we have pv equals rt now in your chemistry course you would

have written this as pv equals nrt um but we don't use that form we're using the form that's in terms of

of the properties that we have so pressure specific volume this r value is actually an r value for

the substance it's the universal r value divided by the molecular mass of the substance we're talking about so for

instance for air this value r would be 0.287 okay and temperature of course in kelvin

so pv equals rt so we have a relationship between pressure specific volume and temperature

but we need something for internal energy and we need something for h okay so let's look at internal energy

first and most often we're interested in changes in the internal energy

we have standard air tables where you can just simply look up the internal energy the enthalpy

at a particular temperature but those tables are actually developed using the analysis that we're going to talk about

here so let's look at internal energy as a function of temperature and specific

volume so two independent intensive properties that's enough to define the state

let's take the total derivative so we have du dt by times dt plus du dudv

times dv now for an ideal gas dudv going to zero actually defines the region when it's a vapor

and this is a characteristic of an ideal gas so this term actually gets wiped out and we're left with d u equals d u dt

dt we recognize du dt as being the specific heat of constant volume which is very different from cp for an

ideal gas for a compressible substance just taking this what's left over this means that

changes in internal energy are equal to cv times changes in temperature which if we integrate a change in the

energy level from from u1 to u2 is the integral t1 to t2 of cv

which can be a function of t in general it's not necessarily a constant so we have to keep it inside the

integral by dt for cases where cv is constant now four cases so cv

is not a constant relative to temperature in fact it's graduated as a function of temperature

but there's lots of situations where you can assume that cv is constant over the range of temperatures that we're

integrating over so if we assume that cb is a constant then this integral just comes out to be

delta u is equal to cb delta t this is a very useful um expression because most often when

you're using the first law of thermodynamics you're looking at changes in the energy

level not necessarily the value of the energy level at a particular state so changes in u c v times changes in t

let's look at h the enthalpy for an ideal gas so once again h is a combined property

its internal energy plus this flow work pv and because this is an ideal gas we can

actually replace pv by rt just using the ideal gas law so h is u plus pv which is equal to u

plus rt now by some argument since r is a constant r is the gas constant

and it's developed from the universal gas constant divided by the molecular mass both of which are constants so

r is still a constant and we've also shown that u can be characterized as u of t

now for a compressible substance clearly when you compress them they warm up but we've characterized the changes in t

or in due to pressure by integrating this rt into it so we can characterize

u as being a function of t then that means that since then h can also be characterized

as a function of t so from the definition then of cp which is dh by dt

dh is cpdt right dhdt is cp so dh is cpdt which we can write as a change in the enthalpy

is the integration across two temperatures cp which is a function of t times dt that's a very general statement

if cp is a constant for the same rationale as c b could be a constant then the

change in enthalpy is simply cp times t2 minus t1 so for ideal gases we have two fairly

simple statements we have that u2 minus u1 is cv delta t h2 minus h1 is cp

delta t cv and cp are different values for compressible substances so it's really important to use the right one

if you're wondering which is the right one always refer back to the definitions of cv and cp so cp

is dhdt cv is du dt these are things to remember one other interesting note for ideal gases

if we take this form right here and we take the derivative of it so dh by dt is d u dt plus d of rt by dt

and since r is a constant we can take it out so we have du dt plus r dt by dt which is just one

and we recognize that dhdt is cp equals du dt which is cv plus this gas constant

r so cp is equal to cv plus the gas constant quite often tables won't show you both

cp and cv they'll give you cp and the gas constant and you can derive cv from that but you

have to remember this expression right here we've talked a lot about properties of

substances now and we've considered substances that have phase changes

solid liquid vapor liquid vapor we've looked at ideal gases and just as a quick summary so

water we've looked at solid liquid and vapor series of experiments we're not really going to think too much more

about the solid we've drawn this phase diagram we've drawn

tv diagrams i'm going to show you in the next slide how to how to draw pv diagrams information for all of the

different phases and combinations are available in the appendices in tables a 4 5 6 and 7 for water

if we look at r 134a information for liquid and vapors are in tables a11 12 and 13. very analogous

information to what's found for water obviously much different temperature range uh and

for ideal gases i've introduced the ideal gas law pv equals rt

and also ways of evaluating changes in internal energy changes in enthalpy the connection between the specific

heats and the gas constant for air and so forth now there's a lot of information here

and the only way to learn how to be to be good at evaluating properties is to do many many examples

so we're going to devote one or two tutorials to solving examples of of

evaluating properties at different states so we're going to be looking at water we're going to be looking at

refrigerant we're going to be looking at air and other gases and i'm going to be giving you

lots of practice problems for you to take responsibility to learn how to do this