 So we spent a little time talking about heat and work and more recently figured out how to relate heat and work to internal energy and it turns out that doing things under constant volume conditions is quite convenient because the heat is the same thing as the internal energy regardless of whether we're talking about the differential amount of heat and internal energy or finite change in the internal energy or finite amount of heat. On the other hand experimentally it's more convenient usually to think about things in a constant pressure environment. If you want to do an experiment at constant volume you've got to be very careful, have a rigid container, make sure it's fully sealed, make sure that it's sturdy enough that the volume can't change so it's a lot more difficult to do an experiment at constant volume at constant pressure on the other hand. This room is at one atmosphere or pressure roughly if I did a reaction in this room if something expands it just pushes some of the air out of the room and keeps the air the whole environment at one atmosphere pressure so it's quite convenient to do things at constant pressure rather than constant volume but then we lose the advantage of not being able to not having to worry about the PV work in a process. So that's the next step we can think about is we can define a quantity that has all the advantages of the internal energy and not having to think too hard about PV work when we do a process but also the experimental advantages of being more convenient at constant pressure rather than constant volume and the name for that new quantity that we're going to define is called the enthalpy something that quite possibly you've heard of in earlier chemistry classes so we define in fact I'll use a triple equal sign to say that for the very first time at least in this class I'm defining the enthalpy to be equal to the quantity internal energy plus pressure times volume so that's just the definition we choose to use with that definition we can ask ourselves what is the change in you or what is the differential change in you or in this case what is the change in enthalpy so if I take the differential of both sides of this equation the differential change in my new quantity the enthalpy I should give this a name dh on the left on the right I have du and then the differential of p times v I can use the product rule to write that as pdv plus vdp so that's the differential of both the left and the right but we know some things about the terms in this expression du we know from the first law is equal to dq plus dw I'll just leave the pdv and vdp in there we also know at least if the only type of work we're concerned with is pv work we know how to write down the differential change in work dq I can write as minus p external dv if I leave the rest of these terms unchanged so now we've got two terms here a minus pdv and a plus pdv but the difference is this one is the external pressure and this one is the internal pressure the system's pressure we know that the external pressure and the internal pressure are the same as each other p is equal to p external if we have a reversible process where the pressure and the external pressure remain equal to each other throughout that whole reversible process so in that circumstance p external and p are the same thing this negative term cancels this positive term and we're left with enthalpy is heat plus change in enthalpy is change is the amount of heat plus vdp and now we're at the point where we can do what I promised at the beginning if we're under constant pressure conditions so that any change in the pressure is zero because the pressure is not changing if dp is equal to zero because we're at constant pressure that second term goes away and what we're left with is the statement that the change in the enthalpy is equal to the change the amount of heat transferred so that's the equivalent of this statement that we've had earlier internal energy is equal to heat if we're under constant volume conditions if we're under constant pressure conditions this new quantity that we've defined the enthalpy the change in the enthalpy is equal to the amount of heat when we're at constant pressure so we can write that either in differential form or if we prefer for finite size changes in this enthalpy so we've got completely analogous expressions to these two that we had for the internal energy but now for the enthalpy and really if you get confused about what's the difference between energy and enthalpy all enthalpy is is a slightly tweaked version of the internal energy the energy of a system tweaked in such a way that when we do things under constant pressure conditions the enthalpy is the same thing as the heat that makes it a convenient quantity to use under constant pressure conditions because under those conditions we don't have to worry too much about the PV work we can ignore the the work contribution and it's only the heat that's related to this new quantity we've defined called the enthalpy so we can use the enthalpy in an example in fact we'll do the same example we considered previously we'll talk about heating of an ideal gas so let's take exactly the same problem as in the prior video lecture let's say we've got one mole of an ideal gas at one atmosphere and we're going to change it from condition one to condition two but we're going to do it at constant pressure so the initial pressure the final pressure are both the same number what we are going to do though is is heat the gas up room temperature 298 Kelvin increase the temperature by 50 Kelvin to 348 Kelvin so what that's the main thing we're doing is we're heating this gas up when we do it at constant pressure of course the volume is going to change using PV equals on RT with these numbers will tell us that the initial volume has to be 24 and a half liters at least if we're talking about an ideal gas and the volume after we've heated it up because the temperature is increased but the pressure hasn't changed that volume is going to have increased again PV equals on RT as we considered previously tells us the final volume of 28.6 liters okay so that's the setup the question then is what is the enthalpy change for this process and we can use the equations we've just determined one very nice thing about the enthalpy similar to what was nice about internal energy when we're comparing it to heat and work heat and work are sometimes inconvenient because their path dependent processes enthalpy because we've defined it using only state variables energy pressure and volume enthalpy is also a state variable so I don't actually have to tell you whether what path I took to get from these initial conditions to these final conditions doesn't matter whether reversible path irreversible path what changed what didn't change the enthalpy change is just the enthalpy under conditions to minus the enthalpy under conditions one because enthalpy is a state variable so initially when the gas is under these conditions we can calculate the enthalpy as U plus PV so we need to know the energy of the gas and the pressure and the volume for an ideal gas at least one that we can treat like a 3d particle in a box we've seen previously that enthalpy is three halves and RT so we can plug numbers into this expression we know moles pressure temperature and volume so the enthalpy of the gas before we've heated it up is three halves and which is one mole gas constant times the initial temperature 298 Kelvin to that we need to add P times V so I'm going to add initial pressure one atmosphere multiplying by the initial volume 24 and a half liters all right so now that's numbers plugged into this expression if I stop and look at units however moles will cancel moles and Kelvin will cancel Kelvin three halves and RT gives me units of joules which is fine because we're interested in something that behaves like an energy a tweaked version of the energy is what the enthalpy is the second term the PV term I've got liters times atmospheres that's also an energy but it's not in units of joules so I need to convert from joule from liter atmospheres to units of joules in order to get rid of these slightly inconvenient units of liter atmospheres convert them to joules and then my final answer will be in units of joules and when I do that calculation three halves times one times gas constant 298 added to the second term 24 and a half times this conversion factor we find that the enthalpy of the gas under this initial set of conditions turns out to be 6200 joules 6.2 kilojoules if you prefer is the enthalpy of the gas under this set of conditions so again enthalpy is a state function for a given system under some thermodynamic state if I supply pressure temperature volume we can calculate the enthalpy of the gas at that state likewise we can calculate the enthalpy of the gas under the final state of the system where I've increased the temperature and also increased the volume but left the pressure the same and I won't work through all the arithmetic again but if we use the same expression 3 F's NRT with the final temperature and then P2 times V2 all that's going to change here is instead of multiplying by 298 Kelvin I'm going to multiply by 348 Kelvin and then instead of one atmosphere times 24 and a half liters for the PV term I'm going to have one atmosphere times 28.6 liters for the PV term so both of these terms have actually increased if we do the arithmetic what we find you're welcome to double check and make sure you agree with this result is that we get an enthalpy of 7,240 Kelvin so the main thing we're interested in the change in the enthalpy when we go from condition one where we have 6200 joules worth of enthalpy to condition two where we have 7,240 joules of enthalpy the change in the enthalpy is a little over a thousand joules a thousand and forty joules which if you recall the answer from the previous example we did when we considered this exact same heating gas under constant pressure turns out that the heat required to heat that gas at constant pressure was the same number of thousand and forty joules because as expected the change in the enthalpy is equal to the heat under conditions of constant pressure so we can notice a couple things about that number one we didn't have to explicitly think about PV work that was the the promise I made when we defined the enthalpy is we wouldn't have to think about the heat and the work separately and add them together turns out we didn't actually save ourselves that much arithmetic because the PV work is really hiding in this this p1v1 and p2v2 the difference between those two terms but at least we were able to do the calculation without specifically having to ask ourselves about path dependent functions like heat and work and instead we've now got a state function that we can use to calculate the change in enthalpy or if we prefer the amount of heat required to get from state one to state two so that's that's really the biggest advantage of the enthalpy is it's a state function and we don't have to burden ourselves with questions about path functions one more thing that we can do with the enthalpy before we move on is because we've been talking about ideal gases in particular if I go back to my definition of enthalpy as being energy plus p times v just like we've seen here if we have an ideal gas that behaves like a particle in a box in other words an ideal gas whose internal energy is three halves times nrt then the enthalpy will be three halves nrt plus pv one additional shortcut we can take that we didn't explicitly use here is if it's an ideal gas pv is of course equal to nrt so I've just used the ideal gas expression to rewrite pv as nrt and then I've got three halves nrt plus an additional one nrt so all together I've got five halves nrt so there's an expression that we didn't use explicitly but that we can put in a box and use from this point on if we have an ideal gas that behaves like a 3d particle in a box then the enthalpy of that gas will be five halves nrt so that's a comparable expression to the result we've had previously that gives us the internal energy of an ideal gas now we have an expression for the enthalpy of an ideal gas so now that we know about enthalpy what we can do is see what that tells us in particular about heat because enthalpy is the same thing as heat when we do things at constant pressure so we'll focus a little more carefully on heat in the next video lecture