 So at the end of the last example problem, I had posed the question why is knowing the specific gas constant of our mixture useful? And it's because we can use our ideal gas law to describe the properties of the mixture. Well, that leads us into a broader conversation about modeling the interaction of properties of mixtures. When we talk about mixtures of gases, there are two ways that we can model the mixtures and their behavior. The first is called Dalton's Law, which, like all models, is inaccurate but is useful. Dalton's Law treats each of the components or species of the mixture as though they occupy the entire volume. Then each of them contributes part of the pressure to the entire pressure of the mixture. So if you have a volume of a container, Dalton's Law says that each of the substances inside that mixture takes up the entire volume. And they contribute a partial pressure to the pressure of the mixture. The other modeling method assumes that each of the substances is at the same pressure but only takes up part of the volume. That's called Amigut's Law. Amigut's Law assumes that each of the species in the mixture contributes part of the volume but exists at the mixture pressure. For real gases, each of these models is close. They're more or less accurate in different circumstances, but as a general rule, Dalton's Law is more useful. For ideal gases, they produce equivalent results. They are identical. So because we're modeling ideal gases, it doesn't really matter if we use Dalton's Law or Amigut's Law because they will produce identical results. However, because Dalton's Law is more useful in the general case, we will use primarily Dalton's Law to describe our mixtures. Furthermore, note that when we're analyzing ideal gases, we can write out the relationship between the partial pressure in Dalton's Law or the partial volume in Amigut's Law to the pressure and volume of the mixture in terms of the ideal gas law, which would simplify down to the number of moles in each of the species divided by the number of moles of the mixture. So the molar fraction in both cases represents the proportion of the partial pressure to the pressure of the mixture in Dalton's Law or the partial volume to the volume of the mixture in Amigut's Law. But again, this is only valid for ideal gas relations. Furthermore, note that when we talk about properties of the mixture, we will also be involving the other thermodynamic quantities we use, like internal energy and enthalpy and entropy, in which case we can write out the properties of the mixture in terms of the specific properties of each of the individual species and the amount of each of the individual species. So if we have a mass-specific quantity, we're going to multiply by the mass to figure out how much total internal energy, for example, each species contributes to the mixture and we add them all together. Also remember, the horizontal line represents a molar-specific quantity. We can write out our total properties, our changes in properties, we can even write out the relationship between our specific heat capacities in terms of those quantities. Anyway, on to our next example problem. Say you had a big ol' box containing 50 kilograms of nitrogen and 50 kilograms of carbon dioxide at 400 Kelvin. I want us to figure out the specific enthalpy of the mixture on a mass basis and a molar basis. So we are looking for lowercase h and lowercase h bar. And to figure out the specific enthalpy of the mixture, we're going to be using the specific enthalpy of each of the individual species, which means we're going to have to go into our property tables. As we peruse our property tables, we will come across the specific enthalpies of nitrogen and carbon dioxide in table A23. So on table A23, I have specific enthalpy, specific internal energy, and zero point entropy as a function of temperature. So I want the enthalpy for carbon dioxide and nitrogen at 400 Kelvin, which means I'm going to be using this quantity here and this quantity here. And note that those are the molar-specific enthalpies. So at 400 Kelvin, the molar-specific enthalpy of nitrogen is 11,640. And that's kilojoules per kilomole. And the molar-specific enthalpy for carbon dioxide at 400 Kelvin is 13,372. Now that we have the molar-specific enthalpies, we can calculate the specific enthalpy on a mass and molar basis for the mixture. But before you get too sidetracked trying to find the right equation, it's easiest to just think through what it is you actually want. The mass-specific enthalpy of the mixture is going to be the total enthalpy of the mixture divided by the mass of the mixture. While we know the mass of the mixture, it's 100 kilograms. We can figure out the total enthalpy of the mixture by multiplying the molar-specific enthalpy by the number of moles. So we can take the number of moles of CO2 times the molar-specific enthalpy of CO2 and add to that the number of moles of diatomic nitrogen times the molar-specific enthalpy of diatomic nitrogen and divide that in entire quantity for the mass of the mixture. And just like earlier, if we wanted to, we can manipulate these quantities and write it out in terms of a proportion. To do that, I could write, instead of the mass of the mixture, I could write the number of moles of the mixture times the molar mass of the mixture, at which point I would be left with the molar fraction times the molar-specific enthalpy divided by the molar mass of the mixture for each of these quantities. But doing that would require that I determine the molar mass of the mixture. Well, for the molar mass of the mixture, I can just take the molar fraction and multiply that by the molar mass of each of the individual species and sum that together. But again, I don't think that it's useful to get too caught up in trying to simplify these algebraically. Instead, I think that it's more useful just to actually calculate how many moles there are, multiply them, and then divide by 100 kilograms. So let's figure out how many moles of nitrogen and carbon dioxide are in 50 kilograms. To do that, I'm going to divide by mass, 50 kilograms, by the molar mass of each of the individual species, which means I have to go back to table A1. And for nitrogen, I recognize that my molar mass is 28.01, so calculator, if you would please. 50 kilograms divided by 28.01 gives me a number of moles of 1.7858. And for CO2, I take 50 kilograms divided by 44.01, and I get 1.13611. So now I can calculate the quantity by taking the number of moles of CO2, which is 1.13611 times the kilo moles, times the molar specific enthalpy of CO2, which is 13,372 kilojoules per kilo mole. Because these are kilo moles of CO2 that cancel, then I do the same thing with nitrogen. I have 1.7858 multiplied by the molar specific enthalpy of nitrogen, which is 11,640 kilojoules per kilo mole. And again, because it's the kilo moles of nitrogen, in both cases they cancel, leaving me with a total enthalpy in kilojoules in the numerator, which I can divide by 100 kilograms. That will give me kilojoules per kilogram of mixture. So 1.13611 times 13,372 plus 1.7858 times 11,640. Divide that quantity by, divide that quantity calculator by 100, and we get 359. So the mass specific enthalpy of the mixture at 400 Kelvin is 359.7 kilojoules per kilogram. For part b we're looking for the same quantity, but on a molar specific basis instead of a mass specific basis. There are a number of ways that we could approach that. We could figure out the molar mass of the mixture and then use that molar mass to convert this. That would allow us to do a direct conversion from mass specific to molar specific. We could take our total enthalpy here, h-mix, and divide it by the number of moles of the mixture, which we could calculate by taking 1.7858 plus 1.13611, and that would give us the molar specific enthalpy. I could take the molar fraction of each species and multiply it by the specific enthalpy on a molar basis for each of the species, and that would represent an average molar specific enthalpy. There are a number of ways that I can do it, but I think the easiest way is just to step back to the definition again. For the molar specific enthalpy, I'm going to be taking the total enthalpy of the mixture, divided by the number of moles of the mixture, and the total enthalpy of the mixture is still the number of moles times the molar specific quantity. I just happened to be dividing by the number of moles of mixture this time around, and the number of moles of the mixture is just nCO2 plus nN2. So we take that same numerator from earlier, and we're just dividing by 1.7858 plus 1.13611, and that gives us 12,313. And again, we could have figured out the molar fraction and taken the molar fraction of each of the species times the molar specific enthalpy of each of the species and summed those all together. We could have determined the molar mass of the mixture and used that with this quantity to do a direct conversion, but I'm of the opinion the easiest way to solve these problems is just to think about what you have and what you're trying to get to.