Okay in this video I'm going to continue on with my tutorials on thermodynamics and statistical mechanics. This is video number 43 and I got to discuss the chemical potential of an ideal gas. I'd like to draw your attention to my website, universityphysicstorial.com. The previous video to this is in where 42 were discussed the relationship between the Gibbs free energy and the chemical potential. And what we saw is that the Gibbs free energy per particle is equal to the chemical potential. So just a bit of a revision to include the video number 42. What we saw was that DG is equal to mu DN. So of course when we integrate that we find that the Gibbs free energy totally gives free energy is the chemical potential times the number of particles. So that's what we say is that the Gibbs free energy per particle is the chemical potential. Alright so going back to our thermodynamic identity for the Gibbs free energy. We find the anhydrousal change in the Gibbs free energy is minus S DT plus P DV plus mu DN. So let's look at some of the partial derivatives we have here. So we know of course that del G del N okay is going to give us the chemical potential but if we look at del G del P we're going to get the volume and you'll see in a moment while we're going to use that. So going back to our relationship between the Gibbs free energy and the chemical potential or the chemical potential is rolled as the Gibbs free energy per particle. We saw that G is equal to mu times N. So what we're going to do now is take the partial's with respect to P. So we're going to get del G del P and we're going to get del mu del P. Of course N is not a function of pressure. Alright? Cannot be a function of pressure. But we we already have those partial's here. So what we're going to have is follows that the volume is equal to the number of particles capital A times del mu del P. Alright? del mu del P. But the next thing we need to know is that the ideal gas is that PV is equal to N kT or V over N is equal to k times T over P. So that's our that's our little aside or how we get from what we have at the moment to the ideal gas law. That means that we can rewrite what we have at the moment as the following. We're going to have del mu del P is going to be equal to V over N. It's going to be equal to k times T over P. Alright? Now what we're going to do is we're going to integrate all of this. So this is a pretty straight forward and pretty usual integral and it turns out to be follows. We're going to get del mu integrated between mu zero and mu. That's assumed that the economic of potential can change. Then we're going to get kT integrated from P zero to P and we're going to have Dp over B. Notice it's kind of total derivative at this time because we can't as in there or there are no there are no passions to suppose on these sides. So I'm going to just bear with me and rub out all this and go back to the start. I'm sure that's integral your way to able to do it this stage. So that integral anyway turns out to be mu minus mu zero is equal to kT outside the natural logarithm of P minus the natural logarithm of P zero or is kT log P over P zero. Alright? So that's what we have. So another way of writing this is that the chemical potential that we have at the moment is the chemical potential which we started with plus kT the natural logarithm of the ratio of the current pressure and the initial pressure. Okay? Now that's it. That's the chemical potential of an ideal gas. But I'd just like to do a small bit of analysis of this. First of all, if we look, what? Look at what the chemical potential is a function of. So we say that the chemical potential is a function of temperature and pressure. Okay? We'll say that mu zero is also a function of temperature and pressure. And it's equal to plus, we'll say kT the log of P over P zero. Okay? So we know of course pressure is just a function of pressure. Now let's look at a small bit of mass. We know the pressure is force per unit area. Okay? Aside to that, we know that mass is equal to density times volume. It's equal to density times the area times the length. All we can say is that the area is equal to the mass divided by rho times the length. Alright? So if we sub this back in up here, what we're going to get is that the pressure is equal to the force divided by the mass multiplied by the density. The conclusion here is that pressure is proportional to the density. That's the area conclusion here. So going back to the equation that I have in red, this particular equation indicates how the chemical potential varies with pressure. Or because pressure is proportional to density, it also indicates how the chemical potential varies with density. So this is the chemical potential of an ideal gas. Thanks for watching. Please pass it on to your friends and subscribe to my channel. Any adults who visit universityphysicsTorales.com.