Wrap up Chapter 16 and start Chapter 17 Lecture 23 Goals: Wrap up Chapter 16 and start Chapter 17 Assignment HW-9 due Tuesday, Nov 22 Wednesday: Read Chapter 17 Third test on Thursday, December 1 1

R=8.31 J/mol K: universal gas constant Ideal Gas Law Assumptions that we will make: “hard sphere” model for the atoms density is low temperature not too high P V = n R T n: # of moles R=8.31 J/mol K: universal gas constant We have discussed the ideal gas law, so here basically under certain assumptions, all gases whether it is O2, H2, neon, or air, satisfy a universal law, where PV=n R T. Here P is the pressure of the gas, V is the volume, n is the number of moles, and T is the temperature. 2) Now, why all gases satisfying these assumptions should have the same universal relationship with the same value of R is not obvious at all, it requires some thinking, and we will talk about this more when we discuss chapter 18. 3) However, there are also a lot of things about this relationship that intuitively makes sense. Consider a container at a fixed volume V. You would expect that the more gas you put in the container, the more pressure you would have, which is exactly what this equation tells you. With V fixed, as n gets larger, P gets larger as well. Similarly, the higher the temperature of the has, you would expect more pressure, which is exactly what this equation tells you. 4) Discuss the microscopic theory of pressure: i.e. individual molecules colliding with the wall.

P V = constant if n, T fixed Suppose that we have a sealed container at a fixed temperature. If the volume of the container is doubled, what would happen to pressure? Double Remain the same Halved None of the above 1) We will use the ideal gas law frequently over the next few weeks. Here is a quick problem. P V = n R T P V = constant if n, T fixed

kB: Boltzmann’s constant P V = n R T n: # of moles n=N/NA P V = n R T=(N/NA) R T =N (R/NA) T P V= N kB T kB: Boltzmann’s constant There is also a slightly different way to write the ideal gas law where instead of the number of moles we use the number of particles. So once again, we have PV=nRT, where n is the number of moles. Now, the number of moles is the number of particles N divided by the avagadros number, 6.02 x 10^23. So if we have avagadro’s number of particles, we have 1 mole, if we have twice the avagadro’s number of particles, we have two moles and so on. Now if we just plug-in this expression for the number of moles, in the ideal gas equation, we get….. Boltzmann’s constant can be thought as the gas constant per unit particle.

PV diagrams Pressure 3 atm 1 atm 1 Liter 3 Liters Volume Now, it is very useful to represent ideal gas processes on diagrams called PV diagrams. Here, we plot pressure versus volume. Each point on the graph represents a state of the gas. For example, here we have a gas that starts at this point, with a pressure of 1 atm, and the volume of the gas is 1 liter. Now although, we only show pressure and volume, other state variables of the gas can typically be found by using the ideal gas law. If we know the number of moles for example, or the number of particles, for each pressure and volume point, we can find the temperature by using the ideal gas law. In this example, the gas ends up at this point, and this line shows the particular trajectory that defines the process. Not only the end-points but also the trajectory is critical. As we will see in a moment, the work done on the gas is different for different trajectories even though the end points may be identical. 1 Liter 3 Liters Volume

PV diagrams: Important processes Isochoric process: V = const (aka isovolumetric) Isobaric process: P = const Isothermal process: T = const Now, a process that an ideal gas goes through can be quite complicated, so that trajectory can be quite complicated. There are, however, a number of special cases, that are very frequently encountered, and for these processes, the PV diagrams look particularly simple. Many important gas processes take place in a container of constant, unchanging volume. Such constant volume processes are called isochoric processes. Iso is a prefix that means constant or equal. We frequently encounter processes where the pressure of the gas does not change, these processes are called isobaric. So P is constant Now, processes where the temperature remains fixed are called isothermal processes.

Which one of the following PV diagrams describe an isobaric process (P=constant) Volume Pressure 1 2 Volume Pressure 1 2 Volume Pressure 1 2 Discuss how you could implement each process Discuss mixed processes V constant, isochoric PV constant, isothermal P constant, isobaric

ΔW=-P ΔV Work done on a gas Δx ΔW = Fext Δx = P A Δx = P ΔV Fext=P A ΔV=Vfinal-Vinitial Fext=P A Now, we will discuss a very important concept, the concept of work done on a gas. This is a pretty difficult concept, so I will try to go through this argument slowly, please interrupt me if you have questions. Now, here the issue is when a gas goes through a particular process, how much work do we do on the gas, how much do we increase the energy of the gas. When we discussed Bernouilli’s principle, we already argued that PV has the units of work. So not surprisingly, our final answer will be related to the PV diagrams that we have been discussing. The final result that we will get will be valid for arbitrary processes, but to make the discussion simpler, let’s consider this simplified geometry where we have an external force pushing on a piston. The movement of the piston compresses the gas. Now if the piston moves by an amount deltax, the work done by the external force will be, Fext deltax. Now if we assume the ideal case that the piston does not have any mass and that there is no friction, then the external force must balance the force applied by the gas on the piston, which is PA. Without worrying about the sign for the moment, Adx is the volume change, we therefore get, P delta V. Now, what about the sign, the work done by the external force will be positive, when the dx is aligned with the force, which means when the gas is compresssing. But when the gas is compressing, its volume will decrease, which means that dV is a negative number. ΔW=-P ΔV

W= - the area under the P-V curve W= the area under the F-x curve Work done on a gas ΔW=-P ΔV W= - the area under the P-V curve W= the area under the F-x curve Volume Pressure 1 2 1) For an isobaric process, the area is especially easy to evaluate since this is just a rectangle.

For an isochoric process (V=const), W=0 2 Pressure 1 1) This is analogous to saying that when we apply a force and if the object does not move then there will be no work done since there is no displacement. Volume

A) |W1|>|W2| B) |W1|=|W2| C) |W1|<|W2| Pressure How does the work done on the gas compare for the below two processes? A) |W1|>|W2| B) |W1|=|W2| C) |W1|<|W2| Volume Pressure 1 2 1) Work depends on the path!!

Work and Energy Transfer When you do work on a system, the energy of the system increases: Potential Energy (ΔU) Kinetic Energy (ΔK) Thermal Energy (ΔEthermal) Give examples for each case where work is done increasing the potential energy, kinetic energy, or the thermal energy. examples: accelerating a mass, increasing the height of an object, rubbing hands ΔU+ΔK+ΔEthermal=ΔEsystem=Wexternal

Q: Thermal energy transfer This description is incomplete. Energy can be transferred without doing any work. Q: Thermal energy transfer ΔU+ΔK+ΔEthermal=ΔEsystem=Wexternal+Q Q>0, environment is at a higher temperature Q<0, environment is at a lower temperature This is not obvious, it took a couple of centuries for people to realize the connection that both heat and work are a form of energy Joule’s experiments.

First Law of Thermodynamics ΔU+ΔK+ΔEthermal=ΔEsystem=Wexternal+Q For systems where there is no change in mechanical energy: ΔEthermal =Wexternal+Q Last time, we discussed the first law of thermodynamics. We argued that, there are two ways that you can change the total energy of a closed system. You can either do work on the system, or you can exchange heat, which is the thermal energy transfer. The total energy of the system is its potential energy+kinetic energy+thermal energy, which is proportional to the temperature. For thermodynamic discussions, we concentrate on systems where there is no change in mechanical energy. This equation therefore reads, the change in thermal energy of a system is the work that you do + the heat that you provide. Heat and work both has units of energy, in SI units, Joules, but historically, a different unit calories has been used for heat. 1 calorie=4.186 Joules 1 food calorie=1000 calories

Work on system, W>0 Work by system, W<0 Environment System Energy in Energy out Heat from system, Q<0 Heat to system, Q>0

Isochoric process (const V) Pi W=0 Work is: -(the area under the curve) Pressure Using ideal gas law: PV=nRT Tf < Ti Pf Volume From first law of thermodynamics we conclude: ΔEthermal =W+Q<0 Q < 0 Now, we are going to go through several processes in detail to make sure that we understand all the concepts that go into the first law of thermodynamics. Let’s first consider an isochoric process, a constant volume process. In this particular example, the volume is fixed, and the pressure of the gas drops from Pi to Pf. We are also considering a sealed container, the amount of gas does not change, the number of moles does not change. Now, first observation is that the work done on the gas, is the area under the curve, which as we discussed before, is zero. So there is no work done on the gas. Now the second observation is that, from the ideal gas law, PV=nRT. When we compare these two points, the volume does not change, n does not change, R is a constant, so since Pf<Pi, we must conclude, Tf<Ti. So the thermal energy of the gas decreases, the gas becomes colder. 5) From first law of thermodynamics, the change in thermal energy is W+Q, since W is zero, we conclude that Q must be less than 0. So during this process, heat was transferred from the gas to the environment.

Isothermal expansion (const T) Pi ΔEthermal=0 Temperature does not change Pressure Pf W<0 Work done on the gas is: Vi Vf Volume From first law of thermodynamics we conclude: ΔEthermal =W+Q=0 Q=-W Now, let’s consider another example. Here we will focus on an isothermal expansion process, where the temperature does not change. So we start at an initial point Pi Vi, and the gas expands to a final point Pf Vf. Now the first observation is that because there is no change in temperature, the thermal energy of the gas does not change. Now, because the gas is expanding, the work done on the gas is negative. When we integrate this curve from Vi to Vf, it is just the area under this curve, so it is a positive number, and the work is – the area, so we conclude that Work done on the gas is a negative number Now, finally, from first law of thermodynamics, we conclude that Q>0. Which means that there was heat energy transferred from the environment to the gas, so we need access to thermal energy for this to happen. Note that heat energy transfer does not necessarily result in a change in temperature.