1. Chapters 16 and 17, thermodynamics
Lecture 23
Goals:
Chapters 16 and 17, thermodynamics
Assignment
HW-9 due Tuesday, Nov 23
Wednesday: Read through Chapter 18.2
Third test on Thursday, December 2 1

2. Temperature Three main scales Farenheit Celcius Kelvin 212 100 373.15
Water boils 32
273.15
Water freezes
A measure of thermal energy
When we say something is hot, we mean that It has more thermal energy when compared to something cold.
Absolute zero means, when the thermal energy is zero. If we have a gas for example, absolute zero would be the point where the pressure drops to zero. Or where the motion of the atoms/molecules that form the gas come to a stop.
At room temperature, the average velocity of the molecules around us are about 500 m/s. When we go to absolute zero temperature, the average velocity goes to zero.
Nowadays, we can cool atoms to a temperature of pK using laser cooling techniques, we can basically create the coldest places in the universe.
Absolute Zero

3. Phase-diagrams Recall “3” Phases of matter: Solid, liquid & gas
All 3 phases exist at different p,T conditions
Triple point of water:
p = 0.06 atm
T = 0.01°C
Triple point of CO2:
p = 5 atm
T = -56°C
1) Note that you can condense stuff by changing the pressure, without changing the temperature.

4. 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.

5. 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

6. 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.

7. PV diagrams Pressure 3 atm 1 atm 3 Liters 1 Liter 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.
3 Liters
1 Liter
Volume

8. 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.

9. 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

10. 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, 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

11. 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.

12. For an isochoric process (V=const), W=02
Pressure 1
Volume

13. 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!!

14. 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

15. 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.

16. First Law of Thermodynamics
ΔU+ΔK+ΔEthermal=ΔEsystem=Wexternal+Q
For systems where there is no change in mechanical energy:
ΔEthermal =Wexternal+Q

17. 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