Quiz 7 8:30-8:50am TODAY Have your calculator ready. Cell phone calculator NOT allowed. Closed book Quiz 3 Re-evaluation Request Due this Thursday, 2/28. Quiz 4 Re-evaluation Request Due next Thursday, 3/6. Turn in you original Quiz along with the Re-evaluation Request Form. Note: It is possible for your grade to be lowered after the re-evaluation. Quiz 4 rubrics posted on the website. Quiz 4 info (average score, grades) will be posted this afternoon. Quiz 5,6 graded, grades being recorded. Next lecture March 4 Quiz 8 (Last Quiz!) will cover the material from today’s lecture and material from DLM12 and 13, excluding FNTs for DLM14.

Measuring Heat capacity Thermometer Stirring rod Heating element Substance of interest Insulating material

All measurements at 25 C unless listed otherwise (-100 C) (0C)

All measurements at 25 C unless listed otherwise (-100 C) (0C)

All measurements at 25 C unless listed otherwise (-100 C) (0C)

All measurements at 25 C unless listed otherwise (-100 C) (0C) k B N A = R = 8.31 J/K  mole : Gas constant or Ideal gas constant k B (Boltzman constant) = 1.38 x Joule/Kelvin N A (Avogadro’s number) = 6.02 x 10 23

6 modes All measurements at 25 C unless listed otherwise (-100 C) (0C) k B N A = R = 8.31 J/mole  K : Gas constant or Ideal gas constant k B (Boltzmann constant) = 1.38 x Joule/Kelvin N A (Avogadro’s number) = 6.02 x 10 23

All measurements at 25 C unless listed otherwise

6 modes All measurements at 25 C unless listed otherwise For polyatomic substances, the values of molar specific heat of liquids are greater than the values for solids. Limitation of our model… For monatomic substances, the value of molar specific heat of liquids is similar to the values for solids. Our model works well here!

(100 C) All measurements at 25 C unless listed otherwise (500 C)

diatomic (no vibrations) (100 C) All measurements at 25 C unless listed otherwise (500 C) monatomic

diatomic (no vibrations) (100 C) All measurements at 25 C unless listed otherwise (500 C) monatomic Oops! What’s going on??!

The discrepancy explained… Closed box of gasOpen box of gas

The discrepancy explained… Closed box of gasOpen box of gas Closed box: all heat goes into the gas’s energy Open box: Some heat goes into pushing air out of the way

The discrepancy explained… Closed box of gasOpen box of gas Closed box: all heat goes into the gas’s energy Open box: Some heat goes into pushing air out of the way C V measurement C P measurement

The discrepancy explained… Closed box of gasOpen box of gas Closed box: all heat goes into the gas’s energy Open box: Some heat goes into pushing air out of the way C V measurement C P measurement Question So then does it take More energy to raise the Temperature of Closed box of gas Or Open box of gas?

(100 C) All measurements at 25 C unless listed otherwise (500 C) monatomic Whew, Now that makes sense.

diatomic (no vibrations) (100 C) All measurements at 25 C unless listed otherwise (500 C) monatomic Whew, Now that makes sense.

“Constant volume”“Constant pressure” “Process” seems to matter… => Chapter 4 Models of Thermodynamics (definition of heat capacity) In this example of thermal phenomena (i.e.,measuring heat capacity),

Equipartition tells us that each (active) mode has the same amount of energy, and that the temperature is the measure of energy per mode Our mode counting works well for solids and gasses. For gasses, we need to distinguish between work and heat carefully (Chapter 4). Our mode counting does not work well for liquids. Summary of Equipartition of Energy

What is Thermodynamics? Sadi Carnot ( ) Father of Thermodynamics James Joule ( ) Lord Kelvin ( ) In a nutshell, we study the transfer of energy between systems and how the energy instills movement, i.e., how the system responds. Ex. If we heat something, it expands. Carefully distinguish between heat and work Learn what a state function is Think about things that depend on the process, and ask about processes. (e.g. ice melts/water freeze, fry egg) Why doesn’t the fried egg turn into raw egg again? Why can carbon exist as a diamond as well as graphite but not ice and vapor?

State functions Depends only on properties of the system at a particular time Work, heat LHS: depends only on i and f Q,W depend on process between i and f Not a property of a particular object Instead a property of a particular process or “way of getting from the initial state to the final state” Ex. E thermal of a gas For an ideal gas, E thermal depends only on * Temperature * Number of modes Remember conservation of energy?

∆U : Internal energy Energy associated with the atoms/molecules inside the body Of material A comment Remember conservation of energy? ∆E total must include all changes of energy associated with the system… ∆E total = ∆E thermal + ∆E bond + ∆E atomic + ∆E nuclear + ∆E mechanical Energy associated with the motion of a body as a whole So then, if there’s no change in ∆E mechanical ∆ U First law of Thermodynamics

Depend only on what the object is doing at the time. Change in state function depends only on start and end points. Examples: T, P, V, modes, bonds, mass, position, KE, PE... State functions Depend on the process. Not a property of an object Examples: Q, W, learning, Process- dependent

Work initial final P VV P Along any given segment

initial final P V First section: W 1 < 0 (volume expands) Second section: W 2 = 0 (volume constant) Third section: W 3 > 0 (volume contracts) Work

initial final P V Here are two separate processes acting on two different ideal gasses. Which one has a greater magnitude of work? The initial and final points are the same. initial final P V A) Magnitude of work in top process greater B) Magnitude of work in lower process greater C) Both the same D) Need more info about the gasses.

initial final P V Is the work done in the process to the right positive or negative? A) Positive B) Negative C) Zero D) Impossible to tell.

Heat and the first law of Thermodynamics initial final V P We can read work directly off this graph (i.e. don’t need to know anything about modes, U, T, etc.) If we know something about the gas, we can figure out U i, U f and U f - U i

initial final V P Heat and the first law of Thermodynamics

Example 5 moles of a monatomic gas has its pressure increased from 10 5 Pa to 1.5x10 5 Pa. This process occurs at a constant volume of 0.1 m 3. Determine: * work, * change in internal energy * heat involved in this process. initial final P V

Heat depends on the process Work depends on the process only depends on initial and final W = 0 for all constant volume processes A comment

Enthalpy Is a state function: - U depends only on state of system - P depends only on state of system - V depends only on state of system => H depends only on state of system (Hess’s law) Who cares?!?!

initial final P V initialfinal P V W = 0 Constant volumeConstant pressure Note: nothing about gasses used - works for solids and liquids too! Enthalpy *Derivation in P.84

Below boiling point Freezing of modes: Nitrogen (per molecule)

Below boiling point California temps Freezing of modes: Nitrogen (per molecule)

Closed Book Don’t forget to fill in your DL section number!