Announcements 9/16/11 Prayer Still at least three unregistered clickers: , 16488CD2, 1DAE9D2E “Real” thermodynamics (more unified, fewer disjointed topics): a. a.Today – – PV diagrams – – work – – isothermal contours – – Internal energy – – First Law of Thermodynamics b. b.Continues for the next 4 lectures after today. Then one more lecture. Then exam! Pearls Before Swine

Reading quiz (graded) Which of the following is NOT true of the work done on a gas as it goes from one point on a PV diagram to another? a. a.It cannot be calculated without knowing n and T. b. b.It depends on the path taken. c. c.It equals minus the integral under the curve. d. d.It has units of Joules. e. e.It is one of the terms in the First Law of Thermodynamics.

Work done by an expanding gas 1 m 3 of an ideal gas at 300 K supports a weight in a piston such that the pressure in the gas is 200,000 Pa (about 2 atm). The gas is heated up. It expands to 3 m 3. Plot the change on a graph of pressure vs. volume (a P-V diagram) How much work did the gas do as it expanded? a. a.How do you know it did work? = 400,000 J

More on Work… PV diagrams What if pressure doesn’t stay constant? Work done on gas vs work done by gas

Thought question (ungraded) A gas in a piston expands from point A to point B on the P-V plot, via either path 1 or path 2. Path 2 is a “combo path,” going down first, then over. The gas does the most work in: a. a.path 1 b. b.path 2 c. c.same work

Quick Writing Describe with words how you could actually make a gas (in some sort of container) change as in path 2.

Internal Energy, E int (aka U) E int = Sum of all of the microscopic kinetic energies. (Also frequently called “U”.) Return to Equipartition Theorem: a. a.“The total kinetic energy of a system is shared equally among all of its independent parts, on the average, once the system has reached thermal equilibrium.” b. b.Each “degree of freedom” of a molecule has kinetic energy of k B T/2 c. c.Monatomic molecules  3 d.o.f. d. d.At room temperatures, diatomic  5 d.o.f. (3 translational, 2 rotational)

Internal Energy Monatomic: E int = N  3 k B T/2 = (3/2)nRT Diatomic: (around room temperature) E int = N  5 k B T/2 = (5/2)nRT

Thought question (ungraded) The process in which  E int is the greatest (magnitude) is: a. a.path 1 b. b.path 2 c. c.neither; it’s the same

Isothermal Contours A gas changes its volume and pressure simultaneously to keep the temperature constant the whole time as it expands to twice the initial volume. What does this look like on a PV diagram? What if the temperature is higher? Lower?

“First Law”  E int = Q added + W on system What does that mean? You can add internal energy, by… a. a.…adding heat b. b.…compressing the gas Possibly more intuitive version: Q added =  E int + W by system When you add heat, it can either …increase internal energy (temperature) …be used to do work (expand the gas)

Three Specific Cases Constant pressure, “isobaric” a. a.Work on = ? Constant volume, “isovolumetric” a. a.Work on = ? Constant temperature, “isothermal” a. a.Work on = ? 0 –P  V

Worked Problems For each problem, draw the process on a P-V diagram, state what happens to the temperature (by visualizing contours), and calculate how much heat is added/removed from gas via the First Law. a. a.A monatomic gas (1.3 moles, 300K) expands from 0.1 m 3 to 0.2 m 3 in a constant pressure process. b. b.A diatomic gas (0.5 moles, 300K) has its pressure increased from 100,000 Pa to 200,000 Pa in a constant volume process. c. c.A diatomic gas (0.7 moles, 300K) gets compressed from 0.4 m 3 to 0.2 m 3 in a constant temperature process. T increases, Q =  E int + P  V = 8102 J added T increases, Q =  E int = 3116 J added T stays constant, Q = –W on gas = –1210 J (i.e., 1210 J of heat removed from gas)

Quick Answers From Students  E int will be positive if ______________ Q added will be positive if ______________ W on system will be positive if ______________