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Announcements 1/24/11

Join us for a session on PV diagrams, internal energy, and the First Law of Thermodynamics. Learn about work done by gases, internal energy calculations, and isothermal contours. Engage in thought-provoking questions and quick writing activities.

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Announcements 1/24/11

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  1. Announcements 1/24/11 • Prayer • Substitute today: Dr. Stokes • Today’s topics: • PV diagrams • Work • Isothermal contours • Internal energy • First Law of Thermodynamics • Some specific state changes

  2. 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? • It cannot be calculated without knowing n and T. • It depends on the path taken. • It equals minus the integral under the curve. • It has units of Joules. • It is one of the terms in the First Law of Thermodynamics.

  3. Work done by an expanding gas • 1 m3 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 m3. • Plot the change on a graph of pressure vs. volume (a P-V diagram) • How much work did the gas do as it expanded? • How do you know it did work? = 400,000 J

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

  5. 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: • path 1 • path 2 • same work

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

  7. Internal Energy, Eint (aka U) • Eint = Sum of all of the microscopic kinetic energies. (Also frequently called “U”.) • Return to Equipartition Theorem: • “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.” • Each “degree of freedom” of a molecule has kinetic energy of kBT/2 • Monatomic molecules  3 d.o.f. • At room temperatures, diatomic  5 d.o.f. (3 translational, 2 rotational)

  8. Internal Energy • Monatomic: Eint = N  3 kBT/2 = (3/2)nRT • Diatomic: (around room temperature) Eint = N  5 kBT/2 = (5/2)nRT

  9. Thought question (ungraded) • The process in which Eint is the greatest (magnitude) is: • path 1 • path 2 • neither; it’s the same

  10. 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?

  11. “First Law” • DEint = Qadded + Won system • What does that mean? You can add internal energy, by… • …adding heat • …compressing the gas • Possibly more intuitive version: Qadded = DEint + Wby system • When you add heat, it can either • …increase internal energy (temperature) • …be used to do work (expand the gas)

  12. Three Specific Cases • Constant pressure, “isobaric” • Work on = ? • Constant volume, “isovolumetric” • Work on = ? • Constant temperature, “isothermal” • Work on = ? –PDV 0

  13. Worked Problems (as time permits) • 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 monatomic gas (1.3 moles, 300K) expands from 0.1 m3 to 0.2 m3 in a constant pressure process. • A diatomic gas (0.5 moles, 300K) has its pressure increased from 100,000 Pa to 200,000 Pa in a constant volume process. • A diatomic gas (0.7 moles, 300K) gets compressed from 0.4 m3 to 0.2 m3 in a constant temperature process. T increases, Q = DEint + PDV = 8102 J added T increases, Q = DEint = 3116 J added T stays constant, Q = –Won gas = –1210 J (i.e., 1210 J of heat removed from gas)

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