1. AP Physics B Ch. 12: Laws of Thermodynamics

2. Internal energy (U) Sum of the kinetic energy of all particles in a system. For an ideal gas: U = N K ave U = N (3/2 k B T) U = n (3/2 R T) Since k B = R /N A

3. Zeroth Law of Thermodynamics Two objects placed in thermal contact will eventually come to the same temperature. Heat always flows from hot to cold until the objects reach thermal equilibrium.

4. Heat transfer Conduction Transfer of heat energy by collisions between particles inside a material Occurs if there is a difference in temperature between two parts of the conducting material Rate of heat transfer: H = Q/t (units are J/s – just like power, which is rate of energy transfer) H = kA (T hot –T cold )/L k = thermal conductivity T = temperature A = cross-sectional area L = thickness

5. Heat transfer

6. Other methods of heat transfer Convection: transfer of heat energy by the movement of a substance Heated fluid expands, decreases density, rises and replaces cooler fluid http://virtualskies.arc.nasa.gov/weather/3.ht ml http://virtualskies.arc.nasa.gov/weather/3.ht ml Radiation: transfer of heat energy by electromagnetic waves

7. Problem Find the energy transferred in 1.00 h by conduction through a concrete wall 2.0 m high, 3.65 m long, and 0.20 m thick if one side of the wall is held at 20 degrees C, and the other at 5 degrees C. k concrete = 1.3 J/s*m*°C

8. Methods of Energy Transfer Energy can be transferred by a system via: Heat (symbolized by letter Q) Work Work can be done on a system by an outside force Work can be done by the system on the environment Heat is energy transfer between a system and its surroundings resulting from random motion Work is energy transfer between a system and its surroundings resulting from organized motion

9. Work done by/on gas: Derivation

10. First law of thermodynamics Δ U = Q + W Where: Δ U = internal energy of system Q = energy transferred to system by heat W = work done on the system by the environment W = -P∆V ** remember that the work done by the system will have the opposite sign!

11. Problem A system absorbs 200 J of heat energy from the environment and does 100 J of work on the environment. What is its change in internal energy?

12. Sample Problem Calculate the work done by a gas that expands from 0.020 m 3 to 0.80 m 3 at constant atmospheric pressure. How much work is done by the environment when the gas expands this much?

13. Sample Problem What is the change in volume of a cylinder operating at atmospheric pressure if its internal energy decreases by 230 J when 120 J of heat are removed from it?

14. Thermal Processes State properties: depend on the physical state of the gas (pressure, volume, temperature, internal energy) Gas process: describes how a gas gets from one state to another Isothermal: process occurs at constant temperature Isobaric: process occurs at constant pressure Isometric/Isochoric: process occurs at constant volume Adiabatic: process is insulated, no heat energy enters or leaves system (Q = 0)

15. P-V Diagrams

16. Isothermal process

17. Isothermal Process Because P is inversely proportional to V at constant temperatures, the graph of P vs. V is hyperbolic for an isothermal process

18. Adiabatic process System is insulated from surroundings, or process occurs rapidly – thermally isolated No exchange of heat between system and surroundings Δ U = W If gas expands, the work done on the gas is negative and internal energy decreases (not an isothermal process) Examples: internal combustion engine, air being released from a punctured tire

19. Adiabatic Process In an adiabatic process, the temperature is not constant, resulting in a steeper curve than that of an isothermal process.

20. Isobaric Process In an isobaric process, the pressure is constant. The temperature and volume change. Since pressure is constant, the work done on the gas by the surroundings equals -P∆V, and the work done by the gas is P∆V. If the gas expands, it does positive work; the surroundings do negative work. If the gas is compressed, it does negative work; the surroundings do positive work Example: expansion of a gas in a cylinder with a freely- moving piston

21. Isobaric Process The area under the graph is equal to the work done by the gas (this is true for any process, but it is particularly easy to calculate on an isobaric P-V diagram).

22. Isochoric/Isometric/Isovolumetric Volume is constant No work is done by the gas. Therefore Δ U = Q So the change in the internal energy is equal to the amount of energy transferred into or out of the system by heat. Examples: heating a gas in a closed-volume container

23. Isochoric Process NO WORK IS DONE IN AN ISOCHORIC PROCESS

24. General Processes Occasionally a process will not follow any of the special models, but we can still analyze the PV diagram using the following facts: The area under the P-V curve is equal to the work done by the gas; the negative of this area is equal to the work done on the gas by the surroundings. The internal energy of the system can be determined using a combination of the ideal gas law and U = n (3/2 R T), where R = 8.31 J/K*mol

25. Sample Problem One mole of a gas goes from state A (200 kPa and 0.5 m 3 ) to state B (150 kPa and 1.5 m 3 ). A) Draw this process, assuming the smoothest possible transition (straight line) for the process. B) Calculate the work done by the gas during the process. C) Calculate the work done by the environment during the process.

27. Things to remember The internal energy of a gas depends on its temperature. On an isotherm, the internal energy remains constant, because the temperature is constant; pressure and volume change The higher the isotherm on the P-V diagram, the higher the temperature The area under the P-V curve is equal to the work done by the gas The amount of work done by a gas depends on the path it takes from one state to another.