1 / 30

Lecture 5

Lecture 5. First Law of Thermodynamics. You can’t get something for nothing. Nothing is for free. First Law of Thermodynamics. We will discuss these statements later…. Thermodynamic System. Consider a closed system (e.g., a parcel of air).

bern
Download Presentation

Lecture 5

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Lecture 5 First Law of Thermodynamics

  2. You can’t get something for nothing. Nothing is for free. First Law of Thermodynamics We will discuss these statements later…

  3. Thermodynamic System Consider a closed system (e.g., a parcel of air). It has internal energy (“u”) = energy due to molecular kinetic and potential energies. Suppose some energy (dq) is added to the system. Example: via radiation from the sun What happens?

  4. Thermodynamic System Some of the energy goes into work done (dw) by the system against its surroundings. Example: expansion What’s left is a change in internal energy. du = dq – dw Conservation of Energy principle “nothing is for free”

  5. Thermodynamic System Heat Environment System System can exchange energy with environment via heat flow.

  6. Thermodynamic System In addition, system can do work on environment or vice versa. Example: expansion

  7. First Law, General Form • dU = dQ – dW • dU = change in internal energy of system • dQ = heat exchanged with environment • dQ > 0  heat flowing into system • dW = work done by or on system • dW > 0  system is doing work

  8. Ideal Gas • Consider a system consisting of an ideal gas in a cylinder. • Cylinder has a piston, which allows volume to be changed.

  9. Cylinder weights more weight, more pressure Walls of cylinder: 1) perfect heat conductors 2) perfect insulators Gas

  10. Cases • Case 1: Walls of cylinder are perfect conductors •  heat can freely flow between system and environment •  in equilibrium, temperature of system must equal temperature of environment • Case 2: Walls of cylinder are perfect insulators • no heat flow between system and environment •  temperatures of system and environment need not be equal

  11. Work (Qualitative) • System does work on environment (dW > 0) if gas expands • (Piston is pushed upward.) • Work is done on system (dW < 0) if gas contracts • (Piston is pushed downward.)

  12. Work (Quantitative) • Suppose piston is pushed upward a distance dx.

  13. Work (Quantitative) • Suppose piston is pushed upward a distance dx. Area of piston = A Pressure force = pA dW = pAdx = pdV dx

  14. First Law, Ideal Gas Usually, we are interested in energy per unit mass Divide both sides by m Define u = U/m; q = Q/m;  = V/m

  15. First Law, Ideal Gas (per unit mass)

  16. Internal Energy • For an ideal gas, u is a function of T only • u is an increasing function of T • Change in internal energy depends only on change in temperature. • Doesn’t depend on the way in which that temperature change is accomplished. • du = kdT, where k is a constant • (Value of k will be determined shortly.)

  17. Heat Capacity, C SI units: JK-1 Amount of heat required to change a “body’s” temperature by a given amount For a gas, C depends on the particular process. Cv = heat capacity at constant volume (dV = 0) Cp = heat capacity at constant pressure (dp = 0)

  18. Specific Heat • Specific heat is heat capacity per unit mass • cv (lower case) = specific heat at constant volume • cp = specific heat at constant pressure • Specific heat is the heat energy needed to raise the temperature of a unit mass of a substance by one degree.

  19. Specific Heat Suppose we add some thermal energy (dq) to a unit mass of a substance like air, water, soil. We expect T(substance) to increase How much? We can define Specific Heat as Heat added Temp change Constant volume Constant pressure

  20. Specific Heat of Dry Air • cv = 717 Jkg-1K-1 • cp = 1004 Jkg-1K-1 • Note: cp – cv = 287 Jkg-1K-1 • Look familiar? • cp – cv = Rd • (Not a coincidence!)

  21. Constant Volume Processes 0

  22. Internal Energy Change • The following is always true: (3.40) W&H

  23. First Law for Ideal Gas, Re-written (1)

  24. Constant-Pressure Processes • Go back to ideal-gas law: Take differential of both sides:

  25. But,…  (2)

  26. Substitute (2) into (1)

  27. Simplify (3) Constant-pressure process 

  28. But, … For a constant-pressure process,

  29. Compare and Result:

  30. Rewrite (3) (4) We will use this in the next lecture.

More Related