1 / 24

First Law of Thermodynamics

First Law of Thermodynamics. Michael Moats Met Eng 3620 Chapter 2. Thermodynamic Functions. To adequately describe the energy state of a system two terms are introduced Internal Energy, U Enthalpy, H Internal Energy related to the energy of a body. Example: Potential Energy

sonya-banks
Download Presentation

First Law of Thermodynamics

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. First Law of Thermodynamics Michael Moats Met Eng 3620 Chapter 2

  2. Thermodynamic Functions • To adequately describe the energy state of a system two terms are introduced • Internal Energy, U • Enthalpy, H • Internal Energy related to the energy of a body. • Example: Potential Energy • Body has a potential energy related to its mass and height (mgh) • To move a body from one height to another takes work, w. w = force * distance = mg * (h2-h1) = mgh2 – mgh1 = Potential Energy at height 2 – Potential Energy at Point 1

  3. Work and Heat • A system interacts with its environment through work, w, or heat, q. w = -(UB – UA) (if no heat is involved) • If w < 0, work is done to the system • If w > 0, work is done by the system q = (UB – UA) (if no work is involved) • If q < 0, work flows out of the body (exothermic) • If q > 0, heat flows into the body (endothermic)

  4. 1 P1 a b Pressure c P2 2 V1 V2 Volume Change in Internal Energy U is a state function. Therefore, the path between condition 1 and condition 2 do not affect the differential value. Heat and work are not state functions. The path taken between condition 1 and 2 does matter, hence the use of a partial differential.

  5. Fixing Internal Energy • For a simple system consisting of a given amount of substance of a fixed composition, U is fixed once any two properties (independent variables) are fixed.

  6. Convenient Processes • Isochore or Isometric - Constant Volume • Isobaric – Constant Pressure • Isothermal – Constant Temperature • Adiabatic – q = 0

  7. Constant Volume Processes • If during a process the system maintains a constant volume, then no work is performed. Recall and Thus where the subscript v means constant volume • Hence in a constant volume process, the change in internal energy is equal to heat absorbed or withdrawn from the system.

  8. Constant Pressure Processes • Again starting with the first law and the definition of work: • Combining them and integrating gives where the subscript p means constant pressure • Solving for qp and rearranging a little gives

  9. Enthalpy • Since U, P and V are all state function, the expression U+PV is also a state function. • This state function is termed enthalpy, H H = U + PV • Therefore qp = H2 – H1 = ΔH • In a constant pressure system, the heat absorbed or withdrawn is equal to the change in enthalpy.

  10. Heat Capacity • Before discussing isothermal or adiabatic processes, a new term is needed to make the calculations easier. • Heat Capacity, C is equal to the ratio of the heat absorbed or withdrawn from the system to the resultant change in temperature. • Note: This is only true when phase change does not occur.

  11. Defining Thermal Process Path • To state that the system has changed temperature is not enough to define change in the internal energy. • It is most convenient to combine change in temperature while holding volume or pressure constant. • Then calculations can be made as to the work performed and/or heat generated.

  12. or Thermal Process at Constant V • Define heat capacity at a constant volume • Recalling that at a constant volume • Leading to

  13. or Thermal Process at Constant P • Define heat capacity at a constant pressure • Recalling that at a constant pressure • Leading to

  14. Molar Heat Capacity • Heat capacity is an extensive property (e.g. dependent on size of system) • Useful to define molar heat capacity where n is the number of moles and cv and cp are the molar heat capacity at constant volume and pressure, respectively.

  15. Molar Heat Capacity - II • cp > cv • cv – heat only needed to raise temperature • cp – heat needed to raise temperature and perform work • Therefore the difference between cv and cp is the work performed. Long derivation and further explanation in section 2.6

  16. Reversible Adiabatic Processes • Adiabatic; q = 0 • Reversible; • First law; • Substitution gives us; • For one mole of ideal gas; • Recall that • Leading to

  17. gives constant Reversible Adiabatic – cont. • Rearranging gives • Combining exponents and recalling that cp-cv=R gives • Defining a term, • From the ideal gas law • Finally

  18. and Reversible Isothermal P or V Change • Recall • Isothermal means dT = 0, so dU = 0 • Rearranging first law • Substituting reversible work ideal gas law gives • Integration leaves • Isothermal process occurs at constant internal energy and work done = heat absorbed.

  19. Example Calculation • 10 liters of monatomic ideal gas at 25oC and 10 atm are expanded to 1 atm. The cv = 3/2R. Calculate work done, heat absorbed and the change in internal energy and enthalpy for both a reversible isothermal process and an adiabatic and reversible process.

  20. First Determine Size of System • Using the ideal gas law

  21. Isothermal Reversible Process • Isothermal process; dT = 0, dU = 0 • To calculate work, first we need to know the final volume. • Then we integrate

  22. Isothermal Reversible Process –continued • Since dU = 0, q = w = 23.3 kJ • Recall definition of enthalpy H = U + PV Isothermal = constant temperature

  23. constant Reversible Adiabatic Expansion • Adiabatic means q = 0 • Recall • Since cv = 3/2R, then cp = 5/2R and • Solve for V3 • Solve for T3

  24. Reversible Adiabatic Expansion - continued Text shows five examples of path does not matter in determining DU.

More Related