First Law of Thermodynamics

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means constant volume • Hence in a constant volume process, the change in internal energy is equal to heat absorbed or withdrawn from the system.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is termed enthalpy, H H = U + PV • Therefore qp = H2 – H1 = ΔH • In a constant pressure system, the heat absorbed or withdrawn is equal to the change in enthalpy.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.