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. 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.
Pressure
Volume P1 P2 V2 V1 a b c 1 2

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Hence in a constant volume process, the change in internal energy is equal to heat absorbed or withdrawn from the system.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In a constant pressure system, the heat absorbed or withdrawn is equal to the change in enthalpy.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Heat Capacity, C is equal to the ratio of the heat absorbed or withdrawn from the system to the resultant change in temperature.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. Thermal Process at Constant V
Define heat capacity at a constant volume
Recalling that at a constant volume
Leading to or

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

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

18. 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.
and

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

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