Presentation on theme: "First Law of Thermodynamics Michael Moats Met Eng 3620 Chapter 2."— Presentation transcript:
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 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 * (h 2 -h 1 ) = mgh 2 – mgh 1 = Potential Energy at height 2 – Potential Energy at Point 1
Work and Heat A system interacts with its environment through work, w, or heat, q. w = -(U B – U A ) (if no heat is involved) If w < 0, work is done to the system If w > 0, work is done by the system q = (U B – U A ) (if no work is involved) If q < 0, work flows out of the body (exothermic) If q > 0, heat flows into the body (endothermic)
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 P1P1 P2P2 V2V2 V1V1 a b c 1 2
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.
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v means constant volume Hence in a constant volume process, the change in internal energy is equal to heat absorbed or withdrawn from the system. Constant Volume Processes
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 q p and rearranging a little gives
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is termed enthalpy, H H = U + PV Therefore q p = H 2 – H 1 = ΔH In a constant pressure system, the heat absorbed or withdrawn is equal to the change in enthalpy.
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 Heat Capacity, C is equal to the ratio of the heat absorbed or withdrawn from the system to the resultant change in temperature.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.
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.
Thermal Process at Constant V Define heat capacity at a constant volume Recalling that at a constant volume Leading to or
Thermal Process at Constant P Define heat capacity at a constant pressure Recalling that at a constant pressure Leading to or
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 c v and c p are the molar heat capacity at constant volume and pressure, respectively.
Molar Heat Capacity - II c p > c v c v – heat only needed to raise temperature c p – heat needed to raise temperature and perform work Therefore the difference between c v and c p is the work performed. Long derivation and further explanation in section 2.6
Reversible Adiabatic Processes Adiabatic; q = 0 Reversible; First law; Substitution gives us; For one mole of ideal gas; Recall that Leading to
Reversible Adiabatic – cont. Rearranging gives Combining exponents and recalling that c p -c v =R gives Defining a term, From the ideal gas law Finally gives constant
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
Example Calculation 10 liters of monatomic ideal gas at 25 o C and 10 atm are expanded to 1 atm. The c v = 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.
First Determine Size of System Using the ideal gas law
Isothermal Reversible Process Isothermal process; dT = 0, dU = 0 To calculate work, first we need to know the final volume. Then we integrate
Isothermal Reversible Process – continued Since dU = 0, q = w = 23.3 kJ Recall definition of enthalpy H = U + PV Isothermal = constant temperature
Reversible Adiabatic Expansion Adiabatic means q = 0 Recall Since c v = 3/2R, then c p = 5/2R and Solve for V 3 Solve for T 3 constant
Reversible Adiabatic Expansion - continued Text shows five examples of path does not matter in determining U.