Presentation on theme: "First Law of Thermodynamics"— Presentation transcript:
1 First Law of Thermodynamics Michael MoatsMet Eng 3620Chapter 2
2 Thermodynamic Functions To adequately describe the energy state of a system two terms are introducedInternal Energy, UEnthalpy, HInternal Energy related to the energy of a body.Example: Potential EnergyBody 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 HeatA 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 systemIf w > 0, work is done by the systemq = (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.PressureVolumeP1P2V2V1abc12
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.
7 Constant Volume Processes If during a process the system maintains a constant volume, then no work is performed.Recall andThuswhere the subscript v means constant volumeHence in a constant volume process, the change in internal energy is equal to heat absorbed or withdrawn from the system.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 giveswhere the subscript p means constant pressureSolving for qp and rearranging a little gives
9 EnthalpySince U, P and V are all state function, the expression U+PV is also a state function.This state function is termed enthalpy, HH = U + PVTherefore qp = H2 – H1 = ΔHIn a constant pressure system, the heat absorbed or withdrawn is equal to the change in enthalpy.
10 Heat CapacityBefore 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a new term is needed to make the calculations easier.Heat Capacity, C is equal to the ratio of the heat absorbed or withdrawn from the system to the resultant change in temperature.Note: This is only true when phase change does not occur.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 volumeRecalling that at a constant volumeLeading toor
13 Thermal Process at Constant P Define heat capacity at a constant pressureRecalling that at a constant pressureLeading toor
14 Molar Heat CapacityHeat capacity is an extensive property (e.g. dependent on size of system)Useful to define molar heat capacitywhere 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 > cvcv – heat only needed to raise temperaturecp – heat needed to raise temperature and perform workTherefore the difference between cv and cp is the work performed.Long derivation and further explanationin section 2.6
16 Reversible Adiabatic Processes Adiabatic; q = 0Reversible;First law;Substitution gives us;For one mole of ideal gas;Recall thatLeading to
17 Reversible Adiabatic – cont. Rearranging givesCombining exponents and recalling that cp-cv=R givesDefining a term,From the ideal gas lawFinallygivesconstant
18 Reversible Isothermal P or V Change RecallIsothermal means dT = 0, so dU = 0Rearranging first lawSubstituting reversible workideal gas law givesIntegration leavesIsothermal process occurs at constant internal energy and work done = heat absorbed.and
19 Example Calculation10 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 = 0To 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 kJRecall definition of enthalpyH = U + PVIsothermal = constant temperature
23 Reversible Adiabatic Expansion Adiabatic means q = 0RecallSince cv = 3/2R, then cp = 5/2R andSolve for V3Solve for T3constant
24 Reversible Adiabatic Expansion - continued Text shows five examples of path does not matter in determining DU.