6.3 The First Law of Thermodynamics

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1 6.3 The First Law of Thermodynamics Physics Tool box Thermodynamic System - any collection of objects that is convenient to regard as a unit, and may have the potential to exchange energy with its surroundings. First Law of Thermodynamics the change in internal energy of a closed system is given by: U Q W, where Q is the heat added to the system and W is work done by the system. Isobaric process P=0 (pressure remains constant) Isothermal process T 0 Adiabtatic process Q 0, no heat is allowed to flow into or out of the system. W Fd PAd P V We have studied energy transfer through mechanical work and through heat transfer, we now are going to combine these principles. We always talk about energy transfer to or from some specific system. The system may be a quantity of material, biological, or mechanical in nature. In general, a thermodynamics system is any collection of objects that is convenient to regard as a unit, and may have the potential to exchange energy with its surroundings. Thermodynamic process is one in which there are changes in the state of a thermodynamic system. Signs for Heat and Work in Thermodynamics We describe the energy relations in any thermodynamics process in terms of the quantity of heat Q added to the system and work W done by the system. Both Q and W may be positive, negative, or zero.

2 A positive Q represents heat flow into the system (with a corresponding input of energy), negative Q represents heat flow out of the system. A positive value of W represents work done by the system against its surroundings (like work done by expanding gas) and corresponds to energy leaving the system. Negative W is work done on the system (compressing gas) and represents energy entering the system. Work Done During a Volume change When a gas expands, it pushes outward on its boundary surfaces as they move outward. Hence an expanding gas always does positive work (the same is tru for any expanding solid or fluid material). Looking at the process at the molecular level, we notice that when a molecule collides with a stationary surface, it exerts a momentary force on the wall and in moving the wall does positive work. The Force, F, is the pressure, P, exerted on the area A. This change d and surface area represent a change in volume, V W Fd PA d P Ad P V True equation: W V2 V1 PdV We can represent this relationship as a graph of P (pressure) as a function of V (volume). Such a PV diagram. Work, w is the area under the curve. Remember work is positive when a system expands thus the area under the graph in (a) is positive. The work is negative during a compression thus so is the area in (b) (notice V 2 in to the left of V. 1

3 Internal Energy and the First Law of Thermodynamics We tentatively define the internal energy of a system as the sum of the kinetic energies of all of its constituent particles (plus the sum of all the potential energies of interactions among these particles). We use the symbol U for internal energy (not the same as u in potential energy). During a change of state of the system the internal energy may change from an initial value U1 to a final value U 2, we denote the change in internal energy as U U 2 U1. We know that heat transfer is energy transfer. When we add a quantity of heat Q and it does no work during the process, the internal energy increases by the amount equal to q; that is U Q. When the system does work by expanding against its surroundings and no heat is added during the process, energy leaves the system and so the internal energy decreases, that is U W. When both heat transfer and work occur, the total change in internal energy is: U Q W The total change of energy is independent of path. The change in internal energy of a system during any thermodynamic process depends only on the initial and final states, not the path leading from one to another. Example You are to eat a 1000 calorie (1 food calorie = 1 kcal = 4190 ) piece of cake, and then run up stairs to work off the energy you have taken in. How high do you have to climb? Assume your mass is 50kg. Solution: Eating the cake corresponds to heat flow into the system, and running the stairs means you do work. Since you want to work off the energy, U 0 U Q W kcal kcal Then Q mgh Q h mg m 50kg s 8551m Assuming 100% efficiency in conversion of food energy into mechanical work.

4 Example A series of thermodynamic processes is shown in the PV diagram. In process ab, 150 of heat is added to the system, and in process bd, 600 of heat is added. Find a) the internal energy change in process ab b) the internal energy change in process abd c) the total heat added in process acd Solution: a) No volume change occurs during ab, so Wab 0 Thus U Q W b ab ab b) first we notice that process bd occurs at constant pressure, so the work done by the system during this expansion is bd W P V V Pa m m 240 The total work for process abd is W W W abd ab bd The total heat is Q Q Q abd ab bd Apply the first law, U Q W abd abd abd

5 c) Because U is independent n the path taken, U U 510 The total work for the path acd is: W W W acd ac cd 2 1 P V V Pa m m Now for heat acd abd Q U W acd acd acd Kind of Thermodynamic Processes Adiabatic Process A process with no heat transfer in to or out of the system; Q=0, thus U W Isochoric Process A process with a constant volume. When the volume of a thermodynamic system is constant, it does no work on its surroundings. Thus W=0 and U Q Isothermal Process Is a constant temperature process. For a process to be isothermal, any heat flow into or out of the system must occur slowly enough that thermal equilibrium is maintained.