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Work = dw = F FF F  dL = (Apext )  dL = p pp pext  (AdL) But, AdL = V 2 - V 1 = dV (infinitesimal volume change) dw = p ext (V 2 - V 1 ) = p ext dV.

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Presentation on theme: "Work = dw = F FF F  dL = (Apext )  dL = p pp pext  (AdL) But, AdL = V 2 - V 1 = dV (infinitesimal volume change) dw = p ext (V 2 - V 1 ) = p ext dV."— Presentation transcript:

1 work = dw = F FF F  dL = (Apext )  dL = p pp pext  (AdL) But, AdL = V 2 - V 1 = dV (infinitesimal volume change) dw = p ext (V 2 - V 1 ) = p ext dV dL A T1T1 T2T2 Gas p ext =F / A  Definition of work using calculus: Infinitesimal work done dw by infinitesimal change in volume of gas dV: But sign is arbitrary, so choose dw = -p ext dV (w < 0 is work done by gas, dV > 0) dw = -p ext dV (Note! p is the e ee external pressure on the gas!) Movable Piston Bonus * Bonus * Bonus

2 dw = - pextdV Total work done in any change is the sum of little infinitesimal increments for an infinitesimal change dV.  dw =  - pextdV = w (work done by the system ) Two Examples : ( 1 ) pressure = constant = pexternal, V changes vi  vf w = - p ext dV= - p ext dV = - p ext ( v f - v i ) = -p ext  V  {Irreversible expansion if pext  pgas That is if, pgas = nRT/V  pexternal }

3 Example 2 : dV ≠ 0, but p ≠ const and T = const: P, V not const but PV = nRT = const (Isothermal change) w = - nRT = - nRT w = - nRT ln ( v f / v i ) w = -  pdV is the area under p in a plot of p vs V.] p ext = p gas = w = -  nRT (Called a reversible process.) {Reversible isothermal expansion because p ext = p gas } [Remembering that  f(x) dx is the area under f(x) in a plot of f(x) vs x,

4 Expansion At constantPres- sure p ext =P 2 Graphical representation of ∫ p ext dV V V i = V 1 V f = V 2 P1P1P1P1 P2P2P2P2 no work P P= nRT/V Isothermal reversible expansion V V i = V 1 V f = V 2 P P = nRT/V p ext = nRT/V  constant PV= PV=const PV=const is a hyperbola Compare the shaded area in the plot above to the shaded area in the plot for a reversible isothermal expansion with pext= pgas = nRT/V shaded area = -w P2P2P2P2 P1P1P1P1

5 Work done is NOT independent of path : Change the State of a gas two different ways: Consider n moles of an ideal gas w = 0 for 2nd step since V = const Final condition: T f = 300 K, V f = 1 liter, p f = 4 atm. Initial condition: T i = 300 K, V i = 2 liter, p i = 2 atm. Step 2: Warm at constant V: 2 atm, 1 liter, 150 K  4 atm, 1 liter, 300 K. Step 1 : 2 atm, 2 l, 300K cool at 2 atm, 1 l, 150K w = - p ext ( V f - V i ) for the first step, p ext = const = 2 atm w = - 2 atm ( 1 - 2 ) l = 2 l -atm w tot = 2 l -atm Path 1 consists of two steps:  V=0 for this step  V  0 for this step

6 Since nRT = const = PV = 4 l-atm  Path 2 is a single step reversible isothermal compression: 2 atm, 2 l, 300K  4 atm, 1 l, 300K (T constant) w = - p dV = - nRT = - nRT w = - nRT ln ( v f / v i ) = -nRT ln ( 1/2 ) w = -4 l-atm ( ln 1/2 ) = (.693 ) 4 l-atm = 2.772 l-atm Compare to w for path 1: w = 2 l-atm w for two different paths between same initial and fianl states is NOT the same. Work is NOT a state Function! p ext =p gas = nRT/V= p

7 Heat : Just as work is a form of energy, heat is also a form of energy. Heat is energy which can flow between bodies that are in thermal contact. In general heat can be converted to work and work to heat -- can exchange the various energy forms. Heat is also NOT a state function. The heat change occurring when a system changes state very definitely depends on the path. Can prove by doing experiments, or (for ideal gases) can use heat capacities to determine heat changes by different paths.

8 The First Law of Thermodynamics I) Energy is a state function for any system : State 1 State 2 Path a Path b ∆E a ∆E b ∆Ea and ∆Eb are both for going from 1  2 If E not a state function then: Suppose Ea > Eb - now go from state 1 to state 2 along path a, then return to 1 along path b. Ea  Eb

9 The First Law The energy increase of a system in going between two states equals the heat added to the system plus the work done on the system. Energy change = E = Ea - Eb E > 0. Have returned system to its original state and created energy. Experimentally find no situation in which energy is created, therefore, Ea = Eb and energy is a state function. No one has made a perpetual motion machine of 1st kind.

10 Totally empirical law. The result of observations in many, many experiments. q and w are NOT state functions and do depend on the path used to effect the change between the two states of the system. dE = dq + dw q > 0 for heat added to the system w > 0 for work done on the system (dV < 0) dw = -p ext dV (w < 0 is work done by system, dV > 0) E = q+w (Here is where choice of sign for w is made) E is a state function independent of the path.

11 Taking a system over different paths results in same E but different q, w: qa + wa qa, qb, qc all different, E = E2 - E1 Same for Paths a, b, c 1 2 q a, w a q b, w b q c, w c wa, wb, wc all different, but E2 – E1 = E = qc + wc = qb + wb =

12 Initial State and system: O2 and N2 gas at 25 C and P(O2) = P(N2) = 1 atm. (1 mole each) Measurements of E Suppose we want to measure E for the following change : Final State : 2 moles NO at 25oC, 1 atm. (This is really a conversion of energy stored in the chemical bonds of O2 and N2 into stored chemical energy in the NO bond.)

13 Change in energy for a chemical reaction carried out at constant volume is directly equal to the heat evolved or absorbed. We know E = q + w a) What is w? 1st let us carry the change above out at constant volume :N2 + O2  2NO Then no mechanical work is done by the gases as they react to form NO because they are not coupled to the world --- no force moving through a distance --- nothing moves  w = 0. E = qv If qv > 0 then E > 0 and energy or heat is absorbed by the system. This is called an endoergic reaction. If qv < 0 then E < 0 and energy or heat is evolved by the system. This is called an exoergic reaction.

14 Can we find or define a new state function which is equal to the heat evolved by a system undergoing a change at constant pressure rather than constant volume? i.e. is there a state function = qp? Yes! H  E + pV will have this property Note E, p, V are state fcts.  H must also be a state fct. Let us prove H = qp : (for changes carried out at constant p) E = q + w H = qp + w + p V, since p=const dHp = dqp + dw + + + + pdV ; dw = - pdV w = - p V for changes at const p dH = dqp - pdV + pdV = dqp  H = qp - p V + p V H = E +  (pV) H = qp dH = dqp


Download ppt "Work = dw = F FF F  dL = (Apext )  dL = p pp pext  (AdL) But, AdL = V 2 - V 1 = dV (infinitesimal volume change) dw = p ext (V 2 - V 1 ) = p ext dV."

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