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work = dw = F  dL = ( Ap ext )  dL = p ext  ( A dL) PowerPoint Presentation
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work = dw = F  dL = ( Ap ext )  dL = p ext  ( A dL) - PowerPoint PPT Presentation

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work = dw = F  dL = ( Ap ext )  dL = p ext  ( A dL)
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  1. T1 T2 Bonus * Bonus * Bonus Definition of work using calculus: Infinitesimal work done dw by infinitesimal change in volume of gas dV: dL A pext=F / A Gas Movable Piston work = dw = F dL = (Apext)  dL = pext  (AdL) But, AdL = V2 - V1 =dV (infinitesimal volume change)  dw = pext(V2 - V1) = pextdV But sign is arbitrary, so choose dw = -pextdV (w < 0 is work done bygas, dV > 0) dw = -pextdV (Note! p is the external pressure on the gas!)

  2. w = - pextdV= - pext dV = -pext ( vf - vi ) = -pextV 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 {Irreversible expansion if pext  pgas That is if, pgas = nRT/V  pexternal }

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

  4. Graphical representation of ∫ pext dV Expansion At constantPres- sure pext=P2 P P1 no work P= nRT/V Isothermal reversible expansion P2 shaded area = -w V Vi = V1 Vf = V2 P pext= nRT/V  constant 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 P1 P = nRT/V PV= const P2 shaded area = -w V Vi = V1 Vf = V2

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

  6. w = - p dV = - nRT = - nRT Path 2 is a single step reversible isothermal compression: pext=pgas= nRT/V= p 2 atm, 2 l, 300K  4 atm, 1 l, 300K (T constant) w = - nRT ln ( vf / vi ) = -nRT ln ( 1/2 ) Since nRT = const = PV = 4 l-atm  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!

  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 : ∆Ea Path a State 1 State 2 ∆Eb Path b ∆Ea and ∆Eb are both for going from 1  2 If E not a state function then: Ea  Eb Suppose Ea > Eb - now go from state 1 to state 2 along path a, then return to 1 along path b.

  9. 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. 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.

  10. E = q+w (Here is where choice of sign for w is made) dE = dq + dw q > 0 for heat added to the system w > 0 for work done on the system (dV < 0) dw = -pextdV (w < 0 is work done bysystem, dV > 0) Totally empirical law. The result of observations in many, many experiments. E is a state function independent of the path. 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.

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

  12. Measurements of E Suppose we want to measure E for the following change : Initial State and system: O2 and N2 gas at 25 C and P(O2) = P(N2) = 1 atm. (1 mole each) 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. 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 Change in energy for a chemical reaction carried out at constant volume is directly equal to the heat evolved or absorbed. 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) H = E +  (pV) E = q + w H = qp + w + p V, since p=const w = - p V for changes at const p H = qp  H = qp - p V + p V  dHp = dqp + dw + pdV ; dw = - pdV dH = dqp dH = dqp - pdV + pdV = dqp 