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P Dv = - Nrt = - Nrt Úvi Úvi V Úvi V W = - Nrt Ln ( Vf / Vi ) = -Nrt Ln ( 1/2 )

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P Dv = - Nrt = - Nrt Úvi Úvi V Úvi V W = - Nrt Ln ( Vf / Vi ) = -Nrt Ln ( 1/2 ) Definition of work using calculus: Bonus * Bonus * Bonus Infinitesimal work done dw by infinitesimal change in volume of gas dV: A dL pext=F / A Gas Movable Piston T1 T2 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 Æ dw = -pextdV (w < 0 is work done by gas, dV > 0) dw = -pextdV (Note! p is the exexternaternall pressure on the gas!) dw = - pextdV Total work done in any change is the sum of little infinitesimal increments for an infinitesimal change dV. Ú dw = Ú - pextdV = w (work done by the system ) Two Examples : ( 1 ) pressure = constant = pexternal, V changes vi Æ vf vf vf w = - pextdV= - pext dV = - pext ( vf - vi ) = -pextDV fi Úvi Úvi {Irreversible expansion if pext ≠≠ pgas That is if, pgas = nRT/V ≠≠ pexternal } Example 2 : dV ≠ 0, but p ≠ const and T = const: nRT pext= pgas = (Called a reversible process.) V dV w = - Ú nRT V v vf f dV dV w = - nRT = - nRT Úvi Úvi V V [Remembering that Ú f(x) dx is the w = - nRT ln ( vf / vi ) area under f(x) in a plot of f(x) vs x, w = - Ú 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 } Graphical representation of ∫ pext dV P Expansion At constantPres- P1 sure pext=P2 no work P= nRT/V Isothermal reversible P P2 expansion shaded area = -w V Vi = V1 Vf = V2 P PV=const is pext= nRT/V Compare the shaded ≠ constant a hyperbola area in the plot above P1 P = nRT/V to the shaded area in the PV= plot for a reversible const P2 isothermal expansion with shaded area = -w pext= pgas = nRT/V V Vi = V1 Vf = V2 Work done is NOT independent of path : Change the State of a gas two different ways: 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: DV≠0 for Step 1 : 2 atm, 2 l, 300K cool at 2 atm, 1 l, 150K ææconst æ -ææp Æ this step compress Step 2: Warm at constant V: 2 atm, 1 liter, 150 K Æ DV=0 for 4 atm, 1 liter, 300 K. this step 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 Path 2 is a single step reversible isothermal compression: 2 atm, 2 l, 300K Æ 4 atm, 1 l, 300K (T constant) pext=pgas= nRT/V= p v v v f f dV f dV w = - p dV = - nRT = - nRT Úvi Úvi V Úvi V w = - nRT ln ( vf / vi ) = -nRT ln ( 1/2 ) Since nRT = const = PV = 4 l-atm Æ 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! 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. The First Law of Thermodynamics I)I) EEnnergergyy isis aa stastatete fufunnctioctionn foforr aannyy sysystemstem :: ∆E Path a a State 1 State 1 State 2 ∆Eb Path b ∆Ea and ∆Eb are both for going from 1 Æ 2 If E not a state function then: DEa ≠ DEb Suppose DEa > DEb - now go from state 1 to state 2 along path a, then return to 1 along path b. Energy change = DE = DEa - DEb DE > 0. Have returned system to its original state and created energy. Experimentally find no situation in which energy is created, therefore, DEa = DEb 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. DE = 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 by system, dV > 0) Totally empirical law. The result of observations in many, many experiments. DE 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. Taking a system over different paths results in same DE but different q, w: qa , wa 1 2 qb , wb DE = E - E q , w 2 1 qc , wc Same for Paths a, b, c qa, qb, qc all different, wa, wb, wc all different, but qa + wa = qb + wb = qc + wc = DE = E2 – E1 Measurements of DE Suppose we want to measure DE 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.) We know DE = q + w a) What is w? 1st let us carry the change above out at constant volume : N2 + O2 Æ 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 Æ w = 0. DE = 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 DE > 0 and energy or heat is absorbed by the system. This is called an endoergic reaction. If qv < 0 then DE < 0 and energy or heat is evolved by the system. This is called an exoergic reaction. 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 DH = qp : (for changes carried out at constant p) DE = q + w DH = DE + D (pV) DH = qp + w + p DV, since p=const w = - p DV for changes at const p \ DH = qp - p DV + p DV Æ DH = qp dHp = dqp + dw + pdV ; dw = - pdV dH = dqp - pdV + pdV = dqp Æ dH = dqp.
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