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** dw = -pextdV Bonus * Bonus * Bonus**

Definition of work using calculus: 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 dw = -pextdV (w < 0 is work done by gas, dV > 0) dw = -pextdV (Note! p is the external pressure on the gas!)

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**{Irreversible expansion if pext pgas**

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 = pextdV= - pext dV = - pext ( vf - vi ) = -pextV {Irreversible expansion if pext pgas That is if, pgas = nRT/V pexternal }

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**{Reversible isothermal expansion because pext= pgas }**

Example 2 : dV ≠ 0, but p ≠ const and T = const: pext= pgas = (Called a reversible process.) w = - nRT w = nRT = - nRT [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 }

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**Isothermal reversible**

Graphical representation of ∫ pext dV P Expansion At constantPres- sure pext=P2 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

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**V0 for this step V=0 for this step**

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: V0 for this step Step 1 : 2 atm, 2 l, 300K cool at atm, 1 l, 150K 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 ( ) l = 2 l -atm w = 0 for 2nd step since V = const wtot = 2 l -atm

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**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 = p dV = nRT = - nRT 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 = 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!

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**Heat : Just as work is a form of energy, heat is also a form of**

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.

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**I) Energy is a state function for any system :**

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.

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**The First Law 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.

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**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 by system, 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.

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**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

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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.)

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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.

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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 = E + (pV) H = qp + w + p V, since p=const w = - p V for changes at const p H = qp - p V + p V H = qp dHp = dqp + dw + pdV ; dw = - pdV dH = dqp dH = dqp - pdV + pdV = dqp

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