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Home , Heat, Ideal gas, State function

IdealIdeal GasGas StateState functions:functions: ConsiderConsider 11 molemole ofof anan idealideal gasgas

PV = RT

Suppose begin with:

o P1 = 5 atm V2 = 4.92 liters T1 = 300 K

And end with:

o P2 = 4 atm V2 = 12.3 liters T2 = 600 K ∆∆PP == 44--55 == --11 atmatm →→ ∆∆VV == 7.387.38 litersliters

∆∆TT == 600600 -- 300300 == 300300°° If we got to the condition 4 atm, 12.3 liters, and 600°°K by going as follows: 5 atm, 4.92 liters, 300°°→→4 atm, 6.15 liters, 300°° →→4 atm, 12.3 liters, 600°°or by the path: 5 atm, 4.92 liters, 300°°→→5 atm, 9.84 liters, 600°° →→4 atm, 12.3 liters, 600°° Would get same ∆∆P,P,∆∆V,V,∆∆TT ChangesChanges inin statestate functionsfunctions areare independentindependent ofof path.path. Important State Functions: T, P, V, , and any combination of the above. Important Non State Functions: , . Bonus * Bonus * Bonus 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 ) Two Examples :

( 1 ) = constant = pexternal, →→ V changes vi vf

vf vf ∆∆ ⇒⇒ w = ∫ - pextdV= - pext∫ dV = - pext ( vf - vi ) = -pext V vi vi ≠≠ {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 = - nRT V

v vf f dV ∫ dV w = -∫ nRT = - nRT v vi V i V

[Remembering that ∫ f(x) dx is the w = - nRT ln ( vf / vi ) area under f(x) in a plot of f(x) vs x, 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 } Graphical representation of ∫ pext dV (If done at constant T, P Expansion At pressure follows constantPres- curved P= nRT/V path

P1 sure pext=P2 while pext remains fixed) 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

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 Step 1 : 2 atm, 2 l, 300K  cool  at  → 2 atm, 1 l, 150K const −p this step compress Step 2: Warm at constant V: 2 atm, 1 liter, 150 K → ∆V=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 → 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 ∫ vi vi V vi V 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 ! 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 ) can use heat capacities to determine heat changes by different paths. TheThe FirstFirst LawLaw ofof ThermodynamicsThermodynamics I)I) EnergyEnergy isis aa statestate functionfunction forfor anyany systemsystem ::

∆E Path a a State 1 State 2

∆Eb Path b ∆Ea and ∆Eb are both for going from 1 →→ 2 ∆∆ ≠≠ ∆∆ IfIf 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. ∆∆ ∆∆ ∆∆ 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 machine of 1st kind.

TheThe FirstFirst LawLaw

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. ∆∆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. TakingTaking aa systemsystem overover differentdifferent pathspaths resultsresults inin samesame ∆∆EE butbut differentdifferent q,q, w:w:

qa , wa

1 2

qb , wb

∆∆E = 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 = E = E2 – E1 MeasurementsMeasurements ofof ∆∆EE

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 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 in the NO bond.) We know ∆∆E = q + w a) What is w? 1st let us carry the change above out at →→ constant : 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 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. 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! HH ≡≡ EE ++ pVpV 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 outout atat constantconstant p) ∆∆E = q + w ∆∆H = ∆∆E + ∆∆ (pV) ∆∆ ∆∆ H = qp + w + p V, since p=const w = - p ∆∆V for changes at const p ∴∴ ∆∆ ∆∆ ∆∆ →→ ∆∆H = q H = qp - p V + p V H = qp

dHp = dqp + dw + pdV ; dw = - pdV →→ dHdH == dqdq dH = dqp - pdV + pdV = dqp p H is called the . The enthalpy change for a syste m at constant pressure equals the heat absorbed or released.

∆∆ qp < 0 heat evolved H < 0 ∆∆ Endothermic process qp > 0 heat absorbed H > 0 ∆∆ ∆∆ Are H and E similar ( i.e. are qv, qp similar ) ? ∆∆H = ∆∆E + ∆∆ ( pV ) Obviously when ∆∆ ( pV ) << ∆∆E ∆∆H ≈ ∆∆E For reactions where only and are involved, and where reaction is carried out at constant pressure (usually in the presence of the ) ∆∆V and therefore ∆∆ ( pV ) = p ∆∆V is usually negligible.

For gas reactions, ∆∆ ( pV ) can be large. For a reaction carried out at const T: