Pressure Volume Work 2

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Sabina Patterson
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1 ressure olume Work Multi-stage Expansion 1 3 w= 4 5 ( ) + 3( 3 ) + 4( 4 3) + ( ) Reversible Expansion Make steps so small that hen d 0, d 0 δ w= d ( ) = + d d int d int w= dw= d path 1 int For ideal gases int = nr / 1 wrev = nrln = nrln and at fixed temperature 1 1 aul ercival

2 Heat apacity ransfer of heat to a system may result in a rise in. δ q= d path function, so depends on conditions. Define: δ q at constant volume, no work = d δ q = d at constant pressure, only work From 1 st Law, du =δ q +δ w =δq d du =δ q f o r d = 0 U = assume no other work Similarly dh = du + d d = ( δq d) + d d =δ q for d= 0 H = dh = du + d = du + nrd For ideal gases ( ) d = d + nrd = + R aul ercival

3 he Relation Between and But since H U = U U = + U U du = d + d U U U = + H = U + U = + work needed to overcome U = intermolecular forces internal pressure pansion per degree For ideal gases U nr = 0, = = nr For liquids and solids is so small that aul ercival

4 Adiabatic Expansion Reversible adiabatic pansion of ideal gases: du =δw δ q = 0 nr d = d = d ideal gases only 1 1 d = R d 1 1 ln / = Rln / ( 1) ( 1) = ( ) ln ( / 1) ( ) = ( γ ) ( ) ln / 1 ln / 1 1 = 1 1 γ 1 γ= Also, since 1 1 γ = = or = γ γ 1 1 = 1 1 ( γ 1)/ γ 1 1 γ 1 aul ercival

5 Given he lausius Inequality Substitute into the 1 st Law: All act differentials, so path independent. But Even more generally, onditions for: ds = δ q rev du = ds d ( ) d 0 surr thermal equilibrium and du =δ q +δ w =δq d ds d =δq d δq ds = + ( ) δ w = d ( ) surr dsuniv = dssurr + d d If >, d > 0; if <, d < 0 lausius Inequality ds δq rev Fundamental Equation of hermodynamics mechanical equilibrium In general, i.e. any path Equal for reversible change aul ercival

6 he Fundamental Equation of hermodynamics ombine or du =δq d with du = ds d ds = 1 ds = du + d δ q rev Reversible change but true for all paths since d act his fundamental equation generates many more relationships. U U Example 1: omparison with du = ds + d S S U U = and = S Example : onsider that du is act and cross differentiate. = S S S a Maxwell relation Example 3: S S = S S = = S cyclic rule and again! = = =α κ aul ercival

7 How Entropy Depends on and ompare with 1 ds = du + d U U U du = d + d = d + d 1 U ds = d + d + S S ds = d + d S = S 1 U = + Δ S = d 0 for ideal gases 1 Δ S = d = nr d = nrln / ( ) 1 For any substance, For ideal gases, (eqs 3.7.4, 6.1.6) = α + κ ds d d Δ S = ln + nrln 1 1 assuming is independent aul ercival

8 Entropy Depends on and (not in tt) 1 ds = du + d First problem: replace du ; second problem: replace d. Use But du = dh d d 1 ds = dh d and both are solved! H H H dh = d + d = d + d ompare with 1 H ds = d + d S S ds = d + d unnatural variables S = Δ S = d 1 Δ = = S d nr d ( ) ( ) Δ S = nrln / = nrln / S 1 H = 0 for ideal gases 1 1 aul ercival

9 Entropy of Mixing onsider the mixing of two ideal gases :,, 1 n 1,, n,, 1 + n 1 +n n Δ S = nrln = nrln = nrln χ n1+ n n Δ S = n Rln = n Rln = n Rln χ 1+ n1+ n ( ) ( ) Δ S = nrln χ n Rln χ = n + n R χ ln χ +χ ln χ mix In general Δ S = χ χ mix ntot R iln i his pression applies to the arrangement of objects (molecules) just as well as fluids (gases and liquids). For ample, arrange N identical atoms in N sites in a crystal: Ω= N!/ N! = 1 S = kln Ω= 0 ompare with the arrangement of two types of atom, A and B. Ω= N! S k( ln N! ln NA! ln NB! )!! Δ = N A N B Application of Stirling s approximation ( ) leads to Δ S = kn( χ ln χ +χ ln χ ) config A A B B i ln z! = zln z z aul ercival

10 he arnot ycle A H B D L step A B B D D A w q ΔU nr ln( / ) w 0 H B A AB ΔU 0 ( ) B H L nr ln( / ) w 0 L D D ΔU 0 ( ) DA H L otal nr( ) ln( / ) w H L A B cyc 0 w nr ( )ln( / ) ε= = = 1 q nr ln( / ) out H L A B L in H A B H ( H L) ε= H for best efficiency, maximize H minimize L aul ercival

11 How Free Energy Depends on G = H S Δ G =ΔH ΔS dg = d Sd (1) () (3) definition constant fundamental eqn. But G G H = S = G G H = ( G/ ) 1 G G = from (1) (4) (5) Gibbs-Helmholtz Equation ( G/ ) H = substitute (4) in (5) alternative form: ( G/ ) (1/ ) = H By applying the Gibbs-Helmholtz equation to both reactants and products of a chemical reaction, Δ ( G/ ) ΔH = aul ercival

12 How Free Energy Depends on G = Δ G = G G1 = d 1 For solids and liquids does not change much with, so G ( ) G ( ) + ( ) 1 1 nr For a perfect gas = Δ G = nrln( / ) where G is the standard free energy defined at =1 bar 1 G = G ( ) + nrln( / ) he chemical potential for a pure substance is the molar Gibbs energy: G μ= G = n For a perfect gas μ=μ +R ln( / ) like eq aul ercival

What is Thermo, Stat Mech, etc.?

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