CHEM443 Physical Chemistry

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1 CHEM443 Physical Chemistry Sectin 00, Fall 0 Quiz #, September 7 0. (5 Pints). Place the letter f the crrect answer, fund in the clumn n the right, in the line prvided in the left clumn.. Equilibrium state variables and functins depend n the path a system takes A. Equatin f State t achieve the equilibrium. ( r F) False. Wrk is a state functin. ( r F) False B. zer 3. An adiabatic bundary prevents heat interactins between a system and its C. Fluid surrundings. ( r F) rue 4. A physical mdel that relates state variables in rder t describe a material D. lw density is an example f a(n) A 5. Elements needed fr defining a temperature scale include: E. high temperature C, G, J, and N. 6. he change in a state variable (r functin) assciated with a system that F. 00 feet belw sea underges a cyclic prcess is B. level 7. he ideal gas equatin f state is a suitable apprximatin fr real gases G. Physical prperty under what cnditin(s)? D, L, E_ that varies with temperature 8. hermdynamics directly tells us infrmatin abut the kinetics H. static (timescales) needed t bserve changes. ( r F) False 9. As defined in this curse, the amunt f pressure- vlume wrk is related t I. equilibrium O n a pressure- vlume graph. 0. Prcesses in which a system is cnstantly at equilibrium are referred t as Q. J. Interplatin scheme K. steady- state L. lw pressure M. true N. reference pints O. area under curve P. False Q. reversible. (5 Pints) Derive he expressin fr the wrk needed t expand an ideal gas isthermally frm vlume V t vlume V using a reversible prcess. Shw all wrk in as detailed a manner as pssible. dw = p ext dv = pdv = nr V dv = nrd(lnv ) V w = nrd(lnv ) = nr ln V = nr ln V V V V

2 CHEM443 Physical Chemistry Sectin 00, Fall 0 Quiz #, September 4 0. (5 Pints) Cnsider an isthermal, reversible expansin f an ideal gas frm vlume V t vlume V. A. What is the change in the internal energy f the gas during this prcess? Fr ideal gas, we shwed in class that internal energy is nly a functin f temperature; thus, fr isthermal prcess, ΔU = 0 B. What amunt f heat is exchanged with the envirnment, and what is the directin f the exchange? Applying the First law: ΔU = q + w = 0 q = w hus, nce we find the amunt f wrk fr this prcess, we knw that the heat is equal in magnitude t that. Als, since the gas expands, the heat must be added t the system in rder t maintain cnstant temperature. C. What amunt f wrk is assciated with this prcess, and what is the directin f the exchange f wrk with the envirnment? Wrk is dne by the system n the surrundings. he amunt f wrk is: dw = p ext dv = pdv = nr dv = nrd(lnv ) V V w = nrd(lnv ) = nr ln V = nr ln V V w < 0 V V. (5 Pints) Fr a reversible, adiabatic expansin f an ideal gas f cnstant vlume heat capacity C V = 3 R frm vlume V t vlume V, what is the final temperature? (Prvide an expressin via apprpriate derivatin). his is an adiabatic prcess; thus n heat is exchanged (q=0). find the final temperature, we perfrm ur first law analysis:

3 du = d ʹ w nc V d = pdv = nr dv V = nrd(lnv ) nc V d(ln) = nrd(lnv ) Integrating: C V d(ln) = Rd(lnV ) ln = R ln V C V V = V R C V V

4 CHEM443 Physical Chemistry Sectin 00, Fall 0 Quiz #3, September 0. (3 Pints) he heat (enthalpy) f reactin at cnstant pressure ( bar) and 98.5K is kj fr the cmbustin f ne mle f fructse: C 6 H O 6 (slid) + 6O (gas) 6CO (gas) + 6H O (liquid) What is the standard enthalpy f frmatin f fructse at 98.5K? ΔH rxn = 86.7 kj = ν i ΔH frmatin,i = (6ml) ΔH frmatin,co (ml)δh frmatin,c6 H O 6 (ml)δh frmatin,c6 H O 6 (ml)δh frmatin,c6 H O 6 ΔH frmatin,c6 H O 6 i=reactants,prducts = 86.7 kj + (6ml)ΔH frmatin,co = 86.7kJ + (6ml)ΔH frmatin,co + (6ml) ΔH frmatin,h O + (6ml)ΔH frmatin,h O + (6ml)ΔH frmatin,h O = 86.7 kj + (6ml)( kj /ml) + (6ml)( 85.8 kj /ml) = 49. kj /ml (ml) ΔH frmatin,c6 H O 6 (6ml)ΔH frmatin,o (6ml)(0 kj /ml) (6ml) ΔH frmatin,o. ( Pints) Fr a gas that is described by the Van der Waals equatin f state: P = R V b a V the Jule- hmsn cefficient in the limit f zer pressure is determined t be: µ J = C P a R b In this limit, what expressin fr the inversin temperature des ne btain? µ J = a C P R b = 0 a R b = 0 inversin = a Rb

5 3. (5 Pints) A quantity f nitrgen gas at 98K is cmpressed reversibly and adiabatically frm a vlume f 0.0 dm3 t 5.0 dm3. Assuming ideal gas behavir, calculate the final temperature f the nitrgen gas. Cnsider a mlar cnstant vlume heat capacity f.5r, where R is the gas cnstant. du = d ʹ w nc V d = pdv = nr dv V = nrd(lnv ) nc V d(ln) = nrd(lnv ) Integrating: C V d(ln) = Rd(lnV ) ln = R ln V C V V = V R C V V 5.0dm 3.5 = (98K) = 58.8K 0.0dm 3 he cmpressin under adiabatic cnditins leads t increase in temperature as ne wuld intuit.

6 CHEM443 Physical Chemistry Sectin 00, Fall 0 Quiz #4, September 8 0 A. (3 Pints) Cnsider an ideal gas with cnstant vlume mlar heat capacity f.5r where R is the gas cnstant. Obtain an expressin fr the entrpy change f ne mle f an ideal gas at a cnstant vlume V reversibly cled frm P,, V, t P, V, 4. Recall C P = C V + R du = dq + dw = dq (n wrk fr cnstant vlume prcess) dq = du = nc V d = dq reversible dq reversible = nc d V = nc V d(ln) = ds ΔS = C V ln 4 B.(3 Pints). Obtain an expressin fr the entrpy change when ne mle f the abve gas underges a reversible expansin at cnstant pressure P frm P, V, 4 t P, V,. du = dq + dw = dq p pdv = dq p d( pv ) dq p = du + d( pv ) = dh = nc p d = dq p,reversible dq p,reversible = nc d p = nc p d(ln) = ds ΔS = C P ln 4. (3 Pints) What is the mlar entrpy change fr vaprizatin f water at 98.5K and bar pressure? vapr vaprizatin vapr vapr dq p,reversible ΔH ΔS = = dq vaprizatin p,reversible = dh vaprizatin vaprizatiin p,reversible = vaprizatin liquid = H vapr H liquid vaprizatin = vaprizatin liquid 44.0kJ /ml 98.5K = 0.5 kj /ml K vaprizatin liquid > 0 vaprizatin 3 ( Pint) What can yu say abut the entrpy change fr an adiabatic, irreversible prcess? Entrpy change is greater than 0 (it s irreversible, s adiabaticity des nt matter)

7 CHEM443 Physical Chemistry Sectin 00, Fall 0 Quiz #5, Octber 0, 0. (3 Pints) he mlar cnstant vlume heat capacity f ethene is apprximated as: C V () R =6.405 ( K) ( K ) ver the temperature range 300K < < 000K. Calculate the change in entrpy f ne mle f ethene when heated frm 300K t 600K at cnstant vlume. here is n phase change during this prcess. ΔS = = nr 600K 300K 600K 300K = 38.75J /K nc V d K + 886K d 3. (3 Pints) Given the fllwing data, cnclude whether the value f the entrpy difference between liquid and gaseus water at bar pressure and 98.5 K as given in the Equatins Handbk is valid. Cnsider the data: vaprizatin = 373.5K ΔH vap ( vap ) = 40.65kJ /ml J C P,liquid = 75.3 K ml J C P,vapr = 33.8 K ml he use f a reversible path cnnecting the states at 98.5 K will be useful. Cmment n any differences r equivalences and their pssible rigins. Slutin: Fr this prblem, we need t cnstruct a reversible path (r thermdynamic cycle) frm water at 98.5K and bar t liquid water at the nrmal biling pint at bar (373.5K) t gaseus water at bar and 373.5K t finally gaseus water at 98.5K. We calculate the entrpy changes fr each f thse segments, add them up, and then cmpare t the value in the hand bk. We ll cnsider ne mle f water, but still retain the mlar prperties given abve. Nte that we will use cnstant heat capacities alng the reversible paths.

8 ΔS = 373.5K C P,liquid J 373.5K J d = (75.3 ) ln =6.9 K ml 98.5K K ml 98.5K ΔS = ΔH vaprizatin vaprizatin ΔS 3 = 98.5K C P,gas = 40.65J /ml 373.5K =08.9 J K ml d = J 98.5K J (33.8 ) ln = K ml 373.5K K ml 373.5K J ΔS + ΔS + ΔS 3 =8. K ml his is slightly lwer than the value in the handbk, but therwise ne wuld say that the value in the handbk is accurate. We are nt accurate in assuming that the heat capacities are abslutely cnstant ver the temperature range f 00K cnsidered fr the temperature dependent segments f ur prcess. And f curse,the handbk is ging t be accurate! 3. ( Pints) If ne treats a fluid using the Van der Waals Equatin f State (Page 4-, Equatins Handbk), what is the expressin fr the isthermal reversible wrk fr a prcess frm (P,V ) t (P,V )? dw = pdv = nr V nb n a V dv V w = nr V nb n a V dv = V V V V n a V nr dv V nb w = n a nr ln(v V nb) ln(v nb) = n a nr ln V nb V V V nb [ ] 4. ( Pints) Fill in the crrect answer fr the fllwing: 4A. (0.5 Pints) he natural variables fr entrpy, S, fr a clsed system (pure substance with n change in amunt f material) are U (internal energy) and V (vlume). 4B. (0.5 Pints) If a system is initially in a nn-equilibrium state, what will happen t what thermdynamic state functin under cnditins where the system is maintained at cnstant temperature and pressure, with n change in amunt f substance in the system? Gibbs free energy will decrease until equilibrium is reached, at which pint G is cnstant. 4C. (0.5 Pints) he natural variables fr internal energy, U, fr a clsed system (pure substance with n change in amunt f material) are S (entrpy) and vlume (V). 4D. (0.5 Pints) What are the natural variables fr the Helmhltz Free Energy fr pure substance and clsed system? V (vlume) and (temperature).

9 Physical Chemistry Sectin 00, Fall 0 Quiz #6, Octber 9, 0. (5 Pints) Since we knw that da(,v,n,n ) = pdv Sd + µ dn + µ dn A and that: G(,P,n,n) = A(,V,n,n) V V demnstrate that µ = A n V,,n,n,n = G n P,,n Given : da(,v,n,n ) = pdv Sd + µ dn + µ dn his suggests : µ = A n,v,n G(,P,n,n ) = A(,V,n,n ) V A V,n,n = A(,V,n,n ) + PV dg(,p,n,n ) = da(,v,n,n ) + d(pv ) = da(,v,n,n ) + PdV +VdP Substitute fr ttal differential f A and keep explicit derivatives fr the chemical ptentials dg(,p,n,n ) = pdv Sd + A dn n + A dn n + PdV +VdP,V,n,V,n dg(,p,n,n ) = VdP Sd + A dn n + A dn n,v,n,v,n hwever, since G is a functin f,p,n,n, we see that the ttal differential f G needs t have terms with partial derivative f G with respect t n and n; thus, we see clearly the equality : A n,v,n = G n,p,n

10 . Alternative Slutin (Mr. Sean Herrn) his slutin uses the fact that the Gibbs and Helmhltz free energies are first- rder extensive functins. We have seen in class that: G(,P,n,n ) = µ n + µ n = G n n + P,,n G n n P,,n As we did in class fr G, we can use a similar analysis fr A as fllws, starting frm the fact that A is first- rder extensive in amunt f material (and thus in vlume): λa(,v,n,n ) = A(,λV,λn,λn ) ake derivative f bth sides with respect t lambda: A(,V,n,n ) = d [ dλ A(,λV,λn,λn ) ] = (λv ) A(,λV,λn,λn ) dλv [ ] + dλ (λn ) A(,λV,λn,λn ) [ ] [ (λn ) A(,λV,λn,λn )],n,n,v,n dλn dλ,v,n dλn dλ + = PV + µ n + µ n = PV + A (n ),V,n A = PV + A (n ) n + A (n ),V,n,V,n n + A (n ) n,v,n n A Nw, we use the given infrmatin, G(,P,n,n) = A(,V,n,n) V V t arrive at ur desired demnstratin:,n,n G(,P,n,n ) = G n n + P,,n G n n + P,,n G n n = A (n P,,n ) G n n = A(,V,n,n) + PV = PV + A (n P,,n ),V,n Cllecting terms with n and n tgether: n + A (n ),V,n n,v,n n + A (n ),V,n n + PV G n P,,n A n,v,n n + G n P,,n A n,v,n n = 0 Since we cannt have n and n zer, the nly way the equality hlds generally (as it

11 must) is if the terms multiplying n and n are each zer. With this being the case, we btain the result that we must demnstrate, namely: G n P,,n = A n G = A n n P,,n,V,n,V,n. (5 Pints) Use what yu knw abut the Fundamental Equatin f thermdynamics, it s relatin t the ther majr thermdynamic ptentials, and U Maxwell Relatins, t determine fr a pure Van der Waals fluid. V du(s,v ) = ds pdv at cnstant temperature : U = S p V V btain the derivative f entrpy with vlume, we can resrt t Maxwell Relatins in the apprpriate thermdynamic ptential t give us a mre cnvenient relatin in terms f an equatin f state (EOS). Since we have S,V, as the variables, we can first try the Helmhltz free energy (since the ther bvius ptential is U, which we already started with) da = pdv Sd Maxwell Relatin : p V hus : U V = S V = p V p Nw we use ur Van der Waals EOS t btain the derivative f pressure with respect t temperature at cnstant vlume p = nr V nb n a V p s V U V = nr V nb = p V p = nr V nb nr V nb n a V = n a V

12 Physical Chemistry Sectin 00, Fall 0 Quiz #7, Octber 6, 0. (0 Pints) Knwing that fr the chemical ptential f an ideal gas in a mixture at temperature and pressure P is: µ i mixture (,P) = µ i pure (,P) + R ln(x i ) what is the Gibbs free energy change f mixing f tw ideal gases, A and B, at cnstant temperature and pressure? Slutin: Initially we have NA mles f A and NB mles f B, separated frm ne anther. In this separated state, the ttal Gibbs free energy at this and pressure cmbinatin is simply the sum f the individual mlar Gibbs free energies: G initial = N A µ A pure (,P) + N B µ B pure (,P) After mixing, the Gibbs free energy f the mixture is simply (making use f the relatin given abve fr the ideal gas chemical ptential: G final = N A µ A mixture (,P) + N B µ B mixture (,P) = µ A pure (,P) + R ln(x A ) + µ B pure (,P) + R ln(x B ) At equilibrium after mixing, the difference in Gibbs free energies is: ΔG = G final G initial = N A µ pure A (,P) + N A R ln(x A ) + N B µ pure B (,P) + N B R ln(x B ) N A µ pure A (,P) N B µ pure B (,P) = RN A ln(x A ) + N B R ln(x B ) [ ] = NR x A ln(x A ) + x B ln(x B )

13 Physical Chemistry Sectin 00, Fall 0 Quiz #8, Nvember 9, 0. (5 Pints) Fr the fllwing chemical transfrmatin, the value f KP =.3x0 at =98K: NO(g) + O (g) = NO (g) Cnsider that initially, there are mles f NO(g) and mle f O(g). What extent f reactin ptimizes the ttal Gibbs free energy f this system? Cnsider the gases t be ideal and make necessary apprximatins t arrive at an expressin fr the extent f reactin in terms f the ttal pressure, P, and the equilibrium cnstant, KP given abve. Nte the magnitude f KP in thinking abut what apprximatin t invke in yur slutin. Slutin: At the given temperature, the Gibbs free energy is ptimized at equilibrium. hus, t find the extent f reactin, we nly need t slve fr the mle fractins at equilibrium. NO O NO Initial Mles 0 Equilibrium - ε - ε ε Mles Equilibrium Mle Fractin ( ε) 3 ε KP=.3x0 he equilibrium cnstant is determined as: K P = p NO = x NO = ε (3 ε) p NO p O p x NO x O p ( ε) 3 ( ε) 3 ε ε 3 ε Since Kp is >>>, we can apprximate ε t be CLOSE t, which allws us t write K P = p NO p NO p O = p x NO x NO x O = ε (3 ε) () (3 ) p ( ε) 3 p ( ε) 3 p ( ε) 3 We cannt ignre the epsiln in the denminatr, as this term is the ne that drives the slutin because it is a small number, cubed, and lcated in the denminatr. hus, ( ε) 3 pk P / 3 ε pk P his is an apprximate slutin fr the ptimal equilibrium extent f reactin at this temperature. We see that increasing the pressure will increase the extent f

14 reactin (i.e. cnversin). his is reminiscent f Le Chatelier s principle fr pressure effect n equilibrium.. ( Pint) he Virial expansin is a perturbatin thery that applies t real fluids under cnditins where the density is lw. What is the reference state fr the Virial expansin t describe fluids with attractins and finite mlecular vlumes? he ideal gas EOS is the reference state fr liquids. 3. ( Pints) What is the entrpy change fr an adiabatic, reversible prcess? Slutin: ΔS = final state ds = initial state final state dq reversible = initial state final state initial state 0.0 = 0 4. ( Pint) Fr a binary system (tw distinct chemical cmpnents) in tw- phase equilibrium, hw many intensive degrees f freedm are available? Slutin: his is a prblem fr Gibbs Phase Rule. N= P = # phases = hus, F = + N - P = degrees f freedm Cnsider, if we have chemical species distributed in tw phases, we have tw degrees f freedm (independent, intensive variables we can cntrl, r dial in ). What if we dial in specific values f temperature,, and pressure, P? We have used up all f ur degrees f freedm. Well, des this mean that the cmpsitin f each phase in equilibrium is then determined fr us? Are we stuck?

15 Physical Chemistry Sectin 00, Fall 0 Quiz #9, Nvember 30, 0. (5 Pints) Cnsider a water()-ethanl() binary liquid slutin in equilibrium with the gaseus mixture (als cnsisting f ethanl and water). ake the vapr t be ideal. Determine a relatin between the liquid and vapr mle fractins f either cmpnent at a certain temperature and pressure P where the fllwing cnditin hlds: Slutin: γ sat = y PP sat γ x P hus γ γ = P x sat P sat sat y P = y P sat x P x y = P sat P sat y x x y = y = x y x = y x y = x y = x his is an example f an azetrpic thermdynamic pint.

16 . (5 Pints) Calculate the biling pint (at atm) f a slutin cntaining 6 g f acetne (Mw = 58) and 7 g f water (Mw = 8) by using the fllwing table. Assume ideal slutin and vapr behavir. emperature C Vapr Pressure Acetne (atm) Vapr pressure Water (atm) A liquid starts t bil when its vapr pressure matches the atmspheric pressure ( atm in this case). hus, accrding t Rault's law P = x acetne P acetne + x water P water = atm Frm the given data we can calculate the mlar fractins mles acetne =6/58= mles water = 7/8 = 4 ttal mles = 6 x acetne = /4 = /3 x water = 4/6 = /3 thus P = /3 P acetne + /3 P water= atm By trials, using the table, we can find the values f vapr pressure which satisfies the abve equatin. he best result is btained by using the values at 80 C : P = / /3. =.0 atm then the biling pint is abut 80 C.