FUGACITY. It is simply a measure of molar Gibbs energy of a real gas.

Size: px
Start display at page:

Download "FUGACITY. It is simply a measure of molar Gibbs energy of a real gas."

Transcription

1 FUGACITY It is simly a measure of molar Gibbs energy of a real gas. Modifying the simle equation for the chemical otential of an ideal gas by introducing the concet of a fugacity (f). The fugacity is an effective ressure which forces the equation below to be true for real gases: θ f µ,t) µ (T) + RT ln θ ( where θ atm A lot of the chemical otential for an ideal and real gas is shown as a function of the ressure at constant temerature. The fugacity has the units of ressure. As the ressure aroaches zero, the real gas aroach the ideal gas behavior and f aroaches the ressure.

2 If fugacity is an effective ressure i.e, the ressure that gives the right value for the chemical otential of a real gas. So, the only way we can get a value for it and hence for µ is from the gas ressure. Thus we must find the relation between the effective ressure f and the measured ressure. let f φ φ is defined as the fugacity coefficient. φ is the fudge factor that modifies the actual measured ressure to give the true chemical otential of the real gas. By introducing φ we have just ut off finding f directly. Thus, now we have to find φ. Substituting for φ in the above equation gives: θ µ (, T) µ (T) + RT ln + RT ln φ µ (ideal gas) + RT ln φ θ µ (,T) µ (ideal gas) RT lnφ This equation shows that the difference in chemical otential between the real and ideal gas lies in the term RT ln φ. This is the term due to molecular interaction effects.

3 The equation relating f or φ to the measured ressure Note that as 0, the real gas ideal gas, so that f and φ The chemical otential for an ideal gas and a real gas at two ressures and is Videal,m d d µ ideal µ ideal (, T) µ ideal (, T) RT ln f Vm d d µ µ (,T) µ (,T) RT ln f Subtracting the first equation from the second gives ( m ideal,m ) f V V d RT ln RT ln f f / or ln ( Vm Videal,m ) d f / RT Let 0, then in the initial state the real gas the ideal gas. Thus f f ln ( Vm Videal,m ) d RT 0 3

4 Since for an ideal gas V ideal,m RT and for a real gas Vm RT Z where Z is the comressibility factor f RT RT Z(, T) ln Z d d ln RT φ 0 0 Z(, T) f ex d 0 The fugacity coefficient φ f/ is given by φ ex 0 Z(,T) d Thus the fugacity of a gas is readily calculated at same ressure if Z is known as a function of ressure u to that articular ressure. 4

5 At low to intermediate ressures Z<, attraction dominates and the Z area under the curve of the RHS lot, d is negative. Thus f ex(negative ower). and f <, φ<. Z At high ressures Z>, reulsion dominates and d >0 and f>, φ> 5

6 The Fugacity of van der Waals From the comressibility factor Z of a Van der waals: Z V RT m RT V b a m V m Z + b a RT RT From the fugacity equation ln f 0 b f a RT RT ex b d ln a RT f b Z d 0 RT a RT RT 6

7 Examle: Given that the van der Walals constants of nitrogen are: a.408 L atm mol - and b L mol -. Calculate the fugacity of nitrogen gas at 50 bar and 98 K. ln f b a RT RT ln f ln f ln + ln L. mol b a RT.408L (0.0834L. bar. K RT. bar. mol. mol 50bar )(98K ) (0.0834L. bar. K. mol )(98K ) fugacity (f) 48. bar 7

8 Fundamental equations for oen system An oen system is the one that can exchange matter with its surroundings. Consider a two comonent system at temerature T and ressure which is comosed of n moles of substance and n moles of substance. The Gibbs function can be written as GG(T,, n, n ). It is a function of four indeendent variables. The total differential is G G G G dg dt + d + dn + dn T n n,n,n T,n,n,T,n,T,n For a system of constant comosition ie a closed system: dg -S dt + V d and G T G S V,n,n T,n,n Also from the earlier definition of the chemical otential : G G µ µ n n Gibbs function for a two comonent oen system can be written as,t,n,t,n dg -S dt + V d + µ dn + µ dn -S dt + V d + µ i dn i 8

9 This is the most general exression for the driving force of a system. It is very useful in examining what haens in a system at constant temerature and ressure when the amounts of n and n are changed. This occurs frequently in chemical and biochemical rocesses for examle in the transfer of a solute to a solution. The starting equation in this case where dt 0 d is dg µ dn + µ dn dw e,max This equation is also interesting because we know that the reversible free energy change at constant temerature and ressure is the maximim non V work. This equation shows that non V work is the result of comosition changes in the system. This is exactly what haens in a electrochemical cell. As the electrochemical reaction occurs there are comositional changes and these lead to electrical current. Finally, if a rocess is at equilibrium then dg 0 and µ dn + µ dn 0 or µ dn - µ dn This equation known as the Gibbs-Duhem equation is the starting oint for many develoments where equilibrium comositional changes occur 9 under constant temerature and ressure.

10 Fundamental equations for oen system An oen system is the one that can exchange matter with its surroundings. Consider a two comonent system at temerature T and ressure which is comosed of n moles of substance and n moles of substance. The Gibbs function can be written as GG(T,, n, n ). It is a function of four indeendent variables. The total differential is G G G G dg dt + d + dn + dn T n n,n,n T,n,n,T,n,T,n For a system of constant comosition ie a closed system: dg -S dt + V d and G T G S V,n,n T,n,n Also from the earlier definition of the chemical otential : G G µ µ n n Gibbs function for a two comonent oen system can be written as,t,n,t,n dg -S dt + V d + µ dn + µ dn -S dt + V d + µ i dn i 0

11 This is the most general exression for the driving force of a system. It is very useful in examining what haens in a system at constant temerature and ressure when the amounts of n and n are changed. This occurs frequently in chemical and biochemical rocesses for examle in the transfer of a solute to a solution. The starting equation in this case where dt 0 d is dg µ dn + µ dn dw e,max This equation is also interesting because we know that the reversible free energy change at constant temerature and ressure is the maximum non V work. This equation shows that non V work is the result of comosition changes in the system. This is exactly what haens in a electrochemical cell. As the electrochemical reaction occurs there are comositional changes and these lead to electrical current. Finally, if a rocess is at equilibrium then dg 0 and µ dn + µ dn 0 or µ dn - µ dn This equation known as the Gibbs-Duhem equation is the starting oint for many develoments where equilibrium comositional changes occur under constant temerature and ressure.

12 Phase Equilibrium A Phase (P): It is a art of a system, uniform through out in chemical comosition and hysical roerties, that is searated a uniform comonent art of the system that is searated from the other homogeneous arts (e.g. solid, liquid, or gas) by boundary surface. A hase may of course contain several chemical constituents, which may or may not be shared with other hases. The number of hases is reresented in the relation by P. For examle: If you have some ice floating in water, you have a solid hase resent and a liquid hase. If there is air above the mixture, then that is another hase. The Degrees of Freedom [F] is the number of indeendent intensive variables (i.e. those that are indeendent of the quantity of material resent) that need to be secified in value to fully determine the state of the system. Tyical such variables might be temerature, ressure, or concentration.

13 Phase Rule (F) The number of degree of freedom for descrition of the intensive state of the system. For a system involve V work only: For examle: A system with one comonent F C P + Where C is the number of comonents of the system F P + 3-P Case : one hase is resent (a balloon full of carbon dioxide). F 3 There is two degrees of freedom: temerature and ressure are necessary to define the system and each one of them can be varied indeendently. Case : two hases are resent (liquid and vaor for instance): F 3 Either temerature or ressure (one variable) is necessary to define the system. Case 3: three hases are resent (solid, liquid and vaor for instance): F Such a system is said to be invariant and is reresented by a oint in a lot of hase diagram (T-), where the ressure and temerature are fixed by the equilibrium. 3

14 Phase diagrams It is a grahical way to deict the effects of ressure and temerature on the hase of a substance. A hase diagram shows exactly what hases are resent at any given temerature and ressure. This hase diagram is characterized by area, line and Area trile oint. Trile oint Line The hases will simly be the solid, liquid or vaor (gas) states of a ure substance. Area: it reresents a one hase (F ), where the two variables T and are necessary to define, like areas of solid, liquid, and vaor. Line: it reresents two hases (F ) in equilibrium. The system can be comletely define by T or. Trile oint (T): it reresents three hases (F 0) in equilibrium. The system can be comletely define by T or. It is only one oint in the diagram. If T or change, the two hase will dissaear. 4

15 The three areas Under the set of conditions at in the diagram, the substance would be a solid because it falls into that area of the hase diagram. At, it would be a liquid; and at 3, it would be a vaor (a gas). At constant ressure, as the temerature increases to the oint where it crosses the line, the solid will turn to liquid. If you reeated this at a higher fixed ressure, the melting temerature would be higher because the line between the solid and liquid areas sloes slightly forward. 5

16 Phase diagram of CO It shows that there is equilibrium between solid and gas at atm and -78 o C. Liquid CO is roduced only above 5. atm and -57 o C. The only thing secial about this hase diagram is the osition of the trile oint which is well above atmosheric ressure. It is imossible to get any liquid carbon dioxide at ressures less than 5. atmosheres. That means that at atmoshere ressure, carbon dioxide will sublime at a temerature of -78 C. This is the reason that solid carbon dioxide is often known as "dry ice". You can't get liquid carbon dioxide under normal conditions - only the solid or the vaor. 6

17 Existence of hases in a one-comonent system When a solid is heated, the solid hase changes to liquid and then to gas. To understand these rocesses, a lot of the chemical otential versus T at constant for various hases is studied. µ Solid Gas T m is the melting oint at which solid and liquid at equilibrium at the same chemical otential. Liquid T b is the boiling oint at which liquid and gas hases at equilibrium at the same chemical otential. Below T m, the solid has the lowest µ and is therefore, the stable hase. T m Temerature T b Between T m and T b the liquid is the stable hase. The sloe of the lines giving the molar entroy of the hases, where: Since S is ositive, the sloes are negative. dµ dt S m Since S m (g) > S m (l) > S m (s), The sloe is more negative for a gas than a liquid than a solid 7

18 For a single hase of a ure substance, µ is a function of T and. Thus, it is convenient to consider, µ as a function of T at secified ressures. At lower ressure, the lots of µ vs. T are dislaced. The effect of on µ of a ure substance at constant T is given by: dµ d T V µ decreases as the is decreased at constant T. m µ Solid Liquid Gas Since V m (g) >>> V m (l), V m (s), T m T m T b The effects is much larger for gas than a liquid or solid. Temerature Reducing the ressure lowers T b and T m, and the effect on T b is much larger because of the large difference in molar volume of gas and liquid. T b The range of T over which the liquid is the stable hase has been reduced. At the same articular, the solid, liquid, and gas lines will intersect at a oint referred to as the trile oint. 8

19 The Claeyron Equation Consider one comonent system with two hases α and β. At equilibrium, T, and µ are same in the two hases. When T only changes or only changes, one of the hases will disaear. When both T and change in such a way to kee µ is equal in the two hases, the two hases will continue to co-exist. This is the co existence curve for the system α β. Along the curve, µ α µ β If T changes by dt, will changes by d. At oint : At oint : d µ d µ α µ β α + d µ α µ β + µ β µ d Since dµ dg V d α µ β m m m Vm, α d Sm, α dt Vm, β d Sm, β Thus d dt S V m, β m, β S V m, α m, α S V m m H T V Note that: sub H fus H + va H m m S dt dt d β dt This equation can be alied to vaorization, sublimation, fusion or transition between two hases. T α 9

20 va H of water kj mol -, V m,g x 0-3 m 3 mol -, V m,l 0.09 x 0-3 m 3 mol - all at 00 o C and atm. What is the change in boiling oints of water er Pascal. Answer : - d H m, va kj mol dt T V V ) (373.5K )( ) x 0 (m mol ) ( m, g m, l Note: J Pa d dt m 3 363Pa. K dt d.768x0 4 K. Pa Calculate the change in ressure required to change the freezing oint of water o C. At o C the heat of fusion of ice is J g -, the density of water is g cm -3 and ice is g cm -3. Answer V : m,l / ( g cm -3 ).000 cm 3 g - V m,ice.0908 cm 3 g - V m V m,l - V m,ice.000 cm 3 g cm 3 g cm 3 g x0-6 m 3 g x0-8 m 3 g - d H - m, fus J g 7.348x0 Pa. K dt T( Vm, l Vm, ice ) (73.5K )( 9.06 x 0 m g ) 34.8bar. K 0

21 The Clausius-Claeyron Equation Clausius modified Claeyron equation by neglecting the molar volume of liquid in comarison with molar volume of the gas. d dt va TV H g P va RT H d va RT H dt d va R H dt T ln o va RT H + C By lotting vs. ln, H can be calculated for vaorization or sublimation. T o Also, by limits the integration: ln ( vah T T RT T ) d Defects of this equation: To reresent as a function of T over a wide range of T is necessary to consider as T deendence. The vaor has been assumed to be an ideal gas. va R H T T dt T va H o

22 Trouton s Rule: The molar entroy of vaorization at the standard boiling oint (the b at bar) is a constant and its value is about 88 JK - mol -. vas 88JK mol By using this rule va H of a liquid can be estimated aroximately: va S va T b H Examle: The boiling oint of benzene is 80. o C at atm. Estimate the vaor ressure at 5 o C using Trouton s rule. The vaor ressure at ( )353.3 K is.03 bar According to Trouton s rule: According to Clausius-Claeyron equation va H (88 JK - mol - ) (353.3 K) 3. x 0 3 J mol - ln.03bar (3.x03Jmol )(353.3K 98K ) ln (8.34Jmol K )(353.3K )(98K ) ( vah T T RT T ) P 0.43 bar

23 Vaor-liquid Equilibrium of binary liquids mixtures When a binary liquid mixture is in equilibrium with its vaor at constant T: µ ( g) ( l) i µ i i.e. the chemical otential of each comonent is the same in the gas and liquid hase. Assuming that the vaor is an ideal gas: And for a liquid mixture is: Where a, is the activity µ ( g) µ ( g) + i µ ( l) µ ( l) + i o i o i RT And for a ure liquid (its activity equal unity), ln i o RT ln a o i o µ ( g) + RT ln µ ( l) + RT ln a i o i i * o i o µ ( g) + RT ln µ ( l) i o i * Where i is the vaor ressure of ure liquid i. By subtract the last equation (for ure liquid) from that in a mixture i RT ln * i RT ln a i a i i * i i This is for ideal behavior of the vaor.

24 Raoult s Law For a certain solution (ideal), the artial ressure of a comonent is equal to the mole fraction of that comonent times (x) the vaor ressure of the ure comonent. This law is correct when the comonents are quite similar. Since the gas hase assumed to be an ideal gas: i i y x i * i i Dalton s Law Where, y i is the mole fraction of i in the gas hase and is the total ressure. As shown in the figure, change of vaor ressure for benzene-toluene mixture, also with vaor ressure of both benzene and toluene solutions against the comosition of solution. The behavior of mixture follow the ideal solution behavior and Raoult s law. For this solution, it is ossible to calculate the comosition of its vaor which in equilibrium with solution. P (bar) Total Toluene Toluene Mole fraction Benzene Benzene

25 For a mixture of two liquid and : y and y From Raoult s law: x y and * y The relative mole fraction of comonents' in the vaor hase: y y x x * * and x From this diagram: -at any secific vaor ressure it can be determined -the corresonding comosition of liquid mixture. - or at any secific comosition of liquid mixture (a) can be determine the corresonding comosition in vaor hase (b). x y y * * x * P (bar) Liquid (liquid comosition line) (vaor comosition line) a b Mole fraction bubble line Dew line vaor Toluene Benzene

26 Examle: At 60 o C the vaor ressures of ure benzene and toluene are 0.53 and 0.85 bar, resectively. For a solution with 0.6 mole fraction toluene. What are the artial ressures of comonents and what is the mole fraction of toluene in the vaor? Answer: * (benzene) 0.53 bar * (toluene) 0.85 bar x 0.6 y??? and??? x * 0.6 x 0.85 bar 0. bar x - x bar x * 0.4 x 0.53 bar 0.05 bar P total bar 0.36 bar y total 0.35

27 Examle: At 80 o C the vaor ressures of ure benzene and toluene are 00.4 kpa and 38.7 kpa, resectively. a) Calculate the vaor comosition and total ressure for solution at 80 o C consists of 0.5 benzene and 0.5 toluene mole fraction b) Calculate the liquid comosition at equilibrium for the mixture at 80 o C, where the vaor contains mole fraction of benzene Answer: a) y y x x * * (0.5)(00.4kPa) (0.5)(38.7).594 y y (.594) (-y ) (.594) y 0.7 and y 0.78 P total x * + x * (0.5)(00.4 kpa) + (0.5)(38.7 kpa) kpa b) y y x x x x * * x x X and X - X y y (0.75)(38.7kPa) (0.5)(00.4kPa) * *.56

28 NON-IDEAL MIXTURES OF LIQUIDS Deviations from Raoult s Law By lotting the vaor ressure of an ideal mixture of two liquids against their comosition, you get a straight line grah like this: In this case, ure A has the higher vaor ressure and so is the more volatile comonent.

29 Positive deviations from Raoult's Law In mixtures showing a ositive deviation from Raoult's Law, the vaor ressure of the mixture is always higher than you would exect from an ideal mixture, and hence have lower BP. Because the vaor ressure is higher, the liquid is evaorating more easily than would be exected. This means that some of the intermolecular bonds in the liquid must have been broken when the liquids were mixed. i.e., means that molecules are breaking away more easily than they do in the ure liquids, because the intermolecular forces between molecules of A and B are less than they are in the ure liquids. Examle: mixture of ethanol and water. The ethanol molecules have an average of hydrogen bond er molecule, whereas the water molecules have. The ethanol molecules interfere with the hydrogen bonds and therefore make the liquid more volatile. Ethanol and benzene? the hydrogen bonding in the ethanol is reduced by the resence of the benzene.

30 Negative deviations from Raoult's Law In exactly the same way, having mixtures with vaor ressures which are less than would be exected by Raoult's Law. These are cases where the molecules break away from the mixture less easily than they do from the ure liquids. New stronger forces must exist in the mixture than in the original liquids. i.e. the intermolecular forces increase when the liquids are mixed. This is generally because molecules where no hydrogen bonding is resent are mixed to form a liquid with hydrogen bonds. Examle: Trichloromethane and ethoxyethane. CHCl 3 has a olar hydrogen atom but no lone airs and therefore cannot form hydrogen bonds. C H 5 -O-C H 5 has lone airs on the oxygen but no olar hydrogen atom and therefore cannot form hydrogen bonds. When mixed, hydrogen bonds form, decreasing the vaor ressure.

31 Boiling oint - comosition diagrams The BP of liquid mixture is the temerature at which the total ressure of liquid mixture equal to the atmosheric ressure. At constant T, the VP of liquid mixture is changed with the comosition of liquid, as it was shown before. Also, at constant ressure ( atm), the BP of liquid mixture is changed with liquid comosition. Ideal Mixtures The vaor comosition line is over the liquid comosition line where more volatile comonent has higher content in the vaor.

32 The oint on the lower curve at a mole fraction of benzene of about This oint gives the boiling oint of a mixture which is 0.33 mole ercent benzene. Toluene Benzene Mixture what is the comosition of the vaor which is in equilibrium with the liquid at this temerature? The vaor comosition must lie on a horizontal (constant temerature) line going through this oint. At any given temerature and ressure the comosition of the vaor must be richer in the more volatile comonent and benzene is the more volatile comonent. This line a tie line and it intersects the uer curve at the comosition of the vaor in equilibrium with the liquid at the same temerature. The region between the two curves is a two-hase region. In this region two hases, liquid and gas, are in equilibrium with each other.

33 If we heat a liquid mixture at a comosition of about 0.33 mole fraction benzene the liquid will have the comosition shown at oint "a" when the liquid first begins to boil. The first vaor to come off will have the comosition shown at oint "b." As we continue to heat the system the more volatile comonent will evaorate referentially and the liquid hase will become richer in the less volatile comonent (toluene, here). As we continue to heat the mixture the toluene will gradually catch u to the benzene in the vaor hase. The very last bit of liquid to evaorate will have the comosition shown at oint "c," and the vaor will have the original comosition that we started with, "d" (or "a"). Notice that as we heat the system the tie line "tracks" with the lower and uer curves to give the comosition of liquid and vaor in equilibrium.

34 Fractional distillation Boiling diagrams rovide an exlanation for fractional distillation. Suose we start with a liquid mixture with 0.33 mole fraction benzene. As we heat the liquid it will begin to boil when the temerature reaches the temerature of oint "a." The first vaor to come off has the comosition shown at oint "b." Cature the vaor, condense it, and heat it u. The new liquid will boil at oint "c" giving a vaor with comosition at oint "d." Cature this vaor, cool it, and boil it at oint, "e," and so on. by continuing this rocess the vaor can be made as ure in benzene as desired. This is what fractional distillation does. There are more modern ways to searate mixtures, for examle, chromatograhy, but fractional distillation still has industrial and some laboratory imortance.

35 Boiling diagrams for non ideal solutions As was exlained earlier, the solution with ositive deviations from Raoult's law has a higher vaor ressure than would be given by Raoult's law. That means that the solution would boil at a lower temerature than would be exected from Raoult's law. A boiling diagram for such a substance might look like the following: Point "a" denotes a constant boiling mixture called an "azeotroe." (low boiling azeotroe) When a system is azeotroe, its comonents can not be searated by simle fractional distillation.

36 In a mixture with negative deviations from Raoult's law the vaor ressure is lower than would be exected from Raoult's law which will roduce boiling oints higher than exected and a high boiling azeotroe. When a system is azeotroe, its comonents can not be searated by simle fractional distillation.

37 Cooling curves for ure substances If there is ure molten lead and it was allow to cool down until it has all solidified. Plotting the temerature of the lead against time, will givea tyical cooling curve for a ure substance. Throughout the whole exeriment, heat is being lost to the surroundings - and yet the temerature doesn't fall at all while the lead is freezing. This is because the freezing rocess liberates heat at exactly the same rate that it is being lost to the surroundings. Energy is released when new bonds form - in this case, the strong metallic bonds in the solid lead.

38 Cooling curves for tin-lead mixtures If you add some tin to the lead, the shae of the cooling curve changes. If you cool a liquid mixture containing about 67% lead and 33% tin by mass, the cooling curve will be looking like this. Comments: Adding the tin to lead, lowers its freezing oint. Freezing starts for this mixture at about 50 C. You would start to get some solid lead formed - but no tin. At that oint the rate of cooling slows down - the curve gets less stee. The temerature does sto falling at 83 C. Now both tin and lead are freezing. Once everything has solidified, the temerature continues to fall. Changing the roortions of tin and lead If you had less tin in the mixture, the overall shae of the curve stays much the same, but the oint at which the lead first starts to freeze changes. The less tin there is, the smaller the dro in the freezing oint of the lead.

39 For a mixture containing only 0% of tin, the freezing oint of the lead is about 75 C. That's where the grah would suddenly become less stee. BUT... you will still get the grah going horizontal (showing the freezing of both the tin and lead) at exactly the same temerature: 83 C. As you increase the roortion of tin, the first signs of solid lead aear at lower and lower temeratures, but the final freezing of the whole mixture still haens at 83 C. That continues until you have added enough tin that the mixture contains 6% tin and 38% lead. At that oint, the grah changes. This articular mixture of lead and tin has a cooling curve which looks exactly like that of a ure substance rather than a mixture. There is just the single horizontal art of the grah where everything is freezing. However, it is still a mixture. If you use a microscoe to look at the solid formed after freezing, you can see the individual crystals of tin and lead. This articular mixture is known as a eutectic mixture. The word "eutectic" comes from Greek and means "easily melted".

40 The eutectic mixture has the lowest melting oint (which is, of course, the same as the freezing oint) of any mixture of lead and tin. The temerature at which the eutectic mixture freezes or melts is known as the eutectic temerature. The hase diagram Start from data obtained from the cooling curves. Draw a grah of the temerature at which freezing first starts against the roortion of tin and lead in the mixture. The only unusual thing is that you draw the temerature scale at each end of the diagram instead of only at the left-hand side.

41 If there is a mixture of 67% lead and 33% tin. That's the mixture from the first cooling curve lotted above. Suose it is at a temerature of 300 C. if you cool that mixture. Eventually the temerature will dro to a oint where it crosses the line into the next region of the diagram. At that oint, the mixture will start to roduce some solid lead - in other words, the lead (but not the tin) starts to freeze. That haens at a temerature of about 50 C. When the first of the lead freezes, the comosition of the remaining liquid changes. It obviously becomes roortionally richer in tin. That lowers the freezing oint of the lead a bit more, and so the next bit of lead freezes at a slightly lower temerature - leaving a liquid still richer in tin. This rocess goes on. The liquid gets richer and richer in tin, and the temerature needed to freeze the next lot of lead continues to fall. The set of conditions of temerature and comosition of the liquid essentially moves down the curve - until it reaches the eutectic oint.

42 Once it has reached the eutectic oint, if the temerature continues to fall, you obviously just move into the region of a mixture of solid lead and solid tin - in other words, all the remaining liquid freezes. Tin-lead mixtures as solder Traditionally, tin-lead mixtures have been used as solder, but these are being hased out because of health concerns over the lead. This is esecially the case where the solder is used to join water ies where the water is used for drinking. New non-lead solders have been develoed as safer relacements. Tyical old-fashioned solders include: 60% tin and 40% lead. This is close to the eutectic comosition (6% tin and 38% lead), giving a low melting oint. It will also melt and freeze cleanly over a very limited temerature range. This is useful for electrical work. 50% tin and 50% lead. This will melt and freeze over a wider range of temeratures. When it is molten it will start to freeze at about 0 C and finally solidify at the eutectic temerature of 83 C. That means that it stays workable for a useful amount of time. That's helful if it is being used for lumbing joints.

43 Two comonent system (Solid Liquid hases) In these systems, the comonents are comletely miscible in liquid hase and immiscible in the solid hase. So, that the ure solid hase searate out in cooling. Cooling Curves Cooling curve (Temerature vs. time at constant ressure) for the system bismuth-cadmium. When molten solution of Bi or Cd is cooled, the lot has nearly constant sloe. When solid crystallizes out, the lot becomes horizontal results from a heat evolution, like at 73 and 33 oc for Bi and Cd, resectively. When a solution is cooled, there is a change in sloe of the cooling curve at the temerature at which one of the comonents begins to crystalline out. Such changes in sloe are evident in the cooling curve of 0% Cd and 80% Cd. All the cooling curves ofd solutions show horizontal section at 40 o C. At this temerature, both solid Cd and Bi crystallize out. This is an eutectic temerature. The eutectic is not a hase, it is a mixture of two solid hases and has a fine grain structures.

44 Temerature comosition diagram of Bi-Cd In the area above JKL, there is one hase: F P + at constant ressure F 3 P at constant ressure For a single hase, P F Two degree of freedom are necessary, Temerature and mole fraction, to define any oint in this area. Along JK, Bi freezes out; along LK, Cd freezes out, there are two hases: Solid and solution having a comosition given by line JK and LK, resectively F P + at constant ressure F 3 P at constant ressure For a two hases, P F One degree of freedom is necessary, only Temerature or comosition (mole fraction) of liquid can be secified.

45 At the eutectic oint K, there are three hases: Solid, Bi, Solid Cd, and liquid solution with 40% Cd. F P + at constant ressure F 3 P at constant ressure For a three hases, P 3 F 0 Zero degree of freedom, there is only one temerature and only one comosition (mole fraction), at which the tree hases coexist at equilibrium at a given constant ressure of liquid can be secified. It is reresent the lowest freezing oint of whole solution series. The eutectic mixture is a mechanical mixture and not a comound. The area below the eutectic temerature K, is a two hase area: Solid Bi and Solid Cd. F P + at constant ressure For a two hases, P F F 3 P at constant ressure Only one temerature is needed to be secified (comosition may not necessary because both Bi and Cd are resent in ure comonents. The addition of Cd lowers the freezing oint of Bi along JK, and addition of Bi lowers the freezing oint of Cd along LK. OR, JK is the solubility curve for Bi in liquid Cd and LK is the solubility curve for Cd in liquid Bi

46 Calculation of the Solubility To determine the molar mass of a solute B in a solvent A by the deression of the freezing oint of A. Assume the solution is ideal and that ure crystalline A freezes out of solution, Equation for the equilibrium is: o A µ ( S, T ) µ (, T, x ) µ (, T ) + Thus at T at which the two hases are in equilibrium, ln x A o A µ ( S, T ) µ A(, T ) RT o fus G o A RT A ( T ) fus G o A (T ) Is the Gibbs energy of fusion of the solute at T. fus o A G ( T ) fus H o A T Substituting this equation in the above one: fus S o A A o A fus, A o A RT ln x o fus H o T fus H A T fus H A( ) T T A fus,a ln x A fus o G A( T ) R o H A T T R TT fus fus fus R fus, A, A H o A T T fus, A fus R H o A T fus TT, A T fus, A

47 Examle: Calculate the solubility in an ideal solution at 50 and 00 o C. If fus H Bi 0.5 kj mol - at 73 o C and T fus,bi 73 Increasing the temerature from 50 to 00, will increase the solubility. Use the equation: at 50 o C ln x Bi fus R H o A T T TT fus, A fus, A x 0.5Jmol 43K 546K ln JK mol Bi K * 546 K at 00 o C ln x Bi 0.50 x 0.5Jmol 473K 546K JK mol Bi K * 546 K x Bi 0.70 As exected, the solubility increases with increasing temerature.

dn i where we have used the Gibbs equation for the Gibbs energy and the definition of chemical potential

dn i where we have used the Gibbs equation for the Gibbs energy and the definition of chemical potential Chem 467 Sulement to Lectures 33 Phase Equilibrium Chemical Potential Revisited We introduced the chemical otential as the conjugate variable to amount. Briefly reviewing, the total Gibbs energy of a system

More information

Phase transition. Asaf Pe er Background

Phase transition. Asaf Pe er Background Phase transition Asaf Pe er 1 November 18, 2013 1. Background A hase is a region of sace, throughout which all hysical roerties (density, magnetization, etc.) of a material (or thermodynamic system) are

More information

The Second Law: The Machinery

The Second Law: The Machinery The Second Law: The Machinery Chater 5 of Atkins: The Second Law: The Concets Sections 3.7-3.9 8th Ed, 3.3 9th Ed; 3.4 10 Ed.; 3E 11th Ed. Combining First and Second Laws Proerties of the Internal Energy

More information

We now turn to considerations of mixtures. To keep our discussion reasonably simple,

We now turn to considerations of mixtures. To keep our discussion reasonably simple, 143 Lecture 23 We now turn to considerations of mixtures. To kee our discussion reasonably simle, we will limit our discussion to non-reacting systems, and to non-ionic systems. In other words, we will

More information

whether a process will be spontaneous, it is necessary to know the entropy change in both the

whether a process will be spontaneous, it is necessary to know the entropy change in both the 93 Lecture 16 he entroy is a lovely function because it is all we need to know in order to redict whether a rocess will be sontaneous. However, it is often inconvenient to use, because to redict whether

More information

Lecture 8, the outline

Lecture 8, the outline Lecture, the outline loose end: Debye theory of solids more remarks on the first order hase transition. Bose Einstein condensation as a first order hase transition 4He as Bose Einstein liquid Lecturer:

More information

4. A Brief Review of Thermodynamics, Part 2

4. A Brief Review of Thermodynamics, Part 2 ATMOSPHERE OCEAN INTERACTIONS :: LECTURE NOTES 4. A Brief Review of Thermodynamics, Part 2 J. S. Wright jswright@tsinghua.edu.cn 4.1 OVERVIEW This chater continues our review of the key thermodynamics

More information

The Role of Water Vapor. atmosphere (we will ignore the solid phase here) Refer to the phase diagram in the web notes.

The Role of Water Vapor. atmosphere (we will ignore the solid phase here) Refer to the phase diagram in the web notes. The Role of Water Vaor Water can exist as either a vaor or liquid in the atmoshere (we will ignore the solid hase here) under a variety of Temerature and ressure conditions. Refer to the hase diagram in

More information

Chapter 20: Exercises: 3, 7, 11, 22, 28, 34 EOC: 40, 43, 46, 58

Chapter 20: Exercises: 3, 7, 11, 22, 28, 34 EOC: 40, 43, 46, 58 Chater 0: Exercises:, 7,,, 8, 4 EOC: 40, 4, 46, 8 E: A gasoline engine takes in.80 0 4 and delivers 800 of work er cycle. The heat is obtained by burning gasoline with a heat of combustion of 4.60 0 4.

More information

Chemistry 531 Spring 2009 Problem Set 6 Solutions

Chemistry 531 Spring 2009 Problem Set 6 Solutions Chemistry 531 Sring 2009 Problem Set 6 Solutions 1. In a thermochemical study of N 2, the following heat caacity data were found: t 0 C,m d 27.2Jmol 1 K 1 f t b f C,m d 23.4Jmol 1 K 1 C,m d 11.4Jmol 1

More information

STATES OF MATTER UNIT 5

STATES OF MATTER UNIT 5 32 32 CHEMISTRY UNIT 5 After studying this unit you will be able to exlain the existence of different states of matter in terms of balance between intermolecular forces and thermal energy of articles;

More information

not to be republished NCERT STATES OF MATTER UNIT 5 INTRODUCTION

not to be republished NCERT STATES OF MATTER UNIT 5 INTRODUCTION 32 CHEMISRY UNI 5 SAES OF MAER After studying this unit you will be able to exlain the existence of different states of matter in terms of balance between intermolecular forces and thermal energy of articles;

More information

3. High Temperature Gases FPK1 2009/MZ 1/23

3. High Temperature Gases FPK1 2009/MZ 1/23 3. High Temerature Gases FP 9/MZ /3 Terms and Concets Dissociation Diatomic element gases Bond energy, dissociation energy (enthali) Flood s dissociation diagram Vaorization, eaoration, boiling, sublimation

More information

The extreme case of the anisothermal calorimeter when there is no heat exchange is the adiabatic calorimeter.

The extreme case of the anisothermal calorimeter when there is no heat exchange is the adiabatic calorimeter. .4. Determination of the enthaly of solution of anhydrous and hydrous sodium acetate by anisothermal calorimeter, and the enthaly of melting of ice by isothermal heat flow calorimeter Theoretical background

More information

HEAT, WORK, AND THE FIRST LAW OF THERMODYNAMICS

HEAT, WORK, AND THE FIRST LAW OF THERMODYNAMICS HET, ORK, ND THE FIRST L OF THERMODYNMIS 8 EXERISES Section 8. The First Law of Thermodynamics 5. INTERPRET e identify the system as the water in the insulated container. The roblem involves calculating

More information

I affirm that I have never given nor received aid on this examination. I understand that cheating in the exam will result in a grade F for the class.

I affirm that I have never given nor received aid on this examination. I understand that cheating in the exam will result in a grade F for the class. Chem340 Physical Chemistry for Biochemists Exam Mar 16, 011 Your Name _ I affirm that I have never given nor received aid on this examination. I understand that cheating in the exam will result in a grade

More information

Ideal Gas Law. September 2, 2014

Ideal Gas Law. September 2, 2014 Ideal Gas Law Setember 2, 2014 Thermodynamics deals with internal transformations of the energy of a system and exchanges of energy between that system and its environment. A thermodynamic system refers

More information

THE FIRST LAW OF THERMODYNAMICS

THE FIRST LAW OF THERMODYNAMICS THE FIRST LA OF THERMODYNAMIS 9 9 (a) IDENTIFY and SET UP: The ressure is constant and the volume increases (b) = d Figure 9 Since is constant, = d = ( ) The -diagram is sketched in Figure 9 The roblem

More information

ONE. The Earth-atmosphere system CHAPTER

ONE. The Earth-atmosphere system CHAPTER CHAPTER ONE The Earth-atmoshere system 1.1 INTRODUCTION The Earth s atmoshere is the gaseous enveloe surrounding the lanet. Like other lanetary atmosheres, it figures centrally in transfers of energy between

More information

AT 25 C! CH10090 Thermodynamics (part 2) Enthalpy changes during reactions. Let s remember what we did in CH10089

AT 25 C! CH10090 Thermodynamics (part 2) Enthalpy changes during reactions. Let s remember what we did in CH10089 CH10090 hermodynamics (art ) Let s remember what we did in CH10089 Enthaly changes during reactions o o o H98 ( reaction) = νi Hf, 98( roducts) νi Hf, 98( reactants) ν i reresents the stoichiometric coefficient.

More information

PHYS1001 PHYSICS 1 REGULAR Module 2 Thermal Physics Chapter 17 First Law of Thermodynamics

PHYS1001 PHYSICS 1 REGULAR Module 2 Thermal Physics Chapter 17 First Law of Thermodynamics PHYS1001 PHYSICS 1 REGULAR Module Thermal Physics Chater 17 First Law of Thermodynamics References: 17.1 to 17.9 Examles: 17.1 to 17.7 Checklist Thermodynamic system collection of objects and fields. If

More information

General Physical Chemistry I

General Physical Chemistry I General Physical Cheistry I Lecture 12 Aleksey Kocherzhenko Aril 2, 2015" Last tie " Gibbs free energy" In order to analyze the sontaneity of cheical reactions, we need to calculate the entroy changes

More information

Chapter 1 Fundamentals

Chapter 1 Fundamentals Chater Fundamentals. Overview of Thermodynamics Industrial Revolution brought in large scale automation of many tedious tasks which were earlier being erformed through manual or animal labour. Inventors

More information

MODULE 2: DIFFUSION LECTURE NO. 2

MODULE 2: DIFFUSION LECTURE NO. 2 PTEL Chemical Mass Transfer Oeration MODULE : DIFFUSIO LECTURE O.. STEDY STTE MOLECULR DIFFUSIO I FLUIDS UDER STGT D LMIR FLOW CODITIOS.. Steady state diffusion through a constant area Steady state diffusion

More information

Thermodynamics in combustion

Thermodynamics in combustion Thermodynamics in combustion 2nd ste in toolbox Thermodynamics deals with a equilibrium state and how chemical comosition can be calculated for a system with known atomic or molecular comosition if 2 indeendent

More information

Liquid water static energy page 1/8

Liquid water static energy page 1/8 Liquid water static energy age 1/8 1) Thermodynamics It s a good idea to work with thermodynamic variables that are conserved under a known set of conditions, since they can act as assive tracers and rovide

More information

Prediction of Vapour Pressures of Dioxin Congeners. R & D Center, Baoshan Iron & Steel Co., Ltd., Fujin Rd., Baoshan District, Shanghai , China

Prediction of Vapour Pressures of Dioxin Congeners. R & D Center, Baoshan Iron & Steel Co., Ltd., Fujin Rd., Baoshan District, Shanghai , China Prediction of Vaour Pressures of Dioxin Congeners Xian-Wei Li, 1 Etsuro Shibata, 2 Eiki Kasai, 2 Takashi Nakamura 2 1 R & D Center, Baoshan Iron & Steel Co., Ltd., Fujin Rd., Baoshan District, Shanghai

More information

Equilibrium Thermodynamics

Equilibrium Thermodynamics Part I Equilibrium hermodynamics 1 Molecular hermodynamics Perhas the most basic equation in atmosheric thermodynamics is the ideal gas law = rr where is ressure, r is the air density, is temerature, and

More information

They provide us with the knowledge of phase composition and phase stability as a function of temperature (T), pressure (P) and composition(c).

They provide us with the knowledge of phase composition and phase stability as a function of temperature (T), pressure (P) and composition(c). They provide us with the knowledge of phase composition and phase stability as a function of temperature (T), pressure (P) and composition(c). PHASE EQUILIBRIUM one of the most important sources of information

More information

Chapter 2 Solutions. 2.1 (D) Pressure and temperature are dependent during phase change and independent when in a single phase.

Chapter 2 Solutions. 2.1 (D) Pressure and temperature are dependent during phase change and independent when in a single phase. Chater Solutions.1 (D) Pressure and temerature are deendent during hase change and indeendent when in a single hase.. (B) Sublimation is the direct conversion of a solid to a gas. To observe this rocess,

More information

Lecture 28: Kinetics of Oxidation of Metals: Part 1: rusting, corrosion, and

Lecture 28: Kinetics of Oxidation of Metals: Part 1: rusting, corrosion, and Lecture 8: Kinetics of xidation of etals: Part 1: rusting, corrosion, and the surface rotection, all about chemistry Today s toics hemical rocesses of oxidation of metals: the role layed by oxygen. How

More information

arxiv: v1 [physics.data-an] 26 Oct 2012

arxiv: v1 [physics.data-an] 26 Oct 2012 Constraints on Yield Parameters in Extended Maximum Likelihood Fits Till Moritz Karbach a, Maximilian Schlu b a TU Dortmund, Germany, moritz.karbach@cern.ch b TU Dortmund, Germany, maximilian.schlu@cern.ch

More information

δq T = nr ln(v B/V A )

δq T = nr ln(v B/V A ) hysical Chemistry 007 Homework assignment, solutions roblem 1: An ideal gas undergoes the following reversible, cyclic rocess It first exands isothermally from state A to state B It is then comressed adiabatically

More information

Chapter 12 Intermolecular Forces of Attraction

Chapter 12 Intermolecular Forces of Attraction Chapter 12 Intermolecular Forces of Attraction Intermolecular Forces Attractive or Repulsive Forces between molecules. Molecule - - - - - - Molecule Intramolecular Forces bonding forces within the molecule.

More information

THERMODYNAMICS. Prepared by Sibaprasad Maity Asst. Prof. in Chemistry For any queries contact at

THERMODYNAMICS. Prepared by Sibaprasad Maity Asst. Prof. in Chemistry For any queries contact at HERMODYNAMIS reared by Sibarasad Maity Asst. rof. in hemistry For any queries contact at 943445393 he word thermo-dynamic, used first by illiam homson (later Lord Kelvin), has Greek origin, and is translated

More information

MODULE 2: DIFFUSION LECTURE NO DIFFUSION COEFFICIENT: MEASUREMENT AND PREDICTION

MODULE 2: DIFFUSION LECTURE NO DIFFUSION COEFFICIENT: MEASUREMENT AND PREDICTION NPTEL Chemical ass Transfer Oeration OULE : IFFUSION LECTURE NO. 4.4 IFFUSION COEFFICIENT: ESUREENT N PREICTION The roortionality factor of Fick s law is called diffusivity or diffusion coefficient which

More information

I have not proofread these notes; so please watch out for typos, anything misleading or just plain wrong.

I have not proofread these notes; so please watch out for typos, anything misleading or just plain wrong. hermodynamics I have not roofread these notes; so lease watch out for tyos, anything misleading or just lain wrong. Please read ages 227 246 in Chater 8 of Kittel and Kroemer and ay attention to the first

More information

Phase transitions. Lectures in Physical Chemistry 4. Tamás Turányi Institute of Chemistry, ELTE. Phases

Phase transitions. Lectures in Physical Chemistry 4. Tamás Turányi Institute of Chemistry, ELTE. Phases Phase transitions Lectures in Physical Cheistry 4 Taás Turányi Institute of Cheistry, ELTE Phases DEF a syste is hoogeneous, if () it does not contain arts searated by acroscoic surfaces, and () all intensive

More information

Physics 2A (Fall 2012) Chapters 11:Using Energy and 12: Thermal Properties of Matter

Physics 2A (Fall 2012) Chapters 11:Using Energy and 12: Thermal Properties of Matter Physics 2A (Fall 2012) Chaters 11:Using Energy and 12: Thermal Proerties of Matter "Kee in mind that neither success nor failure is ever final." Roger Ward Babson Our greatest glory is not in never failing,

More information

Mixture Homogeneous Mixtures (air, sea water ) same composition, no chemical bond components are NOT distinguishable

Mixture Homogeneous Mixtures (air, sea water ) same composition, no chemical bond components are NOT distinguishable BASIC CONCEPTAND DEFINITIONS1 2 THERMODYNAMICS CHAPTER 2 Thermodynamic Concets Lecturer Axel GRONIEWSKY, PhD 11 th of February2019 Thermodynamics study of energy and its transformation describes macroscoic

More information

REVIEW & SUMMARY. Molar Specific Heats The molar specific heat C V of a gas at constant volume is defined as

REVIEW & SUMMARY. Molar Specific Heats The molar specific heat C V of a gas at constant volume is defined as REIEW & SUMMARY 59 PART Kinetic Theory of Gases The kinetic theory of gases relates the macroscoic roerties of gases (for examle, ressure and temerature) to the microscoic roerties of gas molecules (for

More information

Phase Change DIagram

Phase Change DIagram States of Matter Phase Change DIagram Phase Change Temperature remains during a phase change. Water phase changes Phase Diagram What is a phase diagram? (phase diagram for water) Normal melting point:

More information

Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : ,

Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : , CS SOLUTIONS C C CC Solution : Solution is a mixture of two or more non-reacting ure substances in which comosition may be altered within certain limits. There are two tyes of solution. They are : Homogeneous

More information

GEF2200 vår 2017 Løsningsforslag sett 1

GEF2200 vår 2017 Løsningsforslag sett 1 GEF2200 vår 2017 Løsningsforslag sett 1 A.1.T R is the universal gas constant, with value 8.3143JK 1 mol 1. R is the gas constant for a secic gas, given by R R M (1) where M is the molecular weight of

More information

Paper C Exact Volume Balance Versus Exact Mass Balance in Compositional Reservoir Simulation

Paper C Exact Volume Balance Versus Exact Mass Balance in Compositional Reservoir Simulation Paer C Exact Volume Balance Versus Exact Mass Balance in Comositional Reservoir Simulation Submitted to Comutational Geosciences, December 2005. Exact Volume Balance Versus Exact Mass Balance in Comositional

More information

12.808: Some Physical Properties of Sea Water or, More than you ever wanted to know about the basic state variables of the ocean

12.808: Some Physical Properties of Sea Water or, More than you ever wanted to know about the basic state variables of the ocean 12.88: Some Physical Proerties of Sea Water or, More than you ever wanted to know about the basic state variables of the ocean Salinity Various salt constituents in 1 m 3 of seawater having (t, S) = (2,

More information

REAL GASES. (B) pv. (D) pv. 3. The compressibility factor of a gas is less than unity at STP. Therefore, molar volume (V m.

REAL GASES. (B) pv. (D) pv. 3. The compressibility factor of a gas is less than unity at STP. Therefore, molar volume (V m. SINGLE ORRET ANSWER REAL GASES 1. A real gas is suosed to obey the gas equation ( b) = at STP. If one mole of a gas occuies 5dm 3 volume at STP, then its comressibility factor is (b=.586 L mol 1F) (A)

More information

1. Read the section on stability in Wallace and Hobbs. W&H 3.53

1. Read the section on stability in Wallace and Hobbs. W&H 3.53 Assignment 2 Due Set 5. Questions marked? are otential candidates for resentation 1. Read the section on stability in Wallace and Hobbs. W&H 3.53 2.? Within the context of the Figure, and the 1st law of

More information

CHAPTER 2 THERMODYNAMICS

CHAPTER 2 THERMODYNAMICS CHAPER 2 HERMODYNAMICS 2.1 INRODUCION herodynaics is the study of the behavior of systes of atter under the action of external fields such as teerature and ressure. It is used in articular to describe

More information

Mathematics as the Language of Physics.

Mathematics as the Language of Physics. Mathematics as the Language of Physics. J. Dunning-Davies Deartment of Physics University of Hull Hull HU6 7RX England. Email: j.dunning-davies@hull.ac.uk Abstract. Courses in mathematical methods for

More information

Outline of the Course

Outline of the Course Outline of the Course 1) Review and Definitions 2) Molecules and their Energies 3) 1 st Law of Thermodynamics Conservation of Energy. 4) 2 nd Law of Thermodynamics Ever-Increasing Entropy. 5) Gibbs Free

More information

1 Entropy 1. 3 Extensivity 4. 5 Convexity 5

1 Entropy 1. 3 Extensivity 4. 5 Convexity 5 Contents CONEX FUNCIONS AND HERMODYNAMIC POENIALS 1 Entroy 1 2 Energy Reresentation 2 3 Etensivity 4 4 Fundamental Equations 4 5 Conveity 5 6 Legendre transforms 6 7 Reservoirs and Legendre transforms

More information

arxiv:cond-mat/ v2 25 Sep 2002

arxiv:cond-mat/ v2 25 Sep 2002 Energy fluctuations at the multicritical oint in two-dimensional sin glasses arxiv:cond-mat/0207694 v2 25 Se 2002 1. Introduction Hidetoshi Nishimori, Cyril Falvo and Yukiyasu Ozeki Deartment of Physics,

More information

Statics and dynamics: some elementary concepts

Statics and dynamics: some elementary concepts 1 Statics and dynamics: some elementary concets Dynamics is the study of the movement through time of variables such as heartbeat, temerature, secies oulation, voltage, roduction, emloyment, rices and

More information

Phase Equilibrium Calculations by Equation of State v2

Phase Equilibrium Calculations by Equation of State v2 Bulletin of Research Center for Comuting and Multimedia Studies, Hosei University, 7 (3) Published online (htt://hdl.handle.net/4/89) Phase Equilibrium Calculations by Equation of State v Yosuke KATAOKA

More information

A) sublimation. B) liquefaction. C) evaporation. D) condensation. E) freezing. 11. Below is a phase diagram for a substance.

A) sublimation. B) liquefaction. C) evaporation. D) condensation. E) freezing. 11. Below is a phase diagram for a substance. PX0411-1112 1. Which of the following statements concerning liquids is incorrect? A) The volume of a liquid changes very little with pressure. B) Liquids are relatively incompressible. C) Liquid molecules

More information

Chemical Kinetics and Equilibrium - An Overview - Key

Chemical Kinetics and Equilibrium - An Overview - Key Chemical Kinetics and Equilibrium - An Overview - Key The following questions are designed to give you an overview of the toics of chemical kinetics and chemical equilibrium. Although not comrehensive,

More information

Name: Class: Date: SHORT ANSWER Answer the following questions in the space provided.

Name: Class: Date: SHORT ANSWER Answer the following questions in the space provided. CHAPTER 10 REVIEW States of Matter SECTION 1 SHORT ANSWER Answer the following questions in the space provided. 1. Identify whether the descriptions below describe an ideal gas or a real gas. a. The gas

More information

First law of thermodynamics (Jan 12, 2016) page 1/7. Here are some comments on the material in Thompkins Chapter 1

First law of thermodynamics (Jan 12, 2016) page 1/7. Here are some comments on the material in Thompkins Chapter 1 First law of thermodynamics (Jan 12, 2016) age 1/7 Here are some comments on the material in Thomkins Chater 1 1) Conservation of energy Adrian Thomkins (eq. 1.9) writes the first law as: du = d q d w

More information

Phase Equilibrium: Preliminaries

Phase Equilibrium: Preliminaries Phase Equilibrium: Preliminaries Phase diagrams for two one component systems, CO 2 and H 2 O, are shown below. The main items to note are the following: The lines represent equilibria between two phases.

More information

= = 10.1 mol. Molar Enthalpies of Vaporization (at Boiling Point) Molar Enthalpy of Vaporization (kj/mol)

= = 10.1 mol. Molar Enthalpies of Vaporization (at Boiling Point) Molar Enthalpy of Vaporization (kj/mol) Ch 11 (Sections 11.1 11.5) Liquid Phase Volume and Density - Liquid and solid are condensed phases and their volumes are not simple to calculate. - This is different from gases, which have volumes that

More information

Exam 3 Solutions. ClO g. At 200 K and a total pressure of 1.0 bar, the partial pressure ratio for the chlorine-containing compounds is p ClO2

Exam 3 Solutions. ClO g. At 200 K and a total pressure of 1.0 bar, the partial pressure ratio for the chlorine-containing compounds is p ClO2 Chemistry 360 Dr. Jean M. Standard Fall 2016 Name KEY Exam 3 Solutions 1.) (14 points) Consider the gas phase decomposition of chlorine dioxide, ClO 2, ClO 2 ( g) ClO ( g) + O ( g). At 200 K and a total

More information

Setting up the Mathematical Model Review of Heat & Material Balances

Setting up the Mathematical Model Review of Heat & Material Balances Setting u the Mathematical Model Review of Heat & Material Balances Toic Summary... Introduction... Conservation Equations... 3 Use of Intrinsic Variables... 4 Well-Mixed Systems... 4 Conservation of Total

More information

ANALYTICAL MODEL FOR THE BYPASS VALVE IN A LOOP HEAT PIPE

ANALYTICAL MODEL FOR THE BYPASS VALVE IN A LOOP HEAT PIPE ANALYTICAL MODEL FOR THE BYPASS ALE IN A LOOP HEAT PIPE Michel Seetjens & Camilo Rindt Laboratory for Energy Technology Mechanical Engineering Deartment Eindhoven University of Technology The Netherlands

More information

CHEMISTRY XL-14A PHYSICAL EQUILIBRIUM. August 13, 2011 Robert Iafe

CHEMISTRY XL-14A PHYSICAL EQUILIBRIUM. August 13, 2011 Robert Iafe CHEMISTRY XL-14A PHYSICAL EQUILIBRIUM August 13, 2011 Robert Iafe Chapter Overview 2 Phases and Phase Transitions Solubility Colligative Properties Binary Liquid Mixtures Phases and Phase Transitions 3

More information

Actual exergy intake to perform the same task

Actual exergy intake to perform the same task CHAPER : PRINCIPLES OF ENERGY CONSERVAION INRODUCION Energy conservation rinciles are based on thermodynamics If we look into the simle and most direct statement of the first law of thermodynamics, we

More information

Thermodynamic condition for equilibrium between two phases a and b is G a = G b, so that during an equilibrium phase change, G ab = G a G b = 0.

Thermodynamic condition for equilibrium between two phases a and b is G a = G b, so that during an equilibrium phase change, G ab = G a G b = 0. CHAPTER 5 LECTURE NOTES Phases and Solutions Phase diagrams for two one component systems, CO 2 and H 2 O, are shown below. The main items to note are the following: The lines represent equilibria between

More information

Unit 6 Solids, Liquids and Solutions

Unit 6 Solids, Liquids and Solutions Unit 6 Solids, Liquids and Solutions 12-1 Liquids I. Properties of Liquids and the Kinetic Molecular Theory A. Fluids 1. Substances that can flow and therefore take the shape of their container B. Relative

More information

Introduction: Introduction. material is transferred from one phase (gas, liquid, or solid) into another.

Introduction: Introduction. material is transferred from one phase (gas, liquid, or solid) into another. Introduction: Virtually all commercial chemical processes involve operations in which material is transferred from one phase (gas, liquid, or solid) into another. rewing a cup of Coffee (Leaching) Removal

More information

... g 1. M 1. M n g n. g 3 M 2 M n

... g 1. M 1. M n g n. g 3 M 2 M n Modeling Multi-Element Systems Using Bond Grahs Jürgen Greifeneder Franοcois E Cellier Fakultät Verf-technik und echn Kybernetik Det of Electr & Com Engr Universität Stuttgart he University of Arizona

More information

Chapter 9 Practical cycles

Chapter 9 Practical cycles Prof.. undararajan Chater 9 Practical cycles 9. Introduction In Chaters 7 and 8, it was shown that a reversible engine based on the Carnot cycle (two reversible isothermal heat transfers and two reversible

More information

16. CHARACTERISTICS OF SHOCK-WAVE UNDER LORENTZ FORCE AND ENERGY EXCHANGE

16. CHARACTERISTICS OF SHOCK-WAVE UNDER LORENTZ FORCE AND ENERGY EXCHANGE 16. CHARACTERISTICS OF SHOCK-WAVE UNDER LORENTZ FORCE AND ENERGY EXCHANGE H. Yamasaki, M. Abe and Y. Okuno Graduate School at Nagatsuta, Tokyo Institute of Technology 459, Nagatsuta, Midori-ku, Yokohama,

More information

Chemistry 420/523 Chemical Thermodynamics (Spring ) Examination 1

Chemistry 420/523 Chemical Thermodynamics (Spring ) Examination 1 Chemistry 420/523 Chemical hermodynamics (Sring 2001-02) Examination 1 1 Boyle temerature is defined as the temerature at which the comression factor Z m /(R ) of a gas is exactly equal to 1 For a gas

More information

Atmosphere, Ocean and Climate Dynamics Answers to Chapter 4

Atmosphere, Ocean and Climate Dynamics Answers to Chapter 4 Atmoshere, Ocean and Climate Dynamics Answers to Chater 4 1. Show that the buoyancy frequency, Eq.(4.22), may be written in terms of the environmental temerature rofile thus N 2 = g µ dte T E dz + Γ d

More information

CHAPTER 9: LIQUIDS AND SOLIDS

CHAPTER 9: LIQUIDS AND SOLIDS CHAPTER 9: LIQUIDS AND SOLIDS Section 9.1 Liquid/Vapor Equilibrium Vaporization process in which a liquid vapor open container - evaporation continues until all liquid evaporates closed container 1) Liquid

More information

MATH 2710: NOTES FOR ANALYSIS

MATH 2710: NOTES FOR ANALYSIS MATH 270: NOTES FOR ANALYSIS The main ideas we will learn from analysis center around the idea of a limit. Limits occurs in several settings. We will start with finite limits of sequences, then cover infinite

More information

Intermolecular Forces and Liquids and Solids

Intermolecular Forces and Liquids and Solids Intermolecular Forces and Liquids and Solids Chapter 11 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. A phase is a homogeneous part of the system in contact

More information

Lecture 4-6 Equilibrium

Lecture 4-6 Equilibrium Lecture 4-6 Equilibrium Discontinuity in the free energy, G verses T graph is an indication of phase transition. For one-component system, existing in two phases, the chemical potentials of each of these

More information

Lecture 13. Heat Engines. Thermodynamic processes and entropy Thermodynamic cycles Extracting work from heat

Lecture 13. Heat Engines. Thermodynamic processes and entropy Thermodynamic cycles Extracting work from heat Lecture 3 Heat Engines hermodynamic rocesses and entroy hermodynamic cycles Extracting work from heat - How do we define engine efficiency? - Carnot cycle: the best ossible efficiency Reading for this

More information

Intermolecular Forces and Liquids and Solids

Intermolecular Forces and Liquids and Solids Intermolecular Forces and Liquids and Solids Chapter 11 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1 A phase is a homogeneous part of the system in contact

More information

Last Time. A new conjugate pair: chemical potential and particle number. Today

Last Time. A new conjugate pair: chemical potential and particle number. Today Last Time LECTURE 9 A new conjugate air: chemical otential and article number Definition of chemical otential Ideal gas chemical otential Total, Internal, and External chemical otential Examle: Pressure

More information

Session 12 : Monopropellant Thrusters

Session 12 : Monopropellant Thrusters Session 12 : Monoroellant Thrusters Electrothermal augmentation of chemical rockets was the first form of electric roulsion alied in sace vehicles. In its original imlementation, resistojets were used

More information

Chapter 10. Lesson Starter. Why did you not smell the odor of the vapor immediately? Explain this event in terms of the motion of molecules.

Chapter 10. Lesson Starter. Why did you not smell the odor of the vapor immediately? Explain this event in terms of the motion of molecules. Preview Lesson Starter Objectives The Kinetic-Molecular Theory of Gases The Kinetic-Molecular Theory and the Nature of Gases Deviations of Real Gases from Ideal Behavior Section 1 The Kinetic-Molecular

More information

Chapter 6. Thermodynamics and the Equations of Motion

Chapter 6. Thermodynamics and the Equations of Motion Chater 6 hermodynamics and the Equations of Motion 6.1 he first law of thermodynamics for a fluid and the equation of state. We noted in chater 4 that the full formulation of the equations of motion required

More information

Maximum Entropy and the Stress Distribution in Soft Disk Packings Above Jamming

Maximum Entropy and the Stress Distribution in Soft Disk Packings Above Jamming Maximum Entroy and the Stress Distribution in Soft Disk Packings Above Jamming Yegang Wu and S. Teitel Deartment of Physics and Astronomy, University of ochester, ochester, New York 467, USA (Dated: August

More information

Lecture 13 Heat Engines

Lecture 13 Heat Engines Lecture 3 Heat Engines hermodynamic rocesses and entroy hermodynamic cycles Extracting work from heat - How do we define engine efficiency? - Carnot cycle: the best ossible efficiency Reading for this

More information

Physical transformations of pure substances Boiling, freezing, and the conversion of graphite to diamond examples of phase transitions changes of

Physical transformations of pure substances Boiling, freezing, and the conversion of graphite to diamond examples of phase transitions changes of Physical transformations of pure substances Boiling, freezing, and the conversion of graphite to diamond examples of phase transitions changes of phase without change of chemical composition. In this chapter

More information

Changes of State. Substances in equilibrium change back and forth between states at equal speeds. Main Idea

Changes of State. Substances in equilibrium change back and forth between states at equal speeds. Main Idea Section 4 s Substances in equilibrium change back and forth between states at equal speeds. A liquid boils when it has absorbed enough energy to evaporate. Freezing occurs when a substance loses enough

More information

Towards understanding the Lorenz curve using the Uniform distribution. Chris J. Stephens. Newcastle City Council, Newcastle upon Tyne, UK

Towards understanding the Lorenz curve using the Uniform distribution. Chris J. Stephens. Newcastle City Council, Newcastle upon Tyne, UK Towards understanding the Lorenz curve using the Uniform distribution Chris J. Stehens Newcastle City Council, Newcastle uon Tyne, UK (For the Gini-Lorenz Conference, University of Siena, Italy, May 2005)

More information

Chapter 11. Freedom of Motion. Comparisons of the States of Matter. Liquids, Solids, and Intermolecular Forces

Chapter 11. Freedom of Motion. Comparisons of the States of Matter. Liquids, Solids, and Intermolecular Forces Liquids, Solids, and Intermolecular Forces Chapter 11 Comparisons of the States of Matter The solid and liquid states have a much higher density than the gas state The solid and liquid states have similar

More information

ESCI 342 Atmospheric Dynamics I Lesson 10 Vertical Motion, Pressure Coordinates

ESCI 342 Atmospheric Dynamics I Lesson 10 Vertical Motion, Pressure Coordinates Reading: Martin, Section 4.1 PRESSURE COORDINATES ESCI 342 Atmosheric Dynamics I Lesson 10 Vertical Motion, Pressure Coordinates Pressure is often a convenient vertical coordinate to use in lace of altitude.

More information

Chem 112 Dr. Kevin Moore

Chem 112 Dr. Kevin Moore Chem 112 Dr. Kevin Moore Gas Liquid Solid Polar Covalent Bond Partial Separation of Charge Electronegativity: H 2.1 Cl 3.0 H Cl δ + δ - Dipole Moment measure of the net polarity in a molecule Q Q magnitude

More information

PROCESSING OF LOW-VISCOSITY CBT THERMOPLASTIC COMPOSITES: HEAT TRANSFER ANALYSIS

PROCESSING OF LOW-VISCOSITY CBT THERMOPLASTIC COMPOSITES: HEAT TRANSFER ANALYSIS PROCESSING OF LOW-VISCOSITY CBT THERMOPLASTIC COMPOSITES: HEAT TRANSFER ANALYSIS Dr. Adrian Murtagh, Siora Coll and Dr. Conchúr Ó Brádaigh Comosites Research Unit Det. of Mechanical & Biomedical Engineering,

More information

Chapter 11. Liquids and Intermolecular Forces

Chapter 11. Liquids and Intermolecular Forces Chapter 11 Liquids and Intermolecular Forces States of Matter The three states of matter are 1) Solid Definite shape Definite volume 2) Liquid Indefinite shape Definite volume 3) Gas Indefinite shape Indefinite

More information

Chemistry B11 Chapter 6 Gases, Liquids, and Solids

Chemistry B11 Chapter 6 Gases, Liquids, and Solids Chapter 6 Gases, Liquids, and Solids States of matter: the physical state of matter depends on a balance between the kinetic energy of particles, which tends to keep them apart, and the attractive forces

More information

F = U TS G = H TS. Availability, Irreversibility S K Mondal s Chapter 5. max. actual. univ. Ns =

F = U TS G = H TS. Availability, Irreversibility S K Mondal s Chapter 5. max. actual. univ. Ns = Availability, Irreversibility S K Mondal s Chater 5 I = W W 0 max ( S) = Δ univ actual 7. Irreversibility rate = I = rate of energy degradation = rate of energy loss (Wlost) = 0 S gen for all rocesses

More information

States of matter Part 2

States of matter Part 2 Physical Pharmacy Lecture 2 States of matter Part 2 Assistant Lecturer in Pharmaceutics Overview The Liquid State General properties Liquefaction of gases Vapor pressure of liquids Boiling point The Solid

More information

Efficiencies. Damian Vogt Course MJ2429. Nomenclature. Symbol Denotation Unit c Flow speed m/s c p. pressure c v. Specific heat at constant J/kgK

Efficiencies. Damian Vogt Course MJ2429. Nomenclature. Symbol Denotation Unit c Flow speed m/s c p. pressure c v. Specific heat at constant J/kgK Turbomachinery Lecture Notes 1 7-9-1 Efficiencies Damian Vogt Course MJ49 Nomenclature Subscrits Symbol Denotation Unit c Flow seed m/s c Secific heat at constant J/kgK ressure c v Secific heat at constant

More information