- The value of a state function is independent of the history of the system. - Temperature is an example of a state function.

Size: px
Start display at page:

Download "- The value of a state function is independent of the history of the system. - Temperature is an example of a state function."

Transcription

1 First Law of hermodynamics 1 State Functions - A State Function is a thermodynamic quantity whose value deends only on the state at the moment, i. e., the temerature, ressure, volume, etc - he value of a state function is indeendent of the history of the system. - emerature is an examle of a state function. - he fact that temerature is a state function is extremely useful because it we can measure the temerature change in the system by knowing the initial temerature and the final temerature. - In other words, we don t need all of the nitty-gritty detail of a rocess to measure the change in the value of a state function. - In contrast, we do need all of the nitty-gritty details to measure the heat or the work of a system. Reversibility A reversible rocess is a rocess where the effects of following a thermodynamic ath can be undone be exactly reversing the ath. An easier definition is a rocess that is always at equilibrium even when undergoing a change. Phase changes and chemical equilibria are examles of reversible rocesses. Ideally the comosition throughout the system must be homogeneous. - his requirement imlies that the no gradients, currents or eddys can exist. - o eliminate all inhomogeneities, a reversible rocess must occur infinitely slow! - hus no truly reversible rocesses exist. However, many systems are aroximately reversible. And assuming reversible rocesses will greatly aid our calculations of various thermodynamic state functions. Reversibility during ressure changes ensures that = ex hat is, the ressure on the inside of the container is always equal to the ressure exerted on the outside of the container.

2 heorem of Maximum Work he maximum amount of work that can be extracted from an exansion rocess occurs under reversible conditions. hus the theorem imlies that during an irreversible exansion, some of the energy is lost as heat rather than work. he inhomogeneities (currents, gradients and eddys) in ressure that occur during an irreversible rocess are resonsible for the heat. Initial Definitions of Energy and Energy ransfer Internal Energy - Sum of the kinetic and otential energy in a samle of matter. Microscoic modes of Internal Energy (0 th century view) - degrees of freedom for energy storage translational rotational librational vibrations caused by intermolecular forces vibrational electronic nuclear, etc - more secifics in second semester - unnecessary for understanding of thermodynamics but sometimes helful. Macroscoic view of internal energy (19 th century view) - A reservoir of energy within the samle. - he details of the energy are unknown. Internal energy is a state function. Work Recall the definition of work from classical mechanics. w = F dl Microscoically, work involves the concerted motion of molecules (that is, molecules moving in one direction.) Work is a transfer of energy, not a quantity of energy. Work is not a state function. -calculation of work deends on thermodynamic ath (as seen from definition)

3 A more thermodynamically useful rearrangement (let F = A and d = dl A) is 3 w = d - the negative sign is a sign convention (discussed below) - ex is the external ressure (the ressure outside the container) - ex is used in definition because we really do not know what kind of ressures develo during a thermodynamic change. - i.e., ressure currents and eddies are likely for an arbitrary change. Other forms of work are ossible w = E dq Electrical work ( ) w = E d P Polarization work (e.g., iezoelectricity) w = σ da w = τ dθ Surface tension wisting work Sign conventions for work -w work done by system (exansion for work) +w work done on the system (comression for work) Heat - macroscoically heat is a thermal energy transfer - energy transfer usually characterized by temerature and the zeroth law of thermodynamics. - microscoically heat transfer comes from - inelastic collisions - energy transfer from one molecule to another in multile (every) direction Sign conventions for heat -q heat transferred away from body (heat lost) +q heat transferred into body (heat gained) ex

4 First Law of hermodynamics 4 he hysicists that studied energy changes recognized that the energy of an object could be changed via heat or work. James Joule demonstrated exerimentally the law of conservation of energy, that is, that energy is not gained or lost, but merely transferred from one object to another. he law of conservation of energy is also known as the first law of thermodynamics. Since energy changes can be exressed only as heat or work, the first law of thermodynamics has the mathematical exression U = q + w Subtle oints to make: 1) q and w are energy changes, writing q and w is imroer. - We can t refer to object having an amount of heat or work. he object has internal energy, enthaly, free energy, etc ) he form of the first law of thermodynamics deends on the sign convention chosen for heat and work. Older texts state that U = q w ; however, the sign convention for work is different, Differential Form of the First Law w = + d he differential form of the first law is written as ex du = ñq +ñw he differential forms of heat and work are written with the slash to emhasize that they are inexact differentials. Constant volume heat During a constant volume rocess, w = 0. herefore U = q v hat is, constant volume heat is equal to a change in the free energy.

5 Measureable thermodynamic quantities 5 hermal exansivity he thermal exansivity is the measure of how a material changes its volume as the temerature changes. 1 α = Isothermal comressibility 1 κ = he isothermal comressibility is the measure of how a material changes its volume as the ressure changes. Almost always, volume will decrease with increasing ressure, therefore, a negative sign is included in the definition to allow for the tabulation a ositive values. Both definitions modify an extensive change, or intensive change by dividing the change by the volume., into an Enthaly Background In order to fully exlain his ideas of free energy, Josiah Willard Gibbs needed to construct an energy state function that had the definition H = U + P Gibbs named the quantity the heat content because a change in the quantity corresonded to heat gained or lost by a system at constant ressure rovided no non work is being done, that is. H = q Kamerlingh Onnes eventually named the function, H the enthaly and the name stuck.

6 Heat caacity - Heat is roortional to mass. - Heat is roortional to temerature difference. - Proortionality constant for a secific substance is the heat caacity or secific heat. q = mc 6 Definition of Constant Pressure Heat Caacity C Since constant ressure heat is a change in enthaly, the constant ressure heat caacity can be rewritten as C H = his quantity is one of the easily measurable quantities that is very imortant in thermodynamics. Definition of Constant olume Heat Caacity C v Since constant volume heat is a change in internal energy, the constant volume heat caacity can be rewritten as C v U = v his quantity is another one of the easily measurable quantities that we will use a great deal in thermodynamics studies. Relationshi between C and C One way of examining the difference between internal energy and enthaly is by examining the difference between the constant ressure heat caacity and the constant volume heat caacity. C ( U P) ( P) H U + U U U C = = = + U U = + P

7 U At this oint, we want to find another exression for 7 since it is not an easily U measurable quantity. We can find an alternative exression for based the total differential of the internal energy. he internal energy is a function of temerature and volume, thus the total differential is U U du = d + d Dividing the total differential by d under constant ressure conditions yields U U U U U = + = + wo of the artial derivatives in this exression should be familiar: U = C and = α. Remember the thermal exansivity is 1 α = Let us substitute our result into the original equation for C C, U U U U U C C = + P = + + P U U U = + P = P P + = α + What is the relationshi between C C for an ideal gas? 1 1 nr 1 nr nr α = = = = U nr U nr U C C = α + = + = + nr U nr 1 U = + = nr + 1

8 U Remember that for an ideal gas = 0, therefore 1 U C C = nr + 1 = nr 8 he relationshi can be rewritten in terms of molar heat caacities. Cɶ Cɶ = R his relationshi is very imortant and worth memorizing. Why is C greater than C? - Heat caacity is substance s caacity to store energy. - A substance that can store energy via work has a greater heat caacity than a substance that has its volume ket constant. Heat Reexamined Isothermal Heat Isochoric Heat dq = Cd q = 0 dq C d q C d = = If C is indeendent of temerature, (fair assumtion for small temerature changes), then q = C d = C d = C = C Isobaric Heat ( ) 1 dq C d q C d = = Adiabatic Heat By definition, q = 0 All of the above results are free from assumtions (unless noted).

9 hermochemistry 9 Extent of Reaction o monitor the rogress, a variable known as the extant of reaction is defined as n ξ = n i i,0 ν i where n i is the number of moles of chemical secies i, n i,0 is the initial number of moles of chemical secies i and ν i is the stoichiometric coefficient of the chemical secies i for the secific reaction occurring. When the extent of reaction is defined this way, the value of the extant of reaction does not deend on the chemical secies used to examine the state. Examle: Consider the reaction P 4 (s) + 5 O (g) P O 5 (s). Suose we start with 3 moles of hoshorus and 15 moles of oxygen. a) What is the extant of reaction when 4 moles of P O 5 has been created? n n 4 mol 0 mol ν i i,0 ξ = = = i Note: How much P 4 has been consumed to create 4 moles of P O 5? A: moles. b) What is the extant of reaction when moles of hoshorus has reacted? n n 1mol 3mol ν 1 i i,0 ξ = = = i he definition includes the stoichiometric coefficient to ut the changes of the all the chemical secies on an equal footing. Soon we will be dealing reaction energies and other quantities secific to reactions. he reaction energies need to be intensive quantities and yet each chemical secies may have a different amount. hus reaction energies are defined with resect to the extent of reaction so that the secific chemical secies is not imortant, but the reaction as a whole is imortant.

10 Standard States 10 Standard Pressure - φ hermodynamic quantities are measured with resect to a ressure of 1 bar ( kpa) Standard emerature - φ hermodynamic quantities are measured with resect to a temerature of 5 C (98.15 K) Standard Concentration- c φ hermodynamic quantities are measured with resect to a concentration of 1m (that is 1 molal). - Molal is used as a concentration unit rather than molar because molal is indeendent of temerature. Biological Standard State he biological standard state is a ressure of 1 bar, a temerature of 37 C and H of 7, (that is [H + ] = ).

11 Hess Law 11 Since enthaly is a state function, we can choose more than one thermodynamic ath to calculate a state function. For chemical changes, Hess Law states that the enthaly of a reaction can be calculated from the enthalies of all of the chemical ste rocesses needed for the chemical reaction. In other words, if reactions can be added together to form a resultant reaction, then enthalies of the ste reactions can be added to find a resultant enthaly of reaction. Examle: Calculate the enthaly of reaction for the following reaction: Al(s) + 3 Cl (g) AlCl 3 (s), given the reactions below. Al(s) + 6 HCl(aq) AlCl 3 (aq) + 3 H (g) H = kj/mol HCl(g) HCl(aq) H = kj/mol H (g) + Cl (g) HCl(g) H = -185 kj/mol AlCl 3 (s) AlCl 3 (aq) H = -33 kj/mol Rearrange the equations such that their sum is the reaction of interest. Al(s) + 6 HCl(aq) AlCl 3 (aq) + 3 H (g) H = kj/mol 6[HCl(g) HCl(aq)] H = 6[-74.8 kj/mol] 3[H (g) + Cl (g) HCl(g)] H = 3[-185 kj/mol] [AlCl 3 (aq) AlCl 3 (s)] H = [+33 kj/mol] Al(s) + 6 HCl(aq) + 6 HCl(g) + 3 H (g) + 3 Cl (g) + AlCl 3 (aq) AlCl 3 (aq) + 3 H (g) + 6 HCl(aq) + 6 HCl(g) + AlCl 3 (s) H = kj/mol + 6[-74.8 kj/mol] + 3[-185 kj/mol] + [+33 kj/mol] = kj/mol Al(s) + 3 Cl (g) AlCl 3 (s) H = kj/mol

12 Enthaly of Formation 1 he enthaly of formation is the enthaly of a formation reaction for a articular substance. A formation reaction is that where a comound is formed from elements in the naturally occurring state. C(s) + 3 H (g) + ½ O (g) C H 5 OH (l) ½ P 4 (s) + 5/ O (g) P O 5 (s) 7 Fe(s) + 18 C(s) + 9 N (g) Fe 4 [Fe(CN) 6 ] 3 (s) Enthaly of formation of the elements By definition, the enthaly of formation of an element in its natural state is zero. f H (B(s)) = 0 kj/mol f H (C(grahite)) = 0 kj/mol f H (Br (l)) = 0 kj/mol f H (S 8 (s)) = 0 kj/mol f H (Ag(s)) = 0 kj/mol f H (Xe(g)) = 0 kj/mol Reaction enthalies can be calculated as a stoichiometric sum of enthalies of formation. his technique is an alication of Hess law. Examle: Calculate the enthaly of reaction for the following reaction: CH 3 COOH(l) + C 4 H 9 OH(l) C 4 H 9 OOCCH 3 (l) + H O(l), given the reactions below. C(s) + H (g) + O (g) CH 3 COOH(l) H = kj/mol 4 C(s) + 5 H (g) + ½ O (g) C 4 H 9 OH(l) H = - 38 kj/mol 6 C(s) + 6 H (g) + O (g) C 4 H 9 OOCCH 3 (l) H = kj/mol H (g) + ½ O (g) H O(l) H = kj/mol Note all of the above reactions are formation reactions. Rearrange the equations such that their sum is the reaction of interest. CH 3 COOH(l) C(s) + H (g) + O (g) H = kj/mol C 4 H 9 OH(l) 4 C(s) + 5 H (g) + ½ O (g) H = + 38 kj/mol 6 C(s) + 6 H (g) + O (g) C 4 H 9 OOCCH 3 (l) H = kj/mol H (g) + ½ O (g) H O(l) H = kj/mol CH 3 COOH(l) + C 4 H 9 OH(l) + 6 C(s) + 6 H (g) + O (g) + H (g) + ½ O (g) C(s) + H (g) + O (g) + 4 C(s) + 5 H (g) + ½ O (g) + C 4 H 9 OOCCH 3 (l) + H O(l) H = kj/mol + 38 kj/mol kj/mol kj/mol = - 84 kj/mol CH 3 COOH(l) + C 4 H 9 OH(l) C 4 H 9 OOCCH 3 (l) + H O(l) H = - 84 kj/mol

13 Calorimetry 13 Constant Pressure Calorimeter - coffee cu calorimeter - oen to atmoshere - aroriate for solution chemistry Constant olume Calorimeter - bomb calorimeter - sealed and isolated - aroriate for gas hase chemistry Adiabatic Calorimeter - heat measured by temerature needed to kee thermal energy constant. Process of measuring heat of combustion with bomb calorimeter 1. Mass wick used to start combustion.. Mass water used to absorb heat. 3. Mass standard (benzoic acid) to be combusted 4. Add water to combustion chamber to ensure that water from combustion will be in liquid hase. 5. Load standard and seal 6. Fill combustion chamber with oxygen ( 5 atm) 7. Begin combustion and record temerature change of water. 8. Calculate heat caacity of calorimeter from standard - need to account for heat of wick, and heat caacity of water. 9. Reeat rocess with samle. - with heat from wick, heat from the water and heat from the calorimeter, the heat of combustion of samle can be calculated. Hess law is often used to the enthaly of formation of a substance after its enthaly of combustion has been calculated. Examle: 0.53 g of the military exlosive, cyclotetramethylenetetranitramine (HMX), C 4 H 8 N 8 O 8 is combusted in a bomb (!) calorimeter and an internal energy change of 4.60 kj is measured. Calculate the enthaly of formation for HMX. First write a balanced equation for its comlete combustion. C 4 H 8 N 8 O 8 (s) + O (g) 4 CO (g) + 4 H O(l) + 4 N (g) Next calculate the molar internal energy of reaction from the exerimental data. 4.6 kj g = 616 kj mol 0.53g mol

14 o calculate the enthaly of the reaction, we need to return the definition of enthaly. H = U + ( P) = U + ( nr) = U + n ( R) Reexamining the balanced chemical equation, we see that 8 moles moles = 6 moles of gas has been roduced. C 4 H 8 N 8 O 8 (s) + O (g) 4 CO (g) + 4 H O(l) + 4 N (g) kJ H = U + n ( R) = 616 kj mol K mol K.478kJ = 616 kj mol + 6 = 616 kj mol kj mol = 601kJ mol mol 14 he molar enthaly of reaction can also be written in terms of the molar enthalies of formation. 4 f H (CO (g)) + 4 f H (H O(l)) - f H (C 4 H 8 N 8 O 8 (s)) = r H Rearranged, the equation becomes, f H (C 4 H 8 N 8 O 8 (s)) = 4 f H (CO (g)) + 4 f H (H O(l)) - r H From table, we find the enthalies of formation for CO (g) and H O(l) f H (C 4 H 8 N 8 O 8 (s)) = 4 ( kj/mol) + 4 (-85.8 kj/mol) - (-601 kj/mol) = -116 kj/mol emerature Deendence of Internal Energy and Enthaly Recall that C U = and C H = hese relationshis imly that we can find the internal energy or enthaly at a nonstandard temerature as long as we know the heat caacity. du = C d and dh = C d Integrating both sides of these equations yields U = C d and H = C d

15 For large temerature changes, we can t assume that the heat caacity is indeendent of temerature. hus to calculate the internal energy or enthaly at a nonzero temerature, we need the temerature deendence of the heat caacity. Examle: Calculate the change in enthaly of H (g) from 373 K to 1000 K, given that the constant ressure heat caacity has the form C = d + e + f where d = 7.8 J/K mol, e = J/K mol and f = J K/mol K 373 K ( ) H = H H = C d = d + e + f d = d d + e d + f d J K mol ( 7.8 J K mol)( 67 K) ( K ) e 1000 f 1000 e 1 1 = d = d ( ) + ( ) f = + 1 ( )( ) J K mol K = + + = = J mol 1403J mol J mol J mol 18.51kJ mol 15 Kirchoff s law 0 ( ) ( ) H = H 98 K + ν C d r r i,i 98 i hat is by taking a stoichiometric sum of the heat caacities and integrating over the temerature range, we can find the correction to the reaction enthaly at a non standard state temerature.

16 Examle: 471 kj/mol is the reaction enthaly under standard conditions for the following reaction: Fe O 3 (s) + 3 C(s) 4 Fe(s) + 3 CO (g). What is the reaction enthaly at 1000 K? According to the NIS Webbook Internet site, the constant ressure heat caacities of the chemical secies in the above smelting rocess can be fitted according to the 3 Shumate equation, C = A + B + C + D + E he values for A, B, C, D and E for each secies are given below. A B C D E Fe O C CO Fe Sum K 0 3 ( ) ( ) ( ) H 1000K = H 98K + ν A + B + C + D + E d r r i i i i i i 98K i B C D = H ( 98K) + ν A E i 1000 i i r i i i 98 i kJ ( ) ( ) ( )( ) ( r H 1000 K mol ) 5 8 ( ) ( ) ( ) ( ) = kJ 49.3kJ 1.6kJ 34.1kJ 11.kJ 6.5kJ 413.9kJ rh ( 1000 K) = + + = mol mol mol mol mol mol mol Kirchoff s Law can be stated with in a differential form as well. 16 U Include = C and H = C Bond Enthalies A bond enthaly is the energy needed to searate two atoms. abulated bond enthalies are average values calculated from the dissociation of many different comounds. hus tabulated bond enthalies are aroximate values. *However, they can be useful for aroximating enthalies of reaction because most of the chemical energy of a comound is held in its bonds.*

17 Work Reexamined 17 Isochoric Work work involves a volume change. w = d Since a constant volume rocess has no volume change, w = 0 if there is no other work (no non- work). No other assumtions have been made about the system. Isobaric Work Since the ressure is constant, it has no deendence on the volume. hus the ressure can ulled out of the integral. Subsequently, the integral of d is 1 = ex ( ) w = d = d = = ex ex ex 1 ex he only other assumtion made is that the system does not have any non- work. Isothermal Work In general we need to know the relationshi between ressure and volume to erform the calculation, i.e, we need an equation of state. Let us do the calculation for an ideal gas under reversible conditions. nr d w d d d nr nr ln = ex = = = = 1 Note restrictions on the alicability of the calculation. 1. no other non- work. ideal gas 3. reversible conditions Adiabatic Work No assumtions made! du = ñq + ñw du = ñw w = U

18 Refrigeration and the Joule-homson coefficient 18 Joule coefficient - - internal ressure µ J = U q = 0, w = 0 U = 0 isoenergetic rocess Joule-homson coefficient - µ J = - isoenthalic, adiabatic - µ = 0 for an ideal gas Classic Exerimental Aaratus - measure µ J directly. H 0 U = U U = d d = f f i i f i i f f i 0 U = U H = H f i i i f f i

19 Modern Exerimental Aaratus 19 H - measure µ J via - use isothermal conditions rather adiabatic conditions. isothermal conditions: 1 = H H H 1 1 = = = = H H H H C H H H µ J = C

20 Isoenthals on - lot. - inversion temerature - µ > 0 for substance to be used as a refrigerant - temerature must decrease as ressure decreases. 0

21 Real Gases Reexamined 1 Calculations on van der Waals gas - work w = d ex nr an = b nr an nr an w = ex d = d = d d d = + b b b 1 1 nr ln an = 1 b 1 - Difference in heat caacities C H U Cv = v U U 1 U U du = d + d du = d + d U U U U U = + = + U U U = ( ) H U U H = U + = + = + U U U C Cv = + U U = + = + U U = + = α + = α + π [ ]

22 Internal Energy Reexamined Why is U = q v under isochoric conditions? du = ñq + ñw du = ñq - d du = ñq U = q ex Assuming no non- work. Isobaric Internal Energy du = ñq + ñw du= C d d U = C d d = C d Note: no assumtions about the equation of state. v ex v ex v Isochoric Internal Energy du = ñq + ñw du = ñq d du = ñq U = C d = q Note: assumtion of no non- work ex v v Isothermal Internal Energy du = ñq +ñw = C d d = 0 du = 0 v

23 Enthaly Reexamined 3 Why is H = q under isobaric conditions? ( ) H = U + dh = du + d =ñq +ñw + d + v d dh =ñq + -d + d + v d dh = ñq H = q Assuming no non- work and reversible conditions Isobaric Enthaly ( ) H = U + dh = du + d =ñq +ñw + d + v d dh =ñq d + d dh = ñq H = C d = q Assuming no non- work and reversible conditions Isochoric Enthaly ( ) H = U + dh = du + d =ñq +ñw + d + v d dh =ñq d + d + v d dh = ñq + v d H = C d v d For ideal gas under reversible conditions nr H = Cd vd = Cd d Why not H = Cd nr ln? 1 - temerature has a deendence on the ressure. Isothermal Enthaly H = U + dh = du + d ( ) = Cd H = 0 Imortant: All of the above enthalies assume that the system has no non- work and reversible conditions (so that P = P ex )

24 Summary 4 Isothermal = 0 Isobaric = 0 Isochoric = 0 Adiabatic q = 0 Definition of work: w = d ex Definitions for heat: q = H q = U Sign conventions for heat and work. -w work done by system (exansion for work) +w work done on the system (comression for work) -q heat transferred away from body (heat lost) +q heat transferred into body (heat gained) Reversibility = ex Extent of Reaction n ξ = n i i,0 ν i Definition of Heat Caacities C q H = = C v q U = = v v First Law: du = ñq + ñw U = q + w U = C d ex d Partial Derivatives to be acquainted with: α, κ, C, C v, π, µ J, µ J 1 1 α = κ = C U and internal ressure π = H = C v U = (equals zero for ideal gas) v µ J = U µ J = H

25 Natural variables of U and H and total differentials 5 Internal energy can be exressed as a function of and U(,) Enthaly can be exressed as a function of and H(,) - both of these functional deendences are not unique - will consider other natural variables in the next chater. otal differential can be exressed as U U H H du = d + d dh = d + d Hess law emerature Deendence of Enthaly and Internal Energy U = C d and H = C d Kirchoff s law 0 ( ) ( ) H = H 98 K + ν C d r r i,i 98 i U = C and H = C Miscellaneous Alications of Heat Caacities Cɶ Cɶ = R C γ = C v

Homework: 2, 3, 5, 6 (page 500)

Homework: 2, 3, 5, 6 (page 500) Homework: 2,, 5, 6 (age 500) 2. (age 500) wo constant-volume gas thermometers are assembled, one with nitrogen and the other with hydrogen. Both contain enough gas so that 80 kpa. (a) What is the difference

More information

Simple thermodynamic systems

Simple thermodynamic systems Simle thermodynamic systems Asaf Pe er 1 October 30, 2013 1. Different statements of the second law We have seen that Clausius defined the second law as the entroy of an isolated system cannot decrease

More information

First Law, Heat Capacity, Latent Heat and Enthalpy

First Law, Heat Capacity, Latent Heat and Enthalpy First Law, Heat Caacity, Latent Heat and Enthaly Stehen R. Addison January 29, 2003 Introduction In this section, we introduce the first law of thermodynamics and examine sign conentions. Heat and Work

More information

Thermochemistry. Thermochemistry 1/25/2010. Reading: Chapter 5 (omit 5.8) As you read ask yourself

Thermochemistry. Thermochemistry 1/25/2010. Reading: Chapter 5 (omit 5.8) As you read ask yourself Thermochemistry Reading: Chapter 5 (omit 5.8) As you read ask yourself What is meant by the terms system and surroundings? How are they related to each other? How does energy get transferred between them?

More information

Fugacity, Activity, and Standard States

Fugacity, Activity, and Standard States Fugacity, Activity, and Standard States Fugacity of gases: Since dg = VdP SdT, for an isothermal rocess, we have,g = 1 Vd. For ideal gas, we can substitute for V and obtain,g = nrt ln 1, or with reference

More information

1.3 Saturation vapor pressure. 1.3.1 Vapor pressure

1.3 Saturation vapor pressure. 1.3.1 Vapor pressure 1.3 Saturation vaor ressure Increasing temerature of liquid (or any substance) enhances its evaoration that results in the increase of vaor ressure over the liquid. y lowering temerature of the vaor we

More information

THERMOCHEMISTRY & DEFINITIONS

THERMOCHEMISTRY & DEFINITIONS THERMOCHEMISTRY & DEFINITIONS Thermochemistry is the study of the study of relationships between chemistry and energy. All chemical changes and many physical changes involve exchange of energy with the

More information

a) Use the following equation from the lecture notes: = ( 8.314 J K 1 mol 1) ( ) 10 L

a) Use the following equation from the lecture notes: = ( 8.314 J K 1 mol 1) ( ) 10 L hermodynamics: Examples for chapter 4. 1. One mole of nitrogen gas is allowed to expand from 0.5 to 10 L reversible and isothermal process at 300 K. Calculate the change in molar entropy using a the ideal

More information

OUTCOME 1. TUTORIAL No. 2 THERMODYNAMIC SYSTEMS

OUTCOME 1. TUTORIAL No. 2 THERMODYNAMIC SYSTEMS UNI 6: ENGINEERING HERMODYNAMICS Unit code: D/60/40 QCF level: 5 Credit value: 5 OUCOME UORIAL No. HERMODYNAMIC SYSEMS. Understand the arameters and characteristics of thermodynamic systems Polytroic rocesses:

More information

Basic Concepts of Thermodynamics

Basic Concepts of Thermodynamics Basic Concepts of Thermodynamics Every science has its own unique vocabulary associated with it. recise definition of basic concepts forms a sound foundation for development of a science and prevents possible

More information

Exergy: the quality of energy N. Woudstra

Exergy: the quality of energy N. Woudstra Exergy: the quality of energy N. Woudstra Introduction Characteristic for our society is a massive consumption of goods and energy. Continuation of this way of life in the long term is only possible if

More information

1 Exercise 2.19a pg 86

1 Exercise 2.19a pg 86 In this solution set, an underline is used to show the last significant digit of numbers. For instance in x = 2.51693 the 2,5,1, and 6 are all significant. Digits to the right of the underlined digit,

More information

Problem Set 3 Solutions

Problem Set 3 Solutions Chemistry 360 Dr Jean M Standard Problem Set 3 Solutions 1 (a) One mole of an ideal gas at 98 K is expanded reversibly and isothermally from 10 L to 10 L Determine the amount of work in Joules We start

More information

ENERGY. Thermochemistry. Heat. Temperature & Heat. Thermometers & Temperature. Temperature & Heat. Energy is the capacity to do work.

ENERGY. Thermochemistry. Heat. Temperature & Heat. Thermometers & Temperature. Temperature & Heat. Energy is the capacity to do work. ENERGY Thermochemistry Energy is the capacity to do work. Chapter 6 Kinetic Energy thermal, mechanical, electrical, sound Potential Energy chemical, gravitational, electrostatic Heat Heat, or thermal energy,

More information

Expansion and Compression of a Gas

Expansion and Compression of a Gas Physics 6B - Winter 2011 Homework 4 Solutions Expansion and Compression of a Gas In an adiabatic process, there is no heat transferred to or from the system i.e. dq = 0. The first law of thermodynamics

More information

The First Law of Thermodynamics

The First Law of Thermodynamics Thermodynamics The First Law of Thermodynamics Thermodynamic Processes (isobaric, isochoric, isothermal, adiabatic) Reversible and Irreversible Processes Heat Engines Refrigerators and Heat Pumps The Carnot

More information

Ideal Gas and Real Gases

Ideal Gas and Real Gases Ideal Gas and Real Gases Lectures in Physical Chemistry 1 Tamás Turányi Institute of Chemistry, ELTE State roerties state roerty: determines the macroscoic state of a hysical system state roerties of single

More information

Thermodynamics worked examples

Thermodynamics worked examples An Introduction to Mechanical Engineering Part hermodynamics worked examles. What is the absolute ressure, in SI units, of a fluid at a gauge ressure of. bar if atmosheric ressure is.0 bar? Absolute ressure

More information

7. 1.00 atm = 760 torr = 760 mm Hg = 101.325 kpa = 14.70 psi. = 0.446 atm. = 0.993 atm. = 107 kpa 760 torr 1 atm 760 mm Hg = 790.

7. 1.00 atm = 760 torr = 760 mm Hg = 101.325 kpa = 14.70 psi. = 0.446 atm. = 0.993 atm. = 107 kpa 760 torr 1 atm 760 mm Hg = 790. CHATER 3. The atmosphere is a homogeneous mixture (a solution) of gases.. Solids and liquids have essentially fixed volumes and are not able to be compressed easily. have volumes that depend on their conditions,

More information

Chemical Thermodynamics

Chemical Thermodynamics Chemical Thermodynamics David A. Katz Department of Chemistry Pima Community College Tucson, AZ 85709, USA First Law of Thermodynamics The First Law of Thermodynamics was expressed in the study of thermochemistry.

More information

be the mass flow rate of the system input stream, and m be the mass flow rates of the system output stream, then Vout V in in out out

be the mass flow rate of the system input stream, and m be the mass flow rates of the system output stream, then Vout V in in out out Chater 4 4. Energy Balances on Nonreactive Processes he general energy balance equation has the form Accumulation Inut Outut Heat added = + of Energy of Energy of Energy to System Work by done System Let

More information

Chemistry 433. The Third Law of Thermodynamics. Residual Entropy. CO: an Imperfect Crystal. Question. Question. Lecture 12 The Third Law

Chemistry 433. The Third Law of Thermodynamics. Residual Entropy. CO: an Imperfect Crystal. Question. Question. Lecture 12 The Third Law Chemistry 433 Lecture 12 he hird Law he hird Law of hermodynamics he third law of thermodynamics states that every substance has a positive entropy, but at zero Kelvin the entropy is zero for a perfectly

More information

Chapter 6 Thermodynamics: The First Law

Chapter 6 Thermodynamics: The First Law Key Concepts 6.1 Systems Chapter 6 Thermodynamics: The First Law Systems, States, and Energy (Sections 6.1 6.8) thermodynamics, statistical thermodynamics, system, surroundings, open system, closed system,

More information

ENTHALPY CHANGES FOR A CHEMICAL REACTION scaling a rxn up or down (proportionality) quantity 1 from rxn heat 1 from Δ r H. = 32.

ENTHALPY CHANGES FOR A CHEMICAL REACTION scaling a rxn up or down (proportionality) quantity 1 from rxn heat 1 from Δ r H. = 32. CHEMISTRY 103 Help Sheet #10 Chapter 4 (Part II); Sections 4.6-4.10 Do the topics appropriate for your lecture Prepared by Dr. Tony Jacob http://www.chem.wisc.edu/areas/clc (Resource page) Nuggets: Enthalpy

More information

AP Chem Lab 2 Quiz #1 Calorimetry. Conceptual Understanding. Write complete sentences to show your understanding.

AP Chem Lab 2 Quiz #1 Calorimetry. Conceptual Understanding. Write complete sentences to show your understanding. AP Chem Lab 2 Quiz #1 Calorimetry Name Conceptual Understanding. Write complete sentences to show your understanding. Differentiate between kinetic energy and potential energy. Energy may be transferred

More information

THE CLAUSIUS INEQUALITY

THE CLAUSIUS INEQUALITY Part IV Entropy In Part III, we introduced the second law of thermodynamics and applied it to cycles and cyclic devices. In this part, we apply the second law to processes. he first law of thermodynamics

More information

Final Exam CHM 3410, Dr. Mebel, Fall 2005

Final Exam CHM 3410, Dr. Mebel, Fall 2005 Final Exam CHM 3410, Dr. Mebel, Fall 2005 1. At -31.2 C, pure propane and n-butane have vapor pressures of 1200 and 200 Torr, respectively. (a) Calculate the mole fraction of propane in the liquid mixture

More information

Chemistry B11 Chapter 4 Chemical reactions

Chemistry B11 Chapter 4 Chemical reactions Chemistry B11 Chapter 4 Chemical reactions Chemical reactions are classified into five groups: A + B AB Synthesis reactions (Combination) H + O H O AB A + B Decomposition reactions (Analysis) NaCl Na +Cl

More information

3. Of energy, work, enthalpy, and heat, how many are state functions? a) 0 b) 1 c) 2 d) 3 e) 4 ANS: c) 2 PAGE: 6.1, 6.2

3. Of energy, work, enthalpy, and heat, how many are state functions? a) 0 b) 1 c) 2 d) 3 e) 4 ANS: c) 2 PAGE: 6.1, 6.2 1. A gas absorbs 0.0 J of heat and then performs 15.2 J of work. The change in internal energy of the gas is a) 24.8 J b) 14.8 J c) 55.2 J d) 15.2 J ANS: d) 15.2 J PAGE: 6.1 2. Calculate the work for the

More information

Energy and Chemical Reactions. Characterizing Energy:

Energy and Chemical Reactions. Characterizing Energy: Energy and Chemical Reactions Energy: Critical for virtually all aspects of chemistry Defined as: We focus on energy transfer. We observe energy changes in: Heat Transfer: How much energy can a material

More information

1 Exercise 4.1b pg 153

1 Exercise 4.1b pg 153 In this solution set, an underline is used to show the last significant digit of numbers. For instance in x = 2.51693 the 2,5,1, and 6 are all significant. Digits to the right of the underlined digit,

More information

Heat as Energy Transfer. Heat is energy transferred from one object to another because of a difference in temperature

Heat as Energy Transfer. Heat is energy transferred from one object to another because of a difference in temperature Unit of heat: calorie (cal) Heat as Energy Transfer Heat is energy transferred from one object to another because of a difference in temperature 1 cal is the amount of heat necessary to raise the temperature

More information

Name Class Date. Section: Calculating Quantities in Reactions. Complete each statement below by writing the correct term or phrase.

Name Class Date. Section: Calculating Quantities in Reactions. Complete each statement below by writing the correct term or phrase. Skills Worksheet Concept Review Section: Calculating Quantities in Reactions Complete each statement below by writing the correct term or phrase. 1. All stoichiometric calculations involving equations

More information

Thermodynamics: First Law, Calorimetry, Enthalpy. Calorimetry. Calorimetry: constant volume. Monday, January 23 CHEM 102H T.

Thermodynamics: First Law, Calorimetry, Enthalpy. Calorimetry. Calorimetry: constant volume. Monday, January 23 CHEM 102H T. Thermodynamics: First Law, Calorimetry, Enthalpy Monday, January 23 CHEM 102H T. Hughbanks Calorimetry Reactions are usually done at either constant V (in a closed container) or constant P (open to the

More information

SUPPLEMENTARY TOPIC 3 ENERGY AND CHEMICAL REACTIONS

SUPPLEMENTARY TOPIC 3 ENERGY AND CHEMICAL REACTIONS SUPPLEMENTARY TOPIC 3 ENERGY AND CHEMICAL REACTIONS Rearranging atoms. In a chemical reaction, bonds between atoms in one or more molecules (reactants) break and new bonds are formed with other atoms to

More information

Heat and Work. First Law of Thermodynamics 9.1. Heat is a form of energy. Calorimetry. Work. First Law of Thermodynamics.

Heat and Work. First Law of Thermodynamics 9.1. Heat is a form of energy. Calorimetry. Work. First Law of Thermodynamics. Heat and First Law of Thermodynamics 9. Heat Heat and Thermodynamic rocesses Thermodynamics is the science of heat and work Heat is a form of energy Calorimetry Mechanical equivalent of heat Mechanical

More information

Chapter 20. Thermodynamics p. 811 842. Spontaneity. What have we learned about spontaneity during this course?

Chapter 20. Thermodynamics p. 811 842. Spontaneity. What have we learned about spontaneity during this course? Chapter 20 p. 811 842 Spontaneous process: Ex. Nonspontaneous process: Ex. Spontaneity What have we learned about spontaneity during this course? 1) Q vs. K? 2) So.. Spontaneous process occurs when a system

More information

Entropy and The Second Law of Thermodynamics

Entropy and The Second Law of Thermodynamics The Second Law of Thermodynamics (SL) Entropy and The Second Law of Thermodynamics Explain and manipulate the second law State and illustrate by example the second law of thermodynamics Write both the

More information

The Equipartition Theorem

The Equipartition Theorem The Equipartition Theorem Degrees of freedom are associated with the kinetic energy of translations, rotation, vibration and the potential energy of vibrations. A result from classical statistical mechanics

More information

Bomb Calorimetry. Example 4. Energy and Enthalpy

Bomb Calorimetry. Example 4. Energy and Enthalpy Bomb Calorimetry constant volume often used for combustion reactions heat released by reaction is absorbed by calorimeter contents need heat capacity of calorimeter q cal = q rxn = q bomb + q water Example

More information

Thermodynamics. Thermodynamics 1

Thermodynamics. Thermodynamics 1 Thermodynamics 1 Thermodynamics Some Important Topics First Law of Thermodynamics Internal Energy U ( or E) Enthalpy H Second Law of Thermodynamics Entropy S Third law of Thermodynamics Absolute Entropy

More information

Chapter 5. Thermochemistry

Chapter 5. Thermochemistry Chapter 5. Thermochemistry THERMODYNAMICS - study of energy and its transformations Thermochemistry - study of energy changes associated with chemical reactions Energy - capacity to do work or to transfer

More information

Thermochemical equations allow stoichiometric calculations.

Thermochemical equations allow stoichiometric calculations. CHEM 1105 THERMOCHEMISTRY 1. Change in Enthalpy ( H) Heat is evolved or absorbed in all chemical reactions. Exothermic reaction: heat evolved - heat flows from reaction mixture to surroundings; products

More information

Supplementary Notes on Entropy and the Second Law of Thermodynamics

Supplementary Notes on Entropy and the Second Law of Thermodynamics ME 4- hermodynamics I Supplementary Notes on Entropy and the Second aw of hermodynamics Reversible Process A reversible process is one which, having taken place, can be reversed without leaving a change

More information

HEAT UNIT 1.1 KINETIC THEORY OF GASES. 1.1.1 Introduction. 1.1.2 Postulates of Kinetic Theory of Gases

HEAT UNIT 1.1 KINETIC THEORY OF GASES. 1.1.1 Introduction. 1.1.2 Postulates of Kinetic Theory of Gases UNIT HEAT. KINETIC THEORY OF GASES.. Introduction Molecules have a diameter of the order of Å and the distance between them in a gas is 0 Å while the interaction distance in solids is very small. R. Clausius

More information

Standard Free Energies of Formation at 298 K. Average Bond Dissociation Energies at 298 K

Standard Free Energies of Formation at 298 K. Average Bond Dissociation Energies at 298 K 1 Thermodynamics There always seems to be at least one free response question that involves thermodynamics. These types of question also show up in the multiple choice questions. G, S, and H. Know what

More information

CHEMISTRY STANDARDS BASED RUBRIC ATOMIC STRUCTURE AND BONDING

CHEMISTRY STANDARDS BASED RUBRIC ATOMIC STRUCTURE AND BONDING CHEMISTRY STANDARDS BASED RUBRIC ATOMIC STRUCTURE AND BONDING Essential Standard: STUDENTS WILL UNDERSTAND THAT THE PROPERTIES OF MATTER AND THEIR INTERACTIONS ARE A CONSEQUENCE OF THE STRUCTURE OF MATTER,

More information

ENTROPY AND THE SECOND LAW OF THERMODYNAMICS

ENTROPY AND THE SECOND LAW OF THERMODYNAMICS Chapter 20: ENTROPY AND THE SECOND LAW OF THERMODYNAMICS 1. In a reversible process the system: A. is always close to equilibrium states B. is close to equilibrium states only at the beginning and end

More information

Calorimeter: A device in which the heat associated with a specific process is measured.

Calorimeter: A device in which the heat associated with a specific process is measured. 1 CALORIMETRY p. 661-667 (simple), 673-675 (bomb) Calorimeter: A device in which the heat associated with a specific process is measured. There are two basic types of calorimeters: 1. Constant-pressure

More information

Free Software Development. 2. Chemical Database Management

Free Software Development. 2. Chemical Database Management Leonardo Electronic Journal of Practices and echnologies ISSN 1583-1078 Issue 1, July-December 2002. 69-76 Free Software Develoment. 2. Chemical Database Management Monica ŞEFU 1, Mihaela Ligia UNGUREŞAN

More information

Chapter 15: Thermodynamics

Chapter 15: Thermodynamics Chapter 15: Thermodynamics The First Law of Thermodynamics Thermodynamic Processes (isobaric, isochoric, isothermal, adiabatic) Reversible and Irreversible Processes Heat Engines Refrigerators and Heat

More information

Give all answers in MKS units: energy in Joules, pressure in Pascals, volume in m 3, etc. Only work the number of problems required. Chose wisely.

Give all answers in MKS units: energy in Joules, pressure in Pascals, volume in m 3, etc. Only work the number of problems required. Chose wisely. Chemistry 45/456 0 July, 007 Midterm Examination Professor G. Drobny Universal gas constant=r=8.3j/mole-k=0.08l-atm/mole-k Joule=J= Nt-m=kg-m /s 0J= L-atm. Pa=J/m 3 =N/m. atm=.0x0 5 Pa=.0 bar L=0-3 m 3.

More information

AP* Chemistry THERMOCHEMISTRY

AP* Chemistry THERMOCHEMISTRY AP* Chemistry THERMOCHEMISTRY Terms for you to learn that will make this unit understandable: Energy (E) the ability to do work or produce heat ; the sum of all potential and kinetic energy in a system

More information

Boyles Law. At constant temperature the volume occupied by a fixed amount of gas is inversely proportional to the pressure on the gas 1 P = P

Boyles Law. At constant temperature the volume occupied by a fixed amount of gas is inversely proportional to the pressure on the gas 1 P = P Boyles Law At constant temperature the volume occupied by a fixed amount of gas is inversely proportional to the pressure on the gas 1 or k 1 Boyles Law Example ressure olume Initial 2.00 atm 100 cm 3

More information

Problem Set 4 Solutions

Problem Set 4 Solutions Chemistry 360 Dr Jean M Standard Problem Set 4 Solutions 1 Two moles of an ideal gas are compressed isothermally and reversibly at 98 K from 1 atm to 00 atm Calculate q, w, ΔU, and ΔH For an isothermal

More information

Spontaneity of a Chemical Reaction

Spontaneity of a Chemical Reaction Spontaneity of a Chemical Reaction We have learned that entropy is used to quantify the extent of disorder resulting from the dispersal of matter in a system. Also; entropy, like enthalpy and internal

More information

The Gas Laws. The effect of adding gas. 4 things. Pressure and the number of molecules are directly related. Page 1

The Gas Laws. The effect of adding gas. 4 things. Pressure and the number of molecules are directly related. Page 1 The Gas Laws Describe HOW gases behave. Can be predicted by the theory. The Kinetic Theory Amount of change can be calculated with mathematical equations. The effect of adding gas. When we blow up a balloon

More information

Thermodynamics Worksheet I also highly recommend Worksheets 13 and 14 in the Lab Manual

Thermodynamics Worksheet I also highly recommend Worksheets 13 and 14 in the Lab Manual Thermodynamics Worksheet I also highly recommend Worksheets 13 and 14 in the Lab Manual 1. Predict the sign of entropy change in the following processes a) The process of carbonating water to make a soda

More information

The first law: transformation of energy into heat and work. Chemical reactions can be used to provide heat and for doing work.

The first law: transformation of energy into heat and work. Chemical reactions can be used to provide heat and for doing work. The first law: transformation of energy into heat and work Chemical reactions can be used to provide heat and for doing work. Compare fuel value of different compounds. What drives these reactions to proceed

More information

Chapter 8 Maxwell relations and measurable properties

Chapter 8 Maxwell relations and measurable properties Chapter 8 Maxwell relations and measurable properties 8.1 Maxwell relations Other thermodynamic potentials emerging from Legendre transforms allow us to switch independent variables and give rise to alternate

More information

UNIT 1 THERMOCHEMISTRY

UNIT 1 THERMOCHEMISTRY UNIT 1 THERMOCHEMISTRY THERMOCHEMISTRY LEARNING OUTCOMES Students will be expected to: THERMOCHEMISTRY STSE analyse why scientific and technological activities take place in a variety individual and group

More information

CHEMICAL EQUILIBRIUM

CHEMICAL EQUILIBRIUM Chemistry 10 Chapter 14 CHEMICAL EQUILIBRIUM Reactions that can go in both directions are called reversible reactions. These reactions seem to stop before they go to completion. When the rate of the forward

More information

GAS TURBINE PERFORMANCE WHAT MAKES THE MAP?

GAS TURBINE PERFORMANCE WHAT MAKES THE MAP? GAS TURBINE PERFORMANCE WHAT MAKES THE MAP? by Rainer Kurz Manager of Systems Analysis and Field Testing and Klaus Brun Senior Sales Engineer Solar Turbines Incororated San Diego, California Rainer Kurz

More information

Technical Thermodynamics

Technical Thermodynamics Technical Thermodynamics Chapter 2: Basic ideas and some definitions Prof. Dr.-Ing. habil. Egon Hassel University of Rostock, Germany Faculty of Mechanical Engineering and Ship Building Institute of Technical

More information

Problem Solving. Stoichiometry of Gases

Problem Solving. Stoichiometry of Gases Skills Worksheet Problem Solving Stoichiometry of Gases Now that you have worked with relationships among moles, mass, and volumes of gases, you can easily put these to work in stoichiometry calculations.

More information

87 16 70 20 58 24 44 32 35 40 29 48 (a) graph Y versus X (b) graph Y versus 1/X

87 16 70 20 58 24 44 32 35 40 29 48 (a) graph Y versus X (b) graph Y versus 1/X HOMEWORK 5A Barometer; Boyle s Law 1. The pressure of the first two gases below is determined with a manometer that is filled with mercury (density = 13.6 g/ml). The pressure of the last two gases below

More information

Statistical Mechanics, Kinetic Theory Ideal Gas. 8.01t Nov 22, 2004

Statistical Mechanics, Kinetic Theory Ideal Gas. 8.01t Nov 22, 2004 Statistical Mechanics, Kinetic Theory Ideal Gas 8.01t Nov 22, 2004 Statistical Mechanics and Thermodynamics Thermodynamics Old & Fundamental Understanding of Heat (I.e. Steam) Engines Part of Physics Einstein

More information

Chem 1A Exam 2 Review Problems

Chem 1A Exam 2 Review Problems Chem 1A Exam 2 Review Problems 1. At 0.967 atm, the height of mercury in a barometer is 0.735 m. If the mercury were replaced with water, what height of water (in meters) would be supported at this pressure?

More information

4. Influence of Temperature and Pressure on Chemical Changes

4. Influence of Temperature and Pressure on Chemical Changes 4. Influence of Temerature and Pressure on Chemical Changes Toic: The influence of temerature and ressure on the chemical otential and drive and therefore the behaviour of substances. 4.1 Introduction

More information

k is change in kinetic energy and E

k is change in kinetic energy and E Energy Balances on Closed Systems A system is closed if mass does not cross the system boundary during the period of time covered by energy balance. Energy balance for a closed system written between two

More information

Sample Exercise 20.1 Identifying Oxidizing and Reducing Agents

Sample Exercise 20.1 Identifying Oxidizing and Reducing Agents Sample Exercise 20.1 Identifying Oxidizing and Reducing Agents The nickel-cadmium (nicad) battery, a rechargeable dry cell used in battery-operated devices, uses the following redox reaction to generate

More information

Test Review # 9. Chemistry R: Form TR9.13A

Test Review # 9. Chemistry R: Form TR9.13A Chemistry R: Form TR9.13A TEST 9 REVIEW Name Date Period Test Review # 9 Collision theory. In order for a reaction to occur, particles of the reactant must collide. Not all collisions cause reactions.

More information

In a cyclic transformation, where the final state of a system is the same as the initial one, U = 0

In a cyclic transformation, where the final state of a system is the same as the initial one, U = 0 Chapter 4 Entropy and second law of thermodynamics 4.1 Carnot cycle In a cyclic transformation, where the final state of a system is the same as the initial one, U = 0 since the internal energy U is a

More information

Chapter 3. Chemical Reactions and Reaction Stoichiometry. Lecture Presentation. James F. Kirby Quinnipiac University Hamden, CT

Chapter 3. Chemical Reactions and Reaction Stoichiometry. Lecture Presentation. James F. Kirby Quinnipiac University Hamden, CT Lecture Presentation Chapter 3 Chemical Reactions and Reaction James F. Kirby Quinnipiac University Hamden, CT The study of the mass relationships in chemistry Based on the Law of Conservation of Mass

More information

Name Date Class THERMOCHEMISTRY. SECTION 17.1 THE FLOW OF ENERGY HEAT AND WORK (pages 505 510)

Name Date Class THERMOCHEMISTRY. SECTION 17.1 THE FLOW OF ENERGY HEAT AND WORK (pages 505 510) 17 THERMOCHEMISTRY SECTION 17.1 THE FLOW OF ENERGY HEAT AND WORK (pages 505 510) This section explains the relationship between energy and heat, and distinguishes between heat capacity and specific heat.

More information

Chapter 3. Phase transitions. 3.1 Introduction. 3.2 Thermodynamics of phase transitions

Chapter 3. Phase transitions. 3.1 Introduction. 3.2 Thermodynamics of phase transitions Chater 3 Phase transitions 3.1 Introduction Phase transitions are ubiquitous in Nature. We are all familiar with the different hases of water (vaour, liquid and ice), and with the change from one to the

More information

1 Exercise 3.1b pg 131

1 Exercise 3.1b pg 131 In this solution set, an underline is used to show the last significant digit of numbers. For instance in x = 2.51693 the 2,5,1, and 6 are all significant. Digits to the right of the underlined digit,

More information

1 CHAPTER 10 THE JOULE AND JOULE-THOMSON EXPERIMENTS

1 CHAPTER 10 THE JOULE AND JOULE-THOMSON EXPERIMENTS AER 0 E JOLE AND JOLE-OMON EXERIMEN 0 Introduction Equation 84, γ constant, tells us how to calculate the drop in temperature if a gas expands adiabatically and reversibly; it is expanding against an external

More information

Chapter 10. Can You... 1. draw the Lewis structure for a given covalently bonded molecule?

Chapter 10. Can You... 1. draw the Lewis structure for a given covalently bonded molecule? Chapter 10 Can You... 1. draw the Lewis structure for a given covalently bonded molecule? e.g. SF 6 and CH 3 Cl 2. identify and count the number of non-bonding and bonding domains within a given covalently

More information

Calculating Atoms, Ions, or Molecules Using Moles

Calculating Atoms, Ions, or Molecules Using Moles TEKS REVIEW 8B Calculating Atoms, Ions, or Molecules Using Moles TEKS 8B READINESS Use the mole concept to calculate the number of atoms, ions, or molecules in a sample TEKS_TXT of material. Vocabulary

More information

The Reduced van der Waals Equation of State

The Reduced van der Waals Equation of State The Redued van der Waals Equation of State The van der Waals equation of state is na + ( V nb) n (1) V where n is the mole number, a and b are onstants harateristi of a artiular gas, and R the gas onstant

More information

THE BAROMETRIC FALLACY

THE BAROMETRIC FALLACY THE BAROMETRIC FALLACY It is often assumed that the atmosheric ressure at the surface is related to the atmosheric ressure at elevation by a recise mathematical relationshi. This relationshi is that given

More information

2. The percent yield is the maximum amount of product that can be produced from the given amount of limiting reactant.

2. The percent yield is the maximum amount of product that can be produced from the given amount of limiting reactant. UNIT 6 stoichiometry practice test True/False Indicate whether the statement is true or false. moles F 1. The mole ratio is a comparison of how many grams of one substance are required to participate in

More information

Name Class Date. In the space provided, write the letter of the term or phrase that best completes each statement or best answers each question.

Name Class Date. In the space provided, write the letter of the term or phrase that best completes each statement or best answers each question. Assessment Chapter Test A Chapter: States of Matter In the space provided, write the letter of the term or phrase that best completes each statement or best answers each question. 1. The kinetic-molecular

More information

The Kinetic Theory of Gases Sections Covered in the Text: Chapter 18

The Kinetic Theory of Gases Sections Covered in the Text: Chapter 18 The Kinetic Theory of Gases Sections Covered in the Text: Chapter 18 In Note 15 we reviewed macroscopic properties of matter, in particular, temperature and pressure. Here we see how the temperature and

More information

EDEXCEL HIGHERS ENGINEERING THERMODYNAMICS H2 NQF LEVEL 4. TUTORIAL No. 1 PRE-REQUISITE STUDIES FLUID PROPERTIES

EDEXCEL HIGHERS ENGINEERING THERMODYNAMICS H2 NQF LEVEL 4. TUTORIAL No. 1 PRE-REQUISITE STUDIES FLUID PROPERTIES EDEXCEL HIGHERS ENGINEERING HERMODYNAMICS H NQF LEEL 4 UORIAL No. PRE-REQUISIE SUDIES FLUID PROPERIES INRODUCION Before you study the four outcomes that make u the module, you should be cometent in finding

More information

Sample Problem: STOICHIOMETRY and percent yield calculations. How much H 2 O will be formed if 454 g of. decomposes? NH 4 NO 3 N 2 O + 2 H 2 O

Sample Problem: STOICHIOMETRY and percent yield calculations. How much H 2 O will be formed if 454 g of. decomposes? NH 4 NO 3 N 2 O + 2 H 2 O STOICHIOMETRY and percent yield calculations 1 Steps for solving Stoichiometric Problems 2 Step 1 Write the balanced equation for the reaction. Step 2 Identify your known and unknown quantities. Step 3

More information

INTI COLLEGE MALAYSIA A? LEVEL PROGRAMME CHM 111: CHEMISTRY MOCK EXAMINATION: DECEMBER 2000 SESSION. 37 74 20 40 60 80 m/e

INTI COLLEGE MALAYSIA A? LEVEL PROGRAMME CHM 111: CHEMISTRY MOCK EXAMINATION: DECEMBER 2000 SESSION. 37 74 20 40 60 80 m/e CHM111(M)/Page 1 of 5 INTI COLLEGE MALAYSIA A? LEVEL PROGRAMME CHM 111: CHEMISTRY MOCK EXAMINATION: DECEMBER 2000 SESSION SECTION A Answer ALL EIGHT questions. (52 marks) 1. The following is the mass spectrum

More information

The Gas Laws. Our Atmosphere. Pressure = Units of Pressure. Barometer. Chapter 10

The Gas Laws. Our Atmosphere. Pressure = Units of Pressure. Barometer. Chapter 10 Our Atmosphere The Gas Laws 99% N 2 and O 2 78% N 2 80 70 Nitrogen Chapter 10 21% O 2 1% CO 2 and the Noble Gases 60 50 40 Oxygen 30 20 10 0 Gas Carbon dioxide and Noble Gases Pressure Pressure = Force

More information

IB Chemistry 1 Mole. One atom of C-12 has a mass of 12 amu. One mole of C-12 has a mass of 12 g. Grams we can use more easily.

IB Chemistry 1 Mole. One atom of C-12 has a mass of 12 amu. One mole of C-12 has a mass of 12 g. Grams we can use more easily. The Mole Atomic mass units and atoms are not convenient units to work with. The concept of the mole was invented. This was the number of atoms of carbon-12 that were needed to make 12 g of carbon. 1 mole

More information

Calculations and Chemical Equations. Example: Hydrogen atomic weight = 1.008 amu Carbon atomic weight = 12.001 amu

Calculations and Chemical Equations. Example: Hydrogen atomic weight = 1.008 amu Carbon atomic weight = 12.001 amu Calculations and Chemical Equations Atomic mass: Mass of an atom of an element, expressed in atomic mass units Atomic mass unit (amu): 1.661 x 10-24 g Atomic weight: Average mass of all isotopes of a given

More information

Effect Sizes Based on Means

Effect Sizes Based on Means CHAPTER 4 Effect Sizes Based on Means Introduction Raw (unstardized) mean difference D Stardized mean difference, d g Resonse ratios INTRODUCTION When the studies reort means stard deviations, the referred

More information

CHEM 36 General Chemistry EXAM #1 February 13, 2002

CHEM 36 General Chemistry EXAM #1 February 13, 2002 CHEM 36 General Chemistry EXAM #1 February 13, 2002 Name: Serkey, Anne INSTRUCTIONS: Read through the entire exam before you begin. Answer all of the questions. For questions involving calculations, show

More information

Chapter 2 - Porosity PIA NMR BET

Chapter 2 - Porosity PIA NMR BET 2.5 Pore tructure Measurement Alication of the Carmen-Kozeny model requires recise measurements of ore level arameters; e.g., secific surface area and tortuosity. Numerous methods have been develoed to

More information

Lecture 4: 09.16.05 Temperature, heat, and entropy

Lecture 4: 09.16.05 Temperature, heat, and entropy 3.012 Fundamentals of Materials Science Fall 2005 Lecture 4: 09.16.05 Temperature, heat, and entropy Today: LAST TIME...2 State functions...2 Path dependent variables: heat and work...2 DEFINING TEMPERATURE...4

More information

CLASSICAL CONCEPT REVIEW 8

CLASSICAL CONCEPT REVIEW 8 CLASSICAL CONCEPT REVIEW 8 Kinetic Theory Information concerning the initial motions of each of the atoms of macroscopic systems is not accessible, nor do we have the computational capability even with

More information

Answer, Key Homework 6 David McIntyre 1

Answer, Key Homework 6 David McIntyre 1 Answer, Key Homework 6 David McIntyre 1 This print-out should have 0 questions, check that it is complete. Multiple-choice questions may continue on the next column or page: find all choices before making

More information

Thermochemistry. r2 d:\files\courses\1110-20\99heat&thermorans.doc. Ron Robertson

Thermochemistry. r2 d:\files\courses\1110-20\99heat&thermorans.doc. Ron Robertson Thermochemistry r2 d:\files\courses\1110-20\99heat&thermorans.doc Ron Robertson I. What is Energy? A. Energy is a property of matter that allows work to be done B. Potential and Kinetic Potential energy

More information

Chemical Equilibrium

Chemical Equilibrium Chapter 13 Chemical Equilibrium Equilibrium Physical Equilibrium refers to the equilibrium between two or more states of matter (solid, liquid and gas) A great example of physical equilibrium is shown

More information

CHEMISTRY. Matter and Change. Section 13.1 Section 13.2 Section 13.3. The Gas Laws The Ideal Gas Law Gas Stoichiometry

CHEMISTRY. Matter and Change. Section 13.1 Section 13.2 Section 13.3. The Gas Laws The Ideal Gas Law Gas Stoichiometry CHEMISTRY Matter and Change 13 Table Of Contents Chapter 13: Gases Section 13.1 Section 13.2 Section 13.3 The Gas Laws The Ideal Gas Law Gas Stoichiometry State the relationships among pressure, temperature,

More information