Outline of the Course

Transcription

1 Outline of the Course 1) Review and Definitions 2) Molecules and their Energies 3) 1 st Law of Thermodynamics 4) 2 nd Law of Thermodynamics 5) Gibbs Free Energy 6) Phase Diagrams and REAL Phenomena 7) Non-Electrolyte Solutions 8) Chemical Equilibrium 9) Chemical Kinetics and Rates of Processes

2 Section 3.0. The 1 st Law of Thermodynamics (CHANG text Chapter 4 almost ALL! ) 3.1. Revisiting Heat Capacities 3.2. Definitions and Concepts 3.3. The First Law of THERMODYNAMICS 3.4. Enthalpy 3.5. Adiabatic Expansion of an Ideal Gas 3.6. Thermochemistry 3.7. Bond Energies and Bond Enthalpies

3 3.1. A Closer Look at Heat Capacities Goals: 1. To review definition of heat capacity and molar heat capacity 2. To compare and understand difference between C V and C P 3. To review difference between C for gases and substances in the condensed states (i.e. liquids and solids)

4 3.1. A Closer Look at Heat Capacities..when heat is added to a substance its temperature will rise How much the temperature will rise depends on: (1) the amount of heat delivered (2) the amount of substance present (3) the chemical nature and physical state of the substance (4) conditions under which heat is added to substance

5 Heat Capacity (C) C = U/T = q/t measured in J/K where U is the amount of energy required to raise the T of the substance by T or q is heat required to raise the T of substance by T Molar Heat Capacity ( C ) = heat capacity of one mole of a substance C = C/n = q /(n T) measured in J mol -1 K -1 heat capacity is a quantity that can be measured directly can be measured at constant volume or constant pressure

6 heat capacity can be measured at constant P or constant V At constant volume: C V = U/T = q V /T at constant V the heat absorbed by the system is equal to the change in internal energy At constant pressure: C P = H/T = q P /T at constant P the heat absorbed is equal to the change in enthalpy (H) ****Remember that C = C/n = q /(n T) so we can rearrange to calculate values for U and H for processes at constant V and P, respectively. U = C V T = n C V T H = C P T = n C P T

7 U = C V T = n C V T H = C P T = n C P T for these equations we have assumed that C V and C P are independent of temperature only true at T 300K at T less than 300K only translational and rotational motions make major contribution to heat capacity as we T vibrational motions makes contribution and C C P = a + BT where a and B are constants for a given substance (equation helps to predict temperature dependence of C for a substance)

8 In general C V and C P for a substance are not equal.. Why? because in a constant pressure process volume is changing (expansion!) so work must be done on the surroundings. Constant Volume Initial Final Constant Pressure Expansion More work is done in constant P than constant V process, so C P C V..this is true for gases is it true for liquids and solids???

9 C P C V is true mainly only for gases Why? because the volume of liquids and solids doesn t change much with temperature so work done on surroundings is very small. C P C V for liquids and solids Notice.to this point we have mostly discussed gases..not liquids or solids!!!!

10 How does the heat capacity (C) of a gas differ from that of a liquid????? remember we discussed how to calculate the heat capacity (C) for a gas heat capacity = C = ΔU / ΔT U = E TRANSLATION + E ROTATION + E VIBRATION + E ELECTRONIC heat capacities of liquids > gases and can t be accounted for just by considering translational, rotational and vibrational contributions..there are weak interactions present between molecules in the liquid and solid state.. even though they are weak they still contribute and C H 2 O (g) C P = 33.5 J/mol K (i.e. = 4R) H 2 O (l) C P = 75.3 J/mol K (i.e. = 9R)

11 C P H 2 O (l) > C P H 2 O (g) Why?.because energy must be absorbed into system to break up weak hydrogen bonding interactions present between H 2 O molecules in liquid state.. Hydrogen Bonding between H 2 O Molecules Remember.. due to these weak interactions in liquids and solids we cannot easily predict their heat capacities Gases are simple we can predict their heat capacities using our simple equation!!!!

12 3.1. A Closer Look at Heat Capacities Goals: 1. To review definition of heat capacity and molar heat capacity 2. To compare and understand difference between C V and C P 3. To review difference between C for gases and substances in the condensed states (i.e. liquids and solids) Progress: 1. C V = U/T and C = C/n 2. C V = U/T while C P = H/T 3. For gases C P > C V but for liquids and solids C P C V 4. C for liquids >> C for gases.due to presence of weak interactions between molecules

13 3.2. Definitions and Concepts Goals 1) To understand distinction between System and Surroundings 2) To understand the different types of systems.isolated, closed or open adiabatic vs. diathermic 3) To be comfortable with the following terms..isothermal, adiabatic, state function 4) To define work and heat

14 3.2. Definitions and Concepts We are measuring, predicting properties of the SYSTEM system the SURROUNDINGS is everything around the system We are mostly interested in the system.

15 CLOSED vs. OPEN vs. ISOLATED SYSTEMS OPEN CLOSED ISOLATED Energy & Mass Exchange Energy Exchange No Exchange

16 Systems (1) Diathermic system. allows energy to leave system as heat. (2) Adiabatic system...means no heat exchange between system and surroundings. i.e. an adiabatic container is an isolated system with no heat exchange Figure Physical Chemistry 7 th Edn. Atkins P., de Paula J.

17 Adiabatic Bomb Calorimeter - measure heat of combustion of substances - sample is thermally isolated from surroundings - combustion is started by an electrical discharge - heat released by rxn. is measured by rise in temperature of water filling jacket of calorimeter Page 84 in Chang Text

18 ADIABATIC DIATHERMIC SYSTEM SYSTEM Constant T * Endothermic Exothermic Process Process Endothermic Exothermic Process Process * Isothermal.means Endothermic.heat is absorbed T is held constant Exothermic heat is released Atkins Text 7 th edition page 32

19 In an adiabatic system (no heat exchange). How can energy be transferred to surroundings? in the form of WORK.. Example.. consider a chemical reaction that occurs in a cylinder fitted with a piston that results in production of gases Zn (s) + 2HCl (aq) ZnCl 2 (aq) + H 2 (g) as gas is produced - the container expands.lifts weight energy is transferred to surroundings as work weight piston

20 Work and Heat Work most often defined as force times distance i.e. w = - (force) (distance).but work may also be defined as follows :

21 In Chemistry we often discuss EXPANSION WORK..the reaction between zinc and acid occurring in a cylinder fitted with a weighted piston is an example of this.when the piston is moved a specific distance the force opposing this motion is equal to: F = P external A.the external pressure acts like a weight pushing down on the piston.. Atkins Text 7 th edition page 38

22 Consider a gas in a cylinder fitted with a weightless, frictionless piston. gas expands w = - force x distance = - mass x acceleration x distance = - m g (h 2 h 1 ) = - m g h Initial State P 1, V 1, T Final State P 2, V 2, T h 1 h 2

23 w = - force x distance = - mass x acceleration x distance = - m g (h 2 h 1 ) = - m g h here m is in units of kg; h is in meters; w is in Joules g = 9.81 m s -2 in an expansion process, h 2 h 1 and w is NEGATIVE when system does work on surroundings work is negative quantity in a compression, work is done on system and is a positive quantity

24 When the gas expands (does work!) what is the external pressure that it expands against? i.e. what is the external opposing pressure ( P ex )? P ex = F / (A) = m g /(A) or w = - P ex A h P ex = - P ex V where A is the area of the piston h 1 h 2 A h = V demonstrates that amount of work done depends on the external pressure

25 How can the amount of work done by the gas vary??? 1- have a gas expanding against a vacuum meaning no weight (remove mass from piston) mass P ex = 0 P ex = 0 so - P ex V = 0 thus w = 0 *** not the common case usually we have a mass on the piston

26 2- have a gas expanding against a mass so gas expands against a constant external pressure w = - P ex V and P ex 0 as gas expands P in decreases mass for gas to expand P in > P ex for example if P ex = 3 atm and P in = 10 atm..the gas will only expand until P in = 3 atm.then expansion will stop!!!

27 Sample Problem A quantity of mole of an ideal gas initially at P = 15.0 atm and 300 K is allowed to expand isothermally until its final pressure is 1.00 atm. Calculate the value of the work done if the expansion is carried out (a) against a vacuum, (b) against a constant external pressure of 1.00 atm. Solution w = - Pex V

28 (b) The external opposing pressure is 1 atm so work will be done. The initial and final volumes can be determined using the ideal gas equation: V 1 = (nrt)/p 1 V 2 = (nrt)/p 2

29 Let s convert this into JOULES. we have w = L atm conversion factor is 1 Latm = J So w = Latm x ( J/1 Latm)

30 Question? Is it possible to have the system perform a greater amount of work for the same increase in volume i.e. total volume change is same but different amount of work is done????

31 Question: Is it possible to have the system perform a greater amount of work for the same increase in volume? Answer: YES Suppose that the weight on the piston is made up of an infinitessimal number of small weights each of which contributes to a total external pressure of 10 atm. mass P in = P ex (i.e P in = 10 atm, P ex = 10 atm) Removing one weight will P ex by infinitessimal amount so P in > P ex and gas will expand by infinitessimal amnt. until P in = P ex We can continue to do this by progressively removing one weight at a time.gas expands by infinitessimal amnt. with removal of each weight.

32 remember when P ex was a constant value w = - P ex V in previous discussion P ex is not constant.at every instant P in becomes infinitessimally greater than P ex In this case w = - n R T ln (V 2 /V 1 ) = - n RT ln (P 1 /P 2 ) This equation for w represents the maximum work done! we can maximize work done in expansion by decreasing P ex by only an infinitessimal amount at each stage under these conditions this expansion is a REVERSIBLE process meaning..we can P ex by infinitessimal amnt. halting expansion then again by infinitessimal amnt. causing compression. a reversible process is one in which the system is always infinitessimally close to equilibrium

33 real processes are always irreversible..but we care about reversible processes because it gives us idea of maximum amount of work that can be extracted from a system OK so let s recap and consider difference between work done by a system where P ex is constant and work done by a system where P ex is changed by an infinitessimal amount at each stage of expansion consider CASE A and CASE B

34 Consider now difference between work done in Case A and Case B.. CASE A CASE B Note.. Mass is one weight Initial P in = 10 atm P ex = 1 atm gas expands until P in = 1 atm Mass consists of infinitessimal # of small weights that are removed one at a time.. Initial P in = P ex = 10 atm weights removed until P ex = 1 atm In both cases in the Final State P in = 1 atm and P ex = 1atm Yet work done by system in case A and case B is very different!!!

35 work done is equal to area under P versus V curve in CASE A w = - P ex V and this is an irreversible process in CASE B w = - n RT ln (P 1 /P 2 ) and this is a reversible process Case A Case B

36 Let s Re-visit Previous Sample Problem A quantity of mole of an ideal gas initially at P = 15.0 atm and 300 K is allowed to expand isothermally until its final pressure is 1.00 atm. Calculate the value of the work done if the expansion is carried out (a) against a vacuum, (b) against a constant external pressure of 1.00 atm, (c) reversibly. Solution (c) For an isothermal reversible process..

37 We ve discussed 3 conditions for expansion work 1 - Expansion against no opposing force P ext = 0 NO WORK is DONE w = Expansion against constant pressure P ext 0 P ext = constant so w = - P ext ΔV Irreversible Expansion 3 - Reversible Expansion P ext is changed by infinitessimal amount at each stage of expansion w = - n R T ln (P 1 /P 2 ) - this equation for w represents maximum amount of work

38 Sample Problem Calculate the work done when 50 g of iron reacts with HCl to produce hydrogen gas in (a) a closed vessel of fixed volume, (b) an open beaker at 25 C. Fe (s) + 2HCl (aq) FeCl 2 (aq) + H 2 (g) Solution

39 What have we learned? 1- work is a form of energy transfer another form is heat 2- maximum amount of work is done by a reversible process (when compared to an irreversible process) 3- expansion only occurs when there is a pressure difference i.e. when P external is P internal when P external = P internal expansion stops 4-the amount of work done depends on how the process is carried out this means work is not a state function work is path dependent! A change in any state function depends only on the initial and final states not on the path taken to get from initial to final.

40 State Functions (path independent) examples P, V, T Work is not a state function.it depends on the path taken e.g. CASE A (irreversible expansion) versus CASE B (reversible expansion) PATH Dependence X X Z Z

41 Heat is the transfer of energy between two entities that are at different temperatures heat is transferred from a hot entity to a colder entity and when T of two entities are equal transfer stops heat is not a state function it is path dependent Example how much heat is transferred in raising the temperature of a beaker of water (50 g) from 25 C to 35 C (at const. P = 1 atm)? The amount of heat transferred will depend on process or path taken to increase temperature

42 The amount of heat transferred will depend on the path taken to increase temperature There are many ways to raise the temperature of this system by 10 C.. 1- system may be heated on a bunsen burner or heated by immersing the beaker in a warmed water bath heat = q = m s* T = (50 g)(1.00 cal/g deg)(4.184 J/cal)(10.0 deg) = 2092 J 2- system may be heated by sonicating or mixing on a stir plate in which case mechanical work is used to increase temperature and q = 0 3- or could increase T by combination of heating in water bath and mechanical mixing then q lies between 0 and 2092 J (* s = specific heat)

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44 3.2. Definitions and Concepts Goals 1) To understand distinction between System and Surroundings 2) To understand the different types of systems.isolated, closed or open adiabatic vs. diathermic 3) To be comfortable with the following terms..isothermal, adiabatic, state function 4) To define work and heat Progress System is what we are mostly interested in. Isolated (no flow), Closed (E flow), Open (matter and E flow) Isothermal (const.t), adiabatic (no heat exchange), State function is not path dependent w = - force x distance while q = m s T both are not state functions

45 3.3. The First Law of THERMODYNAMICS Goals: 1- To define the First Law of Thermodynamics 2- To review the sign conventions for heat and work

46 3.3. The First Law of THERMODYNAMICS First Law of Thermodynamics states that energy can be converted from one form to another but cannot be created or destroyed E universe = E system + E surroundings For a given process E universe = E system + E surroundings =0 E system = - E surroundings So if a system undergoes a change in energy the surroundings must go through an equal but opposite change in energy to compensate i.e. energy lost by the system must be gained by the surroundings and vice versa

47 We are mostly only interested in E system (we don t really care about E surroundings ) Total energy for the system = E total = KE + PE + U internal energy (U) includes translational, rotational, vibrational electronic, nuclear contributions as well as intermolecular interactions we assume (to keep things simple) that KE = PE = 0 so that E total = U U is a state function (path independent) so we are only interested in final (U 2 ) and initial (U 1 ) states U = U 2 U 1

48 The First Law of Thermo. may be expressed as follows: U = q + w (we assume all work done is of P, V type) so change in internal energy (U) is equal to heat transferred (q) between system and surroundings and work done (w) by or on system By convention we assign the signs of q and w as follows

49 WORK (w) w 0 Energy w 0 WORK done on system is + WORK done by system on surroundings is

50 HEAT (q) q 0 Energy q 0 HEAT absorbed by system is + (Endothermic) HEAT released from system to surroundings is - (Exothermic)

51 Consider a reaction occurring in a constant-volume adiabatic bomb calorimeter The change in energy as a result of the reaction is U = q v + w = q v -PV = q v So for a constant volume process U = q v

52 3.3. The First Law of THERMODYNAMICS Goals: 1- To define the First Law of Thermodynamics 2- To review the sign conventions for heat and work Progress: 1- The First Law is the conservation of energy law E universe = E system + E surroundings =0 2- Work done on a system or heat absorbed by the system are positive (+) to remember this..when someone gives you something or does work for you..it s a good thing (positive)!!!

53 3.4. Enthalpy Goals: 1. To define enthalpy (H) 2. To understand the relationship between enthalpy and internal energy (U) in reactions involving and not involving gases 3. To review difference between C V and C P for gases

54 3.4. Enthalpy U = q + w = q p -PV.notice here we are talking about q p this is because very few processes occur at constant V more at constant P.. U = U 2 U 1 = q p P (V 2 V 1 ) q p = (U 2 +PV 2 ) (U 1 +PV 1 ) Enthalpy (H) is defined as. H = U + PV H = H 2 H 1 = U + (PV) = (U 2 +P 2 V 2 ) (U 1 +P 1 V 1 ) So when P = P 1 = P 2 then H = q p

55 What is the difference between U and H? Consider the following reaction at constant pressure.. 2Na (s) + 2H 2 O (l) 2NaOH (aq) + H 2(g) H = kj U = H -PV volume of H 2 generated at 1 atm is 24.5L so -PV = Latm = -2.5 kj U = kj So there is a small difference between U and H H slightly U since system did expansion work when H 2 was generated ( so some of the internal energy released is to do work) *conversion factor is 1 Latm = J

56 in general the difference between U and H for rxns. involving gases is PV = nrt = RT n (at constant T) where n is the change in the number of moles of gas. n = n products n reactants In previous example n = 1 mole.thus at 298 K RTn = 2.5 kj 2.5 kj is a very small amount (when compared to 370 kj) but still important for accurate work ***However, in reactions only involving the condensed states (i.e. liquids and solids) the changes in volume are so minimal.that U H ***also in reactions involving gases U H when n = 0

57 More on HEAT CAPACITIES Remember U = C V T = n C V T H = C P T = n C P T more work is done in constant P than constant V process, C P C V C P C V since work is done in expansion (for const. P process) for liquids and solids little expansion so C P C V for an ideal gas what is the actual difference between C P and C V? we know that H = U + PV = U + nrt so C P T = C V T + nrt C P -C V = R

58 Sample Problem Calculate the values of U and H for the heating of g of xenon from 300 K to 400 K. Assume ideal gas behavior and that the heat capacities at constant V and P are independent of T. Solution g of Xe = moles Xe Xenon is a monatomic gas. Remember previously we saw that for a monatomic gas C V = 3/2 R = J mol -1 K -1. Molar Mass of Xe = g/mole

59 3.4. Enthalpy Goals: 1. To define enthalpy (H) 2. To understand the relationship between enthalpy and internal energy (U) in reactions involving and not involving gases 3. To review difference between C V and C P for gases PROGRESS 1. Enthalpy H = U + PV and at constant pressure H = q p 2. U H for liquids and solids for reactions involving gases U = H - RTn 3. For gases C P - C V = R

60 Question When 1 mole of napthalene (C 10 H 8 ) is completely burned in a constant volume bomb calorimeter at 298 K, 5150 kj of heat is evolved. Calculate the values of U and H for the reaction. C 10 H 8 (s) + 12 O 2 (g) 10 CO 2 (g) + 4 H 2 O(l)

61 3.5. Adiabatic Expansion of an Ideal Gas (1) Diathermic system. allows energy to leave system as heat. (2) Adiabatic system...means no heat exchange between system and surroundings. i.e. an adiabatic container is an isolated system with no heat exchange Figure Physical Chemistry 7 th Edn. Atkins P., de Paula J.

62 Adiabatic Expansion of an Ideal Gas Case 1: Reversible Expansion Since q = 0 then T drops in an expansion du = dq + dw = dw = - P ext dv If the process is reversible we can replace P ext with P int Thus: dw = - P int dv = - P dv = - (nrt/v) dv

63 du = dw = P dv = or: nrt V dv du nt dv = R V since: du = C V dt du = nt C V dt = nt dt C dv V = R T V Integrate both sides between initial and final states: T T 2 1 dt V C V = R T V 2 1 dv V

64 If: C V = constant Then: T T 2 1 dt T C V = V T T 2 C = V 1 dt T T C ln T 2 1 V2 = R ln V1 V1 = + R ln V2 Since: R = ( C C P V ) for an ideal gas then: T C V ln T 2 1 V1 = ( CP CV ) ln V2 Divide both sides by: C V to obtain: T ln T 2 1 = CP V1 1 ln C V2 V V1 = (γ 1) ln V2 V = ln V 1 2 ( 1) where: γ = C C P V

65 For monatomic gases: C = V 3 R C P = C V + R = 2 5 R 2 Therefore: γ = C C P V = 3 5 = For diatomic gas molecules (excluding vibrations): C V 5 R C P = C V + R 2 7 R 2 Therefore: γ = C C P V 5 7 = 1.40

66 ln 1 2 T T = ln 1) ( 2 1 V V Take the exponential of both sides of the equation: 1 2 T T = P V V P = 1) ( 2 1 V V since: T V P T P V Therefore: 2 1 V V = 1 2 P P Or: P 1 V 1 γ = P 2 V 2 γ Equation relating P and V for a reversible, adiabatic process

67 Alternatively, divide both sides by nr T PV : P 1 V 1 γ T1 P V 1 1 = P 2 V 2 γ T2 P V 2 2 V 1 (γ 1) T 1 = V 2 (γ 1) T 2 Or: T 1 V 1 (γ 1) = T 2 V 2 (γ 1) Equation relating T and V for a reversible, adiabatic process

68 Alternatively, divide both sides by (nr) γ T PV : P 1 V 1 γ T1 P V 1 1 P 1 (1 γ) T 1 γ = P 2 V 2 γ = P 2 (1 γ) T 2 γ T2 P V 2 2 Raise both sides of the equation to the (1/γ) th power: (P 1 (1 γ) T 1 γ ) (1/γ) = (P 2 (1 γ) T 2 γ ) (1/γ) Or: T 1 P 1 [(1 γ)/γ] = T 2 P 2 [(1 γ)/γ] Equation relating T and P for a reversible, adiabatic process

69 Work done in a reversible, adiabatic expansion is: U w = U 2 1 du = U 2 U 1 = ΔU = T T 2 1 C VdT = C V (T 2 T 1 ) = n C V (T 2 T 1 ) if C V = constant over ΔT range Since: T 2 < T 1 then: w = negative P 1 V 1 T 1 P 2 V 2 T 2 Perform as a 2 step process: (1) Expand Isothermally: P 1 V 1 T 1 P 2 V 2 T 1 [ΔU = 0] (2) Cool gas Isochorically: P 2 V 2 T 1 P 2 V 2 T 2 [ΔU = n C V (T 2 T 1 )]

70 Because U is a state function, ΔU is the same whether the change of a gas from P 1 V 1 T 1 to P 2 V 2 T 2 occurs directly or indirectly.

71 Pressure versus volume plots of an adiabatic, reversible and an isothermal, reversible expansion of an ideal gas. Note that the curve is steeper for an adiabatic expansion because γ > 1.

72 Example: A quantity of mole of a monatomic gas initially at a pressure of 15.0 atm and 300 K is allowed to expand until its final pressure is 1.00 atm. Calculate the work done if the expansion is performed adiabatically and reversibly. Need to calculate T 2

73 A quantity of mole of a monatomic gas initially at a pressure of 15.0 atm and 300 K is allowed to expand until its final pressure is 1.00 atm. Calculate the work done if the expansion is performed adiabatically and reversibly. Since the gas is monatomic: γ = 3 5 and 1 = = 3 5 = 2 5 Evaluate T 2 using: T 1 P 1 [(1 γ)/γ] = T 2 P 2 [(1 γ)/γ] T 2 = T 1 P P 15.0 = (300 K) [(1 ) / ] atm atm 2 / 5 = 102 K (3 significant figures)

74 w = ΔU = n C V (T 2 T 1 ) 3 = (0.850 mol) R (102 K 300 K) 2 3 = (0.850 mol) ( J mol 1 K 1 )( 102 K 300 K) 2 = 2099 J = 2.10 x 10 3 J (3 significant figures) = 2.10 kj (3 significant figures) Note: the work is negative because the process is a reversible, adiabatic expansion of an ideal gas

75 Adiabatic Expansion of an Ideal Gas Case 2: Irreversible Expansion P 1 V 1 T 1 P 2 V 2 T 2 where: P 2 = P ext and q = 0 C V ΔU = w = n (T 2 -T 1 ) = - P 2 (V 2 -V 1 ) but: V 1 nrt1 = and V P 2 = 1 nrt P 2 2 C V therefore: n (T 2 -T 1 ) = - P 2 nrt P2 2 nrt P 1 1 If we know initial conditions and P 2 we can solve for T 2

76 Example: A quantity of 0.27 mole of Ne is confined in a container at 2.50 atm and 298 K and then allowed to expand adiabatically under two different conditions: (a) reversibly to 1.00 atm ; (b) against a constant external pressure of 1.00 atm ; Calculate the final temperature in each case.

77 A quantity of 0.27 mole of Ne is confined in a container at 2.50 atm and 298 K and then allowed to expand adiabatically under two different conditions: (a) reversibly to 1.00 atm ; Calculate the final temperature. Since Ne is a monatomic gas: γ = 3 5 and = = 3 5 = 2 5 Evaluate T 2 using: T 1 P 1 [(1 γ)/γ] = T 2 P 2 [(1 γ)/γ] T 2 = T 1 P P 1 2 [(1 ) / ] 2.50 atm = (298 K) 1.00 atm 2 / 5 = 207 K (3 significant figures)

78 A quantity of 0.27 mole of Ne is confined in a container at 2.50 atm and 298 K and then allowed to expand adiabatically under two different conditions: (b) against a constant external pressure of 1.00 atm ; Calculate the final temperature. q = 0 ΔU = w = n C V (T 2 T 1 ) = P ext (V 2 V 1 ) n 3 2 nrt2 nrt1 R ( T 2 T 1 ) = P ext P2 P1 3 T ( 2 T1 T2 T 1 ) = P ext 2 P2 P1 P ext = P 2 3 T ( 2 T1 T2 T 1 ) = P 2 2 P2 P1 = T 2 + P2T1 P 1

79 Solve for T 2 : 3 2 T 2 + T 2 = P2T P T1 5 2 T 2 2 P2 3 T 2 = T1 5 P1 2 P 3 2 = T1 P atm 3 = (298 K) atm = (298 K) = (298 K) 25 = 226 K (3 significant figures) Note: T 2 for part (b) > T 2 for part (a) since more work is done reversibly than irreversibly.