topic 8-11 with answers
topic 8-11 with answers CHEM3330
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Topic 08 Adiabatic Changes An adiabatic change is one in which no heat is transferred between the system and its surroundings Therefore a 0 whenever the change is adiabatic Calculation of Work for an Adiabatic Process The First Law can be combined with the definition of quotadiabaticquot to derive a simple expression for work during an adiabatic process The First Law states that AU q w By definition a 0 for an adiabatic process Therefore wad AU 1 Here the subscript emphasizes that the process must be adiabatic for the relationship to be correct The relationship is generally valid provided that all forms of work are considered For ideal gases recall that AU vdT evaluated at the upper and lower limits applicable to the problem Although this relationship is rigorously correct only for ideal gases it provides an extremely good approximation for real gases and should be used in this course unless the details provided in a specific problem indicate the necessity of considering the effect of volume changes on AU to be discussed in a later topic When CV can be assumed to be constant with temperature it can be pulled outside of the integral with the result that AU CVAT2 Combining equations 1 and 2 we have another expression for adiabatic work wad CAT 3 Written in differential form this is equivalent to dwad CvdT 3A In many contexts we may not have data that directly identifies the final temperature Under such circumstances how can AT be calculated Relating Temperature and Volume for a Reversible Adiabatic Expansion of an Ideal Gas The equations in this section relate the volumes and the temperatures for reversible adiabatic expansions of an ideal gas when no extra work is done The quantitative results will be close for reversible adiabatic expansions of real gases However it is important to note that the relationships derived below are not valid and cannot be used if the expansion is adiabatic but not reversible For an ideal gas assumed pV nRT 4 For an adiabatic process q 05 For a reversible process pext p 6 For the expansion of any gas we showed in Topic 5 that dw pext dV 7 Substitution of 4 and 6 results in the expression dw nRT dVV8 when the expansion is reversible and the gas behaves ideally Note that we have not made any assumption that temperature is constant in equation 8 Topic 08 Adiabatic Changes Page 1 Equation 8 gives the expression for dw in the reversible expansion of an ideal gas Equation 3A gives the expression for dw in an adiabatic expansion When the expansion meets all three conditions reversible ideal and adiabatic and assuming that all work is expansion work the two expressions for dw can be set equal to each other and then solved for T2 The step by step solution is provided in the table below Explanation set wad eq 3A for adiabatic expansion equal to w 8 for reversible expansion of an ideal gas All work is expansion work Divide both sides by nRT Equation CV dT nRTd V V ELK dV 11R T V ampT2dT V2dV nRTl T V1 V Integrate both sides Assume CV is constant with temperature pulling it and the other constants outside of the integral amp1n21n quotR 71 VI Evaluate the integrals at the limits Basic properly of logarithms nxy nyx Recall that Cvlm CV n Make this substitution Define c C mm R Substitute Exponentiate both sides Raise both sides to the 1c power Rearrange to isolate T2 on the left hand side Flip the volume fraction and change sign of the exponent using the property of exponents analogous to the property of logarithms identified above Equation 9 expresses the final temperature in terms of the initial temperature and volumes final and initial for a reversible adiabatic expansion of an ideal gas when no extra work is done This equation cannot be used for an adiabatic expansion that is not reversible For monatomic ideal gases Cvlm 32 R so 1c 23 For diatomic gases Cvlm 52 R so 1c 25 040 For real gases first look up Cplm and then calculate Cvm Cplm R use the result to calculate C Topic 08 Adiabatic Changes Page 2 Mathematically all quantities in the equation are positive The ratio 1c is equivalent to RCvlm which is always a positive fraction less than one If the system expands to a larger volume V1V2 lt 1 indicating that T2 lt T1 Physically this implies that the temperature must drop during a reversible adiabatic expansion from smaller to larger volume The converse must be true if the system is compressed from larger to smaller volume Equation 9 can be substituted into equation 3 to express reversible adiabatic work in terms of temperature and volume data wad 2 CV AT 2 CV Equation 3 AT expressed in terms of T1 and T2 Substitute the expression for T2 from Equation 9 1 Factor out T1 Relating Pressure and Volume for a Reversible Adiabatic Expansion of an Ideal Gas The results above can be combined and with the ideal gas law to relate the pressure and volume changes during the reversible adiabatic expansion of an ideal gas when no extra work is done l Derived above T1 V2 0 T2 VI 2 p1 V1 Ideal gas law relatIonshIp T2 p2V2 l The left sides of the two equations are equal Set P1 V1 2 the right sides equal to each other p2V2 V1 11 Multiply both sides by VzVl and combine c 11 exponents p2 1 1 R CV m R Cp m Analyze the exponent apply the definition of c 1 2 21 C 2 C 2 C put both terms over a common denominator V V m V m substitute CIDm Cvlm R for an ideal gas C Define gamma y as the ratio of the two molar Pam 7 C heat capacities Vm Topic 08 Adiabatic Changes Page 3 7 Substitute y for 1 1c the exponent in Eqn 11 192 VI p1 VIV 2 p2 V27 12 Rearrange to separate terms for the InItIal and fInal states The relationship in Equation 12 differs from Boyle39s Law because temperature is not constant in an adiabatic expansion Still an increase in volume will be accompanied by a decrease in pressure For monatomic ideal gases Cvm 32 R and CIDm 52 R so y 53 For diatomic ideal gases Cvlm 52 R and CIDm 72 R so y 75 For real gases first look up Cplm and then calculate Cvlm Cplm R use the result to calculate y The results in Equations 12 and 9 can be combined and analyzed together to determine the relationship between temperature and pressure during the reversible adiabatic expansion of an ideal gas when no extra work is done You will complete this analysis as a followup problem Ideal Gas Adiabats and Isotherms constant males For all ideal gases p1V1T1 p2V2T2 For reversible isothermal expansions Tis constant q is not zero and pV is constant For reversible adiabatic expansions T is not constant q O and pW is constant Figure 1 graphically compares the isotherms vs the adiabats for one mole of an ideal monatomic gas y 53 when the initial pressure and volume are the same Note that the adiabat decreases more rapidly than the isotherm Isotherms vs Adiabats ldeal Mlona tomic Gas Pressure atm a m 20 30 40 50 Volume L Figure 1 lsotherm and Adiabat Plots for one mole of an ideal monatomic gas The initial state at 29815 K is the same for both plots Comparing the adiabat to the isotherm the pressure decreases more rapidly as a function of volume Note that temperature decreases along the adiabat as the volume increases Topic 08 Adiabatic Changes Page 4 DISCUSSION QUESTIONS AND FOLLOWUP PROBLEMS 81 82 83 84 85 86 87 88 Identify and explain the differences between isothermal and adiabatic expansions of a gas A system is initially set up at a precisely defined volume pressure and temperature In two separate experiments a In experiment 1 the system expands isothermally and reversibly b In experiment 2 the system expands adiabatically and reversibly Which one of the two expansions performs more work on its surroundings Clearly explain and justify your answer Consider Equations 3 9 10 and 12 Derive each of these equations Explain each step of your derivation and clearly identify all approximations and assumptions Derive the mathematical relationship between p1 p2 T1 and T2 for a reversible adiabatic expansion of an ideal gas when no extra work is done Explain each step of your derivation and clearly identify all approximations and assumptions Calculate the values of the constants introduced in this Topic c and 7 for each of the following gases 1 A monatomic ideal gas 2 A diatomic ideal gas 3 NH3g given cpm 3506 J K391 mol39l 4 PCI3g given CVm 6353 J K391 mol39l 230 moles of a gas identified below are initially contained in a volume of 150 L at 29815 K The gas is expanded reversibly and adiabatically until the final volume reaches 400 L Calculate the initial pressure final pressure final temperature w q AU and AH for the expansion You may assume ideal gas behavior and that no extra work is done Carry out these calculations separately for each of the following gases 1 A monatomic ideal gas 2 A diatomic ideal gas 3 NH3g given Cplm 3506 K391 mol39l 4 PCI3g given Cvlm 6353 J K391 mol39l Use the information in Problem 85 when relevant 325 mol of 02 CIDm 29355 J K391 mol39l originally occupies a volume of 195 L at 27315 K It then undergoes adiabatic expansion against a constant external pressure of 0850 atm until the volume has increased to 550 L at which time the expansion is artificially stopped A Identify the major differences between this process and the process described in Problem 86 B Calculate all of the following quantities for this expansion p1 p2 AT q w AU and AH A gas is expanded to a larger volume via a reversible adiabatic process in which no extra work is done Predict separately whether p and T will increase or decrease Justify your answer on the basis of the relevant equations developed in this Topic Topic 08 Adiabatic Changes Page 5 Topic 09 Enthalpies of Reaction The concept of Enthalpy was introduced in Topic 07 In this Topic we extend the discussion to enthalpy changes that occur during chemical reactions and phase changes Much of the material in this Topic reviews material covered in General Chemistry The last section introduces new concepts that apply to determining enthalpy changes that occur at nonstandard temperatures When the total enthalpy of products is greater than the total enthalpy of reactants the process is endothermic Heat must be transferred from the surroundings into the system in order to convert the reactants to products Conversely when the total enthalpy of products is less than the total enthalpy of reactants the process is exothermic Heat is released by the system into its surroundings when the reactants are converted to products Figure 1 illustrates the enthalpy relationships for endothermic left and exothermic right processes reacta I Tllli39S products Hlt HbU rearta 115 pf udiu cts endothenr c exo henrdc Figure 1 Enthalpy relationships for endothermic and exothermic processes In an endothermic process left heat must be absorbed by the system to produce a higherenthalpy product In an exothermic process right heat is released by the system when higherenthalpy reactants are converted to lowerenthalpy products Standard States Standard thermodynamic data are collected and tabulated for welldefined standard conditions In physical chemistry standard states are defined as follows 1 The pressure of gases is 1 bar exactly Note that General Chemistry tables will typically specify 1 atm as the standard pressure 2 The temperature must be specified for each process Although many data tables are tabulated at SATP with T 29815 K the standard state of a substance may be different at a different temperature For example the standard state of H20 is liquid at 298 K but is gas at 398 K 3 The concentration ofsolutes in solution is 1 m molal mol solute kg solvent Note that General Chemistry tables will typically specify the standard concentration as 1 M Molar mol solute L of solution Liquids and solids are pure Topic 09 Enthalpies of Reaction Page 1 Pure elements are in their thermodynamically most stable form at 1 bar and at the specified temperature It is important to know the stable phases 5 I or g of the pure elements and to know the gases that are naturally found as diatomic molecules Although there are several forms of carbon the most stable form at SATP is graphite solid Hess 395 Law Sometimes AH for a specific reaction cannot be found in a thermodynamic table Sometimes it cannot be directly measured experimentally Hess39s Law provides a method for determining AH corresponding to a process when the overall reaction can be written as a sum of two or more reactions for which the values of AH are known or can be measured Hess39s Law If an overall reaction can be written as the sum of two or more reactions AH for the overall reaction is equal to the sum of the individual reaction AH values This is illustrated in Figure 2 AH State A 39 l 2 g A State B ll ll lFiznall State Initial State K Figure 2 Illustration of Hess39s Law The system is transformed from an initial state to a final state in three steps Hess39s Law states that AH for the overall process is equal to sum of the individualstep AH values AH AH1 AHZ AH3 Hess39s Law is based on the fact that enthalpy is a state function Because AH depends only on the initial and final states its value must be the same regardless of whether the path involves multiple steps or is completed in a single step One important implication is that AH for a reaction written in reverse is equal to AH for the original forwardwritten reaction The application of Hess39s law also depends on the fact that enthalpy is an extensive function if the amount of material moles is increased AH will increase proportionately One important implication is that if all coefficients in the original equation are multiplied by a constant factor c AH for the multiplied reaction equation is c times AH of the original reaction equation The general strategy for applying Hess39s Law is as follows 1 Write the overall reaction as a sum of appropriate separate reactions on the basis of information that is given or can be looked up 2 Modify the appropriate reaction equations so that the results yield the reactions in 1 and modify the values of AH appropriately 3 Add the modified values of AH from 2 to obtain the overall AH Topic 09 Enthalpies of Reaction Page 2 A Worked Hess39s Law Problem Example The following worked example demonstrates an organized systematic process by which a Hess39s Law problem can often be solved Problem Calculate AH for the following reaction 285 3H2g gt BzH5g Given the following four reactions and their standard AH values 1 235 32 029 gt 32035 AH1 1273 k 2 BZH6g 302g gt 32035 3H20g AHZ 2035 k 3 H2g 7202g gt H20I AH3 286 k 4 H20I gt H20g AH4 44 kJ Step One 1 Identify any reactants andor products in the overall target reaction that appear once and only once in the given equations 2 For each such reactant or product determine how the given equation must be modified to guarantee that the reactant or product will show up on the correct side of the equation with the correct coefficient Modification may involve multiplying all coefficients by a constant andor reversing the equation 3 Calculate the modified value of AH accordingly 4 Write the modified equation and its corresponding value of AH The reactant B is found only in given reaction 1 It is already on the correct reactant side of the equation and its coefficient already matches that of the target equation Therefore reaction 1 should be used without any changes 1 285 32 02g gt 82035 AH1 1273 k The reactant H2 is found only in given reaction 3 It is already on the correct reactant side of the equation but its coefficient must be multiplied by 3 to match the coefficient in the target equation Therefore the reaction 3 coefficients must be multiplied by 3 and AH must be multiplied by 3 3 x 3 3H2g 15 02g gt 3H20I 3 x AH3 858 k The product BZHG appears only in given reaction 2 Its coefficient matches the target equation but it is not on the correct product side of the equation Therefore the given reaction must be reversed and the original AH must be multiplied by 1 reverse 2 82035 3H20g gt BZH5g 302g AH2 2035 kJ Step Two Add all equations identified in Step One cancelling identical terms that appear on both sides of the summed equation Add all modified values of AH to obtain AH for the summed equation In this example note that H20 is in two different phases 235 32 02g gt 32935 AH1 1273 k 3H2g 1694 gt 3H20I 3 x AH3 858 kJ BQ 5 3HOq gt BgHgm soggy AH 2035 kJ 235 3H2g 3H20g gt 3H20I BZH5g AHsum 96 kJ Step Three Compare the summed equation in Step Two with the target equation If they are identical you are finished If they are not identical complete the problem with the following steps 1 Identify how any unused equations given or looked up need to be modified to in order to convert the summed equation in Step Two into the final equation 2 Calculate AH corresponding to the modified Topic 09 Enthalpies of Reaction Page 3 equation 3 Add the summed equation from Step Two to the modified equation to verify that the result is the target reaction equation 4 Add AHsum from Step Two to the modified AH identified in 2 to obtain the final target value of AH To eliminate 3H20g gt 3H20I from the summed equation the given reaction 4 must be used Reactants and products are already on the correct side of the equation but coefficients and AH must be multiplied by three to eliminate these reactants and products From Step 2 2Bs 3H2g 3H9g gt 3ng4 B2H5g AHsum 96 k 3 x 4 3ng4 gt 3H9q 3 x AH 132 kJ Final Sum 2Bs 3H2g gt B2H5g AH 36 kl Standard Molar Enthalpies of Formation Theformation reaction of a chemical substance is defined by the chemical reaction that has the following characteristics 1 all reactants must be pure elements in their standard states and 2 the chemical substance that is being formed is the only product When the number of moles of product is exactly one the enthalpy of reaction is defined to be the standard molar enthalpy of formation of the product Here are two examples Cs graphite 029 gt C02g AHm for C02g Aern Lis 12 Br2l gt LiBrs AHfm for LiBrs Aern Both examples reveal the following three characteristics for reactions that define standard molar enthalpies of formation 1 There is exactly one mole of a single product 2 All reactants are in their standard states 3 The reaction equation is balanced Because of 1 it is common for balanced formation reaction equations to contain fractional coefficients The standard enthalpy of formation for a pure element in its standard state must be equal to zero You will prove this statement to be true in the followup problems Calculating Enthalpies of Reaction via Standard Molar Enthalpies of Formation Based on Hess39s Law and the definition of standard formation enthalpy the following relationship can be shown relating the standard enthalpy for any reaction to the standard enthalpies of formation of the relevant reactants and products Mathematically the relationship is given by AHgm 2V1 In this equation the superscript emphasizes that standard formation reactions are being used and that the overall reaction occurs under the same conditions as the formation reactions On the right side of the equation there is a separate term in the sum for each reactant and product AHof mi represents the standard molar enthalpy of the ith component which can frequently be found in a standard thermodynamic table The coefficient in the balanced equation for the ith component is symbolized by vi By Topic 09 Enthalpies of Reaction Page 4 convention these coefficients are taken to be positive for reaction products and are taken to be negative for reactants Given a balanced reaction and all relevant enthalpies of formation the calculation of AH for the balanced reaction is straightforward First multiply each individual enthalpy of formation by the corresponding coefficient in the balanced equation keeping in mind that the coefficients of reactants are taken to be negative Once this is done the results are summed to obtain AH for the reaction Although it may have been presented here in a slightly different manner the process is identical to what you were taught in General Chemistry The Temperature Dependence of AH Kirchhoff39s Law For the method described in the preceding section to be precisely correct the temperature of the reaction for which AH is calculated must be identical to the temperature for which the enthalpies of formation are tabulated which is typically 29815 K What if the reaction temperature is not 29815 K From Topic 7 we know that the enthalpy H is dependent on temperature As a result AH must also be temperature dependent When the temperature does not greatly differ from 29815 K the expectation is that the change in AH is very small and can be neglected In such cases the tabulated value is used without correction In this course you will follow this approach and will use the tabulated values unless information is provided that suggests that you need to account for the temperaturedependent difference The derivation that follows develops and justifies the equations that are used to calculate AH at any temperature that differs from the temperature at which AH is known fAHT1 is a known value at T1 and AHTZ is an unknown value at T2 we define AAH the temperaturedependent change in AH as AHTZ AHT1 Once AAH has been calculated and AHT1 is known AHTZ can be can be calculated via algebraic rearrangement AHTZ AHTl AAH Derivation of the expressions used to calculate AAH Kirchhoff39s Law proceeds as follows Equation Relationship Explanation Justification dH 2 Cp dT Equation derived in Topic 07 for heating a single substance within a single phase The relationship is exact at constant pressure and is usually an excellent approximation in cases where any pressure change is not too large ts validity will be assumed here T2 Integrating both sides from T1 to T2 gives the HT2 HTI ICP dT 1 change of enthalpy that results from heating T1 AHT1 Hm 1 Hmw1 AH for a chemical reaction can be expressed as the total enthalpy of products minus the total AHTz HPTOdaTz Hreacfz 2 enthalpy of reactants This must be true at both temperatures AAH AHT AHT 3 Definition of AAH see text Topic 09 Enthalpies of Reaction Page 5 AAH Z Hmod T2 Hreacjz Hpmd T1 Hreac Tl Substitute expressions from 2 into 3 Rearran e rou in roduct and reactant terms AAH Hp1o112 HPWI HWJZ 11mm g g p g p separately T2 T2 Apply Equation 1 separately to products and A JCppr0d JCp reac reactants T1 T1 T2 Combine the difference of two integrals into a ICpprod CpreacdT Single integral T1 ACP E 211 Cp m 4 Define ACp which is equivalent to Cplpmd Cplreac This definition is analogous to the equation used to calculate AH an on page 4 and should be interpreted in the same manner T2 Substitute ACp into the equation This gives the AAH IACP T 5 general relationship necessary to calculate T1 AAH Equation 5 is known as Kirchhoff39s Law Case 1 When ACp can be assumed to be constant over the temperature range of the problem it can be pulled outside of the integral in Equation 5 The resulting relationship is AAH ACP AT 6 Even though Cp is in general temperature dependent the approximation is often acceptable because Cp changes that occur in the reactants often offset and accidentally cancel Cp changes that occur in the reactants This quotcancellation of errorsquot frequently decreases the absolute error associated with calculated values of ACp making the approximation acceptable Case 2 When the temperature dependence of ACp cannot be neglected the first step is to determine the analytical expression for ACp We will explore this with an example Consider the following reaction R gt 2 P For the reactant R and product P assume that temperature dependence of the molar heat capacity can be expressed via the standard analytical form CIDm a bT cT39z Applying the definition of ACp in this reaction ACp 2Cpmpmd 1Cpmreac b prod react react Thus ACP 2a T cpr0d T z 61mm b T 6 T z Grouping like terms prod in T this rearranges to ACP Zapmd area 2 rod breactT 2 cpmd creactT2 P Substituting this into Equation 5 gives T2 I2aprod areact prrod breactT chrod CreactT 2dT 39 T1 With three different terms in temperature the integral can be split into three separate integrals MM Tm 11 Tm brdr Toe cT2dr T1 T1 T1 As usual constants can be pulled outside of the integral sign before each integral is evaluated Topic 09 Enthalpies of Reaction Page 6 W m 2c am Mr T1 T1 T1 TdT 2b 0 b prod areact Finally the integrals should be evaluated at the limits of integration AAH zaprod areactT2 szrod breactj chrod Creact1i T2 T1 One among many possible ways to simplify the equation is AAH zaprod areactAT WJUEZ T12 chrod Creact 1 ET This equation should not be memorized You should be able to derive it beginning with Equation 5 and the temperaturedependent CIDm expression CIDm a bT cT39z DISCUSSION QUESTIONS AND FOLLOWUP PROBLEMS 91 92 93 94 Review standard thermodynamic conditions pressure concentration of solutes solids liquids and pure elements Know all elements that exist as diatomic molecules at SATP Know which elements are liquids and which elements are gases at SATP Know the standard state of elemental carbon State Hess39s Law Hess39s Law is based on two thermodynamic properties of enthalpy Identify the two properties and explain how they are used in the application of Hess39s Law Given for the reaction CH4g 202g gt C02g 2H20l AH 890 kJ Important M in all thermochemical equations like the one provided above the value of AH is based on the condition that the actual number of moles of each reactant and product is exactly equal to its respective coefficient in the balanced chemical equa on Calculate AH for each of the following four reaction equations or conditions A C02g 2H20I gt CH4g 202g B 3CH4g 602g gt 3C02g 6H20I C 2C02g 4H20l gt 2CH4g 402g D 4000 mol of CH4 are completely combusted in the forward reaction Explain each of the steps used to solve the Hess39s Law example in the text Resolve the problem example without referring to the text Topic 09 Enthalpies of Reaction Page 7 95 96 97 98 99 910 911 912 913 Consider the following four chemical reactions and their standard enthalpies Reaction AH k 1 F gt G 51 2 2A D gt 26 552 3 48 3D gt 2E 2504 4 C3D gt E3F 2105 Using the information provided above determine the standard enthalpy for the following reaction 3A 23 C Show and explain all work including how the individual reactions and their respective AH values are modified Write the balanced formation reaction for the following substances at 29815 K Your balanced equation should have only one mole of the product A C5H5I B Br2l C NaCS D C2H5g E H20g F NH4C Prove with clear explanations that the standard enthalpy of formation must be zero for a pure substance in its standard state Calculate the standard enthalpy of the following reaction If appropriate use data in Table 1 belOW C10H3S 39l39 7 5 39l39 4 Calculate the standard molar enthalpy of formation for C8H18l Use only the following information and data in Table 1 below GlVen C3H13I 8 9 AHO A Assuming Cp values to be constant calculate ACp for the following two reactions Where appropriate use data in Table 1 below 1 NH3g 12 N2g 32 H2g 2 Csgr 02g C02g B Assuming that Cp values are not constant write the correct expression for ACp as a function of T for the same two reactions Where appropriate use data in Table 2 below Consider the following reaction NH3g 12 N2g 32 H2g For the following three questions use the results from Problem 910 and the data in Table 1 andor Table 2 if needed A Calculate AH at 29815 K B Assuming that Cp values are constant calculate AH for the same reaction at 39815 K C Assuming that Cp values are not constant but are given by CIDm a bT cT39Z recalculate AH for the reaction at 39815 K D Comment on the difference between the two AH values at 39815 K How valid is the constant Cp assumption used in Part B for calculating the enthalpy of reaction at 39815 K For additional practice repeat all four parts of Problem 911 for the following reaction Csgr 029 gt C02g Derive Kirchhoff39s Law equation 5 Explain each step Topic 09 Enthalpies of Reaction Page 8 914 Derive the equation used to calculate AH when the temperature dependence of Cp must be taken into account Explain each step Your derivation should begin with Kirchhoff39s Law equation 5 and should end with the final equation in the narrative for this Topic 915 Derive the equation used to calculate AH from standard enthalpies of formation AHgm Zvl As your starting point use Hess39s law and the definition of molar enthalpies of form ation Data for use in this set of Problems formula AH flm kJ mol391 CIDm J K391 mol391 naphthalene C10H8s 7853 carbon dioxide C02g 39351 3711 water liquid H20I 28583 75291 ammonia NH3g 4611 3506 nitrogen N2g 29125 hydrogen H2g 28824 carbon graphite Csgr 8527 oxygen 02g 29335 Table 1 Selected molar enthalpies of formation and constantpressure heat capacities at 1 bar and 29815 K CWquot ll K4 mold a bT cT392 Substance a b10393 K39l c 105 K2 NH3g 2975 251 155 N2g 2858 377 o5o H2g 2728 326 050 Csgr 1686 477 854 029 2996 418 167 C02g 4422 879 862 Table 2 Coefficients for TemperatureDependent Heat Capacities of Selected Substances Topic 09 Enthalpies of Reaction Page 9 Topic 10 Changes in Internal Energy Exact Differentials and State Functions Definitions of Exact and lnexact Differentials In calculus an exact differential is defined as an infinitesimal quantity which when integrated always gives a pathindependent value State functions in thermodynamics must be represented by exact differentials because changes in state functions depend only on the initial and final state and are independent of the path For example the differential of enthalpy dH must be exact because ldH AH where AH H2 H1 regardless of the path taken between the initial and final states Internal energy U is also a state function so dU must be an exact differential An inexact differential is an infinitesimal quantity which when integrated gives a value that depends directly on the actual path taken Path functions in thermodynamics must be represented by inexact differentials For example the differential of work dw is inexact because the amount of expansion work calculated from w lpext dV differs depending on the actual path taken Heat q is also a path function so dq must be an inexact differential Expanding Total Differentials in Terms of Partial Derivatives Let fxy represent a function of two independent variables x and y The total differential off can be expanded in terms of the partial derivatives of U with respect to x and y as follows df df df dx dy 1 Note that the expansion involves a sum Each term in the sum is equal dx y dy x to the partial derivative of the function with respect to one of the independent variables multiplied by the total differential of that variable The same pattern applies when the function depends on more than two variables For example if a function G depends on p T and n it follows that dG a G dp dT dn 6p T 6T M an Tm Expansions of total differentials like these are common in thermodynamics and are important to master Properties of Exact Differentials Because state functions must be represented by exact differentials understanding the properties of exact differentials is critical in the study of thermodynamics Topic 10 Changes in Internal Energy Page 1 Mathematical Test for an Exact Differential Let f be a function that depends on x and y If a total differential df can be expressed as dfxy gxy dx hxy dy 2 df is an exact differential if andonlyif a g a h 3 6y x 6x y Application and Interpretation By direct comparison of Equations 1 and 2 it should be clear that gxy in equation 2 is equal to erax in equation 1 Similarly hxy in equation 2 must be equal to away in equation 1 These expressions can be substituted into equation 3 providing a 61 0x 0y another way of stating the test for an exact differential a y a quot 4 y x x y The expression on the left side of equation 4 represents the second mixed partial derivative of fxy first with respect to x then with respect to y The expression on the right side of equation 4 represents the second mixed partial derivative of f first with respect to y then with respect to x As a result df is an exact differential if and only if the second mixed partial derivatives of the function f are equal regardless or the order in which the partial derivatives are evaluated This is equivalent to the test for an exact differential stated above Properties of Partial Derivatives Two properties of partial derivatives are extremely important in many thermodynamic derivations These two properties are summarized here 6y 1 Inverter If the numerator and denominator in the partial derivative are Z interchanged with the same variable being held constant the result is the reciprocal of the original 6T 1 partial derivative For example if H is a function ofT and p 6H p 6H 6T P a a a Euler Chain Relation Z 1 Note that the variables on the left side of the 6x Z Oz y 6y x equation cycle y in the first partial is replaced by x in the second partial x is replaced by z and z is replaced by y The same cycle holds between the second and third partial derivatives For example if H isafunctionofTandp a T 15 6T p 6p H OH T Topic 10 Changes in Internal Energy Page 2 Partial derivatives can be divided as if they were algebraic terms For example dividing equation5by6p6HTgives a T 1Applyingtheinvertertothe rightsideof 0T p 0P H 0 6H T aHj 6T 6H this equation gives 6T p 6p H 6p T Thermodynamic Coefficients A number of coefficients defined thermodynamically in terms of partial derivatives are commonly tabulated for various substances In many expositions the coefficients are introduced one at a time at the time they are required in a specific derivation We are introducing all of the key coefficients together in this section Two of these Cp and CV have already been discussed in Topics 6 and 7 For each of the coefficients the name is given in the first column of Table 1 the thermodynamic definition is identified in the second column and a brief description of its physical meaning is provided in the third column along with typical units Note that subscripts when included in coefficient symbols identify the variable that is held constant You should memorize the thermodynamic definitions of CV and C Although definitions of the other coefficients need not be memorized you should fully understand their meanings know when you will need to look them up and be able to use them to solve applicable problems Name of Thermodynamic Description What the Coefficient Measures Physically Coefficient Definition Appropriate integration of infinitesimalchange equations yields the value of a measureable laboratory changes ConstantVolume 6U The infinitesimal increase in internal energy U that results Heat Capacity CV V E from an infinitesimal increase in temperature T at constant 6T 1 1 V volume V Typical units are K mol aT infinitesimal increase in temperature T at conStant ConstantPressure 6H The infinitesimal increase in enthalpy H that results from an Heat Capacity Cp Cp in pressure p Typical units are K391 mol39l TET aV from an infinitesimal increase in volume V at constant Internal Pressure 6U The infinitesimal increase in internal energy U that results 7l39T E T temperature T Typical units are atm Expansion 1 6V The infinitesimal relative increase in volume V that results V p Coefficient ocT a T from an infinitesimal increase in temperature T at constant pressure p Typical units are K39l Isothermal 1 6V The infinitesimal relative decrease in volume V that results Compressibility KT KT V T a from an infinitesimal increase in pressure p at constant p temperature T Typical units are atm39l JouleThomson 6T The infinitesimal increase in Temperature T that results Coefficient u U E a from an infinitesimal increase in pressure p at constant H enthalpy H Typical units are K atm39l Table 1 Common Thermodynamic Coefficients Topic 10 Changes in Internal Energy Page 3 Tables 2 collects values of 0c and KT for various condensedphase solid and liquid substances at 20 C Table 3 collects values of u for various gases measured at the temperature identified in the third column Substance 0c 10394 K1 KT 10396 atm39l benzene C5H5l 124 921 carbon tetrachloride CC4I 124 905 ethanol C2H5OHI 112 768 mercury Hgl 182 387 water H20I 21 496 copper Cs 0501 0735 diamond Cdia 5 0030 0187 iron Fes 0354 0589 lead Pbs 0861 221 Table 2 Expansion Coefficients and Isothermal Compressibilities of various condensedphase substances at 20 C Substance u K atm39l temperature airg 0189 323 K carbon dioxide C02g 111 300 K helium Heg 0062 298 K hydrogen H2g 003 298 K nitrogen N2g 027 298 K oxygen 02g 031 298 K Table 3 JouleThomson Coefficients of selected gases The temperature associated with each coefficient is shown in the third column Calculation Examples Involving Thermodynamic Coefficients Before proceeding we will consider two typical problems that involve thermodynamic coefficients Problem 1 A sample of C5H5I occupies 1000 cm3 at 298 K and 1 bar Calculate the volume of the sample at 1000 K and 1 bar assuming that it remains in the liquid phase First identify what the question is asking and what information is required to solve the question The question focuses on how the volume changes when the temperature is increased at constant pressure The expansion coefficient 0c which measures such volume changes is needed 1 6V Write the expression defining the expansion coefficient a E Next obtain the P 6T value of oz for benzene provided in Table 2 CL 124 x 10394 K1 n problems like this you will normally assume that the relevant coefficient is constant over the conditions of the problem You will need to treat the coefficient as changing only if you are provided information indicating that the coefficient value should not be treated as constant Topic 10 Changes in Internal Energy Page 4 Because pressure is constant the definition of oz can be written as a regular derivative 1 dV a Multiply both sides by dT then integrate both sides pulling the constant outside of the 1000K V dV Vf integral a I dT Integrate both Sides and evaluate at the limits aAT 1n 298K V V l V Exponentiate both sides ea 2 f then multiply through by Vi Vf eaAT 139 Finally evaluate the results AT 1000 298 K 702 K ocAT 087048 exp08704g 2388 Multiply by Vi to obtain the final volume Vf 23881000 mL 238 mL Comment The answer is physically meaningless because benzene is not stable in the liquid phase at 1000 K ts standard vaporization temperature at 1 bar is 3532 K The quotmoral of the storyquot is that a correct mathematical calculation may not always result in a physically meaningful result Calculated results should always be evaluated in order to assess whether they are physically reasonable Problem 2 Derive the expression for the expansion coefficient of an ideal gas 1 6V Write the expression defining the expansion coefficient a E Next write the ideal 6T gas expression for V V Substitute this into the definition of a and evaluate I9 nRT a 1 p nR dT nR a Noting that the ideal gas equation rearranges to T pVnR it V 6T pV dT pV P 1 follows that a for an ideal gas Correcting for Volume Changes in the Calculation of AU Consider the internal energy as a function of T and V For a constantmole system the total differential dU can be expanded as discussed on the first page of this Topic dU a U dT a U dV Both of the partial derivatives are thermodynamic coefficients 6T V 6V T defined in Table 1 Substituting these coefficients for the partial derivatives gives dU CVdT 7erV The first term CVdT was discussed in Topic 6 The second term Tth is an additional term that accounts for changes in internal energy that result from volume changes at constant temperature Topic 10 Changes in Internal Energy Page 5 Physically the internal pressure m measures the strength of cohesive forces in the system If TET gt 0 it means that the internal increases when the volume increases attractive forces dominate and must be overcome when the molecules move farther apart If TET lt 0 it means that the internal energy decreases when the volume increase repulsive forces dominate and the system becomes more stable when the molecules move further apart Neither attractive nor repulsive forces exist in ideal gases so TET 0 for ideal gases To summarize the relationship AU CVdT is exact for ideal gases For real gases the additional term Tth is typically very small compared to CVdT and can normally be neglected In this course you will neglect the anV term unless information provided in the problem indicates otherwise Changes in U at Constant Pressure 6U The constantvolume heat capacity is defined as C E As is true for all partial V derivatives the identity of the variable being held constant is of fundamental importance In this section we explore the difference if pressure instead of volume is the variable being held constant We will evaluate the differential increase in internal energy resulting from a differential 6U increase in temperature at constant pressure and then will relate the result to CV P Equation Explanation dU CVdT 39TdV Expression for dU derived above 6T a T a T respect to T at constant p The left side is the target expression On the right side the coefficients CV and TET are assumed constant and are pulled outside of the partial derivatives de aTj aVJ Take the partial derivative of both sides of the equation with V T P P P 611 C av aTaT 1 regardless of what variable is being held constant 2 V T 6T M p dU 1 j CV UNIT V Rearrange the expression for 0c from Table 1 a E aV to 6T p V M 6V get 05V 2 Substitute 05V into the previous equation P for 6T P 6T 3T V 2 UNIT V the equation 611 611 Substitute the definition of CV and subtract it from both sides of P If pressure is constant instead of volume the partial derivative of U with respect to T is different The correction factor is given by omTV This result will be used in the next Topic Topic 10 Changes in Internal Energy Page 6 For an ideal gas we have established that TET 0 page 5 of this Topic Therefore for ideal gases I only 6T p CV DISCUSSION QUESTIONS AND FOLLOWUP PROBLEMS 101 102 103 104 105 106 107 108 Explain clearly why state functions in thermodynamics must be represented by exact differentials Explain the difference between exact and inexact differentials What thermodynamic functions covered up to this point are represented by exact differentials Let A be a state function that depends on S and V Write out the total differential of S in terms of the partial derivatives of S and of V A Following the pattern required in 102 write out the total differential of the following function 2 x3 3y2 4xy x 2y 3 B Is dz an exact differential Prove your answer A State two equivalent ways of testing whether a differential is exact B Apply the appropriate tests to determine whether or not the following functions are exact 1 sinxv 2 4v3e394x 3 sinXv expx2 4 X2v3 5 xx ey v A Derive the expression for the expansion coefficient of a Virial gas B Derive the expression for the isothermal compressibility of an ideal gas C Derive the expression for the isothermal compressibility of a Virial gas using the pressurebased expansion At what total pressure would the molar volume of Hgl be 100 less relative to its molar volume at 1 atm Assume a constant temperature of 20 C At 1 atm the volume of a certain liquid varies with liquid as described by the following 4 6 2 relationship V 2 V0 076 38x10 150x10 where V0 represents the volume at 299 K Calculate the expansion coefficient at 315 K Consider NH3g for which Cvm 2732 J K391 mol39l and TET 840 Pa Initially at 298150 K and 1000 L one mole of the sample is simultaneously heated to 300150 K and compressed to 0900 L A Calculate AU for the process B Compare the magnitude of the temperaturechange contribution to AU to the magnitude of the volumechange contribution Topic 10 Changes in Internal Energy Page 7 109 For a van der Waals gas TET asz Consider 02 treated as a van der Waals gas Initially at 200 L and 2870 K one mole of the gas is isothermally and reversibly expanded to 300 L Calculate AU for the expansion 1010 Review all of the definitions of thermodynamics coefficients Discuss what they physically measure and how they are used to perform relevant calculations Topic 10 Changes in Internal Energy Page 8 Topic 11 The Relationship Between C2 and C In a previous Topic you learned that Cp CV nR or equivalently CVm Cvlm R for an ideal gas This relationship is also an excellent approximation for a real gas and should be used for gas systems unless wording in a problem suggests otherwise Later in this Topic we will explore the exact thermodynamic relationship between Cp and CV The exact relationship applies to all real substances even those in a condensed solid or liquid phase The Temperature Dependence of Enthalpy at Constant Volume 6 The constant pressure heat capacity is defined as Cp E In this section we explore P how the partial derivative of enthalpy with respect to temperature changes when volume instead of pressure is held constant The following derivation makes use of the thermodynamic coefficients and the properties of partial derivatives introduced in Topic 10 Equation Explanation H HTp Let H be expressed as a function ofT and p In this discussion we assume that moles are constant 6H 6H Expand the total differential of H dH 2 dT dp 6T 6p T 5H Substitute based on the thermodynamic definition dHdeT dp opr 61 T 5H 5T 5H 5p Take the partial derivative of both sides with j C respect to temperature holding volume constant 6T p 6T 6p 6T 39 V V T V The left side of the equation is what we are trying to evaluate The thermodynamic coefficients and partial derivatives are considered constant and are pulled outside of the new partial derivatives 6H 6H 6 Simplify 6T6TV 1 2 Cp P 1 6T V 6p T 6T V at 57 5V 1 With the eventual goal of substituting for 6p6TV Z write the related Euler chain relation 6T V 6V T 6p T a j 1 Isolate 6p6TV by division 6T v 6V 6V T 6p T Topic 11 The Relationship Between Cp and CV Page 1 Apply the inverter to 6T6Vp i 6T v 61 6p T a 05V j Because a E i a V substitute ocV for 5T V 01 V aT 6 T avam a 05V j Because KT E i a V substitute KTV for 6T V KT V 6p T 6V6Tp 1 Simplify the right side of the equation 6T V KT 6H 0 6H Substitute OLKT into Equation 1 in place of Cp 2 zapam 6T V KT T 61 With the eventual goal of substituting for 6H6pT 6p T 6T H 6H p write the related Euler chaIn relatIon 6H 1 Isolate 6H6pT by division 619 6H 6T 611 Apply the inverter to both partial derivatives in the denominator K 6p T 6p H 6T p 6H aT Cp Because C E 6 H substitute Cp for aHaTp K 6p T 6p H p 6T p 6H IuCp Because u E substitute u for 5175le K 6p T 6p H 6H a Substitute the right side in place of 6H6pT in a T Cp K J Cp Equation 2 V T Factor out Cp and simplify The subtracted term ocuKT represents the difference between Cp and 6H6TV ie when volume instead of pressure is held constant This result will be used in the second derivation that follows Topic 11 The Relationship Between Cp and CV Page 2 Relating CD and C In this section we explore the difference between Cp and CV First we derive the relationship for an ideal gas We then explore the relationship for real substances including liquids and substances C9 CE for an Ideal Gas The following derivation applies only to ideal gases It proves the relationship for ideal gases introduced in Topic 7 6H 6U Using the definitions of CV and Cp Cp CV 6T 6T p V H U pV Definition of Enthalpy H U nRT ideal gas only pV nRT for an ideal gas Make this substitution 31 nR TU 6U Substitute this expression into the original C C p V equation 6T p 6T V 6U 5nRT 6U Expand the first partial using property of addition C C P V M p 6T M V 6U 6U Relationship derived in Topic 10 for an ideal gas a a V 6T Simplify canceling equal terms and pulling Cp CV quotR a T constants outside of the remaining partial Cp CV nR Simplify given that the partial is equal to 1 CRM CVM R Dividing by moles gives the molar relationship Cn CE for Real Substances Including Liquids and Solids 6H 6U Using the definitions of CV and Cp Cp CV 6T 6T V H U pV Definition of Enthalpy 5U pV 6U Substitute the definition of enthalpy Cp CV 6T 6T p V 6U a V 6U Split the first partial into the sum of partials Cp CV E 6T p 6T 6T V Topic 11 The Relationship Between Cp and CV Page 3 av Because p is constant in the second partial pull it Cp CV UNIV l P outside From the last derivation in Topic 10 P 6T de de substItutefor a7rTV 6T p 6T V Cp CvzarrVapV Because a E i a V substitute ocV for V 6T p 6V6Tp Cp CV 05V p 7 3 Factor out ocV The goal of the following steps is to eventually HT 2 T P substitute for p TE in Equation 3 The relationship V shown to the left will be derived in a future Topic For now we will assume that it is correct a p Rearrange the preceding equation 9 7rT T 6T V aT From the first derivation in this topic p 7rT 61 a KT Substitute this 6T V KT aT Substitute the expression for p TE into Equation 3 Cp CV 05V KT 052 V T Combine terms and simplify Equation 4 gives the thermodynamic relationship between Cp and CV for real substances including liquids and solids First we will evaluate the relationship in Equation 4 for ideal gases to verify that it gives the expected result In a problem example in Topic 10 we demonstrated that CL 1T for an ideal gas In one of the follow up problems you proved that KT 1p for an ideal gas Substituting these into 1 2 VT T V 1 Simplifying gives Cp CV For an ideal gas pVT 19 Equation 4 gives Cp CV T nR Therefore Cp CV nR which is the expected ideal gas result General comments regarding liquids and solids For most liquids and solids under ambient conditions Cp z CV This is because both 0c and KT are small such that the difference XZVTKT is typically also very small However though only infrequently KT can be extremely small for some solids and liquids resulting in a Cp CV difference as large as 30 of the Cp value Topic 11 The Relationship Between Cp and CV Page 4 DISCUSSION QUESTIONS AND FOLLOWUP PROBLEMS 111 Review the derivations in this Topic Be able to clearly explain each step 112 For each of the five substance in the following chart calculate Cplm Cvlm at 200 C Assume that all values in the table are accurate at 200 C Substance MW g mol39l p g cm393 0c 10394 K1 KT 10396 atm39l benzene C5H5l 7812 08565 124 921 ethanol C2H5OHI 4607 0789 112 768 water HZOI 18015 0998 21 496 diamond Cdia 5 12011 351 0030 0187 lead Pbs 20719 11342 0861 221 Topic 11 The Relationship Between Cp and CV Page 5 Numerical Answers to Selected Problems Problem Set 8 Problem Sets 811 85 C CVMR 7 Cpm CVm monatomic ideal gas 32 53 diatomic ideal gas 52 14 NH3g 3217 1311 PCI3g 7641 1125 86 Monatomic Diatomic NH3g PC3g ideal ideal 91 atm 3751 3751 3751 3751 TZ K 15504 20139 21980 26223 AT K 14311 9676 7835 3592 q k 0 0 0 0 AU kJ 4105 4626 4821 5248 w k 4105 4626 4821 5248 AH kJ 6842 6476 6319 5935 p2 atm 0732 0950 1037 1237 87 W 3057 kJ AU 3057 kJ AT 4471 K Tr 22844 K AH 4266 kJ q 0 k P1 3736 atm Problem Set 9 93 890 kl 2670 kl 1780 kl 3560 kl 95 178 k 98 318940 k 99 250 kl mol391 910 A1 227335 K39l A2 0752 K39l Bl Aa 2546 A b 1832 X 10393 Kquot1 A C 205 X 105 K2 B2 Aa 260 Ab 016 X 10393 K39 L AC 159 X 105 K2 911 A 4611 kl B 4838 kl C 4819 kl 912 A 39351 kl B 39359 kl C 39364 kl Problem Set 10 106 261 atm 107 129 x 10393 K391 108 AU 5456 J The temperature change contribution is approximately 650 times larger than the volume change contribution 109 645 J Problem Set 11 112 C2H50HI H20 Cdia s Pbs C5H5I JKmol 4523 2833 048 000 182
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