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# Topic Handouts 1-7 CHEM3330

Texas State

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This 43 page Bundle was uploaded by mythili venkateswaran on Friday September 4, 2015. The Bundle belongs to CHEM3330 at Texas State University taught by Dr. Easter in Fall 2014. Since its upload, it has received 81 views. For similar materials see Physical Chemintry 1 in Chemistry at Texas State University.

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Topic 01 States of Gases A state of matter consisting of particles that does not have a defined volume or a defined shape In a closed container particles in the gas phase will fill the entire volume and the collection will assume the shape of the container Compared to liquids and solids of the same material gas particles are spread far apart have much lower density and have much weaker intermolecular interactions Ideal Gas A gas in which there are no intermolecular interactions between the particles An ideal gas is perfectly described by the ideal gas equation of state pV nRT State A condition of the system in which all essential variables state variables are specifically defined Equation ofState An algebraic equation that relates the values of the state variables in a fixed state For an ideal gas the equation of state is given above pV nRT State Variables for Gases The state of a gas can be described by four state variables p pressure V volume n number of moles and Tabsolute temperature R is the universal gas constant Values of only three state variables are required to define a state the value of the fourth variable can be determined using the equation of state 1 n number of moles of gas specifies the chemical amount of substance Two common calculations used to determine the number of moles include a the total number of particles N divided by Avogadro39s number NA n NNA b the mass m divided by the molar mass M n mM 2 V volume is defined by the physical threedimensional volume of the container Sometimes an imaginary container can be used for example when discussing 1 liter of air in the atmosphere 3 p pressure is defined in physics as the magnitude of the force F acting on a surface per unit area of the surface on which the force acts A p FA Units Units of pressure must be consistent with units of force divided by units of area There are many valid units for pressure that meet this requirement The specific problem or situation will determine which units are best Common units of pressure include pascal Pa bar atmospheres atm torr Torr millimeters of mercury mmHg and pounds per square inch psi It is extremely important to be able to correctly convert between various units of pressure whenever necessary Mechanical equilibrium and pressure Consider a moveable wall with two pressures acting on it from either side pL from the left side and pR from the right side If the pressures are not equal the wall will be pushed by the side with higher pressure toward the side with lower pressure The wall will not quit moving until the pressures on both sides are equal When the wall is no longer pushed in one direction or other it is said to be in mechanical equilibrium and the opposing pressures must be equal Topic 01 States of Gases Page 1 O P E N SYSTEM Figure 1 Cartoon depicting the key elements of a manometer See text for details Measuring pressure via a manometer The manometer is a laboratory device used to measure pressure by difference A cartoon of a typical manometer is shown above The manometer is filled with a liquid of known density typically water or mercury It is open to the atmosphere on one side left side above and is attached to the closed system on the other On the side where the liquid level is higher the liquid will contribute a net downward pressure due to gravitational force The net pressure m the liquid is given by pliq pgAh where p is the liquid39s density 9 is earth39s gravitational constant 9 980665 m 5392 exactly and Ah is the difference between the two liquid heights If the liquid is mercury and the difference in height is measured in mm the pressure contributed by the mercury is simply equal to Ah in units of mmHg or Torr When the liquid in the manometer reaches mechanical equilibrium ie is not moving the pressure on the left side must be equal to the pressure on the right side The pressure contributed by the liquid is always positive and is included on the side where the liquid level is higher In the above illustration peft patm the external atmospheric pressure pright psys pliq These must be equal therefore patm psys pliq Measuring pressure via a gauge A tire gauge does not measure the actual air pressure in the tire it measures the difference between the actual pressure and the ambient pressure The true air pressure in the tire is the sum of the gauge pressure plus the ambient pressure 4 T absolute temperature is the property that reflects the amount of kinetic energy within the component particles The direction of heat flow between two objects is defined by differences in temperature heat always flows from higher temperature to lower temperature Diathermic and Adiabatic Boundaries Diathermic boundaries permit heat to flow and to be transmitted through them Adiabatic boundaries do not permit the flow or transfer of heat Topic 01 States of Gases Page 2 Thermal Equilibrium exists when there is no flow of heat between two objects that are separated by a diathermic boundary The objects must be at the same temperature The Zeroth Law of Thermodynamics states that ifobject A is in thermal equilibrium with object B and ifobject B is simultaneously in thermal equilibrium with object C then object A must be in thermal equilibrium with object C Common Temperature Scales in Chemistry Fahrenheit Originally 0 F represented the freezing point of liquid brine and 100 F represented the average human body temperature Nowadays the scale has been redefined via the freezing point of water 32 F and the boiling point of water 212 F Body temperature is now 986 F instead of 100 F One degree is exactly 1180 the difference between the freezing and boiling points of water Because negative temperatures are permitted the Fahrenheit scale is not an absolute scale Celsius 0 C represents the freezing point of water and 100 C represents its boiling point One degree is exactly 1100 the difference between the two values Because negative temperatures are permitted the Celsius scale is not an absolute scale Standard Conversions C 59 x F 32 F 95 C 32 Absolute zero refers to the lowest possible temperature limit in the physical universe as we know it An object39s temperature can theoretically be lowered extremely close to absolute zero but will never quite get there Temperatures corresponding to absolute zero are 27315 C and 45967 F Absolute Temperature Scales Because most scientific equations require the use of absolute temperature it is necessary to define a scale where 0 corresponds to absolute zero Only absolute temperature values may be used in such equations M The Kelvin scale is based on the Celsius scale the interval spacing between degrees is identical However the Kelvin scale is offset by 27315 degrees so that 0 K corresponds to absolute zero Note that the degree sign is not used with Kelvin temperatures Standard Conversions K C 27315 C K 27315 DISCUSSION QUESTIONS AND FOLLOWUP PROBLEMS 11 Express the pressure unit Pascal Pa in terms of the fundamental SI units kg m 5 Topic 01 States of Gases Page 3 12 In each row of the table below a value of pressure is given Convert the pressure to each of the other pressure units identified in the column headings atm bar psi Pa mmHg atm 1234 atm bar 0865 bar psi 972 psi Pa 507300 Pa mmHg 184 Torr 13 Consider the cartoon of the manometer in Figure 1 A pressure reading is taken in which the liquid used was water p 0998 g cm393 The ambient pressure in the laboratory was 745 Torr and difference between water heights right minus left was 363 inches 100 inch 254 cm Use this information to calculate the system pressure Express your answer in units of atm bar psi Pa and mmHg 14 The pressure of a tire is measured by a gauge to be 35 psig If the atmospheric pressure is 10 atm what is the actual air pressure in the tire 15 In each row of the table below a value of temperature is given Convert the temperature to each of the other temperature units identified in the column headings F C K F 986 F C 750 C K 1983 K 16 Determine the temperature at which the numerical value observed on the Celsius scale is identical to the reading observed on the Fahrenheit scale Topic 01 States of Gases Page 4 Topic 02 Ideal Gases Individual Gas Laws Like almost all major discoveries in science one piece was discovered at a time Finally when enough pieces of information were available they were combined to give us the big picture 8 gt 7 I DC 7 125atm E 5 39 d1 40 E a g 4 30 100 atrn V g g Q 10 I 0 39 39 39 300 400 500 600 o 10 20 30 40 Tm t K em era urex Volume L P 4 200 GayLussac39s Law Avogadro39s Law 3 150 p 1 bar T a 29815 K pressure atm N 300 400 500 600 Temperature K moles n Figure 1 The relationships between state variables described by four individual gas laws The four plots represent Boyle39s Law top left Charles39 Law top right GayLussac39s Law bottom left and Avogadro39s Law bottom right Boyle s Law When the amount of chemical substance moles of gas n and the temperature T are constant the pressure of a gas is inversely proportional to its volume p 0c 1 V where 0c represents proportionality Relating an initial state defined by p1 V1 to a final state defined by p2 V2 Boyle39s law requires that p1V1 pZVZ when n and Tare constant The conditions must apply and must be My m when using Boyle39s Law to solve a problem Because the product pV is constant the plot of pressure vs volume yields a hyperbola Figure 1 top left is a graphical representation of Boyle39s Law at three different temperatures 0 C 100 C and 200 C pV is constant at constant temperature but the product increases with an increase in temperature The three constanttemperature plots are called isotherms Topic 02 Ideal Gases Page 1 Charles Law When the number of moles n and the pressure p are held constant the volume occupied by a gas V is directly proportional to the absolute temperature T Voc T Relating initial and final states Charles39 law requires that VlTl VZTZ when n and p are constant The use of Charles39 Law requires that the temperature be expressed on an absolute temperature scale Kelvin Figure 1 top right is a graphical representation of Charles39 Law at three different pressures 100 125 and 150 atm AVAT is constant at constant pressure but the ratio decreases with an increase in pressure The constantpressure plots are called isobars GayLussac s Law When the number of moles n and the volume V are held constant the pressure of a gas p is directly proportional to the absolute temperature T p 0c T Relating initial and final states GayLussac39s law requires that plTl pZTz when n and Vare constant The use of Gay Lussac39s Law also requires that the temperature be expressed on an absolute temperature scale Kelvin Figure 1 bottom left is a graphical representation of GayLussac39s Law at three different volumes 14 22 and 30 L ApAT is constant at constant volume but the ratio decreases with an increase in volume The constantvolume plots are called isochores Pressure Volume and Relative Temperature Scales It should be clear from the discussion above that an absolute temperature scale K must be used when applying either Charles39 Law or Gay Lussac39s Law Failure to do so will yield invalid results With that reminder you may be surprised to know that pressure and volume both have a linear relationship with relative temperature scales eg C or F when the other state variables are held constant This is verified in Figure 2 where volume V is plotted vs temperature in C with p and n constant left and pressure p is plotted vs temperature in C with Vand n constant right These are similar to two of the plots in Figure 1 with the exception of the temperature scale used for abscissa The darker line segments represent experimental data and the lighter extended lines represent an extrapolation of the data to zero volume left and to zero pressure right 125 aim Q1 A c on quotl 00 atm Volume L e e 1 50 atml pressure atm N 4 D EZDO 0 200 400 2cm o 200 400 Temperature 3 Temperature DC Figure 2 The relationship between volume left and pressure vs temperature in C The plots do not pass through the origin because temperature values are not based on an absolute scale The Celsius temperature equivalent of absolute zero is given by the Celsius temperature that corresponds to zero volume left or zero pressure right This is represented by the point where all three extended lines meet on the temperature axis Topic 02 Ideal Gases Page 2 The information shown in Figure 2 can be used to determine the value of absolute zero for the relative temperature scale being used Based on the gas laws the volume must go to zero as the absolute temperature goes to zero left likewise the pressure must go to zero as the absolute temperature goes to zero right This means that absolute zero corresponds to the temperature reading where V is extrapolated to zero left or where p is extrapolated to zero right For an ideal gas the value of the Xintercept will be the same regardless of the constant pressure value left or the constant volume value right In both plots the xintercept is observed at 27315 C Avogadro s Law When the pressure p and the temperature T are held constant the volume occupied by a gas V is directly proportional to the number of moles n Voc n Relating initial and final states Avogadro39s law requires that V1n1 Vznz when p and Tare constant Figure 1 bottom right is a graphical representation of Avogadro39s Law when p 1 bar and T 29815 K Molar Volume The molar volume Vm is given by the total volume occupied V per mole of gas present n Vm Vn Avogadro39s Law requires that the molar volume of an ideal gas remain constant at constant Tand p The molar volume of a gas is an important thermodynamic quantity and is often substituted into thermodynamic equations Note that the units of molar volume are L mol39l The reciprocal of molar volume nV has units of mol L39l and represents the concentration of a gas The Combined Gas Law The four individual gas laws can be combined into a unified ideal gas law One way of stating the combined law is that the product of pressure and volume pV must be proportional to the product of moles and absolute temperature nT pVoc nT Rearranged the combined law requires that the ratio pVnT be constant pVnT R The constant R is a physical constant known as the gas constant Its value is tabulated in many units Two of the most commonly used values are R 831447 J K391 mol391 and R 820574 x 10392 L atm mol391 K39l The best value to use is determined by the data that is provided in any given problem The ideal gas equation of state can be written pV nRT The molar volume can also be substituted in the equation to yield me RT Relating any pair of initial and final states the law requires that p1V1n1T1 pZVZnsz You will note that the combined law does not require any assumptions about one or more of the state variables being held constant Can the Ideal Gas Law be used for Real Gases Recall that an ideal gas is one in which there are no intermolecular interactions between gas particles Although real gases do have intermolecular interactions real gases are well described by the ideal gas laws as long as the interactions are minimal This will generally be true at low pressures andor at high temperatures For most problems and calculations in this course you will assume that a given gas behaves ideally unless information is provided in the problem that suggests otherwise Standard Temperature and Pressure Conditions for Gases There are two common conventions for defining standard conditions Note that the defined pressures should be considered exact STP Standard Temperature and Pressure T 27315 K 0 C and p 1 atm SATP Standard Ambient Temperature and Pressure T 29815 K 2500 C and p 1 bar Topic 02 Ideal Gases Page 3 Relating Mass Density 0 to the Molecular Mass M of an Ideal Gases The mass density of a gas is given by 0 mV where 0 represents density m represents mass and V represents the occupied volume To apply this in the ideal gas equation we must relate moles to mass n mM Substituting the rightside expression for moles into the ideal gas law pV mMRT Isolating the molecular mass on the left side of the equation M mRTpV Finally the density can be substituted for mVin the rightside expression M pRTp This result which you should be able to m makes it possible to calculate the molecular mass of a pure gas sample given its density absolute temperature and pressure The average molecular mass will be calculated if a mixture of gases is present Dalton 5 Law Partial pressures in gas mixtures Partial Pressures in Gas Mixtures Dalton39s Law states that in a mixture of nonreacting gases the total pressure of the mixture is given by the sum of the partial pressures of the individual gases In a N system with N different gases Dalton39s Law is represented by the equation 9T 2 2p where pr is the i1 total pressure and p is the partial pressure of the ith gas Dalton39s law assumes ideal gas behavior and will not be perfectly followed in systems having significant intermolecular interactions In this course you will use Dalton39s Law for all mixedgas problems It follows that the partial pressure of the ith gas will follow the ideal gas equation of state p nRTV Note that Tand V are systemwide variables only n is specific to the individual gas for which p is calculated Mole Fractions y in Gas Mixtures Dalton39s law can be used to express partial pressures in terms of mole fractions The total gas pressure is given by pr nTRTV where nT represents the total number of all gas moles present Combining the last three equations N N RT RT N RT N N pT 2L 1 Zni 2 nT Z 3 pTZEyiM 2 2p 1 Substitution is i1 V V i1 V i1n T i1 made for p 2 The constant values R T and V are pulled outside of the summation 3 The right side is multiplied by nTnT one inside and one outside of the summation 4 The gasphase mole fraction y is substituted for its definition nnT as is pr for the expression outside the summation 5 The Dalton39s Law equation is used for pr Because the expressions in 4 and 5 must be equal term by term it follows that p ypT Therefore the partial pressure of a gas is equal to its mole fraction in the gas mixture multiplied times the total pressure DISCUSSION QUESTIONS AND FOLLOWUP PROBLEMS 21 Review each of the four individual gas laws 1 Name the law 2 Identify the two active state variables 3 Describe the relationship between the two active variables both qualitatively and quantitatively 4 Specify the state variables that must be held constant Topic 02 Ideal Gases Page 4 22 Starting with the combined ideal gas law derive each of the four individual gas laws 23 Calculate the molar volume Vm of an ideal gas at SATP 24 What fundamental characteristic must be true for a gas to behave ideally Identify and explain the conditions under which real gases are expected to behave ideally 25 A sample of an ideal gas is initially in a state described by V 227 L p 214 atm T 300 K If the gas is heated to 500 K and the volume is expanded to 253 L what is the final pressure of the gas 26 On the planet Geekos temperature is measured in degrees Gekiwidth G A fixed amount of an ideal gas was studied at constant volume and the following pressure in units of Gekibars gb vs temperature observations were recorded T G 500 230 120 200 pgb 442 472 511 543 1 Based on the data alone is it possible to determine whether or not the Gekiwidth scale is absolute Justify your answer 2 Graphically determine the Gekiwidth temperature that corresponds to absolute zero 27 A gas mixture having a total pressure of 125 bar was analyzed and found to have the following composition expressed in mass percent 197 He 305 N2 and 498 Ar Calculate the partial pressure of each gas in the mixture 28 The density of a pure gas was measured to be 104 g L391 at 4200 K and 273 atm 1 Calculate the molecular weight of the gas 2 Identify the gas on the assumption that it is either a pure element or a common gas Topic 02 Ideal Gases Page 5 Topic 03 Real Gases Intermolecular Interactions and The Compression Factor Z Ideal gas molecules are free of intermolecular interactions Real gas molecules do interact A typical particleparticle interaction is illustrated in Figure 1 in which the energy of interaction is plotted vs the distance between two molecules The horizontal dashed line indicates where the interaction energy is zero ideal When molecules are too close to each other the interaction is repulsive The strongest attraction occurs where the energy is at a minimum At larger distances the interaction energy remains attractive but decreases in magnitude as the distance increases Repulsive forces dominate Energy i I 39 Attractive forces dominate 39 Distance between molecules Figure 1 Intermolecular Interaction Energy as a Function of Distance The dashed line represents where the energy is zero separating attractive energies below from repulsive energies above The ideal gas equation of state is given by 9V 2 nRT 1 me 2 RT 2 where 2 Wquot Wquot substitutes Vm for Vn in 1 The equations can be rearranged to give 2 quot 1 Here the nRT RT superscript 0 emphasizes that Vand Vm are ideal in these relationships V V For real gases we use the same ratio to define the compression factor Z Z E p h nRT RT It should be clear that Z 1 for an ideal gas For real gases the value of Z depends on whether the primary interactions in a gas are attractive or repulsive Z gt 1 indicates that repulsive interactions dominate If Z gt 1 it implies that Vm gt Vm0 at the same temperature and pressure When the real molar volume is larger than ideal the gas particles are spreading out due to mutual repulsion Z lt 1 indicates that attractive interactions dominate IfZ lt 1 it implies that Vm lt Vm0 at the same temperature and pressure When the real molar volume is smaller than ideal the gas particles are coming closer together due to mutual attraction Topic 03 Real Gases Page 1 Compression factors plotted as a function of pressure are shown in Figures 2 and 3 Figure 2 shows Zfor four different gases at a fixed temperature of 273 K As pressure increases Zfirst decreases to a minimum for C02 02 and N2 and then increases thereafter At very high pressures the repulsive forces dominate Z gt 1 H2 is different from the other three gases in that Z gt 1 at all pressures Eomlpreeseion Factor varioue gases at 2T3 H 939 l39 in HT ID Elli 4m EUU BEND Preseurel Ear Figure 2 Compression Factor of Four Gases at 273 K httpwwwch mguidecoukphvsicalktrealgaseshtml 14 August 2014 In Figure 3 the compression factor is plotted as a function of pressure for the same substance air at six different temperatures The dashed horizontal line represents Z 1 The Boyle temperature TB is defined as the temperature for which Z 1 initially at low pressures At the Boyle temperature the initial value of the slope must be zero dZdp 0 Among the temperatures plotted in Figure 3 300 K is closest to the Boyle temperature for air P 0 Compression Factor of Air Z 3 3 0 100 200 300 400 500 pmm Figure 3 The Compression Factor of Air plotted vs pressure Each of the six plots represents a different air temperature Topic 03 Real Gases Page 2 Virial Equations of State The Virial Equation describes the states of real gases One form of the equation uses a power B C series expansion in Vm39lz me 2 RT 1 V W Comparing this to the ideal gas law it should m m B C be clear that the terms V 2 represent a correction to the Ideal gas apprOXImatIon It should m m also be clear that the compression factor for this Virial expression is Z 1 V 2 The Virial Equation can also be expressed as a power series in p me 2 RT 1 B39p C39p2 In this form B39p C39p2 represents a correction to the ideal gas approximation and the compression factor is given by 1 B39p C39p2 In these equations the Virial coefficients B and C or B39 and C39 must be empirically determined in the laboratory The constants are specific to a fixed temperature for a given substance Although the volumebased coefficients differ from their pressurebased counterparts they are related as follows i B I C B2 B E and C The Second and ThIrd VIrIal coeffICIents of several gases at 29815 K are collected in Table 1 The first Virial Coefficient is always equal to 1 The data available in a given problem will determine whether the volume or pressurebased coefficients should be used Table 1 Second and Third Virial Coefficients of Several Real Gases at 29815 K M mol L391 Gas B 10393 M391 C10396 M392 B3910394 atm391 C3910398 atm392 H2 141 350 576 253 He 118 121 482 305 N2 45 1100 184 180 02 161 1200 658 157 Ar 158 1160 646 152 CO 86 1550 352 247 Volume Expansion Pressure Expansion When Z is plotted vs pressure as in Figures 23 the slope is given by the derivative dZdp evaluated at the pressure of interest Although the Virial power series expansion does not provide an intuitive understanding regarding the nonideal behavior of gases it does provide very accurate analytical results Topic 03 Real Gases Page 3 The van der Waals Equation of State The van der Waals Equation of State for real gases is not quite as accurate as the Virial approach but it is much more intuitive in terms of describing the nature of nonideal interactions The 2 nRT n RT a equation can be expressed In two forms 19 a or p 2 van der Waals V nb V Vm b m coefficients are different for each gas but are independent of temperature van der Waals coefficients are collected for several gases in Table 2 The coefficient b in the denominator of the first term can be interpreted as the effective molar volume that is occupied by real gas molecules The ideal gas approximation views individual molecules as being point masses that occupy zero volume Subtracting b from Vm corrects for this This correction has the effect of increasing the calculated pressure relative to the ideal Stated in different terms Vm b represents the effective available molar volume The second term represents attractive interactions which have the effect of decreasing the calculated pressure Because concentration is the reciprocal of molar volume the attractive term is proportional to the square of the concentration which is in turn proportional to the frequency of collisions between particles Table 2 Van der Waals Coefficients of Several Real Gases Gas aatm M392 blO392 M391 Ar 1337 320 CH4 2273 431 C2H4 4552 582 C02 3610 429 Hz 02420 265 He 00341 238 NH3 4169 371 N2 1352 387 02 1364 319 Xe 4137 516 The van der Waals equation can be rearranged into two forms convenient for determining the RT molar volume given pressure and temperature The rearrangement Vm b can be used for a 19 solving for Vm via the method of successive approximations The cubic equation pVn3 bp RTan aVm ab 0 can be used in a polynomial curve fit using a programmable calculator Topic 03 Real Gases Page 4 Isotherms of Real Gases it D THE F V Figure 4 Isotherms of Real Gases and van der Waals Gases httpenwikipediaorgwikiReal gas Isotherms of a real gas are depicted in Figure 4 The isotherm temperatures increase as pressure increases At the highest temperature top light blue curve the isotherm mimics that of an ideal gas As temperature decreases nonideal behavior begins to set in The critical temperature is the highest temperature at which the liquid can exist as a distinct phase At this temperature the isotherm red has an inflection point through the critical point K but never condenses to a liquid At temperatures below the critical temperature the gas begins to condense at the point labeled G the molar volume of the vapor point G decreases to the molar volume of the liquid point F at constant pressure Therefore the horizontal line connecting G to F in the lower two isotherms identifies the transition pressure and the distinct molar volumes of the vapor and of the liquid When plotting the same isotherms using the van der Waals equation quotloopsquot are observed in the region of the phase transition The quotloopsquot consist of the segment of the curve connecting the points GCBAF The quotloopsquot are an artifact of the van der Waals equation and are nonphysical Even so the phase transition can be identified by connecting the outside points of the loop G F with a straight horizontal line Critical Values and the Principle of Corresponding States The critical temperature TC is the highest temperature at which the liquid can exist as a distinct phase At this temperature the liquid will be in equilibrium with the vapor at a specific pressure known as the critical pressure pc The molar volume of the equilibrium mixture is called the critical molar volume lec The equilibrium state defined by TC pc and lec is called the critical point Topic 03 Real Gases Page 5 Sometimes it is convenient to use what are referred to as reduced variables Reduced variables are defined as the ratio of the variable value divided by the corresponding critical value For example the reduced temperature T is defined as Tdivided by the critical temperature T TTC In like manner p ppc and V Vmlec When expressed in terms of reduced variables all gases in the same state behave identically This is an expression of the Principle of Corresponding States The implication of the principle is that the equation of state for all gases is identical when reduced variables are used ST The reduced equation of state for a van der Waals gas is given by pr 2 3V 1 W Note r that the equation is general and does not depend on a set of gasspecific coefficients Math Pointers The Method of Successive Approximations Many algebraic equations can be solved via the method of successive approximations Before using the method the relevant equation must be rearranged such that the target variable is isolated The right side of the equation will also include the target variable For solving the molar volume of a van RT a V2 m der Waals gas the rearranged equation is Vm b p Success of the method often depends on selection of a reasonable guess of the true value In this example a reasonable starting guess would be the molar volume of an ideal gas Vm RTp Calculation Cycle 1 Substituting your guess value on the right hand side calculate a new hopefully improved value of the target variable on the left hand side 2 Compare the two values If the new calculated value is equal to the guess value you are finished You have determined the value of the variable 3 If your new value is not equal to the old guess value run another calculation cycle beginning at 1 In the new cycle use the variable value that you just calculated as your improved guess value Repeat the cycle until the guess input value and the calculated output value are equal to the appropriate number of significant digits Topic 03 Real Gases Page 6 DISCUSSION QUESTIONS AND FOLLOWUP PROBLEMS 31 32 33 34 35 36 37 38 39 310 311 312 Starting with the definition of Z explain why a gas with Zlt 1 will be dominated by attractive forces A Which one of the gases in Figure 2 has the strongest intermolecular interactions Explain your answer B Compared to C02 would the minimum value onfor NH3 be lower higher or the same Justify your answer What would the fourth term of the expansion look like in the Virial equation expressed in terms of p A Using the data in Table 1 calculate p for CO when Vm 500 L mol391 at 29815 B Calculate Vm for CO at 29815 K when the pressure is equal to the value you calculated in A Is the result what you expected What is the initial slope of a sz p plot for Ar treated as a Virial gas A Using the Virial equation calculate Vm for Ar at 29815 K when p 300 atm B Use the result to calculate Z under the same conditions Discuss the differences between the Virial Equations of State and the van der Waals Equation of State Discuss the physical interpretation of the two terms in the van der Waals equation A Using the van der Waals equation calculate p for Ar at 29815 K when Vm 602 L mol39l B Use the result to calculate Z under the same conditions The pressure and temperature of a Xe sample are 9000 K and 1100 atm Calculate Vm for Xe in the sample treating it as a van der Waals gas Do this two ways 1 by the method of successive approximations and 2 using a polynomial solver on your calculator What are the smallest and largest permitted values for the compression factor Z Beginning with the van der Waals equation derive the two rearranged forms of the equation used in problem 310 Topic 03 Real Gases Page 7 Topic 04 The First Law of Thermodynamics System Surroundings and Universe System The system is the specific part of the physical universe that is of interest It is the focus of study Surroundings Everything that exists outside of the system Of particular interest in thermodynamics is the subset of the surroundings that interacts with the system Universe System plus Surroundings In principle this includes the entire cosmos Figure 1 The thermodynamic universe consists of the system and its surroundings Kinds of Systems Three kinds of thermodynamic systems are common They are defined in terms of whether matter andor energy can be transferred between the system and its surroundings These properties are summarized in Table 1 Table 1 Three kinds of thermodynamic systems defined in terms of their ability to exchange matter andor energy with their surroundings Kind of Can Matter be Transferred Can Energy be Transferred between System and between System and Example System Surroundings Surroundings Open YES YES open coffee cup Closed NO YES closed can of soda Isolated N0 N0 perfect thermos bottle Kinds of Boundaries Separating System from Their Surrounding Two kinds of system boundaries are common They are defined on the basis of whether heat transfer is possible between system and surroundings Diathermic Boundary Permits the transfer of heat between the system and its surroundings Topic 04 The First Law of Thermodynamics Page 1 Exothermic process heat is released from the system and sent into the surroundings Endothermic process Heat is absorbed by the system from its surroundings Adiabatic Boundary Does not permit My transfer of heat between the system and its surroundings Work Heat and Internal Energy Energy In classical physics energy is a property of matter that can be transferred between objects and converted in form but can neither be created nor destroyed In elementary physics energy is defined as the quotcapacity to do workquot Internal Enerqy U n thermodynamics the internal energy is the total energy of a system The absolute internal energy cannot be measured in the laboratory but changes in internal energy AU can be experimentally measured Internal energy is both a state function and an extensive function see below Energy Transfer between System and Surroundings Energy can be transferred between a system and its surroundings either in the form of heat q or in the form of work w Heat g Heat is the disordered transfer of energy On a microscopic level the energy transfer results in an increase of disordered molecular motion Work w Work is the orderly transfer of energy that is results in orderly molecular motion There are many forms of work mechanical work electrical work and expansion work to name only three n physics work a scalar is defined as the vector dot product between force and the distance through gt gt which the force operates w F x The equation can also be expressed as w F xcos6 where F and X represent the magnitude of the respective vectors and 9 is the angle between the force and distance vectors The negative sign reflects that positive work requires displacement against an opposing force Classifications of Thermodynamic Functions Thermodynamic functions can either be state functions or path functions State functions For a given state of the system the value of a state function is unique and fixed Changes in state function values depend only on the initial and final states and do not depend on the path that transforms the system from the initial state to the final state The internal energy U is a state function Therefore AU Uf U is always the same regardless of the path that is taken to transform the system between the two states Path functions Thermodynamic functions that depend directly on the path that is taken are called path functions Work w and heat q are both path functions Different paths between the same initial and final states will result in different values for w and q Topic 04 The First Law of Thermodynamics Page 2 State functions and variables can also be classified as either extensive or intensive An intensive variable or function is independent of the amount of substance present Temperature is an example of an intensive state function When volume is defined in terms of the physical space occupied as it is for gases volume is also an intensive variable An extensive variable orfunction depends on the amount of substance present The internal energy U is an extensive variable if the number of moles present is doubled the result is to double the internal energy For systems of gases the pressure is also an extensive variable because doubling the number of moles of gas in a fixed volume at constant temperature will double the system pressure The First Law of Thermodynamics The first law of thermodynamics states that whenever energy is transferred between a system and its surroundings the change in the system39s internal energy AU must be equal to the sum of all work that is done on the system w plus the heat that is transferred into the system q In equation form the first law says AU w q Sign Conventions in Chemical Thermodynamics A positive sign is used when the energy increases in the system A negative sign is used when the energy decreases in the system AU gt 0 the internal energy of the system has increased Energy was transferred from the surroundings into the system ggt0 energy was transferred into the system in the form of heat wgt0 energy was transferred into the system in the form of work Work was done on the system by the surroundings AU lt 0 the internal energy of the system has decreased Energy was transferred from the system into the surroundings glt0 energy was transferred out of the system in the form of heat wlt0 energy was transferred out of the system in the form of work Work was done on the surroundings by the system Conservation of Energy and the First Law of Thermodynamics Energy is neither created nor destroyed The energy change in the universe must be zero AUumv 0 This is equal to the sum of AU in the system and AU in the surroundings AUumv AU AUSW Therefore the change in the internal energy of the surroundings must be equal in magnitude and opposite in sign to the change in internal energy of the system AUSU AU Note that a subscript is used to designate changes in the surroundings and in the universe no subscript is used for changes in the system The work done on the surroundings must be equal in magnitude and opposite in sign to the work done on the system wsu w Topic 04 The First Law of Thermodynamics Page 3 The heat transferred into the surroundings must be equal in magnitude and opposite in sign to the heat transferred into the system qsu q DISCUSSION QUESTIONS AND FOLLOWUP PROBLEMS 41 Review all of the terms and definitions in this topic A System Surroundings and Universe B Open Closed and Isolated Systems C Diathermic and Adiabatic Boundaries D work heat and Internal Energy E State and Path functions F Intensive and Extensive variables and functions G The First Law of Thermodynamics 42 Relate the following A AU q and w B AU and AUSW C q and qsw D w and WW 43 Review the sign conventions for AU q and w When are the signs positive and when are they negative 44 For each of the following scenarios identify the correct values of the following six quantities AU q w AUSW qsw and WW A A machine system does 17 of work on the surroundings and gains 12 J of heat in the process B A gas system is compressed by the surroundings via 17 J of work and gains 12 J of heat in the process C A gas system does 17 J of work on the surroundings and in the process transfers 12 J of heat to the surroundings 45 A system undergoes a twostep process SteL1 The system starting in State A is transformed into State B The work and heat in the process are given by W1 and q1 Step The same system starting in State B is transformed into State A The work and heat in the process are given by w and q2 In the overall twostep process what relationships between W1 wz q1 and q2 must always be true 46 Is it practical to talk about the surroundings given that the surroundings extend to the outer reaches of the cosmos Justify your answer 47 Identify the following as state or path variables In all three cases a hiker starts at location A and hike up a mountainside until reaching location B 1 The distance traveled by the hiker 2 The change in altitude from start to finish 3 The geometrical distance between the starting and ending locations Topic 04 The First Law of Thermodynamics Page 4 Topic 05 Expansion Work Expansion of a Gas against an External Pressure The General Case Pressurevolume work is one of the most important varieties of work that can be obtained from systems of gases Consider a sample of gas that is initially confined inside a cylinder fitted with a moveable piston The circular area of the cylinder s base is given by A The external pressure pext pushes against the piston to hold it in place Let us first consider work done when the external pressure is constant and A2 is macroscopic and measureable Let the system initially be set up as shown below left The gas inside the system then pushes the piston out by a distance A2 against the external pressure doing work on its surroundings pext Pent Recall that work is defined by the vector dot product of the opposing force and the distance moved Let us initially consider the work done on the surroundings in this example First recall that pressure is force per unit area p FA therefore F A pext Because the external pressure provides the opposing force the external pressure must be used in this expression Work on the surroundings is given by wsu FAz where F represents the magnitude of the force that opposes expansion Substituting the first expression for force into the equation for work wsu pextAAz Because A412 is equivalent to the change in the cylinder39s volume AV we can write wsur pext AV Finally we express the result in terms of work done on the system Recalling that w wsu the work on the system is w pext AV Note carefully this result is not general It is valid only when pext is constant for the entire volumechange process Topic 05 Expansion Work Page 1 The General Expression for PressureVolume Work To generalize the results we must drop the assumption that pext is constant This means that we must visualize the process as a sequence of continuous infinitesimal changes To do this we must rewrite the equation for work in terms of differentials dw pext dV n words the infinitesimal change in work on the system dw is equal to the infinitesimal change in volume dV multiplied by pext the external pressure specifically corresponding to the infinitesimal volume change This result is general For expansion pressurevolume work dw pext dV This equation should be your starting point for all problems that involve pressurevolume expansion work Three Special Cases of PressureVolume Expansion To determine the value of work done on the system the differential equation must be V2 integrated and solved Idw w Ipm dV First note that the integral of dw is simply equal to the V1 total work w The volume integral must be evaluated over the limits from V1 initial to V2 final If the value of pext is dependent on Vfor the process in question a substitution expressed in terms of V must be made for pext in the integrand before the integral can be evaluated and solved NOTE CAREFULLY to correctly solve problems involving expansion work it is essential that you first identify and understand the nature of the physical process This understanding will determine what expression you will use for pext in the integral Examples of three specific expansion processes are given below Free Expansion Free Expansion refers to the expansion of a gas into a perfect vacuum Stated differently the external pressure against which the gas expands is equal to zero pext 0 V2 To solve for work begin with the general relationship derived above w I pm dV Next VI V2 substitute the correct expression for pext for free expansion w dV 2 0 Because the VI integrand is zero the integral is equal to zero The work associated with a free expansion is equal to zero Expansion against a Constant External Pressure Expansion against constant external pressure means that the external pressure is fixed and is independent of changes in the volume of the system Topic 05 Expansion Work Page 2 V2 Start with the general expression w J39pm dV When external pressure is constant with VI V2 respect to volume pext can be pulled outside of the integral w pm dV Evaluating the integral is VI elementary w pm AV where AV V2 V1 The work associated with expansion against a constant external pressure is equal to pext AV Warning Do NOT memorize this equation All equations used to calculate work should be derived for every single problem beginning with the general expression Reversible Isothermal Expansion of an Ideal Gas Before discussing how the expression for work is derived in this kind of expansion it is important to introduce andor review the meaning of each specific condition There are three conditions all of which must be met for the derived equation to be valid Reversible For a reversible process the consequence of an infinitesimal change in the expansion conditions will be to reverse the direction of expansion Suppose that a system is expanding reversibly against its surroundings If the external opposing pressure is then increased infinitesimally the external pressure will be sufficient to reverse the expansion effecting compression of the system A reversible expansion requires the condition of mechanical equilibrium at each and every infinitesimal point of the expansion Equivalently stated the external pressure must be equal to the pressure of the system at all points of the expansion pext p Therefore a valid expression for the system pressure p can be substituted for pext whenever the expansion is reversible Isothermal The temperature T is constant Ideal Gas The system obeys the ideal gas equation of state pV nRT equivalently p nRTV To derive the expression for expansion work always begin with the general relationship V2 Idw w I pm dV 1 As explained above the external pressure must be equal to the system V1 V2 pressure at all times for a reversible expansion Therefore substitute p in place of pext w I p dV VI 2 For an ideal gas assumed the equivalent ratio nRTV can be substituted in place of p V2 nRT w V dV 3 For an isothermal process Tis constant Both n and R are also constant V1 Topic 05 Expansion Work Page 3 V 2 dV 4 Pull the constants outside of the integral w nRTI7 5 Evaluate the definite integral at its VI V limits Generally ldxx nx C Evaluated at the limits the final expression is w nRTln 1 The work associated with expansion of a reversible isothermal ideal gas is nRTInV2V1 Warning Do NOT memorize this equation All equations used to calculate work should be derived for every single problem beginning with the general expression Graphical Representation of Expansion Work In calculus an integral evaluates to give the area under a curve For example if dz y dx then 9 2 z Iydx If y is plotted as the ordinate vs X as the abscissa z is equal to the area under the curve 7 1 between x1 S X S x2 V2 The application to expansion work is straightforward When w I pm dV plot pext vs V VI The area under the curve between V1 and V2 is equal to minus the work done on the system ie is equal to the amount of work done by the system on its surroundings This is illustrated in Figure 1 where external pressure though not subscripted is plotted vs volume for an ideal gas Two different paths are shown the initial and final volumes of which are 20 L and 35 L The top path represents a reversible isothermal expansion The lower path represents expansion against a constant external pressure which is equal to the final pressure of the first expansion The work done on the surroundings is equal to the area under the relevant curve for 20 L S V S 35 L When the final external pressures are the same the figure demonstrates that the reversible process accomplishes more work on the surroundings than what is accomplished by the irreversible constantpressure process 14 12 10 08 05 04 02 00 pressure I atrn is 20 25 30 35 volume l L Figure 1 PressureVolume Plot for a Gas The external pressure though not explicitly subscripted is plotted vs volume of the gas The area under the curve is equal to the amount of work done by the system on its surroundings wsur Topic 05 Expansion Work Page 4 Principles for Solving Problems Involving Work Do NOT use memorized equations or formulas Work is a path function You cannot calculate work correctly if you do not first identify the conditions path of the expansion Evaluate and understand the conditions of the expansion Identify whether or not pext varies with V If so how If you do not correctly identify the conditions all of your work will be invalid V2 Starting with the general relationship Idw w pm dV derive the appropriate Vl expression for expansion work based on the relevant conditions Evaluate the integral after making the appropriate substitution for pext and after identifying all variables that are constant DISCUSSION QUESTIONS AND FOLLOWUP PROBLEMS 51 52 53 54 55 A Review the derivation of the general differential expression for expansion work dw pext dV B Explain each step C Explain why the external pressure must be used in the equation For one of the specific expansion conditions discussed the result was w pext AV If pressure is in units of atm and volume is given in L the resulting units will be L atm Convert L atm to units of J the normal units for work Show all conversion factors For each of the following three expansion conditions start with the general differential equation Derive the final expression for work by applying the specific characteristics of the expansion A Free expansion B Expansion against constant external pressure C Reversible isothermal expansion assuming ideal gas behavior For each of the following three processes identify the specific conditions of the expansion Starting with the general differential equation derive the expression for work Then calculate the work done on the system expressing your answer in standard energy units of or k A Three moles of gas initially occupy 200 L at 298 K The sample is then freed to expand into vacuum B Three moles of gas initially occupy 200 L at 298 K The gas then expands isothermally against a constant ambient pressure of 0988 atm until the system attains mechanical equilibrium with the external pressure C Three moles of an ideal gas initially occupy 200 L at 298 K The gas then expands reversibly and isothermally until it reaches a final pressure of 0988 atm For the reversible isothermal expansion of an ideal gas prove that w nRTln p1 Justify your answer Topic 05 Expansion Work Page 5 56 Derive the equation needed to calculate work for a reversible isothermal expansion of a nRT n 2 van der Waals gas The van der Waals equation of state IS 19 V b a V n 57 Calculate the work done on the system expressing your answer in standard energy units of or k for the following process Three moles of Ar treated as a van der Waals gas initially occupy 200 L at 298 K The gas then expands reversibly and isothermally until it reaches a final volume of 7425 L The van der Waals constants of Ar are Gas aatm M392 b10392 M391 Ar 1337 320 Compare your results with 54 C Topic 05 Expansion Work Page 6 Topic 06 ConstantVolume Calorimetry Constant Volume Processes with no quotExtraquot Work The first law of thermodynamics states that AU q w We will now explore how this relationship is affected in a special case when the volume is constant and no quotextraquot work is done By quotextraquot work we mean any work that is different from pressurevolume expansion work In the preceding topic we derived the general differential equation for expansion work dw pext dV In a constant volume process there are no changes in volume Therefore AV 0 and dV 0 making the right side of the equation for dw equal to O The conclusion is straightforward In a constantvolume process the expansion work is equal to zero When no other kinds of work are permitted the fact that w 0 implies that AU q We use a subscript to emphasize that the process must be carried out at constant volume AU qv This is a very important result and is the starting equation for many calculations You should spend enough time to completely understand this relationship In general q is a path function However note that qv is equal to the change in a state function AU This implies that there is one and only one possible value of qv for a process that occurs at constant volume when no extra work is done The Measurement of AU via Adiabatic Bomb Calorimetry The relationship AU qv implies that AU can be directly measured via an experiment in which the system volume is held constant The laboratory apparatus used for making such measurements is called an adiabatic bomb calorimeter A calorimeter is a laboratory device used to measure the amount of heat transferred in the course of a chemical reaction or a physical change By definition heat cannot flow through an adiabatic boundary In this case the adiabatic boundary separates the calorimeter39s universe containing its own system and surroundings from the external laboratory surroundings This guarantees that no heat is lost into the laboratory surroundings and that all heat transferred out of the reaction system is completely captured by the water bath surroundings The bomb refers to the rigid constantvolume vessel typically made of steel in which the system reaction occurs Figure 1 contains a schematic of an adiabatic bomb calorimeter Inside of the calorimeter is the system contained within a rigid constantvolume bomb The bomb39s immediate surroundings typically contain a water bath Adiabatic walls surround the water bath ensuring that heat will not be lost into the laboratory surroundings The system reaction occurring inside the bomb is exothermic transferring heat to the water bath and causing a temperature increase in the water bath Topic 06 ConstantVolume Calorimetry Page 1 adiabatic we is 533mm l litml calorimeter Figure 1 Schematic of an Adiabatic Bomb Calorimeter Using Calorimetry Data to Determine AU for a Reaction or Physical Change The heat transferred into the water bath surroundings is given by qsu CsurATsur where ATM is the temperature change measured in the water bath and CM is its heat capacity CM is referred to as the calorimeter constant and the subscript is commonly omitted For the system q qu therefore q 39 CATsur In principle q includes all sources of the heat that is transferred For any reaction under study qv AU because the reaction is carried out at constant volume If any other substances are present and undergo combustion the heat they produce must also be included Typically a small length of ignition wire must be supplied for the purpose of initializing the combustion reaction In that case the total heat transferred with respect to the system is given by q AU qW 2 where qW represents the heat generated via combustion of the ignition wire Combining equations 1 and 2 AU qW C ATsur Rearranging for AU gives AU C ATS qw 3 Note that AU as calculated via equation 3 is extensive The calculated value is proportional to the amount of sample that was combusted To convert the result to a molar change in internal CAT energy AUm AUn divide both sides by the number of moles n AUM 4 The n signs of AUM and qW will always be negative because heat is released goes out of the system into the surroundings when combustion occurs Calibration of the Calorimeter Before equation 4 can be used the value of the calorimeter constant C must be determined This process is referred to as calibration Calibration involves the combustion of a substance for which AUm is accurately known The normal standard is benzoic acid C5H5COOHs MW 1221213 for which AHm is 3228 i 4 kJ mol391 httpwebbooknistgov accessed 28 August 2014 As we will show in the next topic at 29815 K this is equivalent to a AUm value of 3227 i 4 k mol39l Topic 06 ConstantVolume Calorimetry Page 2 AT sur AU Equations 1 and 2 can be combined and rearranged to solve for C C qw 5 Note that AU is extensive and is equal to the number of moles times the molar AU value AU n AUm The signs of AU and qW will always be negative the signs of ATM and Cwill always be positive Mathematical Tools Partial Derivatives Before discussing heat capacity further we must digress to discuss partial derivatives in mathematics When a function y depends only on one variable x the normal derivative dydx directly measures how an infinitesimal change in xwill affect the infinitesimal change in the function y When the dependent variable y depends on more than one independent variable X and z the a partial derivative directly measures how an infinitesimal change in xwill affect the x infinitesimal change in the function y when the independent variable 2 is held constant This course uses the partial derivative notation shown above A curly quot6quot is used in place of a regular quotdquot Subscripts are used outside of parentheses to explicitly specify the variables that are being held constant In this course it is mandatory that the variables being held constant must be explicitly be identified via subscripts Partial derivatives denoted without the correct subscripts are incorrect To evaluate a partial derivative identify all variables being treated as constant These can be pulled outside of the derivative Then follow the procedure for evaluating a normal derivative d377rx3 X Example Taking a normal derivative Consider Pull the constants outside of the 61W X derivative 37 7t Evaluate the derivative in x 37 763x2 Combine and simplify the results 1117rx2 8136z2 tany x3 8x Pull the constants Example Taking a partial derivative Consider yaZ outside of the partial derivative including the variables being held constant write the result as a m Evaluate the derivative in x 136Z2 tan 3x2 d x regular derivative in x 136zz tany Combine and simplify the results 39 ez2 tanyx2 Topic 06 ConstantVolume Calorimetry Page 3 The Heat Capacity Heat capacity is defined as the amount of heat that must be absorbed to raise the temperature of a sample by 1 OC Note that a temperature change of 1 0C is identical to a change of 1 K for changesI the temperature can be measured in 0C or in K The results will be the same The definition can be written in equation form C qAT For the bomb calorimeter the heat capacity was referred to as the calorimeter constant C is an extensive quantity because its value depends directly on the amount of substance present Illustration as the amount of water in a soup kettle is increased from 1 cup to 3 gallons the amount of heat that is required to raise the water temperature by 1 C increases proportionately Intensive Expressions for Heat Capacity The heat capacity can be expressed on a permole basis or on a permass basis The molar heat capacity Cm is the total heat capacity divided by the number of moles Cm Cn The specific heat capacity often just called the specific heat C5 is the total heat capacity divided by the mass usually in grams C5 Cm n standard data tables specific heats are much more common than molar heat capacities Both are examples of intensive variables Their value is unique and is constant for a given substance under specific conditions of temperature and pressure Heat Capacities Depend on Conditions The amount of heat required to raise the temperature of a system by 1 OC depends on conditions The heat capacity for constant volume processes CV is smaller in comparison to the heat capacity for constant pressure processes Cp Stated in different terms a given amount of heat will increase the temperature of a constantvolume system more in comparison to the temperature increase in a constantpressure system The constant volume heat capacity CV is defined as the partial derivative of internal energy U 6U with respect to temperature when volume is held constant CV To obtain the molar V constantvolume heat capacity divide by the number of moles Cum CVn For a monatomic ideal gas Cvm 32 R where R is the universal gas constant For a diatomic ideal gas Cum 52 R The constant pressure heat capacity Cp is defined as the partial derivative of the enthalpy H to be defined in the next topic with respect to temperature when pressure is held constant 8H Cp 2 The molar constant pressure heat capacity is Cpm Cpn For both CV and C note that P the subscript identifies the variable being held constant For an ideal gas Cpm Cum R Implications for the First Law U The thermodynamic definition of CV CV can be rearranged to give an important V result that is exactly true for ideal gases When the volume is constant the partial derivative can be Topic 06 ConstantVolume Calorimetry Page 4 U written as a regular derivative CV This can be rearranged to isolate dU dU 2 CV dT 6 This result is exact for an ideal gas it is very close for a real gas For problems in this course you will assume that the relationship is correct unless information in the problem suggests otherwise The method by which corrections are made for real gases will be discussed later The differential equation is solved by integration Assuming that the system goes from state 1 2 T2 to state 2 the solution is Id U 2 Q d T The left side of the integral is equal to AU If the heat 1 T1 capacity is temperaturedependent the expression for CV in terms of T must first be substituted on the right side before completing the integration This will be discussed in the next topic If C can be assumed constant with temperature it is pulled outside of the integral with the result AU CV AT Unless information is provided in a problem to suggests otherwise you will assume that C is constant and will use this relationship Note that that equation 6 applies to a system that is heated while remaining in the same phase The heat capacity is dependent on the phase of the system The equation does not apply to phase transitions which will be explored in a later topic One important implication of this relationship is that AU 0 for an isothermal process AT 0 in which there is no chemical reaction or phase change DISCUSSION QUESTIONS AND FOLLOWUP PROBLEMS 61 A Identify the specific conditions under which the amount of heat transferred is equal to AU for the process B Beginning with the First Law of Thermodynamics apply the conditions you have identified in A to derive the relationship AU qv Explain each step of your derivation 62 A Explain why an adiabatic bomb calorimeter can be used to measure AU for a chemical reaction or for a physical change B Derive with full explanation equation 5 which is used to determine the calorimeter constant C Derive with full explanation equation 4 which is used to determine AUM 63 In a bomb calorimeter 08028 g of benzoic acid was combusted To effect ignition 00041 g of ignition wire was also combusted The heat generated by combustion of the ignition wire is 1400 calg Note that 1 cal 4184 exactly The initial and final temperatures of the water bath were 2187 0C and 2403 0C respectively A Calculate qW the heat transferred from combustion of the wire in units of k B Calculate AT C Calculate AU for the actual process using AUm provided in the text for benzoic acid and the moles of benzoic acid combusted D Calculate Cfor the calorimeter Topic 06 ConstantVolume Calorimetry Page 5 64 An adiabatic bomb calorimeter was calibrated and the calorimeter constant C was measured to be C 1000 kJ OC39l 04711 g of benzoin C14H1202s MW 2122439 was combusted The amount of ignition wire combusted see problem 63 was 00078 g The final and initial temperatures of the water bath were 2616 0C and 2432 0C respectively A Calculate qW the heat transferred from combustion of the wire in units of k B Calculate AT C Calculate AU for the actual process D Calculate the experimental value of AUm for benzoin 65 Consider the functionfxyz TE cosx ezy 22 Evaluate the following three partial af derivatives simplifying your final answer by combining like terms A B y Z X r fl r fl 8y xz az xy 66 For a certain gas the internal energy U is given by U 4nRT Determine the expression for the molar constant volume heat capacity Cum for this gas 67 When heat is transferred between two objects the heat transferred into object A must be equal in magnitude and opposite in sign to the heat transferred into object B qA q3 Based on the definition of heat capacity C qAT it follows that q CAT This equation must apply separately to both objects involved in the heat transfer 5000 mL of water at 890 0C was mixed with 3000 mL of water at 195 0C Calculate the final temperature of the mixture after it has reached thermal equilibrium Although not quite exact you may assume that the density and specific heat of water are constant over the temperature range and have values of 0 0995 g cm393 and C5 418 K391 g39l 68 A 9000 g block of Fes at 900 0C is placed in 2500 mL of water at 250 0C Assume that the specific heat of Fe is constant and equal to C5 0449 J K391 g39l Calculate the final temperature after thermal equilibrium has been reached For water use the information provided in question 67 69 250 moles of an ideal gas Cum 32 R are initially confined in a volume of 500 L at 29815 K The sample is then expanded reversibly and isothermally until the final volume is 150 L 1 Describe the conditions of the expansion and derive the equations necessary for calculating w q and AU 2 Calculate w q and AU for the process 610 250 moles of an ideal gas Cum 32 R are initially confined in a volume of 500 L at 29815 K The sample is then allowed to expand isothermally into vacuum 1 Describe the conditions of the expansion and derive the equations necessary for calculating w q and AU 2 Calculate w q and AU for the process Topic 06 ConstantVolume Calorimetry Page 6 611 612 613 614 250 moles of an ideal gas Cum 32 R are initially confined in a volume of 500 L at 29815 K The sample is then expanded isothermally against the external ambient pressure of 0985 atm until it achieves mechanical equilibrium with the surroundings 1 Describe the conditions of the expansion and derive the equations necessary for calculating w q and AU 2 Calculate w q and AU for the process 250 moles of an ideal gas Cum 32 R are initially confined in a volume of 500 L at 29815 K The sample is then expanded against the ambient pressure of 0985 atm until its it reaches mechanical equilibrium with the surroundings In the process the temperature of the sample decreases to 28815 K 1 Describe the conditions of the expansion and derive the equations necessary for calculating w q and AU 2 Calculate w q and AU for the process Explain why temperature can either be in 0C or K when you need to calculate AT the change in temperature 250 moles of an ideal gas Cum 32 R are initially at a pressure of 1223 atm and a temperature of 29815 K The sample is then expanded reversibly and isothermally until the final pressure is 0985 atm 1 Describe the conditions of the expansion and derive the equations necessary for calculating w q and AU 2 Calculate w q and AU for the process Topic 06 ConstantVolume Calorimetry Page 7 Topic 07 Enthalpy The Thermodynamic Definition and Properties of Enthalpy The thermodynamic definition of enthalpy H is H E U pV where the quotEquot sign represents equality based on definition You should memorize this relationship General properties of enthalpy are similar to those of internal energy 1 Enthalpy is an extensive function Its value increases in proportion to the amount of substance moles present 2 Enthalpy is a state function Its value depends solely on the state of the system and not on how the system was prepared Changes in enthalpy between two states AH depend only on the initial and final states and not on the path between the two states 3 The absolute enthalpy of a system cannot be measured However changes in enthalpy AH can be directly measured in the laboratory The thermodynamic basis for the laboratory measurement of AH is developed below Mathematical Digression Total Differentials You are familiar with the concept of a derivative eg dydx which measures how much a dependent variable y in this example changes infinitesimally as a response to an infinitesimal change in the independent function x The total differential is similar to a derivative except that the denominator is not present dH designates the total differential of enthalpy and represents an infinitesimal change in enthalpy Rules for evaluating total differentials are similar to the rules for evaluating regular derivatives Here are two important examples For the regular derivative examples below y and z are functions of x For the total differential examples Yand Z are functions that contribute in some way to the function X Regular derivative of a sum function is equal to the sum of the individual derivatives f fx yx zx then dfdx dy zdx dydx dzdx Total differential of a sum function follows the same rule fX YZ then dX dYZ dY dZ Regular derivative of a product function follows the product rule If fx yx zx then dfdx dyzdx z dydx y dzdx Total differential of a product function follows the same product rule fX YZ then dX dYZ ZdY YdZ Topic 07 Enthalpy Page 1 Note also that the integral of a pure total differential is the change in the function itself For a 2 definite integral IdH Hll2 2 H2 H1 2 AH For an indefinite integral IdH H C where C is 1 a constant of integration Mastery of these concepts is essential because total differentials are found extensively in the study of thermodynamics ConstantPressure Pressure Processes Carried Out in Mechanical Equilibrium In Topic 6 it was shown that AU qV the amount of work transferred in a constantvolume process when no extra work is done This makes it possible to measure AU directly for a reaction or phase change when the process is carried out in an adiabatic bomb calorimeter The following derivation will establish the thermodynamic basis for measuring AH directly The derivation will make three important assumptions 1 The system pressure p is constant 2 the system is at all times in mechanical equilibrium with its surroundings p pext and 3 no extra work is done Equation Explanation Justification H U pV Definition of Enthalpy dH dU pV Take the total differential of dH dH dU dpV Use the sum rule dH dU pdV Vdp Use the product rule dp 0 therefore Vdp O ASSUMPTION 1 PRESSURE IS CONSTANT dH dU pdV Simplify since Vdp 0 ldH ldU pldV Integrate both sides Because p is assumed constant it can be pulled outside of the integral AH AU pAV Evaluate integral results for any system going from some initial state to a final state AH q w pAV Substitute AU q w from the First Law w pext AV Recall the expression for expansion work w when a gas expands against constant external pressure and NO EXTERNAL WORK S DONE ASSUMPTION 2 Review Topic 4 if necessary AH q pext AV pAV Substitute this expression for work in the preceding equation P pext ASSUMPTION 3 The system is at all times in mechanical equilibrium with its surroundings AH q pAV pAV Substitute p for pext is the previous equation AH q Simplify by removing terms that cancel AH qp Add subscript p to the notation to emphasize that the process must be at constant pressure Summary the amount of heat transferred in a reaction or process is equal to AH when the following three conditions are met 1 the process occurs at constant pressure 2 the system and its Topic 07 Enthalpy Page 2 surroundings are in mechanical equilibrium so p pext and 3 no extra work is done Enthale is emphasized in elementary chemistry because many physical and chemical processes occur in open systems where p pext The Relationship between AU and AH for an Ideal Gas In General Chemistry a common laboratory experiment involves use of a coffeecup calorimeter The calorimeter is prefilled with a measured volume of water Then a reaction is initiated By measuring the temperature change in the open calorimeter the amount of heat transferred can be determined Because all the conditions are met AH q in the coffee cup calorimeter experiment In Topic 6 we discussed the experimental determination of AU via adiabatic bomb calorimetry It is normal to wonder how AH is related to the AU value measured in a bomb calorimeter The following derivation assumes 1 that the reaction or process is isothermal 2 that all gases behave ideally and 3 that because the volumes of solids and liquids are very small compared to the volume occupied by a gas they can be neglected and the pV product can be well approximated by pVgas Equation Explanation Justification H U pV Definition of Enthalpy pV pVgas ASSUMPTION 1 The volume occupied by gases is so much larger than the volume occupied by liquids and solids that the latter can be quotignoredquot pV is well approximated by pVgas with little error pVgas nRT ASSUMPTION 2 ALL GASES BEHAVE IDEALLY H U nRT Substitute the ideal gas equivalent of pV into the first equation AH AU nRT Consider the macroscopic change in enthalpy AH AU AnRT Apply the rule of addition same as for derivatives T is constant ASSUMPTION 3 THE PROCESS IS ISOTHERMAL AH AU RTAn Pull constant values outside of the change A expression AH AU RTAng Consistent with Assumption 1 add a subscript g to indicate that only gas components should be considered What is Ana To use the derived equation Ang must be correctly determined Begin with a balanced equation identifying the phase 5 I or g of each reactant and product On the product right side add the coefficients of all products in the gas phase to get ngprod do not include solid or liquid products Do the same thing for the reactant left side of the reaction to get ngxeac The difference between the two values is Ang Ang nglpmd nglreac As an example consider the combustion of one mole of benzene C5H5l Start by writing the balanced combustion reaction C6H6I M 02g gt 6 C02g g H20I Note that water is a liquid at 29815 K Because the combustion reaction is for mole of benzene fractional coefficients are common There are 6 moles of gas C02 on the product side H20 is a liquid and is not counted There are 75 moles of gas 02 on the reactant side C5H5 is a liquid and is not counted For this example Ang 6 mol product side 75 mol reactant side therefore Ang 15 mol The sign is important Topic 07 Enthalpy Page 3 The AH value for the combustion of benzene can be looked up in standard thermodynamic tables AH 3267 kl mol39l Now we will calculate AU for the combustion reaction at 29815 K The derived equation is AH AU RTAng This can be rearranged to isolate AU AU AH RTAng Evaluate the last term RTAng 8314471 K391 mol39129815 K15 mol 37184 J In the final calculation this must be converted to k to match the units of AH AU 3267 kl 37184 J1 kJlOOO J 3263 kl Note you will assume that gases in similar reactions behave ideally in this course However whenever you use the equation AH AU RTAng always state the assumptions underlying the equa on The ConstantPressure Heat Capacity dU You will recall that the constantvolume heat capacity is defined as CV The V dH constant pressure heat capacity Cp is defined as Cp 2 Note that the partial derivative of the P enthalpy is taken with pressure being held constant Recall that H U pV Because all three variables on the right side are positive the absolute enthalpy is always greater than the absolute internal energy H gt U This is not generally true for changes in enthalpy It is also generally true as we shall prove later that Cp gt CV for the same substance in the same state For ideal gases Cp CV nR equivalently CIDm Cvlm R In this course whenever you have data giving CIDm for a gas and need a value for Cvlm you should assume the ideal gas approximation to hold and use the above relationship Any exceptions will be made clear by the wording of the question Note that this equation is only used for gases For condensed phases solids and liquids CV and Cp are approximately equal We will explore the exact relationship between CV and Cp in a future Topic AH Calculations for Specific Processes Heating of a Substance in a Single Phase Under conditions of constant pressure the definition of Cp can be converted to a regular derivative Cp dHdT This can be rearranged to give dH deT This expression is exact for ideal gases and is a very good approximation for real gases In this course you will assume that the expression is good to use unless information provided in the problem indicates otherwise Integrating both sides gives AH ledT When the heat capacity can be assumed constant over the relevant temperature range it can be pulled outside of the integral AH CpldT This integrates to give AH Cp AT Unless given information suggesting that Cp is not constant with temperature in this course you should use this relationship for AH Later in this topic we will discuss what to do if Cp cannot be assumed constant as a function of temperature Topic 07 Enthalpy Page 4 When Cp can be assumed constant the change in enthalpy that occurs when a substance is heated in a single phase is given by AH Cp AT Note that a different value of Cp applies to each phase Phase Changes The above discussion does not apply to phase changes For example H20s gt H20l melting or fusion or H20l gt H20l vaporization The molar enthalpy changes associated with phase transitions AHfusm and AHvaplm can be looked up in standard tables Tabulated values correspond to the equilibrium transition temperature of the substance in question at a pressure of 1 bar The Temperature Dependence of ConstantPressure Heat Capacity C2 Unless conditions or information provided in a problem indicate otherwise you will usually assume that Cp is constant over the temperature range of interest However there are some cases in which this assumption cannot be made A One way to deal with the temperaturedependence is to express the heat capacity in the following functional form CIDm a bT cT39Z Three constants a b and c are tabulated for individual substances within a specified phase They are used to calculate the actual value of CIDm at a specified absolute temperature T For such problems we still assume the validity of AH ledT as derived above However because Cp is not constant it must be kept inside the integral AH l a bT cT392dT This can be split into a sum of three separate integrals which can be individually evaluated at the limits AH la dT le dT lcT39sz B At very low temperatures near absolute zero Debye observed that the heat capacity is directly proportional to T3 CIDm aT3 where a is a constant that depends on the substance The same procedure described n A is followed to calculate AH for heating First assume that AH ledT Next substitute the temperaturedependent expression for Cp inside the integral AH l aT3dT Finally integrate and evaluate at the temperature limits to solve for AH Expected Behavior of C2 as a Function of Temperature For real substances when heated in a single phase Cp is expected to increase with an increase in temperature The amount of increase will depend on the substance and the phase When undergoing a phase transition Cp is infinite and discontinuous This is because the heat capacity is generally defined as the differential heat absorbed dq divided by the differential change in temperature dT at constant pressure Cp dqpdT In an equilibrium phase transition heat is absorbed dq gt 0 but the temperature remains unchanged dT 0 Division by zero in the denominator makes Cp infinitely large and therefore undefined Note that the initial value of Cp following a phase transition is generally different from the final value of Cp preceding the phase transition Topic 07 Enthalpy Page 5 These concepts are demonstrated in Figure 1 When heated within a single phase Cp increases as a function of temperature The increase is not generally linear as depicted in the simplified figure This applies to a heating the solid c heating the liquid and e heating the vapor At the phase transitions Cp is infinite and therefore undefined This applies to b fusion and d vaporization Following each phase transition note that the initial value of Cp in the new phase is different from the final value in the previous phase Also the rate of change slope of Cp vs T varies among substances and depends on the phase i c heating llquLIlCll m L E Cp E e heating gas ELJ e o g 395 1 E ahea ngso d Tr T T Figure 1 Sketch of the Expected Behavior of Cp as a Function of Temperature See text for details Graphical Determination of AH In Topic 5 it was emphasized that the area under a pext vs V plot can be interpreted as the amount of work done by the system on its surroundings Because AH ledT the area under the Cp vs T plot between two specific temperatures gives the value AH for heating between those two temperatures provided that the heating occurs within a single phase at constant pressure DISCUSSION QUESTIONS AND FOLLOWUP PROBLEMS 71 The Gibbs Energy is defined by the following equation G E HTS where H is the enthalpy T is the absolute temperature and S is the entropy to be introduced later Expand the total differential dG in terms of the total differentials of H T and S 72 Beginning with the thermodynamic definition of enthalpy derive the expression AH qp Explain each step of the derivation and state all assumptions clearly 73 For each of the following scenarios state whether AH qp is a valid relationship Clearly justify each of your answers A A gas sample initially occupying 300 L at 298 K and 400 atm is allowed to expand isothermally against the constant external pressure of 100 atm B 500 mL of liquid water are heated from 25 C to 95 C in an open beaker C 100 mol of water at 1000 C is completely vaporized at 1 atm pressure Topic 07 Enthalpy Page 6 74 75 76 77 78 79 710 711 712 Write the chemical reaction for complete combustion of each of the following substances all of which are solids at SATP For each substance identify Ang Note for substances containing nitrogen assume that the only nitrogencontaining product is N2g A C12H10N2 azobenzene AHCm 6460 k mol39l B C13H100 benzophenone Aug 6503 k mol39l C C12H10biphenyl Aug 6244 k mol39l Using results from 74 calculate AUCm for azobenzene then calculate AHCm for both benzophenone and biphenyl The combustion occurs at SATP 04903 g of naphthalene C10H8s MW 1281705 was combusted in an adiabatic bomb calorimeter The calorimeter constant was measured beforehand to be 1028 kJ K39l To initiate the combustion 00034 grams of ignition wire were combusted 1400 cal g39l The initial and final temperatures in the calorimeter water bath were 2290 C and 2482 C respectively Calculate the molar enthalpy of combustion AHclm and the molar internal energy of combustion AUCm for naphthalene consistent with the data A Derive the relationship AH AU RTAng explaining each step and clearly stating all assumptions B Derive the analogous relationship for a process in which there is no reaction Ang 0 but the gas is heated AT gt 0 Assume that all gases behave ideally A For ideal gases state the relationship between Cvlm and Cplm B How are the two heat capacity values related for a substance in the solid phase What method is used to calculate AH under each of the following circumstances A Heating a gas from T1 to T2 when Cp is constant over that temperature range B Condensation of a gas to liquid at the equilibrium vaporization temperature C Heating a substance from T1 to T2 when Cp varies with temperature Three moles of an ideal gas CVm 52 R are heated from 250 C to 1000 C at a constant pressure of 1 bar Calculate AH for the heating process Calculate AU q and w for the same heating process Three moles of NH3 g are heated from 1250 C to 2000 C at a constant pressure of 1 bar The temperaturedependent is described by CpmJ K391 mol39l a bT cT39z where a 2975 b 251 x 10393 K and c 155 x 105 K39z Calculate AH for the heating process Calculate AU q and w for the same heating process Three moles of water MW 180153 undergo the liquid vapor phase transition at 1000 C and exactly 1 atm For water AHV 40756 kJ mol39l and the density of liquid water is 095840 g cm393 under these equilibrium phase change conditions Calculate AH for the phase change assuming that the vapor behaves ideally Note that the process occurs in an open system Calculate q w and AU for the phase change Topic 07 Enthalpy Page 7 713 175 moles of an ideal gas Cum 32 R are initially confined in a volume of 350 L at 29815 K The sample is then expanded isothermally against the external ambient pressure of 1020 bar until it achieves mechanical equilibrium with the surroundings What is the final volume of the gas Calculate w q AU and AH for the process 714 175 moles of an ideal gas Cum 32 R are initially confined in a volume of 350 L at 29815 K The sample is then expanded against the ambient pressure of 1020 bar until its it reaches mechanical equilibrium with the surroundings In the process the temperature of the sample decreases to 27815 K Calculate w q AU and AH for the process 715 175 moles of an ideal gas Cum 32 R are initially at a volume of 200 L and a temperature of 29815 K The sample is then expanded reversibly and isothermally until the final pressure is 1020 bar Calculate w q AU and AH for the process Topic 07 Enthalpy Page 8 Problem Set 1 12 13 14 15 16 Problem Set 2 23 25 26 Problem Set 3 34 35 36 39 310 Problem Set 4 44 Numerical Answers to Selected Problems Problem Sets 17 atm bar psi Pa mmHg atm 1234 1250 1813 1250E06 9378 bar 0854 0865 125 865E04 649 psi 661 670 972 670E05 503503 Pa 5007 5073 7358 5073E05 3805 mmHg 00242 00245 0356 245E03 184 atm bar psi Pa mmHg atm 0891 0903 131 903504 677 50 psi F c K F 9860 3700 31015 c 16700 7500 34815 K 10273 7485 19830 40 24790 L 320 atm 1145 G A 488 atm B 500 L moI391 646 x 10394 atm391 A 00769 L mol39l B 0943 A 405 atm B 0995 671 L moI391 AU C W AUsur CIsur Wsur A 51 121 171 51 121 171 B 291 121 171 291 121 171 c 291 121 171 291 121 171 Problem Set 5 52 54 57 Problem Set 6 63 64 67 68 69 610 611 612 614 Problem Set 7 74 75 76 710 711 712 713 714 715 101325 JLatm A w 0 B w 543 k C w 975 k 973 k A o024 k B 217 K C 212 k D 983 k K391 A 0046 k B 184 K C 1835 k D 827 x 103 kJ mol391 629 c 432 c w 681 k Au o q 681 kJ w0AUOq0 w 570 k Au o q 570 k w 549 k Au o312 kJ q 518 k w 156 k AU 0 q w 156 k 15 2 25 6456 k mol39l 6508 k mol39l 6250 kJ mol391 AuM 515 x 103 kJ mol39l AH 516 x 103 kJ mol391 AH 655 k Au 468 k q 655 k w 187 k AH 897 k q 897 k Au 710 k w 187 k AH 12227 kJ q 12227 k w 9302 k Au 11297 k w 398 k Au 0 AH o q 398 k w 369 k Au 04365 kJ AH 6727 kJ q 325 k w 133 k Au 0 AH o q 133 kJ

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