= 1906J/0.872deg = 2186J/deg

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1 Physical Chemistry Homework assignment 2 Problem 1: he heat of combustion of caffeine was determined by first burning benzoic acid and then caffeine. In both cases the calorimeter was filled with 466 g of distilled water. When g of benzoic acid, C 7 H 6 O 2 (s), were burned as well as 1.1 cm of the iron wire used to ignite the sample, the temperature increased from C to C. When g of caffeine, C 8 H 10 O 2 N 4 (s), were burned, along with 1.8 cm of the wire, the temperature increased from C to C. What is the heat capacity of the calorimeter with and without water, and what is the energy change, U, per mol for the combustion of caffeine? (he energy change for combustion of benzoic acid is U = kj/g, the combustion of the wire gives 9.62 J/cm, and the specific heat of water is 1 cal/g deg). A. Find heat capacity of the calorimeter: Burn g of benzoic acid and 1.1 cm of wire. his gives temperature rise of = deg = deg. he heat released is q = g 26434J/g + 1.1cm 9.62J/g = 1895J + 11J = 1906J he total heat capacity is C = C bomb + C prod + C water = 1906J/0.872deg = 2186J/deg Subtract heat capacity of the water, C water which means that C bomb + C prod = 466 g 1cal/g deg J/cal = 1950 J/deg, = ( ) J/deg = 236 J/deg B. Heat of combustion for caffeine: Burn g of caffeine as well as 1.8 cm of wire. his gives temperature rise of = deg = deg. Heat released due to combustion of coffeine is U = U tot - U wire = 0.632deg 2186 J/deg - 1.8cm 9.62J/cm = ( )J = 1365 J from this sample. Per gram of caffeine, the heat released is U = 1365 J/ g = kj/g. Per mol of caffeine, the heat released is kj/g 194 g/mol = 4243 kj/mol where 194 is the molecular weight ( = 194). Problem 2: Methane gas is heated from 25 until the volume has doubled. he pressure is constant at 1 bar. he variation in molar heat capacity with temperature has been measured and the results of those experiments can be summarized by C P = a + b 1

2 where a = JK 1 mol 1 and b = JK 2 mol 1. Calculate H and Ū per mol, (you can estimate the final volume temperature assuming ideal gas behaviour). he final volume is twice the initial volume, so 2 = 2 1. he pressure is constant, so P 1 = P 2. he initial temperature is 1 = 298K. he final temperature can be found assuming ideal gas behaviour, 2 / 1 = P 2 /P 1 = 2. So, 2 = 2 1 = 596 K and = 298 K. H = ( ) H P d = 2 1 C p ( )d = 2 1 (a + b )d = [a + b2 2 ] 2 1 Find change in internal energy, = kj/mol kj/mol = kj/mol U = H P = H R = ( )kJ/mol = kj/mol Problem 3: (a) Show that the first law expression for heat flow (assuming only P-work), δq = du + P d, can be rewritten as δq = C d + R d ln for an ideal gas. (b) (c) Using the expression above, show that δq is not an exact differential. Show that by multiplying δq by 1/, an exact differential is formed, i.e. show that is an exact differential. his shows that δq function is defined as entropy. δq = C d ln + R d ln is the differential of a state function. his state (a) he first law du = δq + δw becomes du = δq pd when the work is only P-work. For ideal gas du = ( ) U d = C d and P d = nr d/ = nr d ln. ogether, this gives δq = C d + nr d ln (b) For δq to be an exact differential, the derivative of the coefficient of d with respect to, ( C ) would need to be equal to the derivative of the coefficcient of d with respect ( ) to, (nr/ ). But, ( ) C = 0 for ideal gas since volume does not affect the energy of an ideal gas (or ) any derivative of the energy) due to the lack of molecular interactions, while = nr/. So, the the condition for an exact differential is not satisfied. ( (nr/ ) 2

3 (c) For δq/ to be an exact differential, ( ) (C / ) ( would need to be equal to (nr/ ) Both are equal to zero, so δq/ is an exact differential. his illustrates for the particular case of an ideal gas that the entropy, which is defined by ds = δq/ for a reversible process, is a state function, while δq is not. ). Problem 4: (a) One mole of a perfect, monatomic gas expands reversibly and isothermally at 25 C from an initial pressure of 5 atm to 1 atm. Determine the value of q, w, U and H for this process. (b) One mole of a perfect, monatomic gas expands reversibly and adiabatically from 5 atm to 1 atm. he initial temperature is 25 C. Determine the final temperature of the gas, as well as q, w and U. (a) Since the temperature does not change (isothermal) and the gas is ideal, the internal energy does not change, i.e. U = 0. By definition of the enthalpy H = U + P so H = U + (P ) = U + (nr ) = U since neither n nor changes. So, H = 0. he work done in a reversible isothermal process is w = 2 1 P ext d = nr ln 2 / 1 = nr ln P 2 /P 1 = 1 mol J/Kmol 298 K ln 1/5 = 3988 J. By the first law, q + w = U = 0 so q = 3988 J. (b) For an adiabatic process, q = 0. As the gas expands against the external pressure it does work and looses energy and, since no heat flows in, the temperature drops. he final temperature can be found from the equation ( ) (γ 1)/γ P2 2 = 1 P 1 where γ = C P /C. For an ideal gas C P = C + nr and for one mol of monatomic ideal gas C = 3R/2. herefore, γ = 5/3 and (γ 1)/γ = 2/5. he final temperature is 2 = 298 K (1/5) 2/5 = 156 K. he change in internal energy gan then be obtained, U = (3/2)R = J/Kmol ( ) K = 1770 J. By the first law, w = U q = = 1770 J. Finally, the enthalpy change is H = U + nr = 1770 J J/Kmol ( ) K = 2950 J. Problem 5: (a) One mol of a perfect, monatomic gas expands reversibly and isothermally at 300 K from a pressure of 10 atm to a pressure of 2 atm. Determine the value of q, w H and U for this process. (b) Now assume the gas expands by the same amount again isothermally but now irreversibly against 1 atm pressure (instead of reversible expansion) and calculate again q, w, 3

4 U and H. Compare with the values obtained in part (a) and explain qualitatively the difference. (c) Compare now the isothermal expansion in (a) to an adiabatic, reversible expansion. Sketch how the pressure changes as a function of volume in the two cases (label the curves clearly) and calculate the final temperature of the gas assuming the initial temperature is 300 K. Problem 6: he enthalpy change is H vap = 40 kj/mol when water boils at 100 C and 1 atm. his accounts for both the increase in internal energy as the molecules go from the liquid to vapour phase, as well as the expansion of the volume when water vapor forms. What fraction of the enthalpy change is due to the volume expansion? he enthalpy change is H = U + (P ). he second term is the contribution due to the volume expansion. Since the pressure is constant, (P ) = P P vap. It is a good approximation here to simply neglect the volume of the liquid. Assuming the ideal gas equation of state is a good enough approximation for the vapor, this can be evaluated as = nr/p and this gives (P ) = P nr/p = nr = 1 mol J/Kmol 398 K = 3.3 kj. he relative contribution of the volume expansion term to the enthalpy change is 3.3/40 = 0.08 or 8%. Problem 7: he constant volume heat capacity of a gas consisting of P 4 (g) molecules has been measured to be J/Kmol. he P 4 molecules are non-linear. (a) Assuming ideal gas behaviour, what is the expected heat capacity ratio, γ = C P /C, based on the experimental data? (b) What value of the heat capacity ratio, γ, would one expect from the high temperature approximations to translational, rotational and vibrational partition functions (the equipartition theorem results)? (c) Often, the vibrational degrees of freedom are not active at room temperature and are therefore skipped when the heat capacity is estimated. What is the heat capacity ratio, γ, one would expect from the high temperature approximations if vibrational degrees of freedom are skipped? Which is more accurate in this case: (i) including vibration or (ii) skipping it? Explain briefly the results you obtain for P 4 (g). (a) For an ideal gas C P = C + R so γ = C P /C = 1 + R/C = /67.15 = (b) he molecule is non-linear, so the total number of vibrational degrees of freedom 4

5 is 3N 6 = 6 and the high temperature approximation (equipartition theorem) predicts C = C,tr + C,rot + C,vib = (3/2)R + (3/2)R + 6R = 9R and a heat capacity ratio of γ equip = 1 + 1/9 = (c) Skipping the vibrational degrees of freedom in the high temperature approximation gives C = C,tr +C,rot = (3/2)R+(3/2)R = 3R and a heat capacity ratio of γ equip vib = 1 + 1/3 = In this case it is more accurate to include the vibrational contribution to the heat capacity even though the high temperature approximation is used. his is because the vibrational frequencies of P 4 are very low, the molecule is floppy. At room temperature the vibrational degrees of freedom are to large extent thermally excited and contribute to the heat capacity. 5

Homework Problem Set 8 Solutions

Homework Problem Set 8 Solutions Chemistry 360 Dr. Jean M. Standard Homework roblem Set 8 Solutions. Starting from G = H S, derive the fundamental equation for G. o begin, we take the differential of G, dg = dh d( S) = dh ds Sd. Next,