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25 de Dec de 2015•0 recomendaciones•372 vistas

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Chm3410hwk02 soln.252145237

Fox solutionkuhlblitz

Advance thermodynamicsiimt college meerut

2 the first law of thermodynamicRanny Rolinda R

Exerc cios resolvidos_-_cap._01-atkins_(a)Alisson Moreira

IB Chemistry Ideal Gas Equation, Kinetic Theory and RMM determination of gasLawrence kok

Capítulo 1 - Mecânica dos Fluidos Ana Flávia Prata

- 1. CHM 3410 – Problem Set 2 Due date: Wednesday, September 7th Do all of the following problems. Show your work. "Thermodyna mics is a funny subject. The first time you go through it, you don't understand it at all. The second time you go through it, you think you understand it, except for one or two small points. The third time you go through it, you know you don't understand it, but by that time you are so used to it, it doesn't bother you any more." - Arnold Sommerfeld 1) The isothermal compressibility of a substance, κ, is defined as κ = - (1/Vm) (∂Vm/∂p)T (1.1) Find an expression for κ for substances that obey the following equations of state. a) pVm = RT (ideal gas) (1.2) b) p = RT – a (van der Waals equation) (1.3) (Vm – b) Vm 2 [Hint: For b, it is easier to find (∂p/∂Vm)T, and then use one of the partial derivative relationships from Handout 1, namely (∂Vm/∂p)T = 1 ] (1.4) (∂p/∂Vm)T 2) Consider one mole of an ideal monatomic gas (recall that for a monatomic gas CV,m = (3 /2) R = 12.472 J/mol. K, Cp,m = (5 /2) R = 20.786 J/mol. K). The gas is initially at a pressure p = 2.000 atm and a temperature T = 400.0 K. Find q, w, ∆U, and ∆H for each of the following processes: a) An isothermal reversible compression of the gas to a final pressure pf = 10.000 atm. b) An reversible cooling of the gas from Ti = 400.0 K to Tf = 300.0 K, at a constant pressure p = 2.00 atm. c) An adiabatic reversible compression of the gas to a final pressure pf = 10.000 atm. d) An adiabatic irreversible expansion of the gas against a constant pressure pex = pf = 1.000 atm. 3) Find a general expression for w (work) when 1.000 mol of a gas obeying the van der Waals equation of state undergoes an isothermal reversible expansion from an initial volume Vi to a final volume Vf. Show that your answer reduces to the result for an ideal gas ( w = - nRT ln(Vf/Vi) ) by setting a = b = 0. This serves as a check of your answer. 4) An unsaturated hydrocarbon contains at least one carbon-carbon double or triple bond. Such compounds can be saturated by reaction with hydrogen (H2), a process called hydrogenation. Using the data in Table 2.6 and 2.8 of Atkins, find the enthalpy change for the hydrogenation of ethene, propene, and acetylene (ethyne) at T = 25.0 °C. For acetylene also find the enthalpy of hydorgenation at T = 100.0 °C. You may assume that the heat capacities of the reactants and products are approximately independent of temperature over the range T = 25.0 °C to T = 100.0 °C. 5) For an adiabatic reversible expansion or compression of an ideal gas with constant heat capacity, we derived the following relationship in class piVi γ = pfVf γ γ = Cp/CV = Cp,m/CV,m (5.1) Starting with equn 5.1 derive the following relationship, which also applies to these processes Tf/Ti = (pf/pi)γ-1/γ (5.2)
- 2. Also do the following from Atkins: Exercises 2.19b From the following data determine ∆H°f for diborane, B2H6(g), at 298. K. (1) B2H6(g) + 3 O2(g) → B2O3(s) + 3 H2O(g) ∆H°rxn = - 2036. kJ/mol (2) 2 B(s) + 3 /2 O2(g) → B2O3(s) ∆H°rxn = - 1274. kJ/mol (3) H2(g) + 1 /2 O2(g) → H2O(g) ∆H°rxn = - 241.8 kJ/mol Problems 2.10 From the enthalpy of combustion data in Table 2.6 for the n-alkanes methane (CH4) to n-octane (C8H18), test the extent to which the relationship ∆H°comb = k Mn holds (where M is the molecular mass of the n-alkane, in g/mol), and find the numerical values for k and n. Predict the value for ∆H°comb for n-decane (C10H22() ), and compare it to the literature value ∆H°comb = - 6772.5 kJ/mol. 2.30 a) Write expressions for dV and dp given that V is a function of p and T (V ≡ V(p,T) ) and p is a function of V and T (p ≡ p(V,T) ). b) Deduce expressions for d lnV and d ln p in terms of the coefficient of thermal expansion and the isothermal compressibility. α = (1/V) (∂V/∂T)p (coefficient of thermal expansion) κ = - (1/V) (∂V/∂p)T (isothermal compressibility) Solutions. 1) a) For an ideal gas pVm = RT , and so Vm = RT/p
- 3. So (∂Vm/∂p)T = ∂/∂p)T RT/p = - RT/p2 And so κ = - (1/Vm) ( -RT/p2 ) = RT/p2 Vm. But pVm = RT, and so κ = RT/(pRT) = 1/p b) For a van der Waals gas p = RT - a (Vm - b) Vm 2 It is difficult to solve the van der Waals equation for Vm. We will instead use one of our partial derivative relationships (∂V/∂p)T = 1/(∂p/∂Vm)T So (∂p/∂Vm)T = ∂/∂Vm)T RT - a = - RT + 2a = [ 2a (Vm - b)2 – RTVm 3 ] (Vm - b) Vm 2 (Vm - b)2 Vm 3 (Vm - b)2 Vm 3 And so (∂V/∂p)T = (Vm - b)2 Vm 3 [ 2a (Vm - b)2 - RTVm 3 ] κ = - (1/Vm) (Vm - b)2 Vm 3 = (Vm - b)2 Vm 2 [ 2a (Vm - b)2 - RTVm 3 ] [ RTVm 3 - 2a (Vm - b)2 ] As a check, note that if we set a = b = 0, we get κT = Vm 4 = Vm = 1 RTVm 3 RT p the result previously obtained for an ideal gas. That doesn’t prove our result for the van der Waals gas is correct, but it does show that it is consistent with expectations. 2) a) The gas is ideal and the process is isothermal, and so ∆U = ∆H = 0. The process is reversible and the gas is ideal, pex = p = nRT V w = - ∫i f pex dV = - ∫i f p dV = - ∫i f nRT dV = - nRT ∫i f dV = - nRT ln(Vf/Vi) V V Since this is an ideal gas and an isothermal process, we know (from Boyle's law) that (Vf/Vi) = (pi/pf), so w = - nRT ln(Vf/Vi) = - nRT ln(pi/pf) = + nRT ln(pf/pi) = (1.000 mol) (8.3145 J/mol.K) (400.0 K) ln(10.000/2.000) = + 5353. J Finally, since ∆U = q + w = 0 , q = - w = - 5353. J b) The process is carried out at constant pressure, and the gas is ideal, and so ∆H = ∫i f n Cp,m dT ∆U = ∫i f n CV,m dT
- 4. Since the gas is an ideal monatomic gas, Cp,m is constant, and so may be taken outside the integral, to give ∆H = ∫i f n Cp,m dT = n Cp,m ∫i f dT = n Cp,m (Tf - Ti) ∆U = ∫i f n CV,m dT = n CV,m ∫i f dT = n CV,m (Tf - Ti) So ∆H = (1.000 mol) (20.786 J/mol. K) (300.0 K - 400.0 K) ∆U = (1.000 mol) (12.472 J/mol. K) (300.0 K - 400.0 K) ∆H = - 2079. J ∆U = - 1247. J The process is constant pressure, and so q = ∆H = - 2079. J Finally, from the first law ∆U = q + w , and so w = ∆U - q = ( - 1247. J) - (- 2079. J) = + 832. J c) The process is adiabatic, and so q = 0 For an adiabatic reversible expansion or compression of an ideal monatomic gas, we may say (as proved in class) pi Vi γ = pf Vf γ γ = Cp,m/CV,m= (5R/2)/(3R/2) = 5/3 While we could use the above equation (since we know pi and pf and have enough information to calculate Vi from the ideal gas law) it is convenient to use the relationship we will derive in problem 5. Tf/Ti = (pi/pf)(1-γ)/γ Tf = Ti (pi/pf)(1-γ)/γ Now, (1-γ)/γ = (1 - 5 /3) / (5 /3) = - (2/5) = - 0.400 So Tf = (400.0 K) (2.000/10.000)-0.400 = 761.5 K We may use the expressions previous found (in part b) for ∆U and ∆H ∆U = n CV,m (Tf - Ti) = (1.000 mol) (12.472 J/mol. K) (761.5 K - 400.0 K) = + 4509. J ∆H = n Cp,m (Tf - Ti) = (1.000 mol) (20.786 J/mol. K) (761.5 K - 400.0 K) = + 7514. J Finally, from the first law, ∆U = q + w. Since q = 0, w = ∆U = + 4509. J d) The process is adiabatic, and so q = 0 The process is irreversible against a constant external pressure pex = pf, so w = - ∫i f pex dV = - pex ∫i f dV = - pf (Vf - Vi) The gas is ideal and monatomic, and so (since CV,m is constant) ∆U = ∫i f n CV,m dT = n CV,m ∫i f dT = n CV,m (Tf - Ti) Since q = 0, it follows that ∆U = w, and so
- 5. n CV,m (Tf - Ti) = - pf (Vf - Vi) = pf Vi - pf Vf If we multiply the first term on the right by pi/pi, then n CV,m (Tf - Ti) = (pf/pi) (pi Vi) - (pf Vf) But pi Vi = nRTi, pf Vf = nRTf, so n CV,m (Tf - Ti) = n CV,m Tf - n CV,m Ti = (pf/pi) nRTi - nRTf Divide both sides by n CV,m Tf - CV,m Ti = (pf/pi) RTi - RTf Collect terms in Tf on the left, and in Ti on the right CV,m Tf + R Tf = (pf/pi) RTi + CV,m Ti Recall that CV,m + R = Cp,m, then Cp,m Tf = [ (pf/pi) R + CV,m ] Ti Tf = [ (pf/pi) R + CV,m ] Ti = [ (1.000/2.000) (8.3145 J/mol. K) + (12.472 J/mol. K) ] (400.0 K) = 320. K Cp,m (20.786 J/mol. K) As we did in part c, we may use our previous result from part b to find ∆U and ∆H ∆U = n CV,m (Tf - Ti) = (1.000 mol) (12.472 J/mol. K) (320.0 K - 400.0 K) = - 998. J ∆H = n Cp,m (Tf - Ti) = (1.000 mol) (20.786 J/mol. K) (320.0 K - 400.0 K) = - 1663. J Finally, since q = 0 it follows from the first law that w = ∆U = - 998. J 3) For a van der Waals gas p = nRT - a n2 (V - nb) V2 In general w = - ∫i f pex dV. Since the process is reversible, pex = p, so w = - ∫i f [ (nRT)/(V - nb) ] - (an2 /V2 ) dV The process is isothermal, and so T is constant. If we do the integral, we get w = - nRT ln[ (Vf - nb)/(Vi - nb) ] - an2 [ (1/Vf) - (1/Vi) ] We may manipulate the terms on the right to get rid of the - signs, to get w = nRT ln [(Vi - nb)/(Vf - nb) ] + an2 [ (1/Vi) - (1/Vf) ] As a check, if we set a = b = 0, we get w = nRT ln(Vi/Vf) = - nRT ln(Vf/Vi) , the ideal gas result. That doesn't prove our result is correct, but does show it is consistent with what we expect. 4) The reactions of interest are ethene C2H4(g) + H2(g) → C2H6(g) propene C3H6(g) + H2(g) → C3H8(g) acetylene C2H2(g) + 2 H2(g) → C2H6(g)
- 6. The data we need are in Table 2.5 of Atkins (recall that ∆H°f = 0 for an element in its standard state). Compound ∆H°f (kJ/mol) Cp,m (J/mol. K) C2H2(g) + 226.73 43.93 C2H4(g) + 52.26 43.56 C2H6(g) - 84.68 52.63 C3H6(g) + 20.42 63.89 C3H8(g) - 103.85 73.5 H2(g) 0.0 28.824 The two equations we need are as follows ; At T = 25.0 °C ∆H°rxn = [ Σ (∆H°f (products)) ] - [ Σ (∆H°f (reactants)) ] At T = 100.0 °C ∆H°f(T2) = ∆H°f (T1) + ∫T1 T2 ∆Cp dT ≅ ∆H°f (T1) + ∆Cp (T2 - T1) ethene C2H4(g) + H2(g) → C2H6(g) T = 25.0 °C ( - 84.68) - ( + 52.26) = - 136.94 kJ/mol propene C3H6(g) + H2(g) → C3H8(g) T = 25.0 °C ( - 103.85) - ( + 20.42) = - 124.27 kJ/mol acetylene C2H2(g) + 2 H2(g) → C2H6(g) T = 25.0 °C ( - 84.68) - ( + 226.73) = - 311.41 kJ/mol T = 100.0 °C ( - 311.41) + [ ( 52.63) - (43.93 + 2 (28.824)) ] x 10-3 ( 100.0 - 25.0 ) = - 315.08 kJ/mol 5) We start with pi Vi γ = pf Vf γ γ = Cp,m/CV,m which applies to an adiabatic reversible expansion or compression of an ideal gas with constant heat capacity. Then Vi = nRTi Vf = nRTf pi pf If we use these relationships to substitute for Vi and Vf, and cancel the common factor (nR)γ , we get pi Ti γ = pf Tf γ Combine the pressure terms pi γ pf γ pi (1-γ) Ti γ = pf (1-γ) Vf γ Take both sides to the 1/γ power
- 7. pi (1-γ)/γ Ti = pf (1-γ)/γ Tf Solve for Tf /Ti Tf/Ti = (pi/pf)(1-γ)/γ = (pf/pi)(γ-1)/γ Exercise 2.19b The formation reaction is 2 B(s) + 3 H2(g) → B2H6(g) We need some combination of processes that sum to this reaction. B2O3(s) + 3 H2O(g) → B2H6(g) + 3 O2(g) ∆H° = (-1) ( - 2036. kJ/mol) = + 2036. kJ/mol 2 B(s) + 3 /2 O2(g) → B2O3(s) ∆H° = (+1) ( - 1274. kJ/mol) = - 1274. kJ/mol 3 H2(g) + 3 /2 O2(g) → 3 H2O(g) ∆H° = (+3) (- 241.8 kJ/mol) = - 725.4 kJ/mol _________________________________ ___________ 2 B(s) + 3 H2(g) → B2H6(g) ∆Hf° = + 37. kJ/mol In the above we have used the fact that reversing the direction of a reaction changes the sign of ∆H°, and multiplying a reaction by a constant value means that ∆H° must also be multiplied by the same constant. Both of these are consequences of enthalpy being a state function. Problem 2.10 We are fitting data to the equation ∆H°c = k Mn where M = molecular mass If we take the logarithm of both sides of this equation, we get ln (∆H°c) = ln(k) + n ln(M) If we plot ln (∆H°c) vs ln(M) we expect a straight line with slope = n and intercept = ln(k). Data are given below, from Table 2.5 of Atkins Compound M(g/mol) ln(M) ∆H°c (kJ/mol) ln (| ∆H°c |) CH4(g) 16.04 2.775 - 890. 6.791 C2H6(g) 30.07 3.404 - 1560. 7.352 C3H8(g) 44.10 3.786 - 2220. 7.705 C4H10(g) 58.13 4.063 - 2878. 7.965 C5H12(g) 72.15 4.279 - 3537. 8.171 C6H14() 86.18 4.456 - 4163. 8.334 C7H16() 100.21 4.607 - 4817. * 8.480 C8H18() 114.23 4.738 - 5471. 8.607 * Calculated from the data in the table, as no value is given in the table for ∆H°c The data are plotted below.
- 8. Based on the data, I get for the best fitting line ln (| ∆H°c |) = 4.206 + 0.927 ln(M) or ∆H°c = - (67.1 kJ/mol) (M)0.927 k = - 67.1 kJ/mol n = 0.927 For C10H22() (M = 142.28 g/mol), the predicted values are ln(| ∆H°c |) = 8.802 ∆H°c = - 6648. kJ/mol (experimental value is ∆H°c = - 6830. kJ/mol) Several comments. For consistency we really should only use data for n-alkanes (straight chain alkanes and not branched alkanes) and for initial states that are all in the same phase (so there is no contribution from the enthalpy for the phase transition). I have done the first of these, but the last three compounds in the fitting are liquids, while the first five are gases. The experimental value for C7H16() was calculated from the enthalpy of formation data in Atkins. There is no value for ∆H°c in Atkins for C10H22(), and so the experimental value quoted above is from another textbook. Finally, there is no particular theoretical justification for the fitting equation used above. Being, naive, I would, if asked to come up with a prediction for how ∆H°c changes with the size of the n-alkane, assume that the value increases by a constant amount for every additional -CH2 group in the n-alkane. There is in fact a common approximation, with ∆H°c increasing in magnitude by about 650. kJ/mol for every additional -CH2 group present. Problem 2.30 a) If we write V ≡ V(p,T), then dV = (∂V/∂p)T dp + (∂V/∂T)p dT Using the definitions of α and κ, we can substitute for the above partial derivatives to get dV = - κV dp + αV dT Now, if we write p ≡ p(V,T) Plot of ln(Hcomb) vs ln(M) 6.5 7 7.5 8 8.5 9 2.5 3 3.5 4 4.5 5 5.5 ln(M) ln(Hcomb)
- 9. then dp = (∂p/∂V)T dV + (∂p/∂T)V dT But (∂p/∂V)T = 1/(∂V/∂p)T = - 1/(κV) and (∂p/∂T)V (∂T/∂V)p (∂V/∂p)T = -1 and so (∂p/∂T)V = - (∂V/∂T)p = α/κ (∂V/∂p)T Substituting, we get dp = - (1/κV) dV + (α/κ) dT b) We may say d ln V = (d ln V/dV) dV = (1/V) dV d ln p = (d ln p/dp) dp = (1/p) dp Since we already have expressions for dV and dp, substituting gives d ln V = - κ dp + α dT d ln p = - (1/κpV) dV + (α/κp) dT There are several ways the above can be reorganized, but the answers given above are fine. What is the use of the expressions for dV, d ln V, dp, and d ln p? Well, they are general equations of state, and so are of greatest importance in modeling systems that do not have simple behavior (such as slightly compressible liquids and gases) where values for α and κ can be measured experimentally.
- 10. then dp = (∂p/∂V)T dV + (∂p/∂T)V dT But (∂p/∂V)T = 1/(∂V/∂p)T = - 1/(κV) and (∂p/∂T)V (∂T/∂V)p (∂V/∂p)T = -1 and so (∂p/∂T)V = - (∂V/∂T)p = α/κ (∂V/∂p)T Substituting, we get dp = - (1/κV) dV + (α/κ) dT b) We may say d ln V = (d ln V/dV) dV = (1/V) dV d ln p = (d ln p/dp) dp = (1/p) dp Since we already have expressions for dV and dp, substituting gives d ln V = - κ dp + α dT d ln p = - (1/κpV) dV + (α/κp) dT There are several ways the above can be reorganized, but the answers given above are fine. What is the use of the expressions for dV, d ln V, dp, and d ln p? Well, they are general equations of state, and so are of greatest importance in modeling systems that do not have simple behavior (such as slightly compressible liquids and gases) where values for α and κ can be measured experimentally.