Unformatted text preview: Chemistry 3615 First Exam Version A October 1, 2008 _________Answer Key__________ Name Page 1 of 19 Chemistry 3615 First Exam October 1, 2008 _____________________________ Name Answer Sheet Circle the Correct Letter. If you change your mind, completely erase the wrong answer if you are using pencil. If you are using pen completely black out the wrong answer and clearly indicate your choice for the correct answer. If you need additional scratch paper, just ask. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. A A A A A A A A A A B B B B B B B B B B C C C C C C C C C C D D D D D D D D D D E E E E E E E E E E 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. A A A A A A A A A A B B B B B B B B B B C C C C C C C C C C D D D D D D D D D D E E E E E E E E E E Score = 5 x ______________ = ______________ Page 2 of 19 Useful Relationships and Definitions 1 V = V T p = 1 V V p T R = 8.314 Jmol1K1 = = 0.08206 Latmmol1K1 1 atm = 760 mm Hg = 760 Torr = 101325 Pa 0.001 m3 = 1 L = 1000 cm3 T JT = p H dW =  Fdz = Cp CV dW = pop dV
W =  p op dV
V1 V2 pV = constant TV 1 = constant
p1 T = constant dU = dQ + dW, U = Q + W H = U + pV a b  2 pV RT b Z= = 1+ + 2 + ... RT V V 1 + 1 b  a p + RT RT pV = Z= RT a a 2 2b  p + ... (RT )3 RT x 1 = y z y x z CV = CV = 3 R + R + (3N  5)R (linear) 2 3 3 R + R + (3N  6)R (non  linear) 2 2 F=2+CP dS = dQ T
T2 T1 x y z = 1 y z z x x y U CV = T V H Cp = T p H (T2 ) = H (T1 ) + C i dT Ci dT T1 T S = k ln k = R/6.02x1023 y y! = = x x!(y  x )! S(T2 ) = S(T1 ) + T2 Page 3 of 19 Useful Info: 1 atm = 101325 Pa, R = 8.314 Jmol1K1 = 0.08206 Latmmol1K1, 1 V 1 V , 1000 L = 1 m3, pV = nRT, = , = V T p V p T dU = dQ + dW, dW = pop dV, W =  p op dV , dS =
V1 V2 dQ , H = U + pV, T U H G = H  TS, A = U  TS, C V = , Cp = T p T V H trs H dp Strs V U p = , p = T  , V = T = + dT Vtrs TVtrs T p p T T V V T p (T, p ) = o (T ) + RT ln po , (T, p ) = o (T ) + RT ln f po p real (T, p )  perfect (T, p ) = k RT(Z  1) dp , dG = SdT + Vdp + i dn i p i =1 0 p l (T, p ) = * (T ) + RT ln l p* T = K f m where K f = G mix = nRT x i ln (x i )
i =1 k M l RT 2 M RT 2 , T = K b m where K b = l H fusion , l H vap, l RT + A 2cs cs M s where A 2 = RTVl
2 2M s Page 4 of 19 Other Useful Information: Table 1.5 contains experimental data. Table 1.6 contains values that are specific to the van der Waals equation of state. Page 5 of 19 1. Using the graphs on the following pages and the provided data tables, what is the value of the molar volume of nitrogen vapor if the pressure is 771.4 atm and the temperature is 378.9 K? (a) 0.064 L/mol (b) 0.64 L/mol
. . . . . (c) 6.4 L/mol
. . (d) 64 L/mol .
. (e) 640 L/mol Hence, Z = 1.6 . / Answer = A 2. Consider the following equation of state: 2 What is Tmax for this equation of state? (a) (b) (c) (d) (e) , Answer = E Page 6 of 19 Page 7 of 19 Note that solid lines represent reduced temperatures. Tr = 1.00 1.05 1.10 1.15 1.20 1.30 1.40 1.50 1.60 1.80 2.00 2.50 3.00 3.50 4.00 5.00 6.00 8.00 10.00 15.00 Page 8 of 19 Note that solid lines represent reduced temperatures. 3. Bromine has a critical pressure of 102 atm and a critical temperature of 311 C, respectively. Assuming the gas behaves as a van der Waals gas, what are the values of the van der Waals constants a and b in SI units? a (Pam6mol2) (a) (b) (c) (d) (e) b (m3mol1)
T = 311 + 273 = 584 K, & 9.37 x 105 0.272 0.962 2.69 9.50 5.80 x 107 3.13 x 105 5.87 x 105
. . . . . 3.13 x 102 5.87 x 102 4. 2.50 mol of substance A existing in the gaseous state at 600 K with an initial molar volume of 1.24 x 104 m3/mol is cooled into the two phase region (T = 300 K) where the molar volume of the liquid is 5.64 x 105 m3/mol and the molar volume of the gas is 1.24 x 103 m3/mol. How many moles of gas are present? (a) 0.0571 (b) 0.143 (c) 0.943 (d) 2.36 (e) 2.50 From the figure and the lever rule:
. . . . . . . Answer = B . Page 9 of 19 5. What is the difference in molar isometric heat capacity between water and carbon dioxide in the limit of infinite temperature? (a) 0 (b) (1/2)R (c) R (d) (3/2)R (e) 2R Water has a bent geometry. Carbon dioxide is linear. & , ,
, , . , 6. Copper sulfate exists as various hydrates [CuSO4xH2O (s)]. If you want dry copper sulfate, you must heat your sample to temperatures high enough to release water (> ~ 150 C) but low enough to not decompose CuSO4 (s) (< ~650 C). You put 73.6 mg of copper sulfate pentahydrate into a 100.0 mL sealed rigid container that is evacuated at room temperature (p = 0), and heat the sample to 400 C what is the pressure inside the container? You may assume ideal gas behavior. (a) 16.5 mPa (b) 48.8 mPa (c) 16.5 kPa (d) 48.8 kPa . . . (e) 82.3 kPa . / . . Answer = E . . . Page 10 of 19 7. Figure I contains two isotherms (T = 600K), one for a perfect gas (PV=nRT) and one for a gas governed by Equation of State 1 (b > 0). Figure II contains two isotherms (T = 300K), one for a perfect gas and one for a gas governed by Equation of State 2 (a > 0). (I) Equation of State 1 (II) Equation of State 2 p p V V Which of the following statements is correct for any isothermal reversible expansion with the same initial and final molar volumes for the ideal and real gas in each case?
(A) The molar work done by the ideal gas is larger for both Figure I and Figure II. (B) The molar work done by the ideal gas is larger for Figure I, but the molar work done by the real gas is larger for Figure II. (C) The molar work done by the real gas is larger for Figure I, but the molar work done by the ideal gas is larger for Figure II. (D) The molar work is done by the real gas is larger for both Figure I and Figure II. (E) There is insufficient information to answer this problem. Work = area under the curve. For Equation of State 1, the upper curve is the real gas as the denominator is smaller for a real gas than an ideal gas undergoing the same compression step. As a consequence, more work is done by the system expandng for the real gas. For Equation of State 2, the upper curve is the real gas as the first term is an ideal gas, and something is added for all molar volumes. Hence, more work is done by the system expanding for the real gas. Answer = D Page 11 of 19 8. Consider the following unbalanced reaction. 1 CH4 (g) + 3 F2 (g) 1 CHF3 (g) + 3 HF(g) If 3.20 grams of CH4 and 30.4 grams of F2 react until the limiting reagent is consumed inside a 0.500 m3 reaction vessel, what is the partial pressure exerted by the products if all gases behave ideally and isothermally at 273 C?
(a) 71.7 Pa (b) 89.6 Pa (c) 5.45x103 Pa (d) 7.26x103 Pa (e) 9.08x103 Pa Moles CH4 = (3.2 g)/(16 g/mol) = 0.200 mol Moles F2 = (30.4 g)/(38 g/mol) = 0.800 mol As one requires 3 mol F2/mol CH4 for the reaction CH4 is limiting (none left) Moles F2 consumed = 0.20 mol CH4 x (3 mol F2/1 mol CH4) = 0.600 mol Moles F2 remaining = 0.800 mol 0.600 mol = 0.200 mol Moles of CHF3 formed = 0.200 mol CH4 x (1 mol CHF3/1 mol CH4) = 0.200 mol Moles of HF formed = 0.200 mol CH4 x (3 mol HF/1 mol CH4) = 0.600 mol As all gases are ideal.
. . . . . Answer = D 9. Assuming one can write the enthalpy as an exact differential of two variables, S and p, yields Making use of the the fact this is an exact differential, indicate which of the following relationships is correct or that none of the above are correct.
(a) (d) (b) (c) (e) None of the above are correct Answer = C Page 12 of 19 10. On earth, the composition of air at sea level (ptotal = 1.000 atm, g = 9.807 m/s2, T = 300.0 K) is 79.00 mol% nitrogen, 20.00 mol% oxygen, and 1.000 mol% argon, what is the ratio of the partial pressure of oxygen to the partial pressure of argon at 100 km if we only consider these three gases and ideal gas behavior? Mg p i (z ) = p i (z = 0) exp  z RT (a) 0.01 (b) 10 (c) 20 (d) 50 (e) 456 . . . . . . . , . . . . . , . . Answer = E 11. Assuming ethene behaves as a van der Waals gas (a = 0.328 Pam6mol2, b = 4.13x105 m3mol1), what is the work associated with 2.00 moles of ethene undergoing isothermal reversible expansion from 1.00 x 104 m3 to 1.00 x 103 m3 at a temperature of 300.0 K? (a) 19.8 kJ (b) 7.97 kJ (c) 11.5 kJ (d) 11.8 kJ (e) 19.8 kJ . . . . . Answer = B .
. . . . . . . . Page 13 of 19 The next three problems use the following information: 2.000 moles of an ideal diatomic gas with a molar heat capacity of (5/2)R is originally at 750.0 K and the volume of the system is 0.3500 m3. The gas is subjected to a three step cycle. First, the gas is compressed from 0.3500 m3 to 0.1000 m3 in an isothermal reversible process. The gas is subsequently expanded isobarically back to its original volume. In a final step, the gas is cooled to its original temperature isometrically? Hint, start by drawing a picture. You may assume that the heat capacity is independent of temperature and the gas behaves ideally for all processes. 12. What is the work for the isothermal step? (a) 8.908 kJ (b) 7.811 kJ (c) 8.908 kJ (d) 15.62 kJ (e) 154.2 kJ 13. What is the energy change in the form of heat for the isobaric step? (a) 109.1 kJ (b) 38.97 kJ (c) 31.18 kJ (d) 77.94 kJ (e) 109.1 kJ 14. What is the internal energy change during the isometric (isochoric) step? (a) 77.94 kJ (b) 38.97 kJ (c) 38.97 kJ (d) 77.94 kJ (e) 769.3 kJ , Page 14 of 19 , , , , , , Cv = (5/2)nR n = 2.000 mol AB BC CA Cycle U
0 (TC TB) (TA TC) UBC + UCA = 0 Q WAB UBC WBC UCA QAB + QBC + QCA (=Wcycle as check) V 0.3500 m3 0.1000 m3 0.3500 m3 Q 15.62 kJ 109.12 kJ 77.94 kJ 15.56 kJ V 0.3500 m3 0.1000 m3 0.3500 m3 W nRTln(VB/VA) pB(VC VB) 0 WAB + WBC n = 2.000 mol A B C Cv = (5/2)nR n = 2.000 mol AB BC CA Cycle n = 2.000 mol A B C p pA = nRTA/VA pB = nRTB/VB pC =pB T 750 K 750K TC = (VC/VB)TB W 15.62 kJ 31.18 kJ 0 15.56 kJ T 750 K 750 K 2625 K U
0 77.94 kJ 77.94 kJ 0 p 35631 Pa 124710 Pa 124710 Pa Page 15 of 19 15. The following equation of state applies at low pressures: If Z = 1.0015 at T = 600.0 K and p = 5.830 x 104 Pa with a Boyle temperature of 190.0 K for the gas, calculate the values of a and b. a /Pam6K2mol2
(A) (B) (C) (D) (E) b /m3mol1 1.326 x 104 2.663 x 105 1.326 x 105 2.663 x 106 1.326 x 106 7559 1519 755.9 151.9 75.59 At TB, Furthermore, Hence, .
.
. . . . . . . and . . Answer = A Page 16 of 19 16. Indicate which of the following statements is false, or that all of them are correct for a perfect gas undergoing the reversible processes depicted in the figure. You may assume that the heat capacities are independent of temperature. (a) (b) (c)  (d)  0 0  (Diatomic) >   =    (Monatomic)
p C = pD p TA = TD C B D A
TB = T C VC VD VB VA p A = pB (e) All of the above are correct V (a) as 0, 0  True. Further explanation the step from B to C is compression. Work is just the area under the curve so this area is positive. (b) as 0, 0  True. Further explanation Area under the curve from A to C is work done on the system during compression and is positive. Area under the curve from C to D to A is work done by the system during expansion so the area is negative. As the negative area is greater than the positive area, the net work is negative. . Isothermal reversible work does not depend on the heat capacity of (c) the gas and pV = nRT does not care how many atoms make up the gas "molecule".  (Diatomic) =   (Monatomic) so Hence atoms count the same as molecules.  this is false. Answer = C (d) U is only a function of T for a perfect gas. TB = TC, both start at TA  True. 17. Indicate which of the following statements is false, or that all of them are true for a onecomponent gas at the critial point. (a) (d) 0 0 (b) (c) 1 (e) All of the above are true (a) True  p vs. V graph: critical point = inflection point: (b) True , , (c) True  At the critical point we form liquid (attractions winning over repulsions) so Z < 1. For a van der Waals gas we showed Zc = 3/8 with experimental values smaller still. (d) False As , a gas becomes ideal. Critical pressures are at relatively high p. Answer = D
Page 17 of 19 18. Indicate which of the following statements is correct, for the phase diagram of sulfur depicted in the figure. Sulfur can exist as a gas, a liquid, or solid. For the solid state there are two distinguishable crystalline solid phases (monoclinic and rhombic). Roman letters in the figure correspond to distinct points (a, b, c, d, e, f, g, h, i, j, k, l, m). Greek letters (, , , ) in the figures correspond to distinct phases. (a) The system has 1 triple point False 3 triple points, c, d, l (b) The system has 3 critical points False 1 critical point, f (c) m has 2 degrees of freedom True, F = 2 + C p = 2 + 1 1 = 2 (d) is a liquid False, = gas, = liquid, , = solid (e) C has 1 degree of freedom False, F = 2 + C p = 2 + 1 3 = 0 Answer = C 19. Indicate which step in the derivation of the relationship between Cp and CV below is incorrect, or that all of the above are correct. Assuming the pressure is constant, H = U + pV, and that U is a function of T and V: (a) (b) (c) (d) (e) All of the above are correct The rest follows (a) through (d) noting that we can divide (a) through by dT to obtain (b), and that we can obtain (c) from (b) by remembering that dp = 0, and that we can obtain (d) from (c) by noting that & Answer = E H = U + pV dH = dU + Vdp + pdV = dU + 0 + pdV dH = dU + pdV Page 18 of 19 20. Which of the following plots is not consistent with perfect (ideal) gas behavior? (a) T = Constant , M = Molecular Weight (b) p M 0 0 = m /V 0 0 1/T (c) Isothermal Reversible from Vi Vf (d) Any Process, CV is T independent
U Q T Cv 0
(e) 0 0
ln (V f /V i) 0 T (a) slope = RT/M decreases as M increases, (a) is false slope = 1, (b) is true (b) (c) slope = RT increases as T increases, (c) is true , slope = (d) decreases as decreases, (d) is true (e) not a straight line but a 0 0 p hyperbola so (e) is true Answer = A Page 19 of 19 ...
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This note was uploaded on 09/15/2009 for the course CHEM 3615 taught by Professor Aresker during the Fall '07 term at Virginia Tech.
 Fall '07
 AREsker
 Chemistry, Physical chemistry, pH

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