Unformatted text preview: CHEM 3615 Answer Key for Problem Set #2 Ideal and Real Gases Due September 8, 2008
Read Chapter 1, the Handout on the Barometric Distribution Law & Chapter 6 (Sections 6.1 and 6.2) regarding the phase rule. Problems 1 through 6 are due at the start of class and will be graded. 1. Show that , , and , are valid expressions for the (recall this problem assumes ideal gas barometric distribution law, behavior). Solution: , From the ideal gas law piV = niRT or pi = (ni/V)RT =
, , From the ideal gas law piV = niRT or pi = (ni/V)RT = ((mi/Mi)/V)RT = iRT/Mi
, or , Note to graders (the students do not need to show the subscript i's which were not in the original problem set). Page 1 of 17 2. The approximate sea level (1 atm) composition of the atmosphere is N2= 78.09%, O2=20.93%, Ar=0.93%, CO2=0.03%, Ne=0.0018% and He=0.0005% (all % values are mole%). Compute the partial pressures (in atm), the total pressure (in atm) and the composition of the atmosphere in mole fractions at altitudes of 6.2 km (Mount McKinley) and 160 km (approximate lowest value for a low earth orbit) assuming constant temperature (25C) with g = 9.807 m/s2 and ideal gas behavior. Solution: Note with respect to sig figs. For this problem I treated the exponential as if it were exact. No simple sig fig rules apply with logarithms and exponentials. Furthermore, I used only the number of sig figs in the composition at sea level for each gas. One exception is helium at 160 km where I kept extra digits because He is almost all of the gas present. The point for the gas at 160 km is that the mole fraction of helium is very close to 1. Values at Sea Level (0 km = 0 m), the po values are obtained from Dalton's Law: pi = xiptotal pN2,0 km = (0.7809)x(1 atm) = 0.7809 atm pO2,0 km = (0.2093)x(1 atm) = 0.2093 atm pAr,0 km = (0.0093)x(1 atm) = 0.0093 atm pCO2,0 km = (0.0003)x(1 atm) = 0.0003 atm = 3 x 10-4 atm pNe,0 km = (0.000018)x(1 atm) = 0.000018 atm = 1.8 x 10-5 atm pHe,0 km = (0.000005)x(1 atm) = 0.000005 atm = 5 x 10-6 atm Note: I will use atm for my units of pressure, but I will use SI units inside the exponential. This mixing of units will not cause any problems because the argument of the exponential is unitless (all units inside the exponential cancel). At 6.2 km = 6200 m
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, . . . . . . . . . ptotal = (0.3928 + 0.09547 + 0.0035 + 1 x 10-4 + 1 x 10-5 + 5 x 10-6) atm = 0.4919 atm xi = pi/ptotal (Dalton for mole percentages, multiply by 100) xN2,6.2 km = (0.3928 atm)/( 0.4919 atm) = 0.7986 (79.86%) xO2,6.2 km = (0.09547 atm) /(0.4919 atm) = 0.1941 (19.41%) xAr,6.2 km = (0.0035 atm) /(0.4919 atm) = 0.0071 (0.71%) xCO2,6.2 km = (1 x 10-4 atm) /(0.4919 atm) = 2 x 10-4 (0.02%) xNe,6.2 km = (1 x 10-5 atm) /(0.4919 atm) = 2 x 10-5 (0.002%) xHe,6.2 km = (5 x 10-6 atm) /(0.4919 atm) = 9 x 10-6 (0.0009%) Page 3 of 17 At 160 km = 160000 m = 1.60 x 105 m
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, . . . . . . . . . . . . . . . . . . . . Page 4 of 17 , . . . . . . . ptotal = (1.557 x 10-8 + 3.340 x 10-10 + 9.7 x 10-14 + 2 x 10-16 + 5 x 10-11 + 3.97 x 10-7) atm ptotal = 4.13 x 10-7 atm (I've ignored sig figs otherwise I would get a mole fraction of 1, where the number was irrelevant Ar, CO2 and Ne, I simply rounded) xi = pi/ptotal (Dalton for mole percentages, multiply by 100) xN2,100 km = (1.557 x 10-7 atm)/(4.13 x 10-7 atm) = 0.038 (3.8%) xO2,100 km = (3.340 x 10-8 atm) /(4.13 x 10-7 atm) = 0.000807 (0.081%) xAr,100 km = (9.7 x 10-14 atm) /(4.13 x 10-7 atm) = 2.3 x 10-7 (0.000023%) xCO2,100 km = (2 x 10-16 atm) /(4.13 x 10-7 atm) = 6 x 10-10 (0.00000006%) xNe,100 km = (5 x 10-11 atm) /(4.13 x 10-7 atm) = 1 x 10-4 (0.01%) xHe,100 km = (3.97 x 10-7 atm) /(4.13 x 10-7 atm) = 0.961 (96.1%) Graders: do not worry about significant figures on this problem. I basically assumed we knew the z values well enough that sigfigs were determined by the prefactor to the exponential and added an extra sig fig to He. 3. Using a computer, generate graphs of p/po vs. z and ln(p/po) vs. z (x-axis ranges of 0 to 50 km) for T = 300K and 1000K (both temperatures should be plotted on the same graph, clearly labeled) assuming the gas is N2 and obeys the barometric distribution law and g = 9.807 m/s2. For 300K, plot p/po vs. z and ln(p/po) vs. z (x-axis ranges of 0 to 50 km) for He, N2 and Ar (all three compounds should be plotted on the same graph, clearly labeled). Which graphs are linear and comment on the significance of the trends, slopes, and intercepts (this problem will count twice as much as the other problems). Solution: , , , : , Page 5 of 17 Part 1: Nitrogen gas at 300 K and 1000 K .
, . . . . . . , . 1.0 0.8 0.6 0.4 0.2 0.0 p/po N2, 1000K N2, 300K 0 10 20 30 40 50 z /km
Graph Analysis: 1. Exponential decay 2. Decreasing T leads to faster decay (reflects T in the denominator of the argument of the exponential) 3. y-axis intercept = 1 (exp (0) = 1). Page 6 of 17 .
, . . . . . . , . 0 -1 -2 -3 -4 -5 0 10 20 ln (p/po) N2, 1000K N2, 300K 30 40 50 z /km
Graph Analysis: 1. Linear representation 2. Smaller T has a more negative slope (bigger magnitude, reflecting the fact that T is in the denominator of the slope). Put another way, 1/300 > 1/1000!!! 3. Intercept is of course zero as ln (p/po) = ln (po/po) = ln (1) = 0!!! Page 7 of 17 Part 2: For He, N2 and Ar .
, . . . . . . . . , . . , . 1.0 0.8 0.6 0.4 0.2 0.0 p/po He, 300K Ar, 300K N2, 300K 0 10 20 30 40 50 z /km
Graph Analysis: 1. Exponential decay 2. As M increases, decay is faster (reflects M in the numerator of the argument of the exponential) 3. y-axis intercept = 1 (exp (0) = 1). Page 8 of 17 .
, . . . . . . . . , . . , . 0 ln (p/po) -2 -4 -6 0 10 20 He, 300K N2, 300K Ar, 300K 30 40 50 z /km
Graph Analysis: 1. Linear representation 2. Larger M has a more negative slope (bigger magnitude, reflecting the fact that M is in the numerator of the slope). Put another way, 39 > 28 > 4!!! 3. Intercept is of course zero as ln (p/po) = ln (po/po) = ln (1) = 0!!! Page 9 of 17 4. A vessel of volume 22.4 L initially contains 2.0 mol H2 and 0.5 mol O2 at 473.15K. The H2 and O2 react to form H2O until the limiting reagent is consumed. Calculate the partial pressures and the total pressure of the final mixture assuming ideal gas behavior. Solution: If all the O2 reacts: . As some H2 remains, O2 is limiting H2 Initial Change Final 2.0 mol -2 (0.5 mol) 1.0 mol . . . . . . . If the students used 1 significant figure because I put 0.5 mol O2 in the problem rather than 0.50 mol, that is fine too. . . . . . . . . If the students give 1.75 x 105 Pa, that is fine too!!! . . . . . . . O2 0.5 mol -(0.5 mol) 0 mol . H2O 0 mol 2(0.5 mol) 1.0 mol . Page 10 of 17 5. What is the volume occupied by 0.500 moles of ethane at 310 K and 45.0 atm assuming (a) ideal gas behavior, and (b) van der Waals behavior (a = 5.507 L2atmmol-2 and b = 0.0651 Lmol-1)? Assuming the van der Waals equation is a good representation of the data, what is the value of the compression factor [Z = pV/(nRT)]? [Hint: This is not trivial for part (b). A solution can be obtained in Microsoft Excel with the Tool Goal Seek (use the help function to find Goal Seek as it is located in different places in different versions of Excel). One can also express the van der Waals equation of state as a polynomial (cubic) set equal to zero, and then graphing the function to find where it crosses the x-axis, thereby satisfying the equation.] Solution: a) for an ideal gas: . . . . / b) for a van der Waals equation of state with a = 5.507 L2atmmol-2 and b = 0.0651 Lmol-1 No matter how hard we try, we cannot rearrange this equation to solve for V!!! How can we solve it? (1) Rearrange the equation to obtain a cubic function: By graphing f(V) vs. V, we can find the solution where the curve intercepts the x-axis. (2) We can rearrange the equation to obtain This equation can be solved iteratively. What do I mean? (a) First I re-write (b) Now I make my first guess using Vguess = V from the ideal gas law and obtain a Vresult. (c) If Vresult is the same as Vguess, we are done. (d) If Vresult is not the same as Vguess, Vresult becomes our new Vguess, and we solve for a new Vresult. Now we go to (c) and compare. We repeat (c) and (d) until the two numbers converge to be the same. (3) A third approach is to use the Goal Seek feature in Microsoft Excel. Page 11 of 17 (a) In this approach, we define a cell say B1 to be our guess of V (again, I would start with the result from the ideal gas law. (b) Next, I would rewrite the van der Waals equation of state as: In cell B2 I would write: Finally, I would use the Function Goal Seek in the What if Analysis (newest version of Excel and would change the precision for iterative calculations to 0.000000000001 under Excel options. Excel has the precision set by default to 0.0001). In the pop-up menu I would Set cell: B2; To value: 0; By changing cell: B1. This is a very useful skill to learn as you can re-arrange any equation to give you a value of zero!!!! You will also do this in PCHEM lab. Here: . . . . . . . The result is that B2 will be zero, and B1 will have the volume in units of m3 (1.69x10-4m3). Note that this number is smaller than for the ideal gas, so the gas is far from ideal. Finally, . . . . . Attractions are very important Page 12 of 17 6. Derive an algebraic expression of the compression factor of a gas (Z = pV/(nRT)) that obeys the equation of state, p(V+nb) = nRT, where b and R are constants. If the pressure and temperature are such that (V/n) = 5b, what is the numerical value of the compression factor? [Hint: Algebraically manipulate the equation of state until you can replace a pV/(nRT) term with a Z before making use of the fact that (V/n) = 5b] Solution: . Page 13 of 17 Practice Problems (These will not be graded). Please note that these problems were taken from various PCHEM textbooks. As a result some of the answers may make reference to equations that do not correspond to equations in your textbook!!! A. (a) Could 131 g of xenon gas in a vessel of volume 1.0 L exert a pressure of at least 20 atm at 25 C if it behaved as a perfect gas? (b) What pressure would it exert if it behaved as a van der Waals gas? Use the data in Table 1.6 (data section of your textbook - p. 993, 8th Edition, or Table 1.5 p. 1075 of the 7th Edition). [(a) Yes, p = 24 atm & (b) p = 22 atm] B. Calculate the pressure exerted by 1.0 mol C2H6 behaving as (a) a perfect gas and (b) a van der Waals gas when it is confined under the following conditions: (i) at 273.15 K in 22.414 L, (ii) at 1000 K in 100 cm3. Use the data in Table 1.6 (data section of your textbook - p. 993, 8th Edition, or Table 1.5 p. 1075 of the 7th Edition). [(ai) p = 1.0 atm, (aii) p = 8.2 x 102 atm, (bi) p = 1.0 atm, & (bii) p = 1.7 x 103 atm] Page 14 of 17 C. In an industrial process, nitrogen is heated to 500 K at a constant volume of 1.000 m3. The gas enters the container at 300 K and 100 atm. The mass of the gas is 92.4 kg. Use the van der Waals equation to determine the approximate pressure of the gas at its working temperature of 500 K. For nitrogen, a = 1.408 L2atmmol-2 and b = 0.0391 Lmol-1. (p = 140 atm) Page 15 of 17 D. The density of water vapor at 327.6 atm and 776.4 K is 133.2 gdm-3. (a) Determine the molar volume of water and the compression factor [Z = pV/(nRT)] from these data. (b) Calculate Z assuming the van der Waals equation is valid with a = 5.536 L2atmmol-2 and b = 0.03049 Lmol-1. [(a) (V/n) = 0.1353 Lmol-1, Z = 0.6957 & (b) Z = 0.649 ] E. Suppose that 10.0 mol C2H6 (g) is confined to 4.860 L at 27 C. Predict the pressure exerted by the ethane from (a) the perfect gas and (b) the van der Waals equation of state. Calculate the compression factor (Z) based on these calculations. For ethane, a = 5.562 L2atmmol-2 and b = 0.06380 Lmol-1. [(a) p = 50.7 atm, Z = 1 & (b) p = 34.8 atm, Z = 0.687] Page 16 of 17 F. A certain gas obeys the van der Waals equation with a = 0.50 m6Pamol-2. Its volume is found to be 5.00 x 10-4 m3mol-1 at 273 K and 3.0 MPa. From this information calculate the van der Waals constant b. What is the compression factor for this gas at the prevailing temperature and pressure? (b = 0.46 x 10-4 m3mol-1 & Z = 0.66) Page 17 of 17 ...
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