This preview shows page 1. Sign up to view the full content.
Unformatted text preview: CHEM 3615 Answer Key for Problem Set #3 –Real Gases Due September 16, 2008 We will be completing Chapter 1 during the week of September 3. Start reading Chapter 2. Problems 1 through 6 are due at the start of class and will be graded. 1. The experimental critical constants for HCl are 374 °C, 81.5 atm, and 81.0 cm3/mol. Calculate the values of a, b, and R using the van der Waals equation. Compare the value of R with the correct value [0.08206 Latm/(mol K)] and note the % difference. Compute the constants a and b only from pc and Tc using the correct value of R. Using these values and the correct value of R, calculate the critical volume and compare it with the experimental value. Solution: For a van der Waals gas: , . . Part 1: . From pc: . From Tc: . . % Error % .% Part 2: Taking the ratio of Tc to pc: . . · . · . · . · · . · · · . . . , , , ° . , · · . Page 1 of 15 From pc: . From : . % Error %
, , , . . . . . . The large errors in the problem arise from two factors. First, the van der Waals equation is not quantitatively correct in the vicinity of the critical point. Second, Alan forgot to change the critical temperature from a previous problem. The actual critical tempertaure is 324.7 K. Below I provide the results for using the correct critical temperature. Notice that the results are a bit better but still suck. Again, the problem is that the van der Waals equation of state has good qualitative predictions for things taking place at the critical point but fails quanitatively. For a van der Waals gas: , , . . % For HCl, pc = 81.5 atm, Tc = 324.7 K, and Part 1: . From pc: . From Tc: . . % Error % .% Part 2: Taking the ratio of Tc to pc: . . . . . . · . · . · · . · · · · · . · . . Page 2 of 15 From pc: . From : . % Error %
, , , . . . . . . .% The large errors arise from the fact that the van der Waals equation is not quantitatively correct in the vicinity of the critical point (even though it has many good qualitative features). 2. If Z = 0.9948 at 327 °C and 1.00 atm and the Boyle temperature of the gas is 1900 K, estimate the values of a and b. (Use the first two terms Z = 1 + B'p from the van der Waals equation of state where B' is a function of a and b as derived in class). Solution: State 1: Z = 0.9948, T = 327 °C + 273 = 600 K, p = 1.00 atm = 101325 Pa State 2: TB = 1900K!!! For a two-term virial expansion in terms of p: At the Boyle temperature: Hence a = RbTB and we can plug this into the equation for state 1: Now we solve for b: Page 3 of 15 . . . . . . . · · . · · . · · · . . · · · · and substituted for Note that we also accepted credit if you used the first equation. Numerically, you obtain the same result because Z is so close to 1. In general, this is not the equation you would use as normally, we start with Z = 1 + B'p + C'p2 + D'p3 + ... and drop p2 and higher order terms. Recall that we tossed Z when we added higher order terms during the derivation. In the future you should use the way I solved the problem. 3. The compression factor for methane is given by Z = 1 + Bp + Cp2 + Dp3. If p is in atm, the values of the constants are as follows: T/K 200 1000 B -5.74 x 10-3 0.189 x 10-3 C 6.86 x 10-6 0.275 x 10-6 D 18.0 x 10-9 0.144 x 10-9 Plot Z as a function of p for the range of 0-1000 atm. Comment on the trends and what they mean. Solution: p /atm
0 100 200 300 400 500 600 700 800 900 Z at 200 K
1 0.5126 0.2704 0.3814 0.9536 2.095 3.9136 6.5174 10.0144 14.5126 Z at 1000K
1 1.02179 1.04995 1.08534 1.12882 1.18125 1.2435 1.31644 1.40093 1.49783 Page 4 of 15 20 200K 1000K 15 Z 10 5 0 0 200 400 600 800 1000 p /atm
At low temperatures, attractive interactions are important resulting in Z < 1, i.e. 200 K lower than the Boyle temperature. At 1000 K, Z > 1 for all p. This result indicates that T > TB. At 1000 K repulsive terms are more important than attractive terms. Note to graders: If students choose to use a smaller Z range to highlight the behavior around Z = 1, that is also acceptable. Page 5 of 15 4. Derive the relationship between the constants a and b of the Dieterici equation: Solution At the critical point Also Or (note that we can divide through by the exponential term as it is not zero) At this point, lets cheat and look at the answer. If we plug into our expression for 2c, we should recover Tc telling us we have not made a mistake so far: !!! Also Page 6 of 15 Note that the first two terms in brackets come from applying the product rule to the first and the three terms in the second set of brackets come from applying the term of product rule to the second term of . Again we can divide through by the exponential term because it is not zero Now let's give everything a common denominator: We can eliminate the denominator by multiplying each side of the equation by We can also divide both sides of the equation by RTc to obtain: Now we make the substitution Such that Page 7 of 15 Next we would plug into as I did above to see that Now all we need to do is find the expression for pc so we use the original expression for the pressure: Now for Zc . . Notice that this value is smaller than the value of 3/8 (0.375) predicted for the van der Waals equation of state and is actually closer to experimental values. For small non-polar gases it is a bit too low, for hydrocarbons it is almost dead on, but it is still on the high side for polar gases. 5. At low pressures the Berthelot equation has the form in which a and b are constants. Find the expression for the Boyle temperature in terms of a, b, and R. Find the expression for α the coefficient of thermal expansion as a function of T and p only. Hint: Try for a final answer of the form that will be a correction of α = 1/T, the result for an ideal gas: 1 ⎛1+ • • • ⎞ α= ⎜ ⎟ where the numerator and denominator are different. T ⎝1+ • • • ⎠ Solution: First consider the Boyle Temperature: So: Page 8 of 15 This is a two-term virial expansion. At TB, B' goes to zero so we want to find: The original equation was actually the two-term expansion of the Berthelot equation of state. Don't memorize the result, understand and be able to reproduce the steps to obtain the Boyle temperature from any general equation. Next, First factor a T out of the denominator: Now, let's multiply the top and bottom of the equation by p/R to get closer to the final form: / / This could also be written as: These results are interesting for several reasons: 1) TB is not always a/Rb (van der Waals Eq. result), it is model dependent. 2) Interactions cause α to deviate from 1/T. Looking carefully at the equation for α, we see that we recover 1/T as T approaches infinity. 3) At the Boyle temperature, the denominator of the correction is 1, but the numerator is not. This means that a Berthelot gas at the Boyle temperature is not ideal with respect to all of its properties. While we may use pV = nRT for (p,V,T) data at the Boyle temperature, but must remember that not all properties of the gas are ideal. We will see this again and again throughout the semester. Page 9 of 15 6. Consider the compression factor, Z, of a real gas (methane, a = 2.27 x10-1 Pa•m6•mol-2, b = 4.28 x10-5 m3•mol-1) in terms of a two-term virial expansion for a van der Waal's gas:
Z= pV a⎞p ⎛ = 1+ ⎜b − ⎟ RT RT ⎠ RT ⎝ Plot Z vs. p for the following temperatures: 1/4TB, 1/2TB, TB, 3/2TB, 2TB, 5/2TB, 10TB, and T → ∞ where TB is the Boyle temperature (let p go from 0 to 100 atm). Solution: For the Boyle temperature: . . . · · . · · · · · · · 0 0 For the initial slope of the two-term virial expansion. Page 10 of 15 1.02
5/2TB 2TB 3/2TB 1.01 10TB Z 1.00 T→∞ TB 0.99
1/2TB 1/4TB 0.98 0 20 40 60 80 100 p /atm
As seen in the figure, T < TB yields Z < 1 in the limit where p approaches 0 indicating that attractive interactions are the most important. At T = TB, the initial slope is zero and the gas exhibits ideal gas behavior with respect to compression factor (Z). You may safely use pV = nRT as long as p is not so large that higher order terms in the virial expansion become important. For T > TB repulsions dominate and the initial slope is positive. However, we know that as T goes to infinity, we must recover ideal gas behavior (Z = 1). As we see in the graph, the initial slope appears to maximize in the vicinity of 2TB. For the van der Waals gas, Tmax is indeed 2TB (something you will show in the next problem set). Later we will also learn that Tmax is known as the inversion temperature (Tinv) for the Joule-Thomson effect. Hence we see that for TB < T < (Tmax = 2TB =Tinv for the van der Waals model), the initial slope is positive and increases with T. For (Tmax = 2TB for the van der Waals model) < T < infinity, the initial slope is still positive, but decreases with increasing T. Page 11 of 15 Practice Problems (These will not be graded). A. The critical constants of methane are pc = 45.6 atm, molar Vc = 98.7 cm3.mol-1, and Tc = 190.6 K. Calculate the van der Waals parameters of the gas and estimate the radius of the "spherical" molecules. (a = 1.33 L2atm.mol-2, b = 32.9 cm3.mol-1, r ~ 0.24 nm) B. Use the van der Waals parameters for chlorine from the data tables at the end of your textbook to calculate approximate values of (a) the Boyle temperature of chlorine, and (b) the radius of a Cl2 molecule assuming it is a sphere (note that this approximation is worse for Cl2 than methane, why?). (TB = 1.4 x 103 K, r ~ 0.28 nm) Page 12 of 15 C. Suggest the pressure and temperature at which 1.0 mol of (a) NH3, (b) Xe, and (c) He will be in states that correspond to 1.0 mol H2 at 1.0 atm and 25 deg. C. Use the critical parameters in data tables in Chapter 1 or at the end of your textbook. (a: p = 8.7 atm, T = 3.64 x 103 K; b: p = 4.5 atm, T = 2.60 x 103 K; and c: p = 0.18 atm, T = 46.7 K) Page 13 of 15 D. Calculate the volume occupied by 1.00 mol N2 using the van der Waals equation of state in the form of a two-term virial expansion with respect to p at (a) its critical temperature using the following data: Tc = 126.3 K, a = 1.408 L2atm.mol-2, and b = 0.0391 L.mol-1. Assume that the pressure is 10 atm throughout. (b) What is the Boyle temperature? Use the (a: V = 0.939 L, b: TB = 439 K, note that this value differs from tabulated values of 327.2 K, why?) Page 14 of 15 E. The mass density of water vapor at 327.6 atm and 776.4 K is 1.332 x 102 g/L. Given that for water Tc = 647.3 K, pc = 218.3 atm, a = 5.536 L2atm.mol-2, and M = 18.02 g.mol-1, calculate (a) the molar volume. Then calculate the compression factor from (b) the data, and (c) the two-term virial expansion of the van der Waals gas with respect to p. (a: molar V = 0.1353 L.mol-1, b: Z = 0.6957, c: 0.71) Page 15 of 15 ...
View Full Document