AnswerKey_ProblemSet_4_Fall2008

AnswerKey_ProblemSet_4_Fall2008 - CHEM 3615 Problem Set#4...

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Unformatted text preview: CHEM 3615 Problem Set #4 –Real Gases Due September 23, 2008 Finish reading Chapter 2. Problems 1 through 7 are due at the start of class and will be graded. 1. In the last problem set, you plotted Z values for methane as a function of temperature expressed in terms of nTB for the two-term virial expansion for a van der Waal's gas: pV a⎞p ⎛ Z= = 1+ ⎜b − ⎟ RT RT ⎠ RT ⎝ Plot the derivative of the compression factor with respect to pressure at constant temperature as a function of temperature (100K < T < 6000K) and make a second plot where the temperature varies between (100K < T < 1x106 K) using a logarithmic scale. In both cases, pick your y axis in such a way as to clearly see any maximum in temperature. At what temperature (Tmax) does the slope maximize in terms of the Boyle temperature? Prove this mathematically. Finally, what is the value of the maximum slope in terms of the van der Waal's constants a and b? {Units for Tmax are K, the maximum slope is in units of Pa-1} Solution: . Where . So we obtain: · · and . . · . · · The derivative of the compression factor with respect to pressure at constant temperature as a function of temperature is plotted as follows: the first plot has the temperature from 100K~6000K, while the second has T from 100 K ~ 1×106 K. Page 1 of 17 2.5 -1 2.0 1.5 1.0 0.5 0.0 -0.5 -1.0 0 TB T = 2TB Figure 1A (∂Z/∂p)T x 10 /Pa 9 2000 4000 6000 T /K Slope vs Temperature (100K~6000K) 2 T = 2TB Figure 1B (∂Z/∂p)T x 10 /Pa -1 1 TB 9 0 -1 -2 10 2 T→∞ 10 3 10 4 10 5 T /K Slope vs Temperature (100K~1×106K) Page 2 of 17 As we can see the graphs clearly have a maximum. I've already assigned the graphs, but how do we know the temperature where the initial slope on a Z vs. p plot reaches its maximum value (Tmax)? We must differentiate our expression for the slope as a function of T with respect to T and set the resulting function = 0 to find the extreme points: To prove it is a maximum, at T = Tmax = 2 TB we need to take the second derivative with respect to is really a maximum, then the second derivative evaluated at should be temperature. If a minimum. As the second derivative is less than zero, the extreme point does correspond to a maximum so we may write for a van der Waals gas that The maximum slope is the value of our expression for the slope evaluated at . . . . · · · . This expression means that big gas molecules with weak attractions have relatively large positive maximum slopes while small molecules with strong attractions have relatively small positive maximum slopes. Page 3 of 17 2. The law of corresponding states predicts two gases have identical states when they have identical reduced variables. Using the critical constants in the data section of your textbook, tell me when oxygen has the same compression factor as helium at 10.0 K, 60.0 mL/mol, and 1.50 atm. {Units for answers are: p in atm, molar V in L/mol, and T in K} Solution: For the Law of Corresponding States, we must first find the pressure, molar volume and temperature reduced by their critical values: , For Helium: . . . . . , · , . . For oxygen: . , . . . · . · , . · . . · · . . . · . . . . · . · · . . · . , . . , · · · · . . . . , · , . . , . · . · . . . · Note that Z is the same in both cases. This will not always be true. Here Zc for He and O2 is essentially the same (0.305 vs. 0.308, respectively). The bigger the difference in Zc, the worse the Law of Corresponding States functions. Page 4 of 17 3. Using the attached graph at the end of the problem set relating compression factors to reduced pressure, estimate the molar volume of ammonia (NH3) at 3450 atm and 1014 K. (Note that the reduced temperatures correspond to the solid lines). {Units are L/mol} Solution: . At this point, Z ~ 2.0 . . . . · · · . / For comparison consider (not part of problem) (a) An ideal gas: . · · · . / %error = [(0.0242 L/mol – 0.0482 L/mol)/(0.0482 L/mol)] x 100 = -50 % (b) For a van der Waals gas (a = 4.224 atmL2/mol2, b = 3.707 x 10-2 L/mol) . · . · · · . · · Use Goal Seek to solve for the molar volume. Here you will need to use the experimental value obtained from the graph as your initial guess (the ideal gas is too far off for Goal Seek to find a solution to the equation). . / %error = [(0.0541 L/mol – 0.0482 L/mol)/(0.0482 L/mol)] x 100 = +12.1 % Page 5 of 17 4. Consider 0.250 mole of ethane (C2H6, pc = 48.20 atm, ρc = 6757 mol•m-3, Tc = 305.4 K) vapor in a 60.0 cm3 sealed container at 400 K. The container is cooled to a temperature of 275K and undergoes partial condensation (ρliquid = 1.9ρc, ρgas = 0.25ρc), what is the mole fraction of liquid in the flask? If the temperature is cooled further to 244K (ρliquid = 2.2ρc, ρgas = 0.15ρc), what is the mole fraction of liquid? {Answers are mole fractions} Solution: & At 400 K, we have 0.250 mol gas in a 60.0 cm3 sealed container, hence . . . (1) at 275 K . Hence . . . (2) at 244 K . Hence . . . . & . . . . & . . . & . . . . & . . Page 6 of 17 and enter the Think about what is happening. We start as a gas (T > Tc) with a certain molar volume two-phase region as we cool the gas. The two-phase coexistence region (L+G) has an anisotropic shape. It is almost parabolic on the liquid side, but is stretched out toward larger molar volume with decreasing temperature on the gas side. As we decrease temperature, becomes closer and closer (relatively) to the corresponding value for the liquid state ( ), i.e. further and further (relatively) from the corresponding gas state ( ). As the lever rule is: we see that the denomintor grows as we cool the gas further, but the numerator grows faster. As a consequence the mole fraction of the liquid grows as we cool the gas to lower temperatures. I hope this makes intuitive sense. The colder we make the gas, the more liquid that forms!!! Page 7 of 17 5. A 7.50 g block of solid carbon dioxide, evaporates inside an initially 100. cm3 piston at a constant temperature of 70.0 °C and behaves ideally. Calculate the work done by the expanding gas against (a) a constant pressure of 1 atm, and (b) the gas expanding isothermally and reversibly to the same volume as in (a) (you may assume that the volume occupied by the solid is negligible). {Units for answers are: (a) J & (b) J} Solution: Consider a piece of dry ice (CO2) in a piston with . . . .° / . . . . a) expansion against a constant pressure of 1 atm, means pinitial = pfinal = 1 atm = pop For an ideal gas we can calculate Vf: . . · · . . . . Expansion the gas is doing work on the surroundings so the sign should be negative. b) for isothermal reversible process . . . Notice that the answer is also negative and that it is a larger negative number. Why? During isobaric expansion, the pressure must be smaller than p we started with (or nothing would expand, see the figure). The "initial pressure" in the figure is the pressure we would have if all the CO2 sublimed and we did not allow the container to expand!!! Then we let the gas expand extremely slowly. In this problem we are saying that sublimation is slow leading to isothermal reversible conditions. . . · · . Page 8 of 17 6. Consider 1.50 mol of carbon monoxide in a 100. cm3 piston at 450 K. If the gas undergoes isometric cooling to half its initial temperature and then undergoes isothermal reversible expansion to twice its initial volume, (a) compute the work done assuming the gas behaves ideally. Next, (b) compute the work done for the entire process assuming the gas behaves as a van der Waal's gas (a = 1.453 atmL2/mol2, b = 3.95 x 10-2 L/mol). (c) Compare and explain your results. {Units for answers are: (a) J & (b) J} Solution: Step 1: Isometric cooling, WAB = 0 as VA = VB. It makes no difference if the gas is ideal or real. Step 2: Isothermal reversible expansion: the result is most likely different as the functional forms of the pressure are different. . . . a) For perfect gas undergoing isothermal reversible expansion: . 2 2 b) for van der Waals gas where a = 1.453 atmL /mol & b = 3.95 x 10-2 L/mol . . · · · . . · · · . . · · . . . · . . · . . · · . . . . . · · . . . c) The work done by the repulsive term for the van der Waals gas is greater than for the ideal gas in part (a) (-3478 J). However, the attractive term requires that some energy (+1656 J) is used to separate the molecules and overcome the attractive interactions. Here we see that the net work is –1821 J (the work done on the surroundings). In this case, the real gas does less work on the surroundings than the ideal gas (attractions are more important than repulsions in this problem). Page 9 of 17 7. Consider 1.50 mol of an ideal monatomic gas at 300K undergoing isobaric expansion from 8.00L to twice its original volume. The gas then undergoes isometric cooling to 200K. Subsequently, the gas undergoes isothermal reversible compression back to its original volume. In the final step, the gas undergoes isometric heating back to 300K. Compute Q, W, and ΔU for each step and the entire process. {Units for answers are: Q, W, and ΔU are in J for all steps and the cycle} Solution: AB – isobaric expansion from 8.00 L to 16.0 L pB = pA, TA = 300 K < TB BC – isometric cooling from TB to 200K, dV = 0, pC < pA, TB < TC CD – isothermal reversible compression back to VA = 8.00 L dT = 0, pD > pC DA – isometric heating from 200K to 300K, V = 8.00 L Graph the steps Now we will set up two additional tables and fill them in with easy information: State (n = 1.50 mol) A B C D Process (n = 1.50 mol) AB BC CD DA Cycle (ABCDA) ΔU p V 8.00L (= 8.00 x 10-3 m3) VB =2VA=16.0 L (= 16.0 x 10-3 m3) VC =VB=16.0 L 8.00 L Q QAB= ΔUAB-WAB QBC= ΔUBC QBC=-WBC QDA= ΔUDA Qcycle=QAB+QBC+QCD+QDA Qcycle=- Wcycle (1st Law) T 300 K pB = pA 200 K 200 K W WAB=-pA(VB-VA) 0 0 Wcycle=QAB+QBC+QCD+QDA Wcycle=- Qcycle (1st Law) 0 ΔUcycle=ΔUAB+ΔUBC+ΔUCD +ΔUDA = 0 (state variable) The things in the table are either given information, or things you just need to know stone cold. Page 10 of 17 Now let's make use of the fact that we have an ideal gas to fill in the rest of these tables with pertinent equations. State (n = 1.50 mol) A B C D Process (n = 1.50 mol) AB BC CD DA Cycle (ABCDA) p pA = nRTA/VA pB = pA pC = nRTC/VC pD = nRTD/VD ΔU ΔUAB=CV(TB – TA) ΔUBC=CV(TC – TB) 0 ΔUDA=CV(TA – TD) ΔUcycle=ΔUAB+ΔUBC+ΔUCD +ΔUDA = 0 (state variable) V 8.00L VB =2VA=16.0 L VC =VB=16.0 L 8.00 L Q QAB= ΔUAB-WAB QBC= ΔUBC QDA= ΔUDA Qcycle=QAB+QBC+QCD+QDA Qcycle=- Wcycle (1st Law) T 300 K TB = (pBVB)/ (nR) 200 K 200 K W WAB=-pA(VB-VA) 0 WCD = -nRTCln(VD/VC) 0 Wcycle=QAB+QBC+QCD+QDA Wcycle=- Qcycle (1st Law) Why (note that the part in parentheses tells when the equation is valied): , , ∆ , , ∆ ∆ , , Page 11 of 17 Now we can calculate for each step: A to B: . With this we can calculate TB: . . . . · · . . . From which we may obtain ΔUAB and WAB: ∆ . . ∆ B to C: . . . · . . . . . · . · . · · . . . · . . . . · · ∆ . ∆ C to D: . . . · . · . · . . . ∆ . . D to A: ∆ . Cycle: ∆ ∆ ∆ ∆ ∆ . . ∆ . . . . . . . . . . . . · · . · Page 12 of 17 We can summarize the results in the following tables State (n = 1.50 mol) A B C D Process (n = 1.50 mol) AB BC CD DA Cycle (ABCDA) p . . . . ΔU . . 0 . 0 . . . . . . . Q . . 0 . 0 . V T 300 K 600 K 200 K 200 K W . Page 13 of 17 Practice Problems (These will not be graded). A. A chemical reaction takes place in a container of cross-sectional area 100 cm2. As a result of the reaction, a piston is pushed out through 10 cm against an external pressure of 1.0 atm. Calculate the work done by the system. (W = -1.0 x 102 J) B. A sample of 4.50 g of methane gas occupies 12.7 L at 310 K. (a) Calculate the work done when the gas expands isothermally against a constant external pressure of 200 Torr until its volume has increased by 3.3. L. (b) Calculate the work that would be done if the same expansion occurred reversibly. You may assume the gas is ideal. (a. W = -88 J & b. W = -167 J) C. In the isothermal reversible compression of 52.0 mmol of a perfect gas at 260 K, the volume of the gas is reduced to one-third its initial value. Calculate W for this process. (W = 123 J) Page 14 of 17 D. Show that the following functions have exact differentials: (a) f(x,y) = x2y + 3y2, and (b) f(x,y) = x cos(xy) (must be exact) E. (a) What is the total differential of z = x2 + 2y2 - 2xy + 2x - 4y -8? (b) Show that z(x, y) is exact. (a: dz = (2x – 2y + 2) dx + (4y – 2x – 4) dy & b: must be exact) Page 15 of 17 F. Starting with the mathematical definition of α and κ, and the total differential of molar V in terms of p and T, show that . Page 16 of 17 Page 17 of 17 ...
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