3.012 PS 3 THERMO MODEL SOLUTIONS Issued: Fall 05 Not due

Transcription

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4 3.012 S 3 HERMO MODEL SOLUIONS Issued: Fall 05 Not due HERMODYNAMICS 1. Why is it hard to make measurements at constant volume? We have discussed in class that experimentally, C v is difficult to measure due to thermal expansion. Let s quantify this difficulty: Suppose you have 1 mole of iron that has a volume of 7.31 cm 3 at 293 K. Determine the pressure that would have to be applied after this material is heated to 298 K (only 5 degrees warmer!) to compress it to the volume it had at 293K- thus maintaining constant volume. Data for Fe: α = 6.3x10-5 K -1-1 κ = 1.10x10-6 atm at 298K We start with the formula for the thermal expansion coefficient, in order to calculate the amount of expansion that occurs when the temperature is raised to 298K: "d = dv V " = 1 $ #V ' & ) V % # ( H # "d = V H # L V L dv V "( H $ L ) = "% = ln V H V L hus the volume after heating is: V H = V L e "# Next, we need to determine the pressure required to compress the material back to V L : " = # 1 % $V ( ' * V & $ ) d = # 1 "V dv + = # 1 " V L, V H dv V = 1 " lnv H V L = 1 " ln V L e-+ V L = -+ " = 286.3atm S 3 1 of 7 10/6/05

5 thus the difficulty in making constant volume measurements. Even this small re-compression requires quite a significant applied pressure. 2. hermal properties of gases. Calculate the thermal expansion coefficient and isothermal compressibility of an ideal gas. " # 1 % $V ( ' * V & $ ) V = nr % $V ( ' * = nr & $ ) " = 1 % nr( % ( ' * = ' V & ) & nr ) * % ' nr & ( * = 1 ) " # $ 1 &%V ) ( + V '% * V = nr &%V ) ( + = $ nr '% * 2," = $ 1 & V $ nr ) & )& ( + = ( + nr ) ( + = 1 ' 2 * ' nr *' 2 * 3. Cooking with a sealed pot. Engel and Reid problem 3.5, p. 59. We make use of the expression derived in class relating temperature/volume changes to pressure changes in a material: d = " # d $ 1 V# dv aking the a and k as approximately constant during this process, we can integrate the first term directly: 333 " # d ' $ = 2.04 %10&4 K &1 * ), 333K & 298K 298 ( 4.59 %10 &5 bar &1 + ( ) =156bar o integrate the second term, we need to estimate the volume change allowed during the expansion. Since the thermal expansion coefficient of the vessel is smaller than that of water, the expansion of the vessel will be the limiting factor. We can calculate the volume change allowed by the vessel: S 3 2 of 7 10/6/05

6 " vessel = 1 $ #V ' & ) V % # ( 4. Irreversibility of a free expansion. Recall in lecture 2 that we discussed irreversible processes, and gave several examples- one of these example irreversible processes was the expansion of a gas to fill a vacuum. All irreversible processes only occur in one direction spontaneously because it is in this direction that the entropy of the universe is increased, in accord with the second law. Let s show this for the irreversible expansion: consider the diagram below. An ideal gas is initially contained in one half of a chamber with adiabatic walls, the volume of the half of the chamber containing the gas is V o. he partition between the right and left halves is suddenly removed, and the gas expands isothermally and adiabatically to fill the entire space (V f = 2V o ). Show that this process fulfills the requirements of the second law for spontaneity by calculating the entropy change in the gas, the surroundings, and the universe, and show that for the process to run in reverse (spontaneous collapse of the gas back to one half of the chamber) will violate the second law. ds universe = ds system + ds surr = ds system = du + dv Because the temperature is held constant and for an ideal gas U = U(), the internal energy is constant: " nr % $ ' # V & ds universe = 0 + dv = nr V dv ()S universe = nrln V f V i = nrln2 > 0 Because the entropy change in the universe is positive, the process is spontaneous. For the process to run in reverse, the entropy change is negative: S 3 3 of 7 10/6/05

7 "S universe = nrln V f V i = nrln 1 2 = #nrln2 < 0 5. Let s revisit the liquid bismuth adiabatic cooling problem we examined on the last problem set. he physical situation is described again below, and the physical data for the system is also given. From your calculations on the last problem set, you showed that the alumina crucible and bismuth equilibrate at a final temperature of 433 K. Demonstrate that this process (cooling of the bismuth, warming of the alumina crucible) obeys the second law of thermodynamics, by calculating the total entropy change of the universe for the process. wenty kg of liquid bismuth at 600 K is introduced into a 10 kg alumina (Al 2 O 3 ) crucible (initial temperature 298K), filling the crucible to the top; the crucible and bismuth are then surrounded by adiabatic walls (illustrated below) and the system is allowed to equilibrate. At equilibrium, according to the zeroth law, the temperatures of the bismuth and alumina crucible must be equal. Use the following thermodynamic data: C p alu min a,solid = C p Bi,solid = C p Bi,liquid = J K " mole J K " mole J K " mole m alu min a = 2327K m Bi = 544K "H m Bi =10,900 J mole o show the process is spontaneous, we must calculate the entropy change in the universe: "S universe = "S system + "S surr = "S system + 0 "S universe = "S Bi + "S Al2 O S 3 4 of 7 10/6/05

8 We know the entropy change in the surrounding is zero because the only energy transfer occurring here is heat transfer, and the samples are surrounded by an adiabatic boundary- thus no heat is moving into the surroundings to change their entropy. f Al 2 O 3 "S Al2 O 3 = n C p # d = # (98.08) i ( ) d "S Al2 O 3 = 4.14 J K 544 "S Bi = n C p # d $ n"h m 600 l 433 s + n C p # d = $2.79 J K 544 "S universe = "S Bi + "S Al2 O 3 =1.35 J K Since the entropy change of the universe is positive, the process is predicted to be spontaneous. 6. Confirming your intuition using the second law. Using your ability to calculate entropy and enthalpy changes at phase transitions and the data given below for nickel, consider the following possible processes and confirm your intuition by a calculation showing whether the process is spontaneous or not according to the second law. he key to this problem is to correctly identify what heat is transferring between the system and the surroundings in these processes, and to carefully calculate the entropy changes for the system. Also critical: pay attention to the number of significant figures you carry in your calculation. a. One mole of solid nickel at K isothermally transforms completely to liquid at this temperature. he surroundings are a heat bath at K, and the process occurs at constant pressure. b. One mole of solid (superheated) nickel at 1736 K isothermally transforms completely to liquid at this temperature. he surroundings are a heat bath at 1736 K, and the process occurs at constant pressure. C p s = J C L p = K " mole J K " mole m =K "H m =17.47 kj mole a. he transformation we are concerned with is shown schematically below on qualitative plots of the entropy and enthalpy of the system near the temperature of interest. Our process is isothermal and isobaric. o test for spontaneity, we need, as in the above problems, to calculate the entropy change in the universe: S 3 5 of 7 10/6/05

9 "S universe = "S system + "S surr he entropy change in the system will be given by the entropy change along the dashed line from A to B on the entropy plot below. We calculate the length of this line segment as illustrated graphically in the plot: "S system = "S m # "S 1 + "S 2 "S m = "H m m =10.12 J K l "S 1 = n C p $ d = $ (1) (38.91) d = 38.91ln = J K s "S 2 = n C p ( ) $ d = $ (1) d =16.49ln ( #) = = J K %"S system =10.18 J K Because the surroundings are at constant temperature, and we assume no work is occurring, the entropy change in the surroundings derives completely from the heat transfer from the system to the surroundings: "S surr = #q sys = #"H sys (K) "H sys = "H m # "H 1 + "H 2 =17,470 # =17.57kJ l "H 1 = $ nc p d = $ (1)(38.91) d = 389.1J s "H 2 = $ nc p d =16.49( #) ( 2 # 2 ) = = 486.7J 2 %"S surr = #10.23 J K Finally, the total entropy change of the universe is: "S universe = "S system + "S surr =10.18 #10.23 = #0.05 J K Since the entropy change is less than zero, this process will not occur S 3 6 of 7 10/6/05

10 b. Now, we follow the exact same scheme to predict the spontaneity of melting a superheated solid nickel sample: We must show that the entropy change of the universe in this process is positive: "S universe = "S system + "S surr 1736 s "S system = "S m # n C p $ d + n C p $ d =10.12 # =10.06 J K "S surr = #q sys (1736K) = #"H sys (1736K) q sys = "H m # 1736 s nc p %"S surr = #10.01 J K l $ d + $ nc p d = 17.37kJ l %"S universe = 0.05 J K which is a spontaneous process S 3 7 of 7 10/6/05