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Unformatted text preview: Name Section Rec TA Ch 1b, Solution Set for Problem Set Four S i d e 1 of 24 Due Friday, Jan. 28, 2011 at 4 PM in the Drop Box 1. va n de r W a a ls P a r ame t e rs (15 poi n ts) Recall van der Waals equation of state for gases, where is pressure, is volume, is the number of moles, is the gas constant, and and are parameters. OGN, p. 118; lecture notes is absolute temperature, pg. 2 of lecture 7), but what happens if you need these difficult. In this problem you will find relationships between the van der Waals parameters and the critical properties of a substance. The critical temperature, , is the temperature above whi ch the vapor of a substance cannot be liquefied. The critical pressure, , is the pressure required to liquefy the gas a t the critical temperature. The critical volume, , is the volume of a fixed mass of fluid at a table of critical values than van der Waals parameters. (a) Express van der Waals equation in terms of the specific molar volume , where . (2 poi n ts) (b) Experimental data on all real substances at their critical temperature indicates that the isotherm passes through a horizontal inflection point at the critical pressure of a substance. Mathematically, this statement means Use this mathematical relationship to express the van der Waals constants and in terms of critical properties. Since tables usually only report critical temperatures and pressures, you should express your answer so that only , and appear. (13 poi n ts) Name Section Rec TA Ch 1b, Solution Set for Problem Set Four S i d e 2 of 24 Due Friday, Jan. 28, 2011 at 4 PM in the Drop Box So l u t i o n 1 a Start with van der Waals equation, as given, Substitute specific molar volume in pressure term directly, Divide both sides by the number of moles and use definition of specific molar volume, (2 poi n ts) Sol u t ion 1b Start with the result of part (a), Solve for pressure, Take the first derivative of zero, with respect to , evaluate at critical point, and set equal to (2 poi n ts)
Take the second derivative of to zero, [1] with respect to , evaluate at critical point, and set equal (2 poi n ts)
Solve for a in Equation [1], [2] [3] Substitute Equation [3] into Equation [2] and solve for b, Name Section Rec TA Ch 1b, Solution Set for Problem Set Four S i d e 3 of 24 Due Friday, Jan. 28, 2011 at 4 PM in the Drop Box (2 poi n ts) Substitute this result into Equation [3], [4] (2 poi n ts) [5] Substitute Equations [4] and [5] into the result for Part (a) evaluated at the critical point, (2 poi n ts) Substitute this results into Equations [4] and [5], and (3 poi n ts) 1.5 pt . e a ch Name Section Rec TA Ch 1b, Solution Set for Problem Set Four S i d e 4 of 24 Due Friday, Jan. 28, 2011 at 4 PM in the Drop Box 2. M or e E qu a t ions of St a t e (15 poi n ts) The virial equation of state is derived from statistical thermodynamics, where is the specific molar volume, is the second virial coefficient, and is the third virial coefficient. The second virial coefficient accounts for twobody interactions, the third virial coefficient accounts for threebody interactions and so on. All of the virial coefficients depend on temperature. (Note: When there are no molecular interactions, and the virial equation of state reduces to the ideal gas law.) (a) Manipulate the van der Waals equation of state so that it resembles the virial equation of state to find as a function of , , , and . You will need to use the fact that for small values of . (4 poi n ts) (b) Using the result from (a), describe what happens at large temperatures. (2 poi n ts) (c) The Boyle temperature is defined as the temperature at which the Boyle temperature of N2 using your result in part (a) and values of the van der Waals parameters presented in class. (3 poi n ts) (d) In lecture you saw a plot of versus , where is a parameter in the van der Waals equation of state, is a parameter in the LennardJones potential, and is the intermolecular distance. Even though the plot showed a reasonable correlation, you might choose to plot versus instead. What value of would you choose for the exponent? (Hint: ; consider what represent.) Explain your answer. (2 poi n ts) (e) Jones potential to find Where the first two terms in the summation, is the gamma function. Keeping only In class you saw a plot of versus that showed reasonable correlation. What combination of parameters is a better choice for the yaxis t answer from part (a); can be ignored; how are and related?) (4 poi n ts) Name Section Rec TA Ch 1b, Solution Set for Problem Set Four S i d e 5 of 24 Due Friday, Jan. 28, 2011 at 4 PM in the Drop Box So l u t i o n 2 a Start with the result in (1a), Rearranging, (2 poi n ts) Using with , Comparing to the virial equation of state, Answer (2 poi n ts) Sol u t ion 2b At high temperatures, , attractive forces between molecules are minimal, and the molecular interactions are dominated by excluded volume interactions and . (2 poi n ts for stating or showing that ) Name Section Rec TA Ch 1b, Solution Set for Problem Set Four S i d e 6 of 24 Due Friday, Jan. 28, 2011 at 4 PM in the Drop Box So l u t i o n 2 c Start with solution for (2a), Boyle temperature means, (1 poi n t) Substituting values for nitrogen (using values from lecture notes), A nswe r (2 poi n ts) No deduction for sig. figs. unless unreasonable
The Boyle temperature for nitrogen is actually 332 K. The van der Waals parameters for nitrogen capture general thermodynamic behavior quite well but sometimes miss specific measurements by wide margins. Be careful when using correlations and understand their limitations. Name Section Rec TA Ch 1b, Solution Set for Problem Set Four S i d e 7 of 24 Due Friday, Jan. 28, 2011 at 4 PM in the Drop Box Sol u t ion 2d The van der Waals parameter, , is a measure of excluded volume of a single molecule. The LennardJones parameter, , is a measure of the intermolecular dist anc e at which the potential energy equals zero. Since the potential curve is so steep at small intermolecular distances, is often viewed as the di ame t e r of the molecule. If we take this latter view, then Therefore, we will take , and a plot of versus should yield a better correlation than versus . (1 poi n t for st a t i n g t h a t n = 3, 1 poi n t for e xpl a n a t i on ) (C2H2, C2H4, C2H6, C3H8, C6H6, CH4, CO, CO2, H2, He, N2, nC4H10, O2, SO2): The graph for vs. is slightly more linear, showing a good direct correlation. So l u t i o n 2 e Take the answer from part (a) and set it equal to the given equation,
(2 poi n ts) Recognize that the parameters, and , will be temperaturedependent: Recognize that and solve for on righthand side, Name Section Rec TA Ch 1b, Solution Set for Problem Set Four S i d e 8 of 24 Due Friday, Jan. 28, 2011 at 4 PM in the Drop Box Since we want to plot a versus some combination of LennardJones parameters, we can ignore temperature and all constants, (2 poi n ts) By using a more appropriate combination of parameters for the yaxis, a much more linear relationship is produced. Name Section Rec TA Ch 1b, Solution Set for Problem Set Four S i d e 9 of 24 Due Friday, Jan. 28, 2011 at 4 PM in the Drop Box 3. I soba r i c C ool i n g (12 poi n ts) Imagine that 2.00 mol of argon is confined by a slowmoving, massless, frictionless piston in a cylinder. The argon is initially at a pressure of 1.00 atm and temperature of 398 K and is cooled to 298 K. You may assume the external pressure remains constant at 1.00 atm throughout the cooling process. You may assume that argon is an ideal gas.
(a) What is the work done on the syst em, w? (4 poi n ts) (b) What is the heat absorbed by the syst em, q? (3 poi n ts) (c) What is the change in the internal energy of the system, U ? (Note that OGN uses E for internal energy while Prof. Heath uses U . You will be asked to use U for internal energy for problems sets, quizzes and exams.) (2 poi n ts) (d) What is the change in the enthalpy of the system, H ? (3 poi n ts) So l u t i o n 3 a In its most general form, work is defined as, where is the wor k done on the syst em, is the resisting pressure, and is the system volume. When we are analyzing the system, the resisting pressure is simply the system pressure, , Since the piston is frictionless, massless, and moving slowly, the system pressure and the external pressure, , are equal at all times; thus, Since the external pressure is constant, we can pull pressure out of the integral, Evaluate the integral, (1 poi n t) where and are the final and initial system volumes, respectively. [1] That was a long way to get to Equation [7.1] in OGN if you just write down the equation an but looking up an equation is not a very intuitive, logical way to solve thermo problems. If you were wondering, these thermo equations will be listed in tables on quizzes and exams but without description. Assume that argon is an ideal gas. Substitute gas law for volume in Eq [1], Name Section Rec TA Ch 1b, Solution Set for Problem Set Four S i d e 10 of 24 Due Friday, Jan. 28, 2011 at 4 PM in the Drop Box The system pressure remains constant and is equal to the external pressure; so pressure divides out of the equation. Since the system is closed, no mass escapes and the mass of argon in the container remains constant. Simplification yields: (1 poi n ts) Substitute known values to solve for work, (1.5 poi n t a n swe r ; 0.5 pt for 3 sig. f igs.) Sol u t ion 3b Since the system is isobaric, we use the heat capacity at constant pressure to find the heat transferred to the system,
(1 poi n t) We know that a n y ide a l gas obeys (OGN, p. 208), and that a monatomic ideal gas has (OGN, Eqtn [7.8]); so (1 poi n t) (1 poi n t a n swe r ; no pts assigned to sig figs) So l u t i o n 3 c From the first law of thermodynamics, the change in the internal energy is defined as (OGN, Eqtn [7.3]):
(1 poi n t) Name Section Rec TA Ch 1b, Solution Set for Problem Set Four S i d e 11 of 24 Due Friday, Jan. 28, 2011 at 4 PM in the Drop Box (1 poi n t a n swe r ; no pts assigned for sig figs) already rounded them to the proper number of significant figures nonrounded answers or develop a new equation, as shown here. you should use the Sol u t ion 3d From the definition of enthalpy (OGN, Eqtn [7.7]),
(1 poi n t) we find the change in enthalpy, Using the ideal gas law for the energy, term and the symbolic result from Part (c) for internal (1 poi n t) (1 poi n t a n swe r ; no pts assigned for sig figs) Notice that for this system, just as we should expect for such a system when the work is entirely pressurevolume work and the external pressure is constant. GRADER: calculations. Name Section Rec TA Ch 1b, Solution Set for Problem Set Four S i d e 12 of 24 Due Friday, Jan. 28, 2011 at 4 PM in the Drop Box 4. E xpa nsion  C ompr ession C yc l e (35 poi n ts) Initially, 3.00 mol of oxygen gas is at a pressure of 1.00 atm and a temperature of 405 K. It expands adiabatically and reversibly until the pressure is reduced to 0.600 atm, and it is then compressed isothermally and reversibly unt i l the volume r e turns to i ts or i gina l va lue. Calculate:
(a) the final temperature (units: Kelvin) of the gas (12 poi n ts) (b) the final pressure (units: atmospheres) of the gas (5 poi n ts) (c) the total heat added (units: Joules) to and the total work done (units: Joules) on the oxygen in this process (14 poi n ts) (d) the energy change, U , (units: Joules) and the enthalpy change, H , (units: Joules) between the final and initial states (4 poi n ts) Take cP(O2) = 29.4 J K1 mol1. Assume that oxygen acts as an ideal gas. So l u t i o n ( G e n e r a l ) : you are given for each state Defining states (and subscripts for variables):
State 1: Initial 2: Intermediate 3: Final Pressure 1.00 atm 0.600 atm Temp. 405 K Volume Note After adiabatic reversible expansion from state 1 After isothermal compression from state 2 to the original volume V1 V2 V3 = V1 T2 T 3 = T2 P3 So l u t i o n 4 a Find the initial volume, , by using the ideal gas law and by recognizing that moles of gas remain constant throughout the entire process since the system is a closed system. (2 poi n ts) For an adiabatic reversible expansion, we know that (OGN, Equation [7.14]) (2 poi n ts) where is the ratio of specific heats, [1] We know that a n y ide a l gas obeys (OGN, p. 208), Name Section Rec TA Ch 1b, Solution Set for Problem Set Four S i d e 13 of 24 Due Friday, Jan. 28, 2011 at 4 PM in the Drop Box and we are given that cP(O2) = 29.4 J K1 mol1, (2 poi n ts; keep many sig. figs. since this is an intermediate calculation) Solve for using : Substitute Equation [1], (2 poi n ts) gas next undergoes an isothermal expansion and we need to know the final temperature. Substitute the ideal gas law into the above equation and solve for T2, (1 poi n t for c a l c) [2] (keeping more sig figs since this is an intermediate calculation) Name Section Rec TA But we know that Ch 1b, Solution Set for Problem Set Four S i d e 14 of 24 Due Friday, Jan. 28, 2011 at 4 PM in the Drop Box (due to isothermal expansion), so (2 poi n ts for a n swe r ; 1 pt for sig f igs) Sol u t ion 4b Calculate the final pressure using the ideal gas law, (1 poi n t) Isothermal compression ( ) to the original volume ( ), (2 poi n ts) Substitute Equations [1] and [2], [3] (2 poi n ts a n swe r ; no pts assigned to sig figs) Hopefully, in Parts (a) and (b) you can see why you want to perform symbolic mathematics quite simple relationships that are very easy to evaluate. Notice that we never used the value of the gas constant (nor the number of moles) to find the final temperature and pressure. The number of moles is an extensive property and will affect energy ( U , H ), work and heat transfer. Temperature and pressure are intensive properties. If you used moles to find your answer, you probably used it twice and its effect canceled out. Name Section Rec TA Ch 1b, Solution Set for Problem Set Four S i d e 15 of 24 Due Friday, Jan. 28, 2011 at 4 PM in the Drop Box So l u t i o n 4 c Find the total thermal energy transferred to the system,
(1 poi n t) where is the thermal energy added to the system during adiabatic expansion and is the thermal energy added to the system during isothermal compression. Find the total work done on the system, (1 poi n t) where is the work done on the system during adiabatic expansion and work done on the system during isothermal compression. is the E va l u a t e t h e a d i a b a t i c e x p a n s i o n to f i n d a nd Since the first process is adiabatic, by definition. (1 poi n t) Write the first law of thermo for the first process,
(1 poi n t) but (adiabatic), The internal energy, , only depends on temperature for an ideal gas (1 poi n t), Substitute Equation [2] and simplify, Recall that a n y ide a l gas obeys (OGN, p. 208), and substitute, (1 poi n t; intermediate calc, so keep sig. figs.) Since , the system is doing work on the surroundings. The sign for the answer should not be a surprise since the volume increased. Name Section Rec TA Ch 1b, Solution Set for Problem Set Four S i d e 16 of 24 Due Friday, Jan. 28, 2011 at 4 PM in the Drop Box E va l u a t e t h e isot h e rm a l compr ession to f i nd Write the first law of thermodynamics for the second process, a nd
(1 poi n t) Since the internal energy, , only depends on temperature for an ideal gas and this process is isothermal, we can say that (1 poi n t) and In its most general form, work is defined as, where is the wor k done on the syst em, is the resisting pressure, and is the system volume. When we are analyzing the work done on oxygen, the resisting pressure is simply the system pressure, , The compressed air is an ideal gas that undergoes an isothermal compression. Substitute the ideal gas law into the above equation, and simplify, (2 poi n ts) We know that since the gas is compressed back to its original volume. From our earlier work in Part (a), we also know that: Substitute both of these equations into , Name Section Rec TA Ch 1b, Solution Set for Problem Set Four S i d e 17 of 24 Due Friday, Jan. 28, 2011 at 4 PM in the Drop Box using the ideal gas law. Substitute Equation [2], (1 poi n t) Since , Calculate C ombi n e r esu l ts to f i n d : a nd (1 poi n t a n swe r ; 1 poi n t si g f igs) Calculate : (1 poi n t a n swe r ; no points assigned to sig figs) Name Section Rec TA Ch 1b, Solution Set for Problem Set Four S i d e 18 of 24 Due Friday, Jan. 28, 2011 at 4 PM in the Drop Box Sol u t ion 4d ( V i ewpoi n t #1) Write the first law of thermo for the overall process,
Use the results from Part (c), (1 poi n t) From the definition of enthalpy (OGN, Eqtn [7.7]), we find the change in enthalpy, The initial and final volume is the same, (2 poi n ts) Substitute the ideal gas law and Equation [3], and into the equation for , (1 poi n t) Name Section Rec TA Ch 1b, Solution Set for Problem Set Four S i d e 19 of 24 Due Friday, Jan. 28, 2011 at 4 PM in the Drop Box Sol u t ion 4d ( V i ewpoi n t #2)
. The change in the internal energy of an ideal gas can be written, Substitute Equation [2] and use the fact that during the isothermal compression, into the equation for : We know that a n y ide a l gas obeys (OGN, p. 208), and we are given that cp(O2) = 29.4 J K1 mol1, The change in the enthalpy of an ideal gas can be written, Make same substitution for T 3 as for internal energy, advantage of the fact that energy is a state function. Name Section Rec TA Ch 1b, Solution Set for Problem Set Four S i d e 20 of 24 Due Friday, Jan. 28, 2011 at 4 PM in the Drop Box 5. U si n g C ompr essed A i r to L a u n c h a D isc f rom a C yl i nde r (23 poi n ts) A hollow cylinder (height: H = 7.5 cm; inside diameter: D = 2.5 cm) is closed at the bottom and open at the top. A disc with mass m = 1.44 kg is held in place by removable latches inside the cylinder at a given height (h = 1.5 cm). Air is contained in the volume below the disc at pressure, P = 12 atm. Since the cylinder is open to the atmosphere at the top, P ext = 1.0 atm. You may assume the disc is frictionless, but you cannot neglect its mass. When you release the latches, you may assume the air below the disc undergoes isothe rma l expansion. You may also assume that air behaves as an ideal gas.
How high will the disc be shot into the air? (a) Find the work (units: Joules) done by the compressed gas during the isothermal expansion. You can assume the final volume is the total volume of the cylinder. (6 poi n ts) (b) Find the work (units: Joules) required to move the disc and external air to the top of the cylinder. (10 poi n ts) (c) 2 poi n ts) (d) Find the highest point (units: centimeters) that the disc will reach after being ejected from the cylinder. Calculate this height relative to the top of the cylinder. You may neglect the drag force on the disc as it moves out of the cylinder assume all kinetic energy is converted to potential energy. (5 poi n ts) Name Section Rec TA Ch 1b, Solution Set for Problem Set Four S i d e 21 of 24 Due Friday, Jan. 28, 2011 at 4 PM in the Drop Box So l u t i o n 5 a In its most general form, work is defined as, where is the wor k done on the syst em, is the resisting pressure, and is the system volume. When we are analyzing the work done on the compressed gas below the disc, the resisting pressure is simply the system pressure, , and initially (when the latches are in place), But we w a n t t h e wor k don e by t h e syst em! The compressed air is an ideal gas that undergoes an isothermal expansion. Substitute the ideal gas law into the above equation, and simplify, (2 poi n ts) Substitute symbols for final and initial volume, constant throughout the isothermal expansion of a closed system. Use the ideal gas law again using the initial state since pressure and volume are given for that state. (1 poi n t) Name Section Rec TA Ch 1b, Solution Set for Problem Set Four S i d e 22 of 24 Due Friday, Jan. 28, 2011 at 4 PM in the Drop Box (2 poi n ts a n swe r ; 1 pt sig f igs) Sol u t ion 5b ( V i ewpoi n t #1 : E f f e c t ive e x t e r n a l pr essu r e) Some people prefer to think about eve r ythi ng in terms of work done against some pressure (PV work). In its most general form, work is defined as, where is the wor k done on the syst em, is the resisting pressure, and is the system volume. When we are analyzing the work done by the system to move the disc from one height to another, where the resisting pressure is simply the pressure of the surroundings, , since the disc is frictionless. Note that is not just the external pressure, but it represents the effective pressure of all forces acting on the disc (the boundary between the ambient and the compressed gas). The pressure of the surroundings is constant through this process. The atmospheric The effective pressure, , is the sum of the external pressure and the pressure caused by the weight of the disc, (5 poi n ts for recognizing that both pressure and disc weight must be included, regardless of which viewpoint the student uses) Substitute this result into the equation for and simplify, Name Section Rec TA Ch 1b, Solution Set for Problem Set Four S i d e 23 of 24 Due Friday, Jan. 28, 2011 at 4 PM in the Drop Box (2 poi n ts for an equation that is similar to Eq [1], regardless of method) [1] (3 poi n ts a n swe r ; no pts assigned for sig figs) Sol u t ion 5b ( V i ewpoi n t #2 : A P  V con t r i bu t ion a nd a P E con t r ibu t ion) If you followed the derivation closely, you probably noticed that we divided by the area early and multiplied by the area later for the term that includes the weight of the disc. bout the contributions in different ways. The system does PV work only against the external pressure and does mechanical work by increasing the potential energy of the disc.
In its most general form, PV work is defined as, where is the wor k done on the syst em, is the resisting pressure, and is the system volume. The resisting pressure is the external pressure, . And we want to evaluate the work done by the system, Since the external pressure remains constant throughout the entire process, Name Section Rec TA Ch 1b, Solution Set for Problem Set Four S i d e 24 of 24 Due Friday, Jan. 28, 2011 at 4 PM in the Drop Box This term of work accounts for the work required to move a massl ess, f r i c t ionl ess disc from its initial height to the top of the container. The mechanical work required to move a disc with mass, m, is, assuming the disc is frictionless. Substituting for height, Combining these two contributions, This is the same equation as we found with the other viewpoint, Equation [1]. Choose equally correct. So l u t i o n 5 c When the gas expands it does not expand reversibly (2 poi n ts). The gas does work to raise the disc to the top of the cylinder and causes the disc to accelerate! Sol u t ion 5d The additional work changes the kinetic energy of the disc,
(1 poi n t) where is the velocity of the disc as it leaves the cylinder. energy since we are neglecting friction/drag forces of air on the disc, (2 poi n ts) where is the highest point (relative to the top of the cylinder) that the disc will reach. Solving for , (2 poi n ts a n swe r ; no pts assigned to sig figs) ...
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This note was uploaded on 03/21/2011 for the course CHEMISTRY 1b taught by Professor Reisman;heath during the Winter '11 term at Caltech.
 Winter '11
 Reisman;Heath
 Chemistry

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