Chemistry 456A (10:30AM Bagley 154)

Transcription

1 Winter 0 Chemistry 456A (0:0AM Bagley 54) Problem Set B (due 9PM Friday, /0/) Q) In the previous homework we compared isothermal one-step, irreversible work with reversible isothermal work. We also compared a one-step isothermal process and a one-step adiabatic process. Now we compare a one-step adiabatic, irreversible process with a reversible, adiabatic process. Let s consider that both of these processes start from the same P,,, called P, for both processes and end at the new pressure is P = P for both the reversible and irreversible adiabatic expansions.. Consider this gas to be an ideal gas with heat capacity C = nr. [As a practice of C your algebraic skills you might try leaving all the quantities in terms of c = before you nr evaluate final answers.] he questions we consider are will the two systems end up at the same final state? And will the processes give us different amounts of heat, work and energy change? Qa) o help organize thinking about this we want to compare the equations: EoS, First Law (Part A), Energy Change (First Law part B), Work. hese expression are fundamental to all problems and are very general, and are the same as done for the isothermal case, or any other problem involving the ideal gas. Relation EoS P = nr First Law (Part A) U = q + w Work w = Pext Internal Energy Change (First Law Part B) U = C Qb) Now adapt the general rules to the specific cases of adiabatic expansion: Relation Reversible Adiabatic Irreversible (Step) Adiabatic EoS P = nr same First Law (Part A) U = q + w same Work w = P d w = Pd = Pd Heat q = 0 same Internal Energy Change (First Law U = C Same

2 Winter 0 Chemistry 456A (0:0AM Bagley 54) Part B) Relation of to C d = Pd C = P Qc) Of the above relations, for the reversible and irreversible processes which one is the key one that distinguishes between reversible and irreversible adiabatic process? he work is the only fundamental equation that makes the two cases different. Qd) Explain why the work cannot be greater than an upper bound: w < C. For the I.G. C is the total internal energy in the gas; this is the only source of work (as q=0) so you can t get more work than the energy you have. Qe) Using the fact that P < P, which means the pressure along the reversible path is always greater than the one on the irreversible path, to explain, qualitatively, why one will get more work from one case than the other. Work is less because P<P, so the temperature drop is less for the reversible step. Qf) Use the premise that the work obtained from the system following the reversible Reversible Irreversible path is greater than that from the irreversible path (i.e. wadiabatic > wadiabatic ) to explain qualitatively why the final temperature of the irreversible expansion must be larger than that of the reversible expansion. he temperature drop must be larger for the reversible case because the reversible work is a larger magnitude, so being negative is a larger negative number, hence smaller. Qg) Qualitatively explain why the volume of the irreversible expansion will be different from that of the reversible expansion. Will the volume expansion for the reversible process be larger than that of the irreversible process? [You might check with the quantitative results below.] Hint: o reason through this assume the reversible path does more work so that the temperature drop must be larger for the reversible process and use the EoS to tell you whether this means the final volume of the irreversible process must be larger or smaller than that of the reversible process. o reason through this assume the reversible path does more work so that the temperature drop must be larger for the reversible process and use the EoS to tell you whether this means the final volume of the irreversible process must be larger or smaller than that of the reversible process.

3 Winter 0 Chemistry 456A (0:0AM Bagley 54) Qh) Quantitatively determine the temperature ratio and the volume ratio for P the reversible case. It is always true that for the I.G.: = P P P ( + c) = ( + c) C d = Pd = nrd ln C c = nr c d ln = d ln c ln c ln = ln = P P c ln = ln = c ln + c ln P P P 0 = ln + ln P c γ = c P 0 = ln + γ ln P P = P = 5 =.9 γ c = ln P P P P c ln = ln = ln = ln ln P P ( c) = P + ln = ln P 5 ( c P + ) = = = P Qi) Quantitatively determine the temperature ratio and the volume ratio for the irreversible case.

4 Winter 0 ( c) Chemistry 456A (0:0AM Bagley 54) C = P = P P P P = nr = nr P P P c = P + = + = + = = 0.7 P c P = = Qj) Summarize the quantitative results and discuss in qualitative terms what you expected for the relative temperature and volume ratios for the two processes. Use the temperature ratios to qualitatively describe the relative amount of work done by both processes. o reason through this assume the reversible path does more work so that the temperature drop must be larger for the reversible process and use the EoS to tell you whether this means the final volume of the irreversible process must be larger or smaller than that of the reversible process. Qk) Draw a P- diagram (which may require redoing your old one) for these two cases, taking into account what you learned above. Label for the reversible and irreversible cases and compare with for isothermal expansion. Block out the areas under the curves that correspond to the work in both adiabatic cases. In your diagram show that rev < and irr-rev>, and the irr-rev-work is just the part of the area for Pext = P from to, so the Reversible work has larger area but does not go out as far. You can think of the reversible work as more efficient, it got more work done with less volume expansion. And if it did more work the temperature had to drop further (but in so doing this makes the temperature ratio a smaller number, which makes the temperature drop larger, which also makes the volume ratio a smaller number, so less expansion.) Q) C Compare the irreversible (one-step) adiabatic expansion with the reversible adiabatic expansion. Assume the system starts at pressure P and drops to pressure P in both processes and that P = P and that the gas is an ideal monatomic gas

5 Winter 0 Chemistry 456A (0:0AM Bagley 54) C = nr. o show that the irreversible work is less than the reversible work prove wone step that = for these two processes independent of the values of or, or n. wreversible 4 he work ratio only depends on the pressure ratio and no other quantity (excluding the fraction from the heat capacity). he ratio of the irreversible work to the reversible work is independent of the values of or, or n. P a wone step U C irr P irr = = = = a wreversible U rev C rev P rev P wone step 0.7 == = a = = w 5 reversible + c he work ratio only depends on the pressure ratio and no other quantity (the fraction from the heat capacity is taken as a constant). Q) We stated a general relation between and the state variables, called the P.EoS (hermodynamic Equation of State) = P. (Eqn.9 in ext) Qa) Evaluate For the van der Waal s gas in terms of P, and. nr an P = nb P nr an π vdw = P = P = nb Qb) Compare this result with that of the ideal gas. For the ideal gas you must get the same answer as the vdw gas with a=0, which is zero. Q4) a) Starting with the two parts of the first law of thermodynamics show that = C for a gas or any other system where the work done is only Pd work.

6 Winter 0 Chemistry 456A (0:0AM Bagley 54) du = q + w = d + d d = 0, w = 0, q = q q q = d = d And the general definition of heat capacity is heat (transferred) divided by temperature rise. b) Using the relation in part A and assuming Cv is independent of temperature show that we recover the result of the ideal gas expression that, for a constant volume process, relates U and. [For the I.G. is this relation restricted to just isochoric processes?] du = C d U = C his relation is valid all the time for any processes for an ideal gas. c) ) If C depends on temperature, how do you in general find U for a constant volume process? U = du = C d) Consider another form of energy, H=U+P, and again assume only Pd work, then show that the heat transferred under constant pressure conditions must be q sub p and H qp it must be dh, and therefore = Cp =. P H H dh = q + w + d ( P ) = d + dp P dp = 0, w = P d = d P, q = q H ext H P qp = d = = C p P P d Q5) his is closely related to problem P.0 of your text. Because U is a state function we know Euler s test for a state function (also called a Maxwell Relation): = P q P

7 Winter 0 Chemistry 456A (0:0AM Bagley 54) a) Using this relationship, show that the heat capacity, C, is independent of C volume for an ideal gas; that is show that = 0. C 0 = = = 0 C b) Using the same relation, show that = 0 for a van der Waals gas. an C π = = = = 0 c) Can you suggest (from your text) another type of non-ideal gas where it is not the case that the heat capacity is independent of the volume of the gas? P P C = = = P P P P = + = So we need to find a case where the pressure will survive derrivatives. For example the Redlich-Kwong EoS does just that: R a P = b b P m m m = a ( ) m ( m b) he RK EoS does predict the heat capacity depends on, and only the a term contributes. Yes, b is in there but b=0 gives a satisfactory answer also, see below.

8 Winter 0 Chemistry 456A (0:0AM Bagley 54) = = b C P a.5 4 m m m m ( m ) b d = d = ln ( b) b ( b) b b m m m m m m m m m ( m b) m m C a a.5.5 ln m C = d = d = 4 4 m m m m b b m b m