Exam 2, Chemistry 481, 4 November 2016

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1 1 Exam, Chemistry 481, 4 November 016 Show all work for full credit Useful constants: h = J s; c (speed of light) = m s 1 k B = J K 1 ; R (molar gas constant) = J K 1 mol 1 van der Waals equation of state: P = R/( b) a/ All these problems were covered either in class, on the homework, or in the office hours and help sessions held by me or by Say (the A). Some of the wording was slightly different. So, my advice: (1) come to class prepared, () come to help sessions and office hours, (3) read the problems carefully. 1. One mole of an ideal gas at temperature 1 expands isothermally from, 1 to 3, 1. Determine, in terms of R and 1, a. he maximum value of w irrev if the gas undergoes a one-step expansion. (8 pts) he gas can expand only if P ext P, thus max w irrev = P d = P ( ). But = 3 so that P = R 1 / = R 1 /3. hus max w irrev = R 1 ( )/(3 ) = R 1 /3. Many people took P ext to be equal to P 1, the initial pressure. Since the gas expands, the final pressure P is only 1/3 of P 1. hus, the expansion work is considerably less than would be predicted if you had, falsely, taken P ext = P 1. b. If the expansion is reversible, determine i. w rev (5 pts) ii. S (10 pts) Isothermal reversible expansion. w rev = P d = R( 1 / )d/ = R 1 ln( ) =3 = R 1 ln 3. Since the expansion is isothermal, U = 0, so that q rev = [ R 1 ln 3] = R 1 ln 3. Since is constant S = path δq rev/ = (1/ 1 ) path δq rev = q rev / 1 = R ln 3 (you can take the 1/ out of the integral sign only because the proccess is isothermal; and you should mention this for full credit!). c. Suppose the ideal gas has a molar heat capacity C = 5 R. Consider a reversible adiabatic expansion (again of one mole of an ideal gas) from 1,, P 1 to, = 3, P. Determine

2 i. he value of the final temperature in terms of 1. (10 pts) Along an adiabatic path, δq = 0, so du = δw rev = P d. Now, if the gas is ideal du = C d so 5Rd = P d = R d/. his can be integrated up to give (5/) ln = ln or ln( 5/ )=constant. Another way of saying this is that /5 is constant along an adiabatic path. Note that the factor is /5, because the heat capacity is 5R. For a monatomic gas, the heat capacity is 3 R, so that the exponent is /3. hus, 1 /5 1 = /5 or = 1 ( / ) /5 = 1 (1/3) /5 = ii. he value of U in terms of R and 1 (10 pts) Since the gas is ideal, the energy depends only on the difference in temperature U = C ( 1 ) = 5 R( ) = R 1 = 0.890R 1. Many people tried to use expressions for an isothermal expansion, which is, obviously, not valid here. iii. he value of S (5 pts; justify your answer). Along an adiabatic path δq = 0, therefore ds = 0.. he r H values for the following reactions are Fe(s) + 3 O Fe O 3 (s) r H = 06 kj mol 1 3Fe(s) + O Fe 3 O 4 (s) r H = 136 kj mol 1 Use these data to calculate the value of r H for the reaction (5 pts) 4Fe O 3 (s) + Fe(s) 3Fe 3 O 4 (s) Homework problem take 4 rxn#1+3 rxn# to get r H = 4( 06) + 3( 136) = = 416kJ mol 1. Most people got this, because they ve had thermochemistry before. 3. In Chapter 8 we will derive the formula = ( ) P For the isothermal expansion of a van der Waal s gas from to determine U. P

3 3 (10 pts) For a van der Waals gas P = nr/( nb) n a/, so ( P/ ) = nr/( nb), and ( P/ ) = nr/( nb). hus = ( ) P Integration from to gives U = P = R nb nr nb n a = n a ( 1 = n a 1 ) = n a Many people stated that U/ = 0. his is true only for an ideal gas, but not for a van der Waal s gas. he purpose of this problem was to use the van der Waal s equation of state, given at the beginning of the exam, to derive the dependence of energy on volume for the van der Waal s gas. 4. he constant-pressure molar heat capacity of a molecule from 300 K to 100 K is given by C P ( )/J K 1 mol 1 = A + B C where is in K, and the units of A, B, and C are J K 1 mol 1 ; J K mol 1, and J K 3 mol 1, respectively. Calculate the value of S, in units of J K 1, when one mole of the molecule is heated at constant pressure from 100 K to 1000 K. (10 pts) his is similar to homework problem 6-7. We know ds = δq rev /. At constant pressure, δq rev = C P d. hus at constant pressure ds = C P /. For this problem, then Integration of ds from 1 to gives S = Now, 1 = 100 and = 1000, so ds = A + B C 1 ds = A ln 1 + B( 1 ) + 1 C ( 1 ) S = A ln B 1 C( ) =.306A + 900B C

4 4 5. Consider a two-level system with energies ε and +ε. he lower level (E = ε) is doubly degenerate. he partition function is given by q = /x + x, where x = exp( βε). In solving this problem remember the power series expansions: ln(1 + y) = y 1 y +..., and y = 1 y + 1 y Also, remember that We ve done a similar problem in class! d ln(q) dx = 1 q dq dx a. Determine a general expression for E in terms of ε and x. (7 pts) he partition function can be written as q = x + x = + x x We know that E = d ln(q)/dβ. By the chain rule, we have d ln q/dβ = d ln q/dx dx/dβ. Now dx/dβ = εx. Using the expression for d ln(q)/dx given in the introduction to the problem, we find. E = 1 q dq dx ( εx) = x [ x + x x + ] x ( εx) = ε x x x + You can also solve this problem by directly making use of the definition of E : E = g 1E 1 exp( E 1 β) + g E exp( E β) q = ε(1/x) + εx ( + x )/x b. From physical considerations what should be the value of the entropy at = 0 and at =? (5 pts; justify your answer) At = 0 everything is in the lowest state. his state is doubly degenerate so S = k ln W = k ln. Many people said that lim 0 S = 0. his is true only for a nondegenerate ground state! I went over this in class. Also, both Say and I looked explicitly at this two-level system with degeneracies in office hours. At =, any of the three levels are equally probable, so W = 3 and S = k ln 3. Many people said that lim S =. his is true only if you have an infinite number of levels.

5 5 c. he general expression for the entropy) is S = E / + k B ln q. Using your answer to part [a], determine a power series expansion in x of S(x), keeping all terms up through x. When = 0, x = 0, so lim x 0 S(x) should be equal to the result you obtained in part [b]. (15 pts) his was the hardest problem on the exam. From the answer to part [a] the power series expansion of E is E = ε x (1 + x /) = ε x (1 x / + ) = ε (1 x + ) Also, from the expression for q given in the introduction to this problem, we find that k ln q = k ln( + x ) k ln x = k ln[(1 + x /)] = k ln + k ln(1 + x /) k ln x From the definition of x and the expansion of ln(1 + y) given in the introduction to this problem, we see also that k ln x = kβε = ε/ and k ln(1 + x /) = kx / +. Combining these results gives S(x) = k ln + ε + εx + ε o check lim x 0 = k ln. + kx / = k ln + x ( k + ε )