Final Exam, Chemistry 481, 77 December 2016

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1 1 Final Exam, Chemistry 481, 77 December 216 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 Pressure units: 1 atm = Pa= bar = 76 torr. van der Waals equation of state: P = R/( V b) a/ V 2 Unless indicated otherwise, k will designate the Boltzmann constant k B. 1. a. What is the pressure in Pa corresponding to 1 12 torr? (6 pts) he pressure P in SI units is Pa (1/76) atm/torr torr = Pa. b. What is the number density of molecules (molecules/cm 3 ) in a chamber with pressure P = 1 12 torr at a temperature of 298K? (You can assume the gas is ideal) (11 pts) Use the ideal gas equation of state: n/v = P/R. So, P/R = /( ) = he units are mol/m 3. But 1 m 3 = 1 6 cm 3. So, there are moles/cm 3. Multiplcation by Avogadro s number gives N/V = molecules/cm For a monatomic gas, the second virial coefficient B 2V is defined as [ B 2V = 2πN A e u(r)/k B 1 ] r 2 dr where u(r) is the potential of interaction between two atoms. Determine B 2V gas where u(r) is defined by (13 pts) for a (a) u(r) = when r σ (b) u(r) = when r > σ When u(r) =, e u(r)/k B 1 = 1 1 =. So, the integration can be limited to the range σ. For r σ, [e u(r)/k B 1] = [ 1] = 1, so σ ( ) r B 2V = 2πN A ( r 2 3 σ dr) = 2πN A = πσ3

2 2 3. Consider a rotating diatomic molecule, with quantum number j. he degeneracy of each level is g j = 2j + 1. he rotational energy levels are E j = Bj(j + 1) where B is the rotational constant, B = 2 /2µr 2 e, where µ is the reduced mass and r e is the bond length. Answer the following questions: a. In terms of B, what is frequency of light which will excite the transition j j + 1? (7 pts) E = E j+1 E j = B[(j +1)(j +2) j(j +1)] = 2B(j +1). Now the Bohr frequency condition is hν = E. So ν = 2B(j + 1)/h. b. Give the expression for the rotational partition function q r in terms of B and k. (11 pts) q r ( ) = j g j exp( E j /k ) = (2j + 1) exp[ Bj(j + 1)/k ] j= c. If the spacing between the rotational levels is small compared to k, then the summation can be replaced by an integral, which leads to lim q r( ) = k/b k >>B Determine from q r an expression for the average rotational energy of the molecule. (11 pts) We know that E = k 2 (d ln q/d ) = k 2 (1/q)dq/d = k 2 (B/k )(k/b) = k d. If the temperature increases from 1 to 2, what will be the change in the rotational entropy of the molecule: S = S( 2 ) S( 1 )? Hint: you can assume the hightemperature limit discussed in part [c] immediately above. (15 pts) We know that S = ( E )/ + k ln q. Here, E = k and q = k/b. So, S = k + k ln(k/b) and S = k(ln k 2 /B ln k 1 /B) = k ln( 2 / 1 ). 4. he vibrational energies of a molecule are often approximated by the levels of a harmonic oscillator, given by E n = (n + 1/2)hν he partition function associated with the vibrational motion of a diatomic molecule is q vib = x 1/2 /(1 x) where x = exp( hν/k ). he average energy of the oscillator is E = hν 2 ( ) 1 + x 1 x a. Determine a power series expansion (in x) for E, valid for small x. Keep all terms up through O(x 2 ). (13 pts)

3 3 We know that 1/(1 x) 1 + x + x 2 + O(x 3 ), so (1 + x)/(1 x) (1 + x) (1 + x + x ) = 1 + 2x + 2x hus E = 1 2 hν(1 + 2x + 2x2 ) 5. he Clapeyron equation states that the slope of the coexistence curve between two phases α and β is given by = H α β µ α=µ β V α β where H α β and V α β designate the enthalpy of the phase change and the change in volume associated with the phase change, both at temperature. a. Suppose the transition is from the liquid to the vapor phase. Derive the Clausius- Clapeyron equation, which expresses (1/P ) ( P/ ) µl =µ g as a function of H vap, boil, and R. (13 pts) For a transition for liquid to vapor, since the vapor is so much less dense, V = V g V l V g. We can then use the ideal gas equation of state, with V g = nr/p. his gives, then ( P/ ) µl =µ g P H vap /( nr ). Rearranging gives 1 P = H R 2 b. he heat of vaporization of water ( H vap ) = 4.65 kj/mol. At an atmospheric pressure of.55 atm, at what temperature will water boil? (14 pts) We can rearrange the Clausius-Clapeyron equation as dp/p = ( H/R 2 )d. his can be integrated to give ln(p 2 /P 1 ) = ( H/R)( 1/ 2 + 1/ 1 ). For this problem we know 1 = 373, P 1 = and P 2 =.55P 1. We are interested in 2, the boiling point at P =.55 atm. So log(.55) = ( /8.314) (1/ 2 1/373) or.5978 = 4889(1/ ). his can be rearranged to give 1/ 2 =.5978/ = Inverting gives 2 = 356.7K. 6. he pressure of a gas is given in terms of the partition function Q as P = k ( ln Q/ ). a. Give a similar expression that relates the average energy E to a derivative of the partition function Q with respect to. (7 pts) E = U = k 2 ( ln Q/ ) V b. he quantity ( U/ ) is called the internal pressure. Starting from the result of part (a), derive an expression for ( U/ ) in terms of P,, and the derivative of P with respect to. (13 pts) Statistical mechanical proof: We know from part (a) of this problem that U = k 2 ( ln Q/ ) V. Differentiate with respect to V, keeping constant to get ( ) U = k 2 2 ln Q = k 2 2 ln Q = k 2 ( ) ln Q (Here we use the identity of the mixed partial second derivatives). But from the introduction to this problem we know that ( ln Q/ ) = P/k, so that

4 4 So, finally, ( ) ln Q = ( ) U ( ) P = P k k k = P + V V 7. a. Derive an expression that relates ( S/ P ) to the coefficient of thermal expansion α which is defined as (11 pts) α = 1 ( ) V P We know that dg = Sd + V dp. he applicable Maxwell relation then gives ( S/ P ) = (/ ) P = V α b. Physically, how would you expect the entropy of a gas to change as you increase the pressure? Explain your answer. (7 pts) As you compress a gas, the disorder will descrease, hence the entropy will decrease. c. Explain whether your answer to part [b] is consistent with your answer to part [a]? (9 pts) Suppose we have an ideal gas, then α = (P/nR)(n/P ) = 1/R. hus ( S)/( ) = (nr/p )(1/R) = n/p. his is a negative number, so that S will decrease as P increases. 8. Calculate expressions for U, q rev, w rev and S when one mole of a monatomic ideal gas undergoes the following transformations. a. A reversible isothermal expansion at temperature from V 1 to V 2 = 3V 1. (12 pts) isothermal, and n = 1 therefore U =. w = P dv = R dv/v = R ln(v 2 /V 1 ) = R ln 3. Since U =, q = w = +R ln(v 2 /V 1 ). since constant, S = q rev / = R ln(v 2 /V 1 ) = R ln 3. b. A reversible adiabatic compression from temperature 1 and volume V 1 to volume V 2 = 1 2 V 2. Give any coefficients as decimal numbers, not sums of numbers raised to rational powers. (14 pts) Adiabatic, thus q rev = and since δq =, ds = and, hence S=. For adiabatic compression ( 2 / 1 ) 3/2 = (V 1 /V 2 ) (monatomic gas). hus 2 = 1 2 2/3. Finally, U = C V = (3/2)R 1 [ 1+2 ( 2/3)] =.8811R 1. Since q =, w = U. Work is positive, since the gas is compressed. 9. Consider a two-level system with energies ±ε; where ε is a positive number). he degeneracy of the upper level is 2; that of the lower state is 1. a. From physical considerations, what is (i) the energy and (ii) the entropy of the two-level system at = and = (justify your answers). (12 pts) At =, system in ground state. Energy = ε. Because the ground state is non-degenerate, S = k ln W = k ln 1 =. At =, all levels equally populated. he energy is the weighted average: E = ε( ) = +ε/3. S = k ln W = k ln

5 5 b. Express the partition function q in terms of x = exp( ε/k ) (12 pts) q = j g j exp( E j /k ) = exp(+ε/k ) + 2 exp( ε/k ) = 1 x + 2x = 1 + 2x2 x c. Determine E from q. Hint: the limits at and of your result should equal the answer you obtained in part [a] above. (14 pts) E = d ln q dβ ln q dx = d dx dβ = 1 dq q dx Now x = exp( ε/k ) = exp( βε). So dx/dβ = ε exp( βε) = εx. hus x dq E = εx 2x dx = ε x 2 2x d. Sketch E(x) in units of ε as a function of x. (13 pts) dx dβ 2x 2 1 = ε 2x2 1 x 2 2x (2 x 2-1)/(2 x 2 + 1).4.2 E / units of x 1. he constant-pressure molar heat capacity of a molecule is given by C P ( )/J K 1 mol 1 = A + B C 2 where is in K, and the units of A, B, and C are J K 1 mol 1 ; J K 2 mol 1, and J K 3 mol 1, respectively. Determine an expression for S, when one mole of the molecule is heated at constant pressure from 1 to 2. (12 pts) We know ds = δq rev /. At constant pressure, δq rev = C P d. hus at constant pressure ds = C P /. For this problem, then ds = A + B C

6 6 Integration from 1 to 2 gives S = 2 1 ds = A ln B ( 2 1 ) C 2 ( )