HW Sol#7

HW Sol#7 - Chemistry 131, Fall 2007 Solutions to Homework...

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Chemistry 131, Fall 2007 Solutions to Homework No. 7 Problem 1. See Levine, problem 8.4. Let x = 1 /V m and solve Levine eq. 8.4 to get P = RT ± x + Bx 2 + Cx 3 + ... ² . Next write an expression for Z with eq. 8.5 and substitute the expression for P found from eq. 8.4: Z = PV m RT = 1 + B P + C P 2 + ... = 1 + B RT ( x + Bx 2 + Cx 3 + ... ) + C R 2 T 2 ( x + Bx 2 + Cx 3 + ... ) 2 + ... = 1 + B RT ( x + Bx 2 + Cx 3 + ... ) + C R 2 T 2 ( x 2 + 2 Bx 3 + ... ) + ... = 1 + ( RTB x + RTB Bx 2 + RTB Cx 3 + ... ) + ( R 2 T 2 C x 2 + 2 R 2 T 2 C Bx 3 + ... ) = 1 + RTB x + ( RTB B + R 2 T 2 C ) x 2 + .... Comparing the powers of x in the above expression with the corresponding powers of x in the expression for Z obtained from eq. 8.4, which is Z = 1 + Bx + Cx 2 + ... , yields B = RTB , C = RTB B + R 2 T 2 C = RTB ( RTB ) + R 2 T 2 C = ( B 2 + C ) R 2 T 2 . Problem 3. See Levine, problem 8.17. (a) Use the Redlich-Kwong equation (Levine eq. 8.3) to ob- tain the required partial derivative, then substitute the latter in the expression giving the change in internal energy of va- porization. To evaluate the integral use the formula given on Levine p. 238. ³ ∂P ∂T ´ V m = R V m - b + a/ 2 V m ( V m + b ) T 3 / 2 Δ vap U m = Z V v m V l m µ RT V m - b + Ta/ 2 V m ( V m + b ) T 3 / 2 - RT V m - b + a V m ( V m + b ) T 1 / 2 dV m = Z V v m V l m 3 a/ 2 V m ( V m + b ) T 1 / 2 dV m = µ 3 a 2 bT 1 / 2 ln V m V m + b V v m V l m = 3 a 2 bT 1 / 2 ln V v m ( V l m + b ) V l m ( V v m + b ) . Adding P ( V v m - V l m ) yields the expression for Δ vap H m given in the problem statement. (b) Use the following values in the given expression for Δ vap H m : a = 1 . 807 × 10 8 cm 6 atmK 1 / 2 mol - 2 , b = 62 . 7cm 3 / mol , V l m = 100 . 3cm 3 / mol , V v m = 1823cm 3 / mol , P = 10 . 85atm , T = 298 . 15K , R = 82 . 0575cm 3 atmmol - 1 K - 1 . The result is: Δ vap H m = (131796cm 3 atmmol - 1 ) × ³ 1m 100cm ´ 3 ³ 101325Pa 1atm ´ = 13 . 354kJ / mol . Problem 5. See Levine, problem 8.20. The temperature, pressure, and molar volume at the critical point must satisfy the equation of state and must cause the first two partial derivatives of P with respect to V to vanish (Levine eq. 8.12) (a) P c = RT c V m ,c - b - a T c V 2 m ,c , (1) ³ ∂P ∂V m ´ T = - RT ( V m - b ) 2 + 2 a TV 3 m (2) RT c ( V m ,c - b ) 2 = 2 a T c V 3 m ,c , (3) ³ 2 P ∂V 2 m ´ T = 2 RT ( V m - b ) 3 - 6 a TV 4 m (4) 2 RT c ( V m ,c - b ) 3 = 6 a T c V 4 m ,c . (5) Divide eq. (3) by eq. (5): V m ,c - b 2 = V m ,c 3 b = V m ,c / 3 (6) V m ,c V m ,c - b = 3 2 . (7) Solve eq. (3) for a and substitute for V m ,c / ( V m ,c - b ) from eq. (7): a = RT 2 c V 3 m ,c 2( V m ,c - b ) 2 = RT 2 c V m ,c 2 ³ V m ,c V m ,c - b ´ 2 (8) = RT 2 c V m ,c 2 ³ 3 2 ´ 2 = 9 RT 2 c V m ,c 8 (9) V m ,c = 8 a 9 RT 2 c (10) V m ,c - b = 2 3 V m ,c = 16 a 27 RT 2 c . (11) Substitute eq. (10) and eq. (11) into eq. (1) and solve for
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This note was uploaded on 05/22/2008 for the course CHEM 131 taught by Professor Lindenberg during the Fall '08 term at UCSD.

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HW Sol#7 - Chemistry 131, Fall 2007 Solutions to Homework...

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