I 2 (g) 2I(g) 0.12 g 254 g mol 1 = mol

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1 hermodynamics: Examples for chapter When 0.12 g of solid iodine was vaporized at 1670 K, the resulting gas displaced 20.2 cm 3 of dry air at 298 K and 99.9 kpa pressure. How many iodine molecules were dissociated (i.e. as atomic iodine) in the gas? he gas phase equilibrium reaction is: I 2 (g) 2I(g) Assume that the gas mixture (of I 2 and I) behave according to the ideal gas law. he amount of I 2 before any dissociation takes place is: n = m M I2 = 0.12 g 254 g mol 1 = mol After the equilibrium has been reached, the amount of I 2 is (1 α)n and the amount of I is 2αn. Here the degree of dissociation is denoted by α. he ideal gas law is: PV = n tot R n tot = PV R n tot = (1 α)n + 2αn = PV R α = PV nr 1 = PV M I 2 mr 1 = (99.9 kn m 2 ) ( m 3 ) (254 g mol 1 ) (0.12 g) (8.31 J K 1 mol 1 1 = 0.72 ) (298 K) he amount of I is 2αn = ( mol) = mol. 1

2 2. What is the molar volume of n-hexane at 660 K and 91 bar according to (a) the ideal gas law and (b) the van der Waals equation? For n- hexane, c = K and P c = 30.3 bar. If you obtain an equation that you cannot solve analytically, attempt to solve it numerically. a) b) V = R P = ( L bar K 1 mol 1 )(660 K) = 0.603L mol 1 91 bar a = 27R2 c 2 = 27( L bar K 1 mol 1 ) 2 (507.7 K) 2 64P c 64(30.3 bar) b = R c = ( L bar K 1 mol 1 )(507.7 K) 8P c 30.3 bar P = R V b ā V 2 = L 2 bar mol 2 = L mol 1 est values for V and see when you get 91 bars: V = 0.39 L mol 1. his equation can also be solved by Maxima: a : 24.81; b : 0.174; R : ; : 660; P : 91; r : solve(p = R*/(V - b) - a/(v*v), V); float(r); his shows three roots of which only one is real (two are complex). he last line converts the complicated algebraic form to numerical values. 2

3 3. en grams of N 2 is mixed with 5 g of O 2 and held at 25 C and bar. (a) What are the partial pressures of N 2 and O 2? (b) What is the volume of the ideal mixture? a) n N2 = n O2 = 10 g 1 = mol g mol 5 g 1 = mol g mol mol y N2 = = mol 0.513mol mol y O2 = = mol 0.513mol P N2 = (0.750 bar) = bar P O2 = (0.750 bar) = bar b) V = nr P = (0.513 mol)( L bar K 1 mol 1 )( K) bar = L 4. Calculate dz/dp for a real gas in the limit P 0 using the virial equation. Z = 1 + B P + C P dz ( ) dz dp = B + 2C P +... = B dp P 0 3

4 5. he critical temperature of gaseous ammonia is 406 K and the critical pressure is 113 bar. Use the van der Waals equation to predict its molar volume at 273 K and 1 bar. Hint: Calculate first the coefficients a and b, then write the equation for the molar volume: V = R +b. Guess P+a/ V 2 an initial value for the molar volume and obtain a new value from the equation. Place the obtained value in the equation again to get a new value. Repeat this procedure until the value does not change anymore. his is an example of iterative methods for solving equations. Use the following equations derived in the lecture notes: a = 27R2 2 c 64P c and b = R c 8P c his gives a = 425 L 2 kpa mol 2 and b = L mol 1. Use the above equation and iterate until you get convergence: V = L mol Derive the expressions for V c, c and P c in terms of the van der Waals constants a and b. he equations in the lecture notes can be rewritten as (the definition of critical point and the equation of state): R c ( Vc b ) 2 = 2a V 3 c 2R c ( Vc b ) 3 = 6a V 4 c P c = R c V c b ā V 2 c Division of the first equation by the second (side by side) yields V c = 3b. Substitution of this expression into the first equation above gives c = 4

5 8a. Substitution of the two previous equations into the last equation 27Rb above yields P c = a. Note that the van der Waals constants a and b 27b 2 are usually calculated using the critical temperature and pressure since they are typically known more accurately than the critical volume. 7. he isothermal compressibility κ of a gas is defined as: κ = 1 V P Calculate κ for a van der Waals gas using the method of implicit differentiation (with respect to P). Show that in the limit of infinite volume, it yields 1/P (the same result as the ideal gas law). Based on the lecture notes, the van der Waals equation can be written: ( P + ā V 2 ) ( V b ) = R Expand the left hand side and substitute V = V/n to get: nr = PV npb + n2 a V n3 ab V 2 Implicit differentiation with respect to P at constant gives (i.e., side by side): 0 = V + P Solving for ( ) V gives: P ( ) ( ) ( ) V nb n2 a V P V + 2n3 ab V 2 P V 3 P P = nb V P n 2 a/v 2 + 2n 3 ab/v 3 5

6 he definition of isothermal compressibility now gives (at last stage V is assumed to be large): κ = 1 V P = V nb PV n 2 a/v + 2n 3 ab/v 2 V PV = 1 P 8. he cubic expansion coefficient is defined by α = 1 V P Calculate the cubic expansion and isothermal compressibility coefficients for an ideal gas. For an ideal gas V = nr/p. he require partial derivatives are therefore: P = nr P and P = nr P 2 Using the definitions for α and κ (using PV = nr below): α = 1 V κ = 1 V P P = nr PV = nr nr = 1 = nr V P 2 = PV V P 2 = 1 P 9. A certain gas has the following equation of state: P = nr V nb 6

7 a) Calculate the following partial derivatives: ( ) P V and ( ) P V b) Show that the following relation holds for the mixed partial derivatives for this equation of state: ( ) ( ) 2 P 2 P = V V a) ( ) P V = nr (V nb) 2 and b) ( ) 2 P = nr V (V nb) 2 and ( ) P V = nr V nb ( ) 2 P = nr V (V nb) Calculate the compressibility factor Z for NH 3 (g) at 400 K and 50 bar by using the attached graph and the virial equation. By reading the value for B from the graph, using relation B = B/R, and the virial equation written in terms of P, we get: 1 + Z = 1 + B P +... = 1 + B R P +... ( L mol 1 ) ( bar L K 1 mol 1 (50 bar) = 0.85 )(400 K) 7

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