pv m = RT + Bp The simplest equation for calculating fugacity, given an equation of state, is Z=1 + B RT p

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1 Chem 42/523 Chemical hermodynamics Homework Assignment # 5 1. *Assume O 2 gas obeys the virial equation pv m = R + Bp with B = 12.5 cm 3 mol 1 at K. Calculate the fugacity of oxygen at p = 1. MPa and = K. he simplest equation for calculating fugacity, given an equation of state, is Z f p Z 1 ln =. p p In this case, from the equation of state and the definition Z = pv m /(R ), weget Z=1 + B R p f ln = B Z p p R = Bp R. herefore, paying special attention to units in the exponential, we get f=pe Bp/(R ) =1. MPa exp =9.95 MPa " m 3 mol # Pa JK 1 mol K 2. Fugacity of liquid water at K is approximately 3.17 kpa. ake the ideal enthalpy of vaporization of water as 43,72 J mol 1 and calculate the fugacity of water at 3 K. Using the Clausius-Clapeyron equation, we get f2 ln = vaph 1 1 f 1 R Jmol 1 1 = JK 1 mol =.188 So, given that f 1 =3.17 kpa, we calculate f 2 : f 2 =f 1 e.188 = kpa 1

2 3. * he second virial coefficient of methane is given by B/(cm 3 mol 1 ) = 46 88, /. (a) Obtain an expression for the fugacity of methane as a function of temperature and pressure. (b) Calculate G m for the expansion of methane at =175Kfromp =4. MPa to p =.2 MPa. (c) Determine H m H m for methane gas at = 175 Kan =4. MPa. (a) From the virial equation pv m = R + Bp, we get pv m = R Z = p p R. herefore, fugacity is given by Z f p Z 1 ln = p p = Z p R = p R. So, after accounting for the units in the exponent, we get f = p exp (b) For an isothermal expansion, we have m 3 mol 1 p (Pa) R(J K 1 mol 1 ) (K) G m = R ln From the temperature and the two pressures given, we use the expression derived in part (a) to calculate f 1 and f 2 at =175K, with B = cm 3 mol 1 ; f2 f 1 p 1 =4. MPa; f 1 =3.556 MPa. p 2 =.2 MPa; f 2 =.199 MPa. herefore,.199 G m = JK 1 mol K ln = Jmol 1 (c) From the expression derived in part (a), we write ln f ln p = e (a b )p/r. (1) From Eq. (6.37), ln f = H m H m p R 2. 2

3 Evaluating the left hand side of Eq. (1), (note that p is held constant) we get ln f = a b p p R 2 + b p 2 e (a b )p/r R a 2b = R 3 p 2 e (a b )p/r. herefore, ln f H m H m = R 2 p = a 2b p e (a b )p/r. Substituting the constants (with proper units) and evaluating, we get H m H m = m 3 mol 1 ( Pa) 175 " m 3 mol # Pa exp JK 1 mol K = Jmol hepartialpressureofbr 2 abovea(x 1 CCl 4 + x 2 Br 2 ) solution is kpa. he composition of the solution is x 2 =.25. he vapor pressure of pure bromine at the same temperature is 28.4 kpa. Assume Raoult s law standard state for bromine and calculate the activity coefficient of Br 2 in the solution. he activity coefficient is related to the activity a i and the mole fraction x i as a i = γ R,i x i. he activity a i is defined, with reference to Raoult s law standard state, as a i = f i f i ' p i p. i herefore, a 2 = =.482. γ R,2 = a 2 =.482 x 2.25 = he following data are for various mixtures of isopropanol (I) in benzene (B) at 25 C. x I p I (torr) p tot (torr) (a) Construct a pressure-composition diagram showing the variation of p I, p B,an tot with x I. (b) Does this solution exhibit positive or negative deviation from Raoult s Law? 3

4 (c) From a pressure composition diagram, determine the activities a I and a B and the activity coefficients γ I and γ B at x I =.2,.5, and.8. (Assume Raoult s law standard state.) (a) he pressure-composition diagram is shown below. he partial pressure of benzene is calculated at each composition as p B = p tot p I. P tot P iso P benz Pressure/orr Mole fraction of Isopropanol (b) It is clear that the solution exhibits positive deviation from Raoult s law since at each point, p I >x I p I, and p B >x B p B. (c) he activities are defined with respect to Raoult s law standard state as a i = f i /f i ' p i /p i and the activity coefficients are defined as γ R,i = a i /x i. he pressures for the table below are read from the graph at the given mole fractions. x I p I (torr) a I γ R,I p B (torr) a B γ R,B A certain substance exists in two solid forms A and B and also in liquid and gas states. Construct a p- diagram for this system given the following triple point data and additional observations. /K p/kpa Phases in equilibrium 2 1 A, B, gas 3 3 A, B, liquid 4 4 B, liquid, gas Below 2 K, the gas is in equilibrium with A and B does not exist. Above 4 K, the liquid is in equilibrium with gas and neither solid form exists. Above 3 K, the equilibrium between solids A and B does not exist. Above 3 K and 4 kpa, A is in equilibrium with liquid up to moderately higher pressures, and solid A is denser than the liquid. he experiments did not go high enough in pressure or temperature to establish the critical point of the gas. (In other words, limit the upper limits of your p- diagram to about 5 K.) 4

5 he qualitative phase diagram is shown below. he line representing the boundary between A(s) and liquid has a positive slope because the solid is denser than the liquid. riple points and two-phase equilibria 45 Liquid p (torr) A(s) 15 B(s) Gas (K) 5